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0, and let d n = d(Kn, OC). Show there is a convex, strictly increasing function 0.) 15. Let T (T). If S = (T) = 0, then is a linear isomorphism from A onto B. If T can be represented by a distribution R E £'(] - rr, n[) and S = (T). we have IJ(T)(n)1 = I(T,:-, en~)1 = I(R y , einY)1 ~ C(1 + Inl)D a. If Izl < a is fixed, we have Izi/iul ::: f.J, < 1 for any U E Ya,O' Then on both half-rays we have
0'»
satisfies the inequalities p (z) :::: log IAn ,
=
for all z E k n \Kn - I , n 2: 1. (Define Ko ¢.) Why is p a subharmonic function in O? (iii) For 0 C, (rn)n~I' a strictly increasing sequence of positive numbers, convergent to 00, and (IAn).;,::h a sequence of strictly positive numbers, show there is a subharmonic function p in C such that p(z) 2: log IAn,
=
for all Z E 8(0, rn)\B(O, rn-I), n :::: I. (Define ro = -1.) (iv) Let f E &(0), show that there is a subharmonic function p in 0 such that
i
If1 2 e- P dm <
00.
=
Conclude that there is a locally integrable function g in 0 such that og/oi f in the sense of distributions. Use Exercise 2.1.1 (d) to conclude, without appeal to the results in [8G, §3.2.1], that
o oi: COO(n>
--*
C""(O)
is surjective.
3. Let p be a subharmonic function in an open set 0
t= C satisfying (a) log (1 + Iz12) =
O(p(z», ZEn, and (b) there exist constants C" C 2 , C 3 , C4 > 0 such that if ZEn, ~ E C, and Iz - ~I ~ exp(-C1P(z) - C 2 ), then ~ En and ~ C3 P(z) + C4 • Show that Wp(O) nJe'(O) = Ap(O).
pen
2.2. Interpolation with Growth Conditions In [BG, §3.5] we have introduced the concept of a multiplicity variety V. Let us recall that it consists of a collection of pairs (Zk> mk), Zk distinct points of C, mk E N*, called the multiplicities of the points Zko and either the collection is finite or IZk I --+ 00. We shall assume throughout that we are in the latter case. We say that 1; E V to indicate that 1; is one of the points Zk. Given two multiplicity varieties, V = (Zko mdk~l, V' = (z;, mj)j~l' we say that V' C; V when (z;)j~1 is a subsequence of (zkh~l and, for the corresponding indices, we have mj ~ mk. For f an entire function, f ¢ 0, V (f) is the collection of zeros of f, with their respective multiplicities. More generally, we recall that for a collection of functions (h )jeJ. V «h )jeJ) denotes the multiplicity variety of their common zeros (Zdk~l and common multiplicities mk = inf{m(fj. Zk) :
2.2. Interpolation with Growth Conditions
119
j E Jl. The Weierstrass theorem [BG, §3.3.l] asserts that for every multiplicity variety V, there is I E Jt"(C) such that V = V (f). Given a multiplicity variety V, we also defined the space A(V) of holomorphic functions on V, i.e., doubly indexed sequences of complex numbers (ak,l )k,[, 0 .::s [ < mk. The reason for this notation is that there is a natural restriction map p: Jt"(C) -+ A(V) given by
g(I)(Zk») peg) = ( - [ - I•
. k,1
Given a sequence a E A(V) we say that the entire function g interpolates it, if peg) = a. The interpolation theorem [BG, §3.4.1] states that p is always surjective. The proof uses in an essential manner that variety V satisfies V = V(f) for some IE Jt"(C)\{Oj. In this section we discuss the corresponding interpolation problem for the space Ap (C). We shall assume throughout that the multiplicity variety V satisfies the non triviality condition V S; V(f)
for some I E Ap(C)\{Oj. We follow [BT 1] and we refer the reader to the references therein for other approaches to the interpolation problem. We start first by defining a space of holomorphic functions with growth conditions on a multiplicity variety. This space must be a good candidate to be the image of Ap(C) by the restriction map p: Jt"(C) -+ A(V). It turns out that there are several natural candidates. The first one is given in the following definition: 2.2.1. Definition. Let V = (Zk> mdk;::l be a multiplicity variety. We denote by Ap(V) the space of sequences (ak,{ h,l E A(V) such that there are constants A, B ~ 0 for which
L
lak'!l.::s
A exp(Bp(Zk»
(k ~
1).
O::;:i<m,
2.2.2. Proposition. For any I that
LI
1(I)(z)
- - I-
1;::0
I.
E Ap(C)
there are constants A ~ 0, B ~ 0, such
I .::s Aexp(Bp(z»
(z E C).
In particular, p(Ap(C» S; Ap(V).
Proof, Assume that I/(t)1 .::s A1eB1p(t) for every t E C. Let z E C be fixed. For any t E B(z. 2) there is { E B(z. 1) such that It - {I .::s 1, hence pet)
.::s
Cop(l;)
+ Do .::s cJp(z) + (Co + l)Do.
Taking these inequalities into account, the Cauchy inequalities 1(/)(z)
I-1-'• -
I
1
:::; 21 II-zl::::2 max 1/(t)1
2. Interpolation and the Algebras A p
120
yield the following estimate:
~ If(l)(z) I:5 2 ~ where A
I!
max If(t)1 :5 2 max AleB'P(t) :5 AeBp(z) , II-zl:S2
II-zl!:2
= 2Ale(Co+I)Do, B = BleB.
o
In general, as we shall see in some examples below, Ap(V) #- p(Ap(C». The reason is that the definition of Ap(V) was done basically taking only into account local conditions. In order to facilitate the appearance of global conditions we proceed as follows: Let r > 0 and z E C, define p(z, r) := maxp(z 1'1:5'
+ ~).
The same argument as that in Proposition 2.2.2 leads to the existence of A > 0, B > 0, such that f(l)(z) I L r' :5 A exp(Bp(z, r», I t~O I. - - I-
for any
z E IC
and
r > O.
This inequality leads naturally to a different candidate for a space of holomorphic functions with growth conditions on a multiplicity variety.
2.2.3. Definition. Let V = (Zko mdk~1 be a multiplicity variety, denote by Ap,oo(V) the space of sequences (ak,th" E A(V) for which there are constants A > 0, B > 0, such that for every r > 0 and every k E N*
L
lak,dr':5 A exp(B(p(Zko r».
O:sl<mk
The following result is immediate:
2.2.4. Proposition. The following inclusions hold: p(Ap(C»
S Ap,oo(V) S Ap(V).
We are now ready to state the interpolation problems of interest to us.
2.2.5. Definition. Let V = (Zko mkh~l be a multiplicity variety. We say that V is an interpolation variety for Ap(C) if p(Ap(C) = Ap(V). If p(Ap(C» = Ap,oo(V) we say that V is a weak interpolation variety. These are only two, the simplest two, definitions of interpolation variety. We shall only study the problem of deciding when V is an interpolation variety, and relegate the similar study for weak interpolation varieties to the exercises.
2.2.6. Lemma. Let V = (Zk, mk)k~l be an interpolation variety for Ap(IC). There are two constants, Clo C2 > 0, and functions hk E Ap(C) such that for every k 2: 1:
2.2. Interpolation with Growth Conditions
(i) Ihk(z)1 :::: CI exp(c2P(Z» (z E C); (ii) for every j ::f. k, one has hj/)(zd
(iii) hfl (Zk) = 0 for all 0 :::: I :::: (iv) him.-1)(Zk) = (mk - I)!.
121
= 0 for 0:::: I :::: mk -
I;
2;
mk -
In other words, (p(hk»j,1
= 8j ,k 81,m.-I,
where 8a ,b is the Kronecker delta symbol.
Proof. We leave it as an exercise for the reader to give a simple proof based on the open mapping theorem. We give here a proof using a minimum of functional analysis. For that purpose, let S := {(ak,/)
L
E Ap(V):
laul:::: I for every
k
~ I}.
O:sl<m.
Consider on S the distance induced by the norm
III (ak,/) II I := sup
L
lak,/l.
k:::I O:;:l<m.
It is easy to verify that S is a complete metric space. For n E N°, let Sn := {p(/): f E Ap(C), If(z)1 :::: n exp(np(z», Vz E
q n s.
We claim that Sn is a closed subset of S. Let (gj)j:::1 be a sequence in Ap(
_J_ _
I!
~a
j_oo k.l·
Therefore, peg) = (au),
and (ak,/) E Sn' Since we are assuming that p is surjective. we have
The Baire Category Theorem [Hor] allows us to conclude there is an no E N* such that Sno ::f. 0. It is obvious that each Sno is a balanced convex set, i.e., it is convex and jf a E Sno and A E
to.
122
For k
2. Interpolation and the Algebras A p ~
1, let ak
= (aJ.I)j.1 =
O;.kOI.m,-1 E
S,
and let A be a fixed number. 0 < )... < roo Then Aa k E 8(0, ro). Hence there is IPk E Ap(lC) such that
s
IIPk(Z)1
no exp(nop(z»
and P(IPk)
The choice hk = IPk/)...,
CI
= Aa
k•
= no/A, and C2 = no has the required properties. 0
As we have seen in [BGl, interpolation problems are related to the structure of the ideals in the algebra of holomorphic functions. This is the reason we start by studying the finitely generated ideals of the algebras Ap(iC).
2.2.7. Definition. Let fl,"" fn E Ap(IC). The local ideal generated by h, ... , fm is the set [loc(fl, ... , fm) of all the functions g E Ap(iC) such that for every z E IC there is an open neighborhood Q of z and functions gl, ... ,gm E Jt"(Q) so that glQ =
L
gj(/jIQ).
I::;j::;m
It is easy to verify that [loc(fl •... , fm) is an ideal in Ap(iC). Moreover, if ] is the ideal generated by h, ... , fm in Jt"(IC) we have [loc(!I. ... , fm)
= ] n Ap(iC).
We have shown in the previous volume that any finitely generated ideal in ;f(iC) is closed [BO, Corollary 3.5.8]. Since the injection i: Ap(lC) -+ Jt"(IC) is
continuous, we conclude that [locUI, ... , fm)
= i-Ie])
is closed in Ap(iC). We denote by [(flo ... , fm) the ideal generated by h,.·., fm in Ap(C) and, by l(fl, ... , fm), its closure in Ap(C). In general, we only have the inclusion l(fl, ... , fm) 5:; l)oc(fl, ... , fm), without equality.
2.2.8. Definition. Let fl,"" fm if
E Ap(IC).
We say they are jointly invertible
When m = I, we say that f E Ap(C) is invertible if [(f) = [Ioc(f). For instance, when p(z) = IZI, Lindelof's theorem [BG, §4.5.7J proves that, in the case Ap(C) = Exp(C), every nonzero function f is invertible. Note that f being invertible does not mean I/f E Ap(C). In fact, only if f is invertible and has no zeros then 1/f E A p (IC).
2.2. Interpolation with Growth Conditions
123
2.2.9. Lemma. Assume that II, ... , 1m are jointly invertible. Given A > 0, there are B > 0 and C > 0 such that if h E Iloc(fI, ... ,1m) and Ih(z)1 ~ exp(Ap(z»,
there exists gl, ... , gm E Ap(C) with
L
h=
gJfj
I:;;j:;;m
and Igj (z) I ~ C exp(Bp(z»,
lor
I S j
sm.
Proof. We give a proof similar to that of Lemma 2.2.6. Consider
s
S = {h E I(ft, ... , 1m): Ih(z)1 with the metric d(h, g)
= sup(e-Ap(z)lh(z) -
exp(Ap(z))),
g(z)!),
ZEIC
induced by the norm of the Banach space Ap.A(C), Let us verify that S is a complete metric space. Given a Cauchy sequence (hn)n?:1 in S, it is easy to find a subsequence (h",h?:1 and h E Ap(CC) such that Ih(z)1
s
exp(Ap(z»
and as
k --+
00.
If V = V (fl, ... , 1m), (Zj, mj) E V, then hn~ vanishes at Zj with multiplicity ~ m j. It follows the same is true for h. Hence,
hE Iloc(fI, ... , 1m) = l(fl, ... , 1m).
This shows that S is a complete metric space. Consider now the linear continuous and surjective mapping
8: (Ap(C)m --+ /(/10 ... , 1m), 8(gl, ... , gm):=
L
gJfj.
I:;;j :;;m
As pointed out above, the ideal I (flo ... , 1m) = floc (fl, ... , f m) is a closed SUbset of Ap(C). For n E N*, denote Sn
:= {(gl, ... , gm) E (Ap(c»m: Igj(z)1
s
ne np (z)(1 S j S m)} n S.
We claim that S" is a closed subset of S. In fact, let (hl)I>1 ~ S" and h E S 1--+ 00. Each hi = 8(gl./' .. ~, gm,I), Igj,l(z) I S ne"p(z). We can therefore extract subsequences (gj.hh?:J and functions gj E ,w(C) such that
be such that d(h l , h) --+ 0 as
124
2. Interpolation and the Algebras A p
Hence for all and
L
gj.IJJ -+
t:sj:sm
L
gJj
Z
E
C,
in Jf"(C).
l:sj:sm
Therefore, h = 8(gl, ... , gm). In other words. h E Sn. Since 8 is surjective. S = Un>1 Sn. There is then an integer no such that Sno ¥= 0. As in Lemma 2.2.6. Sn is a balanced convex set. Therefore. for some ro > 0, the ball B(O, ro) = {h E S. d(O, h) < ro} is contained in Sno' Let 0 < A < roo Then for any h E S we have ).h E B(O. ro). This means that there are gt.···. gn E Ap(C). and 8(gl • ...• gm) = )'h.
Letting B = no and C = nol). we conclude the proof of the lemma.
0
Remark. Note that if hE lloc(fl, ...• fm) and h satisfies Ihl:::: De AP • then we can conclude from Lemma 2.2.9 that h = LI:Sj:sm gj/j and Igj I :::: DCe BP • Let us now state the main result of this section. 2.2.10. Theorem. Let !J, ... , fm E Ap(C) and V:= V(fl, ...• fm) be the multiplicity variety of their common zeros. Assume there are e > 0, C > 0, such that for every (Zk. mk) E Vane has
L
l:sj:Sm
limkl(Zk)1 J
,
~eexp(-Cp(Zk».
mk·
Then V is an interpolation variety for Ap(C). Conversely, if V is an interpolation variety for Ap(C) and if the functions fl, ...• fm are jointly invertible, then there are e > O. C > O. such that the inequality (*) holds for every (Zk. mk) E V. Proof. Let us start proving that in case lloc(f" ... , fm) = l(fl, ... , fm). the inequality (*) is a necessary condition for V to be an interpolation variety. If V is an interpolation variety, from Lemma 2.2.6 we conclude there are constants C I , C 2 > 0 and functions hk E Ap(C) such that Ihk(z)1 :::: C I exp(C2P(Z»
and The functions
2.2. Interpolation with Growth Conditions
vanish at every Zj with multiplicity over, for some A > 0 we have
125
~ mj'
Hence
rpk
E Iloc(!I, ... , /m). More-
+ IZkD(1 + Izl)lhk(z)1 ::: C I (1 + IZkD exp(Ap(z», used that 10g(1 + IzD = O(p(z». Hence, we can apply
Irpk(Z)1 ::: (1
where we have the previous lemma and the corresponding remark to conclude that there are constants B, C > 0, and entire functions gl,b ... ,gm,k such that rpk
L
=
k
gj,k/j,
~
I
I~j~m
and (Z E C),
with C2 = CCI. The choice of the fonn
hk
ensures that the Taylor expansion of
rpk
about
Zk
is of
Condition (t) now implies that 1_< ' "
1
()ll.rr*)(Zk)l
~ gj,k Zk l:'Oj:'Om
, mk·
::: C3 exp(C4P(Zk»
L
If(m')(Zk)1
l:'Oj:'Om
mk'
J
,
'
using once more that log( I + IZk I) = 0 (P(Zk». This inequality is clearly equivalent to (*). We shall now prove that the inequalities (*) imply that V is an interpolation variety for Ap(C). We start by reminding the reader that a corollary of the Minimum Modulus Theorem is the following: 2.2.11. Lemma. Let 1/1 be a/unction holomorphic in 8(0, 2eR), assume that
max
1~I:'02eR
II/I(~)I.:::
M
and
11/1(0)1
~
{, > 0,
then there is a value r, R/4.::: r .::: R/2 such that (**)
min (log II/I(~)I) ~ 9 log .5 - 8 log M. 1~I=r
PrOOf of Lemma 2.2.11. Let us recall the Minimum Modulus Theorem as stated in ([BG, Theorem 4.5.14] or [Lev, Chapter I, Theorem 11]): If / is a holornorphic function in a neighborhood of 8(0, 2eR), 1(0) = 1, and '1 E ]0, 3e/2[,
126
2. Interpolation and the Algebras Ap
then there is a finite family of exceptional closed disks, the sum of whose radii does not exceed 4'1R, such that for z E 8(0, R) and outside the exceptional disks, the following inequality holds log II(z)1
~-
(2 +
log
~~) log M(f, 2eR).
Let us choose 1'/ so that 811R < R14, e.g., 1'/ = 1/33. The width of the annulus R/4 S Izl S RI2 is R/4 and the exceptional collection of disks can at most have width equal to the sum of their diameters, i.e., at most 811R. Hence, there exists r E [R/4, R/2] such that these disks do not intersect the circle Izl = r, so that log If(z)1 > -8 log M(f, 2eR) for all z, Izl = r. Let f(z) 1/I(z)/fJ, then
=
min (log 11/I(z)l-log8) > -8(logM(1/I, 2eR) -log8). Izl=r
In other words, min log 11/I(z)1 > 910g8 - 8 log M(1/I, 2eR), Izl='
o
which is the desired inequality (**). Note that the same argument shows that one can take r We also need the following lemma:
E
]3RI4, R[.
2.2.12. Lemma. Let g be afunction holomorphic in I~ I < 1, continuous in I~ I ~ 1, and bounded in absolute value by M. Assume further that g has a zero of order q at a point ~ = a, 0 < lal < I, and a zero of order m at the origin. If Ig(m)(O)1
=--,--'-' m. then
~
8 > 0,
8 lal q ~ M'
Proof of Lemma 2.2.12. Define an auxiliary function h by the identity
g(~) = ~m (I~ --a~) q h(n· The function h will then be in the disk algebra, Ih(~)1 S M, h(O) =1= 0, h(a) =1= O. Differentiating m times the defining equation for h and letting ~ = 0, one finds g(m)(o) = (-a)qh(O).
m! Therefore,
o
2.2. Interpolation with Growth Conditions
127
Let us now return to the proof of Theorem 2.2.10. Let (Zk. mk)
E
V and define
dk := infll, inflzi -zkll. j#
If dk < I, choose some Zj such that dk = IZj - zd. Since we are assuming that (*) holds, there must be an index j, 1 S j S m, such that If(m;)(zj)1 J
mj!
e
~ _ exp(-Cp(Zj».
m
We shall apply now Lemma 2.2.12 to the function max Ig(OI I{I~I
s
max max Ijj(zj
I~j~m 1,1~2,
+ ~)I s
g(~)
:= jj(Zj
+n
Then
M := Mo exp(NoP(zj»
for some constants Mo, No > 0 that depend only on the bounds for the generators jj. We have now la I = dk and q = mk in the statement of the lemma, hence m e Mdk I ~ - exp(-Cp(Zj». m
Since p(Zj)
s
Cop(z.)
+ Do, we obtain
(tt) for some AI > 0, 0< el S I. Clearly the same inequality holds if dk = I. In order to apply the semi local interpolation Theorem 2. 1.4 we need to show that we can choose £0 > and Ao > 0 such that, for every k ~ I, the connected component Uk of S(lfl, eo, Ao) that contains Zk satisfies Uk £ B(Zko dk/2). Let us choose j so that
°
(#)
and, for
RI s
2e, define
1/1 by the equality ~m'1/I(O
We have for some A2 > 0,
11/1(0)1 =
e2
= jj(Zk
> 0,
If(m ll (Zk )d;"1 J
+dk~)'
mk!
~ e2 exp(-A2P(zk»,
as a consequence of (tt). Moreover, max
RI~2e
11/I(~)1
= (2e)-m, max Ijj(Zk +dkOI 1{1=2,
S max Ijj(Zk Irl~2,
+ r)1
S Moexp(NoP(zk».
It would be tempting to conclude that 1/1 has no zeros in B(O, I), but all We know is that there are no other common zeros to all the jj in B (Zk, dd. Nevertheless, thanks to the Minimum Modulus Theorem we can find r, 1/4 S r :::: 1/2, to which the inequality (**) of Lemma 2.2.11 applies. That is, for all {,I~I = r, log
11/I(~)1 ~ 9 log 11/1(0)1 -
Slog (max
Irl~2e
11/1(1')1) .
2. Interpolation and the Algebras A p
128
Translating this inequality into an inequality for jj, we have that there are constants e3 > 0 and A3 > 0 so that Ijj(Zk
+ dk~)1
::: e34-mk exp( -A3P(Zk»,
=
whenever I~ I r. We still need to estimate the multiplicity mk to obtain a good estimate. This will depend once more on the inequality (#). We have
-1-1
I/rk)(zk)1 = mk! 2Jri
jj(z) dz iz-zki=e (z - Zk)mk+1 '
so that
In other words, for some A 4 , B4 > O. We conclude that there are Ao, Bo > 0 and r, 1/4::: r ::: 1/2, such that Ilj(z)1 ::: eoexp(-Aop(z»
when Iz - zkl = rdk. This shows that the connected component Uk of the set S(l/I, eo, Ao)
= {Z EC:
(L
Ijj(Z)12)
1/2 < eoe-AOP(z)} ,
I::;J:<:m
which contains Zk. must be contained in B(zko dk/2). To conclude the proof of the theorem, let (ak,lh,1 E Ap(V) and define a holomorphic function Xin S(I/I. Eo, Ao) by ()..IUk)(z)
=
" " ~
ak,l(z - Zk) I
and j. == 0 on S(I/I. eo, AO)\(Uk>l Uk). Since (ak./h.1 E Ap(V), there are constants A, B > 0 such that
L
lak."::: AeBp(Zk)
(k ::: I).
O::;i<mk
Hence I(XIUk)(z)l:::
E
lak."
(~k)':::
O::;i<m,
L
lak."::: AeBp(z,) ::: A'eB'p(z) ,
O:<:/<mk
for some A', 8' > O. Clearly this inequality is valid in the whole of S(I/I, eo, Ao). Applying the semilocal interpolation Theorem 2.1.4 to X, we conclude there exists I E Ap(C) such that p(f)
= (ak,l)k,l.
This concludes the proof of the theorem.
o
2.2. Interpolation with Growth Conditions
129
In the case of principal ideals there is a simple analytic condition to ensure that f E Ap(C) is invertible. 2.2.13. Definition. A function f E Ap(C) is said to be slowly decreasing if there are positive constants 6, A, and B such that: (i) every connected component Va of SClfl, 6, A) is relatively compact; and (ii) for every z, ~ E Va, p(~) ~ Bp(z) + B.
2.2.14. Proposition. Any slowly decreasing function is invertible.
Proof. Assume that g
E
.Jt"(C) is such that h = gf
E
Ap(C), i.e., there are
constants C, D > 0 such that Ig(z)f(z)1
~
(z E C).
Cexp(Dp(z»
Let Va be a connected component of the set S(lfl, definition. On the boundary of Va we have If(z)1 hence for z
E
= 6e- Ap
6,
A) given by the previous
(z),
aVa we have
Ig(z)1
~
C
- exp«A 6
+ D)p(z»
~
C
- exp«A 6
+ D) ml:!Xp(~». t;eU.
Since Va is bounded, we can apply the Maximum Principle and obtain C Ig(z) I ~ - exp«A S
+ D) ma,x p(~»,
z EVa'
t;eU.
Using condition (ii) from Definition 2.2.13, we can replace this inequality by Ig(z)1
~
E exp(F(p(z»,
for all z E Va, for some E, F > 0 independent of the component VCt • Therefore, this inequality holds in S(lfl, 6, A). On the other hand, for z ¢ S(lfl, 6, A), CeDp(z) C Ig(z)1 ~ If(z)1 ~ exp«A + D)p(z».
e
o
It is now clear that g E Ap(C) and /U) = iIocU).
2.2.15. Proposition. Let p be a radial weight that satisfies the doubling condition p(2z) = O(p(z», then every f E Ap(iC)\{O} is slowly decreasing. Proof. It is enough to show that there exist a point Zo and two constants 6, A > 0 such that for all r » 1 one can find rl, r2 satisfying r
2 ~ rl inf{l/(z)l: Iz - zol
= rl
< r < r2 ~ 2r,
or Iz - zol
= r2}
~
6exp(-Ap(r».
Since, in this case, we can find a strictly increasing sequence of radii R n , Rn -+ 00, Rn+1 ::: 4Rn, such that I/(zo + ~)I ~ exp(-Ap(Rn if RI = Rn.
»
2. Interpolation and the Algebras A"
130
The doubling condition imposed on the weight p implies that if z and z' satisfy Rn ~ Izo + zl :::: Rn+h Rn ~ Izo + z'l :::: Rn+h then p(z) :::: + C;p(z') for some convenient positive constants C;. Thus, the components of the set S(I/I, e, A) satisfy the requirements of Definition 2.2.13. In order to conclude the proof of the proposition, we choose Izol :::: 1 such that I(zo) =1= O. We can now twice apply Lemma 2.2.11 to the function '1/1(1;) == I(zo +~) to obtain '10 '2. 0
Co
Co'
The conditions imposed on p by Proposition 2.2.15 imply immediately that Ip(z)1 = O(lzl")
for some K > 0, while the inequality p(z) :::: Cop(1;) only implies the growth condition
+ Do whenever Iz -
~I~
1,
Ip(z)1 :::: O(e Alzl )
for some A > O. In fact, for a weight like p(z) = e lzl one does not automatically obtain that every I E Ap(IC)\{O) is slowly decreasing, but only that I is slowly decreasing in Aq(C) with a weight q slightly bigger than p (see [Heyl]). The advantage for the interpolation problems of knowing that a function IE Ap(C) is invertible is that then V = V(f) is interpolating if and only if there are e > 0 and C > 0 such that I/(md(Zk)1 > -Cp(z,) ee mk! -
holds for all (Zk. mk) E V. So the interpolation problem has a clear-cut answer in this case. Whence, the interest of Proposition 2.2.15. It is natural to ask whether condition (*) of Theorem 2.2. IO always holds when V = V(f) is interpolating, even if I is no invertible in Ap(iC). The answer is negative as shown by the following example of Squires ([Sq2], see also [Hey2]). Consider the weight p(z) = 11m zl + log(l + Iz12) and the corresponding space Ap(iC), which coincides with the space F(E/(R» of Fourier transforms of distributions of compact support. If IfJ E V(lR), then there is a constant A > 0 such that for every n 2: 0 IFIfJ(z) I < - (1
Cn
+ Izj)n
exp(AI 1m zj)
(z
E
IC),
for some Cn > O. Clearly FIfJ E Ap(iC), but it cannot be slowly decreasing. (It turns out that in this space, slowly decreasing and invertible are two equivalent conditions, hence no such function is invertible.) Since the derivatives of Frp also satisfy similar estimates, condition (*) of Theorem 2.2. IO cannot hold for FIfJ. Nevertheless, Squires has shown that one can find IfJ E 'D(lR) such that FIfJ(z)
=
sin rr z 00
11 n".l
z2) .
(
1-
n 4
2.2. Interpolation with Growth Conditions
131
Hence V(Frp) ~ V(sinJrz) (all the zeros are simple). On the other hand, V(sinJrz) = {k E Z, mk == I} is interpolating for Ap(IC), namely
~ (sin Jrz)lz=k =
JrcosJrk = (-1/,
and condition (*) of Theorem 2.2.10 is satisfied for I(z) = sinJrz. Since any multiplicity variety W contained in an interpolating variety for Ap(1C) is itself interpolating, it follows that V (F rp) is interpolating and condition (*) fails for Frp. In fact, the concept of a slowly decreasing function originated in the work of Ehrenpreis about convolution equations in the spaces £(lR) and 1J'(JR) [Eh3], [Eh4]. He proved that in F(£'(JR» any invertible function is slowly decreasing. Let us mention here that for p(z) = 11m zl + log(l + Iz1 2 ), the condition 1 being slowly decreasing is equivalent to the following: There is A > 0 such that for every x E R. the function 1 satisfies max{l/(x
+ 1)1: t E R It I :::: A log(2 + Ix!)}
2: (A
+ Ixl)-A.
We shall show in the next chapter that every exponential polynomial N
I(z) =
L Cj(z)e
iaj "
j=l
satisfies this type of lower bound condition, and hence is slowly decreasing. More generally, the ideas of this chapter are also well adapted to the Beurling weights, p(z) = I Imzl + w(lzJ), where w: [0,00) -+ [0, 00) is an increasing, continuous, subadditive function such that w(O) 0, w(t) 2: a + b logO + t), for some a E R., b > 0, and
=
1
00
o
w(t) --2
1+ t
dt < 00.
These functions appear when we consider spaces of Coo functions where one introduces conditions on the growth of the derivatives (see [Bj], [BD3], [Chou)). To conclude this section, let us rephrase the interpolating procedure used in Theorem 2.2.11. We used condition (*) to find 8, A > 0 such that for any a E Ap(V), there is a holomorphic function X on S(I/I, 8, A) satisfying the correct growth conditions and p(J...) a. It is clear from the proof that the map a ~ ). is a linear continuous map into Ap(S(I/I, 8, A». It is therefore natural to ask whether the procedure a ~ A of Theorem 2.2.11 can be also made linear and continuous, in other words, whether there exists a right inverse E to p. Or What amounts to the same thing, does it exist as an extension operator,
=
po E
= id,
Where E is linear and continuous, under the assumption that V is an interpolation variety? One way to do this is to define >.. using an interpolation series, typically
2. Interpolation and the Algebras A I'
132
some kind of Lagrange series. For instance, if V = V(sinlTz) = Z, p(z) = Izl, one can see that E(a)(z)
sin IT Z = 2::>k( -Ok _k kEZ IT(z)
( Z ) Ik 1 -k
(where for k = 0 the term (zlk)lk l is considered to be 1) is a right inverse to the restriction operator p. More generally, if V = V(f) = {(Zk. l)), one could try to define E(aHz)
= 2:::ak kEl.
fez) f'(zd(z - Zk)
(.:.)Ilk Zk
for a convenient sequence of integers JLk ~ 0 (see [Ge], [BrK], [Lev]). It turns out that this is not possible for every weight p. In a seminal paper, B.A. Taylor [Ta2] considered the problems of finding a right inverse for p and for the Cauchy-Riemann a-operator in the correct spaces. He found that the situation is very different for p(z) = Izl", ex > 0, and p(z) = 11m zl + log(l + IzI2). In the first case, there is a right inverse for and, consequently, linear continuous extension operators E: Ap(V) ~ Ap(C) exist when V is interpolating. In the second case, the extension operator exists if and only if V satisfies the additional condition of lying in a logarithmic strip about the real axis, i.e., there is A > 0
a
I Irnzkl
~
A log(2 + IZkl)
for all
Zk E V.
Such a variety is V is called hyperbolic. We refer the reader to the work of Taylor, Vogt, Meise, et al. for further ramifications of these two questions (e.g., [MeTl], [MeT2], [MeTV]). EXERCISES 2.2.
=
Throughout these exercises we denote by V (Zk. mdki!l a multiplicity variety, which we assume is infinite and satisfies V <; V(f), for some f E Ap(iC), unless explicitly stated otherwise. I. (a) Show that the failure of the condition IZk I --+ 00 in the definition of a multiplicity variety is equivalent to saying there is no entire function f #- 0 such that V V(f). Prove that in such a case the map p: Jt"(iC) --+ A(V) is injective but cannot be surjective. (b) Assume now that IZkl--+ 00, but that there is no f E Ap(C)\{O} such that V <; V(f). Then: (i) show that p: Ap(iC) --+ A(V) is injective; (ii) let e j E A(V) be defined by eL = 8j,k81.0, prove that no e j E p(Ap(iC).
=
2. (a) Show that if V is an interpolation variety for Ap(C) and W <; V, then W is also an interpolation variety. (b) Prove that W = N (with multiplicity I) is an interpolation variety for the space Exp(C). *(c) Can you find a single function I E Exp(C) such that V(n = N? Can you find two functions I .. 12 E Exp(C) such that VUh h) N? (Hint: Recall from [BG, Exercise 2.5.20] that if (zkh". denotes the zeros of f E Exp(C), repeated according to their multiplicities, then the sums ~O~I'kl
=
133
2.2. Interpolation with Growth Conditions
3. Let p(z) = IzIP, P > 0, compute explicitly the function p(z, r) defined just before Definition 2.2.3. 4. Show that if the multiplicities
mk ~
M < 00 for all k, then Ap.oo(V)
= Ap(V).
=
*5. For p(z) Izl, construct a multiplicity variety V such that Ap.oo(V) =f. Ap(V). (Hint: Use a variety for which mk :::::: IZkl.)
*6. Let p(z) = IzI P. Use Exercise 2.2.3 to show that a sequence (au) and only if for some A > 0 exp(AlzkjP)
(*)
+ IZkDI
lak." ~ A (I
Furthermore, show that Ap.oo(V) mk
= 0
E Ap.oo(V)
if
(0 ~ / < mb all k).
= Ap(V) if and only if
COg(~z~IZkD) ,
as
k -+
00.
Show also that in this case, (*) is equivalent to the existence of B > 0 such that laul ~ *7. Let p(z) = I Imzl someA>O
Thus Ap.oo(V)
Bexp(BlzkI P ) (/!)I/P .
+ log(l + IzI2),
then (ak./)
la k.1 I -<
I!
E Ap.oo(V)
if and only if for
AI+ 1 exp(Ap(Zk» .
= Ap(V) if and only if mk -
0 (
-
p(Zt) ) logp(zk)
as
k -+
00.
What restriction on the multiplicities does this condition imply when Zk is on the real axis? (Hint: Compute p(z, r) for this weight function.) 8. Simplify the proofs of Lemmas 2.2.6 and 2.2.9 using the open mapping theorem, i.e., Proposition 1.5.1, instead of the Baire category argument given in the text. 9. Let V be an interpolation variety for Ap(C), dk := min{l, min Iz}· - zkll. j",k
Show that there are positive constants e, A, B, C > 0 such that for all k
and mk ~ Bp(Zk)
+ C.
(Hint: Use Lemma 2.2.6 and reconsider the proof of Theorem 2.2.10 without assuming that V is of the form V (fl ..... 1m), with !J, ... , 1m jointly invertible.) 10. (a) Use the previous exercise to show that a necessary condition for a sequence (zkh~l of distinct points converging to 00 to be an interpolation variety for the space Exp(C) is the existence of a constant A > 0 such that inf{lz· - zkleA
i;4t
}
O.
134
2. Interpolation and the Algebras A p
(b) Let (zkh~1 be a sequence of distinct real numbers, IZk I -+ 00. A necessary condition for it to be an interpolation variety for the space :F(E'(JR» is the existence of N > 0 such that
11. Let V = V(fl, ... , 1m), It. ... , 1m jointly invertible in Ap(C). Show that V is an interpolation variety if and only if there are e, A > 0 such that: (i) each Zk E V belongs to a component of S(lfl, e, A) of diameter at most I; and (ii) no two distinct points of V lie in the same component of S(lfl, e, A). 12. Let p(z)
= Izl,
V
= {k E Z, mk = 1; ±(k + e(k», mk =
1}, with e(k)
E
JR, 0
<:
IE(k)1 <: ~.
(a) Show that there exists I E Ap(C) such that V = V(f). Find a condition on e(k) so that there cannot exist 8 > 0, such that Vk E Z.
Show that this condition is independent of the choice of f. (b) Choose two functions EI (k), e2(k) as in part (a) with the additional condition that el (k) <: 0 and E2(k) > 0, let flo fz be the corresponding functions of part (a). Show that V(fh fz) = V(sinrrz), which is interpolating. Why is this not a contradiction to Theorem 2.2.10? *13. Find an even entire function f of exponential type such that it vanishes at all the points x + iy E Z + iZ of the form Ix - nJI :S nj' Iyl :S nj, for an increasing sequence of integers nj , such that nj -+ 00 sufficiently fast. Show that V (f) cannot be an interpolation variety for Exp(lC), even though the conditions on dk , mk from Exercise 2.2.9 are satisfied. (Hint: Check whether the value of 1f'(nj)1 is too small. This depends only on the behavior (Z/Zk)2), where Zk lies in the block of zeros surrounding nJ.) of the finite product
no -
= IzI
14. Let).. E JR+ \IQ!, p(z) and
P,
P:::
1,
fez) == sin(rrz) sin(nz/)..),
{I).. - ;; I: m :S n, mEN} ,
n E N°.
Find necessary and sufficient conditions for V(f) to be an interpolating variety in Ap(C) in terms of the behavior of
E
E'CIi..), f(t;) := (Tx' e- ix ,) be the Fouriertransform:FT of T. Let P(D) =
L7~o Gj Dj and be a linear differential operator with constant coefficients.
(a) Show that if there is S E E'(JR) such that peDiS = T, then I(t;)/ P(it;) is an entire function. (b) Let q(z) = amz m + ... + ao be a polynomial of degree m, h a holomorphic function in B(O, 1). Show that lamh(O)1 :S max Iq(z)h(z)l· 1,1';;1
=
=
(Hint: Use that when Izi 1, Iq(z)1 Ir(z)l, r(z) = am + am_Iz + ... + aoz m.) (c) Assume that g(~) := I(r;)/ PUt;) is an entire function. For ~ fixed consider q(z) := P(i (z + r;)) to show that lamllg(~)1
::: max 1/(r; + z)l· Izl!S1
2.2. Interpolation with Growth Conditions
135
Conclude that there is S E £'(IR) such that :FS = g, T = P(D)S, and cv(suppS) = cv(supp T). (This proves that any polynomial is invertible in Ap(e), for p(z) I Imzl + 10g(1 + Iz 12 ). In fact, it proves the same result holds for any weight p.) (d) Show that for some constant C > 0, IP(z)1 ~ Cd(z, v(p»m. Use this to prove that P is slowly decreasing in any Ap(e).
=
16. Prove that I sinzl ~ Cd(z, nZ)e11m'l for some C > O. Use this to show that sinz is slowly decreasing in any Ap(e) such that p(z) ~ 11m zl + logO + IzI2). 17. For B > 0 let us define Yl..I := Yl..I(B) := inf{rl exP(Bp(Zb
r»: r > O},
Yk := Yk(B) := Yk.ml-I(B).
Show that a = (au)
E Apoo(V)
implies that for some A> 0, B > 0, lak.ll :::: AYk,I(B).
*18. Let fl' ... , f m be jointly invertible and V = V (fl, ' .. , f m) a weak interpolation variety. Follow the notation of the previous exercise. Show that for every B > 0 there are constants e, C > 0 such that
~
1.fj(mk)(zk)1
j=1
mk·
L.....-
,~eYi
(B)
-ep(',)
e
'.
(Hint: Consider functions hk like those in Lemma 2.2.6 with condition (iv) replaced by (iv)' hlmk-')(zd = (mk - I)! nCB). Follow the proof of the necessity part in Theorem 2.2.10.) *19. For each k
~
1, let Rk
= R.(B)
~
1 be a value such that
R;;(ml-l) exp(Bp(Zko Rk»
:::: 2n(B).
Assume that fl,"" fm E Ap(e), V = V(fl,"" 1m), are such that for each B > 0 there exist constants e, C lo C2 , C3 > 0 so that for all (Zk> mk) E V the following three properties hold: (i) mk :::: C 1P(Zk) + C 2 : (ii) P(Zk: 2Rd :::: CIP(Z) + C 2 Vz E B(Zk, 2R k ): ...
(Ill)
m
Ilmk)(zk)1
j=1
mk·
L: }
,
~ eYk(B) exp( -C 3P(Zk».
Prove that: (a) V is a weak interpolation variety. (b) There are 8, KO, KI, K2 > 0 such that every Zk E V belong to a bounded component of S(I/I, 8, Ko) with the properties: (i) p(z) :::: KIP(l;) + K2 for any two points z and l; in the same component; and (ii) no two distinct points of V lie in the same component of S(I/I, Il, Ko). (Hint: Apply Lemma 2.2.12 to the function g(l;) = jj (Zk + Rkl;), for a convenient choice of j. Conclude that
I/(z)1 ::: Yk(B)
Iz - zkl mk R mk
1}
exp( -Cp(Zk»,
k
for Iz - Zk I < ~ Rko where of Theorem 2.2.10.)
1},
C > 0 are suitable constants. Fol1ow the rest of the proof
2. Interpolation and the Algebras Ap
136
*20. Let p(z) = Izl P , P > 0, and follow the notation of the preceding exercise. (a) Show that mk = O(P(Zk», Rk = O(p(zd), and condition (ii) always holds for this particular weight. (b) Show that condition (iii) is equivalent to ~
f;;(
1ft,) (zk)1 mk!
> eexp(-Ciz,IP) (I + IZk I)m,
for some convenient e, C > O. (Compare with the previous Exercise 2.2.6.) (c) Conclude that when fI. ... , fm are jointly invertible the condition in part (b) is equivalent to V being a weak interpolation variety. "21. Let f be a slowly decreasing function in F(e'(R» and V = V(f). (a) Show that mk = O(P(Zk», Rk = O(P(Zk» and condition (ii) of Exercise 2.2.19 holds (here p(z) = 11m zl + 10g(1 + IzI2». (b) Use Exercises 2.2.7 and 2.2.19 to show that V is a weak. interpolation variety if and only if there are e, C > 0 so that If (m,)(z )1 > eexp(-Clzkl) k
for every Zk
E
-
+ IZkl)C
(1
V.
22. Show that if for some A, B > 0, lak I ~ Ae B1kl for all k ~
sin7fz
fez) := L."ak(-l)k ( _ k)
keZ
(where (z/ k)lkl == 1 for k ak for every k E Z.
7f
E
Z, then the function
(Z) Ikl k
Z
= 0), is an entire function of exponential type such that f(k) =
23. Let V = hh~I. IZkl - 00, mk == I, and let A(V) be the set of all sequences The object of this problem is to prove there are no linear continuous extension operators E: A(V) _ .1t"(C), poE = id ([Rudl]). The topology on V is that of convergence for each index k. (a) Show that the topology we described is the compact open topology on the space of all functions V-C. (b) Let ek E A(V), ek(Zj) = Ojk. Assume E exists, set gk := E(ek). Show there is a E C such that gk(a) =1= 0 for all k. (c) Let fk: ek!gk(a) E A(V). Show It - 0 in A(V). (d) Use the homogeneity of E to verify Efk(a) = I for all k. Conclude that E cannot be continuous. (aklk~I'
2.3. Ideal Theory in Ap In this section we follow [Sk2], [Gul], [KTl], [KT2] to obtain sufficient conditions which imply that a function belongs to a finitely generated ideal in a space Ap. We start by generalizing the Hilbert space ideas that led to Theorem 2.1.3. Let HI> H 2 , H3 be three Hilbert spaces. We denote by (·I·)j their scalar products and by II . IIj the corresponding norms, j = I, 2, 3.
2.3. Ideal Theory in Ap
137
Consider the following situation:
where (i) Tl is a continuous linear operator and we denote its adjoint Tt: H2 ~ HI; and (ii) T2 is a closed linear operator with dense domain, denoted dom(T2). In this situation, T2 also has an adjoint T2*: H3 with dense domain dom(T2*) and, moreover:
~ HI
which is a closed operator,
(a) (Tn· = T2 ; (b) the conditions x E dom(T2) and y = T2(x) are equivalent to (xIT2*(Z»1 = (ylzh for every z E dom(T2·); and (c) the condition x E kerT2 is equivalent to (xIT2*(Z»1 = 0 for every z E dom(T2*), i.e., ker T2 = (1m TnJ. (see [Hor]).
Now let G 2 be a closed subspace of H2 such that Tl (ker T2 ) the following criterion to decide whether Tl (ker T2) = G2: 2.3.1. Proposition. G 2 that
~ G2.
We have
= Tl (ker T2) if and only if there is a constant C > II Tt"(X2) + T2*(X3) II! ::::
0 such
CIIX2112
for every X2 E G2 and every X3 E dom(Tn. In this case, for every X2 E G2 there is Xl E ker T2 such that Tl (Xl) = X2 and
IIxIII! :::
(l/C)lIx2112 .
•Proof. Let G l = kerT2, it is a closed subspace of HI. Denote still by Tl the operator TtlG l : G l ~ H2. The following lemma shows that Tl(G l ) = G2 if and only if there is C > 0 such that:
2.3.2. Lemma. Let T: E 1 ~ E2 be a continuous linear operator between two Hilbert spaces. In order that T be surjective, it is necessary and sufficient that there is a constant C > 0 such that for every y E E 2 , lIyll2 ::: ClIT*(y)1I1. In this case,for y E E2 there is X EEl such that y T(x) and IIxlll ::: Cllyll2.
=
ProOf of Lemma 2.3.2. Assume T is surjective. Then T is an open map, hence there is a constant C > 0 such that if z E £2, IIzll2 :::: 1, there is X EEl, T(x) = z, IIxll ::: C. Given an arbitrary y E £2, Y ¥= 0, let z = y Illy liz. and let x E El as above, then
lIyl12 = I(zlyhl = I(T(x)lyhl = l(xIT*(Y»d :::: CilT*(y) 111.
2. Interpolation and the Algebras Ap
138
To prove the converse, assume the existence of a constant C > 0 such that for every y E E 2, we have lIyll2 ~ CIIT*(y)llI. Clearly T* is injective, hence for every y E E2 the following map is a well-defined linear map on Im(T*): u: T*(z) ~ (zlyh,
which verifies lu(T*(z»1
= I(zlyhl ~ Iizli211yll2 ~ CIIyll2I1T*(z)II2'
Since EI is a Hilbert space, with the help of the Hahn-Banach theorem, we conclude that there is an x E EI representing the linear form u and IIxlli ~ CIIy III, i.e., (T*(z)lx)1 = u(T*(z» =
(zlyh
for all z E E 2 . Therefore, T(x) = y. This concludes the proof of the lemma. 0 Let us now return to the proof of the proposition. G I is also a Hilbert space, hence G I is isomorphic (including the identity of norms) to its dual G;. On the other hand, Gil ;:; HI / Gt, also with equality of norms. Since G I = ker T2 and, by a previous observation (see [HorD, HI = ker T2 EB Im(Tn,
we have
Therefore, TI (G I) = G2 is equivalent to the existence of a constant C > 0 such that for every X2 E G2 and X3 E dom(Tn we have
II Tt(X2) + T2*(X3)i11
::: CII X2112.
In the case where TI(G I ) = G z, we have IITt(X2)III ::: Cllx2112 which, by Lemma 2.3.2, implies the existence of XI E G 1 with IIxdli ~ Clix2112 and T1(Xl)=X2. 0 The preceding considerations will be applied to the following types of spaces. Let Q be an open set in C, as usual dm denotes the Lebesgue measure on 0, and let rp: Q ~ lR be a continuous function (upper semi-continuous would be enough). Denote by L2(Q, rp) the space of functions in n which are square integrable with respect to the measure e-CP dm. Consider 8/oz as an unbounded operator T from L2(0, rp) into itself with domain dom(T) =
{f
E L 2 (n, rp):
:~
E L 2(n, rp) } ,
where af joz is the derivative in the sense of distributions. The following lemma is essentially included in the proof of Theorem 2.1.3:
2.3. Ideal Theory in Ap
139
2.3.3. Lemma. T is a closed operator with dense domain.
Proof. verify which hence
The subspace dom(T) contains V(Q), hence it is dense in L2(Q, cp). To that T is closed, let {(fn, T(fn»n~d be a sequence in the graph of T converges to (f, g) in the graph norm. In particular, In --+ I in V'(Q), afn/az --+ af/az in V'(Q). But afn/az = T(fn) and T(fn) --+ g in L2(Q, cp) ensures that g af/az in V'(Q). Hence I E dom(T) and T(f) = g.
=
o
n,
Now let CPI, CP2 be two continuous real valued functions in and let gl, ... , gn be holomorphic in a neighborhood of which we assume to be compact. Let
n,
H I =H3:=(L 2(Q,CPI»n,
H2:=L 2(Q,CP2),
and use 11·111 to denote both the norm in U(Q,CPI) as well as that in HI' Consider TI (hi, ... , h n ):= gjhj,
L
l~j9
which is clearly continuous HI -
H2 • Define an operator T2: HI -
H3 by:
(a) h = (hi, ... , h n ) E dom(T2) if and only if hj E dom(T) for every j; (b) T2(h) = (ahl/az, ... , ahn/az).
In this case, the space kerT2 is exactly (L2(Q, CPI) n Je"(Qnn. The operator TI sends ker T2 into a closed subspace G2 of L2(Q, cpz), G 2 := L2(Q, CP2)
n Jt"'(Q).
We are in a situation like that of Proposition 2.3.1, hence, in order that every holomorphic function I in L2(Q, cpz) could be written in the form
1=
L
hjgj ,
I~j~n
with h j E L2(Q, CPI) n Je"(Q), it is enough to prove the estimate in Proposition 2.3.1 . . . We will assume henceforth that Q is an open bounded set with Coo regular ~undary and that CPI, CPi are defined and of class C 2 in a neighborhood of Q. We let cP := CP2 - CPl. For any h = (hi, ... , h n ), denote, as usual, Ihl = (EI~i~n Ih j I2 )1/2. Moreover, we will assume that
Igl > 0
in a neighborhood of
n,
n.
i.e., the functions gj have no common zeros in Let us compute Tt(u) for u E L2(Q, CP2). Tt(u) is defined by the formula (TI(h)lu)Z (hITt(u»I' On the other hand, for h = (hi, ...• h n ) E H], we have
=
(TI(h)lu)Z =
1 (~ Q
I~J~n
hi gj
)
ue-'P2dm
=
1(L Q
I~J=,n
hj(gjUe-
2. Interpolation and the Algebras A p
140
We conclude that Tt(u) = (glue-
Let v = (Vt, ... , Vn) E dom(TD. Since T2*(v) inequality
= (T*(VI), ... , T*(v n»,
the
becomes in this case
II Tt(u) + T2*(v)lIr
=
L
IIgjue-
+ T*(vj)lIi
2: C2I1ull~,
I~j~n
for u E G2 = L2(Q, Cf.J2) n Jf(Q) and Vj E dom(T*), I :::: j :::: n. Rewriting the left-hand side of this inequality as IITt(u)
+ T2*(v)lIt = II Tt(u) IIi + 2 Re(Tt(u)IT2*(v»1 + IIT2*(v)lIt,
it shows we have to study three terms separately. A. Let us start by rewriting the first one
II Tt(u) IIi =
L 15j~n
1Igjue-
Q
B. Estimation oj2Re(Tt(u)IT2*(v»l. Since u is holomorphic we have T(gjue-
(:z
hence gjue-'P E dom(T). Therefore, 2Re(Tt(u)IT2*(v»1 = 2Re
L (gjue-'PIT*(vj»1 t~j~n
= 2Re
L
(T(gjue-'Plvj)h
15j5n
Recall that for any a, b 2: 0, a > 0, we have 1
2ab < _a 2 -a
Hence
+ ab 2 •
(gje-'P») ,
2.3. Ideal Theory in Ap
It follows that for any
141
> 0,
(X
2Re(Tt(u)IT2*(v)h :::
-.!. { IgI 2 Iule- 2'P-'P' dm (X
In
C. Evaluation of IIT2*(v)lIr. We are going to show that IIT2*(v)IIT :::
10 IvI2 :::~e-'P'
dm,
IvI2 =
EI:U:5n Ivje. For that purpose, we note that IIT2*(v)IIT = IIT*(vj)lIi, hence it is enough to obtain this estimate for the case of a single function (i.e., n 1). Let £(Q) be the family of all Coo functions on Q (i.e., in a neighborhood of Q). For f E £(Q) n dom(T*) we claim that
with
EI:5i:5n
=
T*(f) = -e'P'
~(fe-'P').
oz
In fact, we must have for every h
E 'D(Q).
Now, (T(h)If)1
=
1. --=oz oh
n
fe-'P' dm
and, from
we conclude that (T(h)lfh = -
= -
1. L n
0 h <>_(/e-V")dml (}z
+
1
0 -::::(h/e-'P')dm n oz
h [:/fe-'P')] dm,
since the other term vanishes, as one sees by applying the Stokes formula.
2. Interpolation and the Algebras A p
142
Rewriting the resulting identity we obtain
whence
as claimed. We shall admit a technical result whose proof can be found in an even more general setting in [Ho3, Lemma 4.1.3]. 2.3.4. Lemma. (i) The space £(0) is dense in dom(T), endowed with the graph norm
(lIf1it + lIoflozllr)1/2. (ii) The space £(!.1)
norm (II flit
n dom(T*) is dense in dom(T*), endowed with the graph
+ liT" flit)1/2.
Having accepted this lemma, let us give a characterization of the functions in n dom(T*).
£(!.1)
2.3.5. Proposition. A function
o on on.
I
E £(0)
belongs to dom(P)
if and only if I =
Proof. A function I E £(!.1) belongs to dom(T*) if and only if for every g E dom(T) one has (gIT*(f»)) = (T(g)lf)l. Since £(n) is dense in dom(T) (with the graph norm), it is necessary and sufficient to verify that for every g E £(!.1) one has
- (gle911 !...(fe-91I»)
az
I
= (o~ II) . oz 1
In other words, we need to verify the identity 0=
=
i
[g (e911 :Z (fe-91I»)
1. [oza - n
g-::(fe 911»
+ :~
1]
e-91, dm
1. 11 -
og - -] + -;::(fe '1'1) dm = oz
0 - /\ dz = -: 11. --;:(gle-91I)dz az
= -: 21
21
n
We conclude that f
E
an
dom(T*) if and only if
{ gte-911 dz = 0
Jan
0 --;:(gle-91I)dm n dZ
gle-911 dz.
2.3. Ideal Theory in Ap
143
for every g E t'(Q). Let s be the arc length parameter in an, the function dz/ds is Coo on an, hence we can find a function g E t'(Q) such that g = Jdz/ds on We conclude that J E dom(T*) if and only if J = 0 on 0
an.
an.
We can now show that the estimate we needed for II T2* (V ) II i holds.
2.3.6. Proposition. For every
J
IIT*(j)1I7
E
domCT*) one has the inequality
~ In IJI
2 e-\?1
::~~ dm.
Proof. From Lemma 2.3.4 we conclude that it is enough to prove the inequality for f E c(Q) n dom(T*). Let us recall that we have shown in the proof of Proposition 2.1.2 that for any u E C 2 one has
aq;1 12 + e\?1 lui 2 --_ a2q;1 e\?1 1-au +uaz az ilzilz = a_ [e\?1
az
If u
2 (u auilz +IUI2ilCP')] _~ (e\?lu -au) +e\?'l au az az az az I
= 0 on an, when we apply Stokes's formula we obtain { e\?1 1au 12 dm = { IUI 2 e\?1 a2q;~ dm + { e\?1 1au + acp, 12 dm if! az if! az az if! ilz az ~
When
J
E
t'(Q)
( lul2e
ifl
n dom(T*)
we
J =0
have
on
an
and
-e\?'(a/ilz)(e-\?' j), so we can apply the previous inequality to u = obtain
IIT*(f)lIi
= ~
i
2 le'l'l :z(e-
{ le-
if!
=
i
T*U) = -e\?1
f
and
2 e
= ( IJI 2e
as we wanted to prove.
Summarizing the estimates contained in A, B, and C, we see that for ex ~ I, u E G 2 = L 2 (n,cpz) n ~(n), and v = (Vi, •.. , Vn) E dom(Tt), we have the inequality
II Tt(u)
+ T2*(v)lIr ~
(1 - ~) LIgI IuI e2
2
2
dm
144
2. Interpolation and the Algebras A p
In what follows we shall take
cp := log(lgI 2 ) = log
(~ Igj 12) . l:OJ:on
2.3.7. Lemma. The following identities are correct: (i)
a2cp _ _ azai Igl 4
(ii)
"
~
Igk agj
l:ok<j:on
_
Bz
g. Bgk 12 J
Bz
Hence,
Therefore,
and the identity (i) follows. (ii) An elementary computation yields
We can now apply an identity due to Lagrange: 2
L l:oj:on
a/Jj
+
L
lakbj - aj b kl 2 =
which is valid for any choice aj, bj ~~
L
lajl2lbd,
l:Oj.k:On
l:ok<j:on E
C, I ::: j ::: n. This concludes the proof
0
2.3. Ideal Theory in Ap
145
2.3.8. Lemma.Letaj,bj,vj EIC,l:::j :::n,laI2=L:l~j~nlajI2,andsimilarly IvI2 = L:l!U~n IVjI2. Then 2
L Proof. Let S
ak(akbj - ajbk)vj
:::
lal 2 lvI2
L
lakbj - bj akl2.
= L:iik(akbj -ajbk)Vj. We note that
L
2S =
[iik(akbj - ajbk)Vj
+ iij(ajbk -
akbj)vk]
l~j.k~n
L
=
(akbj - ajbk)(iikvj - iijVk)
l~j.k~n
L
= 2
(akbj - ajbd(iikVj - iijVk).
l~k<j~n
From the Cauchy-Schwartz inequality we obtain
ISI 2
::: (
L
lakbj - ajb
L
d) (
l~k<j~n
liikVj - iijVd) .
l~k<j~n
On the other hand, the Lagrange identity used in Lemma 2.3.7 yields 2
L
liikV; - iij Vkl 2 =
lal 21vI 2 -
L
ajVj
:::
lal 21vI2,
which combined with the previous inequality, proves the lemma.
o
These two lemmas allow us to state that
Let us now take CPl :=
with
1/1 a subhannonic
C2
1/1 + acp,
function still to be chosen. Then we obtain 2
a2 cpl - - a Ivi --_ 2 2
az az
Igl
'"' a ~ e'P-(g·e-'P)v-
l~j~n
oz
J
J
Since IgI 2 e- 2'P-'P' = e'P-2I{J, = e-'P2, with cpz := cP fOllowing result:
> -
a2 cp Ivl 2 --_.
oz oz
+ CPl, we have obtained the
146
2. Interpolation and the Algebras A p
2.3.9. Proposition. Let Q be an open bounded subset in the complex plane with Ceo regular boundary. Let 1/1 be any C 2 function in the neighborhood ofQ. which is subharmonic in O. Assume that gj' 1 :::: j .:::: n, are holomorphic functions in a neighborhood ofQ. such that Igl 2 = ~1:5j:5n Igjl2 > 0 on Q. Fix any a 2: 1 and let CPI = 1/1 + a log(lgl2) and CP2 = 1/1 + (a + 1) 10g(lge). Then,for any u E G 2 and v E dom(Tz*) one has the inequality IIT(u)
+ Tz*IIt
2:
(1 -~) f luI 2e-
As a consequence of these rather technical results, we shall now be able to obtain two theorems guaranteeing that a function belongs to a certain ideal. Here is the first one.
2.3.10. Theorem. Let 0 be an open subset of C. 1/1 a subharmonic function 0, and a fixed constant a > 1. Let gl • ...• gn (resp. (gj )j~l) be a system of n holomorphic functions in 0 (resp. a sequence of holomorphic functions such that the series Igl 2 = ~j~l Igj 12 converges uniformly on every compact subset of 0). For every function J. holomorphic in Q, such that
1.
Ifl2 ->/I d -2--2 e m < 00, n Igl a+
(*)
there exist n holomorphicfunctions hi, ... , h n in 0 (resp. a sequence holomorphic functions) such that
J=
L
(hj)j~1
of
hjgj
l:5j:5n
=
(resp. f ~j:~1 hjgj , the series being convergent in ~(O)). Moreover, the following inequality holds:
f ~e->/I dm
in Igl 2a
< _a_ -
f
J.£t..e-Vt dm.
a-I in Ig1 2a+2
Proof. We consider first a simple case and remove the unnecessary restrictions one by one. Assume that 0 is a bounded open set with Coo regular boundary. the gj form a finite family of n functions holomorphic in a neighborhood of Q. with Igl > 0 on Q., and 1/1 is of class C 2 in a neighborhood of Q.. Since 82 1/1 /8z 8z 2: 0, Proposition 2.3.9 provides the inequality
II T(u) + T2*(V) liT 2:
(I - ~) In
luI 2 e-
We can appeal now to Proposition 2.3.1 to obtain the theorem in this case. Let us now eliminate the condition that n is finite. Let (gj)j~1 be a sequence of functions which are holomorphic in the same neighborhood of Q. and 0 < 2:">1 Igjl2 converges uniformly on Q.. Let f be a holomorphic function satisJ_ • fying condition (*) with respect to this sequence. T!!ere IS an integer no » 1 such that for n ::: no we have LISjsn Igj 12 > 0 on Q and condition (*) holds
2.3. Ideal Theory in Ap
147
when we replace the infinite sequence by gl, ... ,gn' Hence, there is a sequence (hj,n)j~1 such that hj,n = 0 for j > n:
1= L
(i)
gjhj,n;
I::=,j
(il)
1( n
L Ihj ,nl 2 's]
I::=,~n
)"'-- dm '" a
,1
~
IgJ12
If I
n
C::;~n Igj 12)
•• , ,--dm '" M
<
00
where the bound M is obtained from the Monotone Convergence Theorem. It follows that for n ~ no
In
L Ihj,n 12 I::;j
e-'"
Ig1 2
dm ::: M.
C1
Hence, if B(z, 2r) ~ Q we have, for any ( E B(z, r),
L
Ihj,n«()1 2 :::
j~1
]l'~2 ~8(z,2r) L
j~1
Ihj ,nl 2dm
(We have used that Lj~1 Ih j ,n«()1 2 is a subharmonic function, being a finite sum of subharmonic functions.) Therefore, we can find a sequence (nkh~ I, nk ~ 00, such that for every j, hj,n, converges to a function hj in Jf"(Q). Given K Cc Q and N an arbitrary integer,
~
15J5N
llhj,n,,2Ig,-2C1e-'" dm ::: cx:
I
K
and letting k ~
L
00
f
1<'
f 1/12 ( L Igjf) -Cl-I e-1/I dm in l::;j5n,
one obtains
Ih j I2 Igl- 2C1 e-'" dm ::: _cx_ a -I
Since K is arbitrary and we can let N ~
00,
f
in
IfI 2Igl- 2(a+l)e-1/I dm.
we finally conclude that
{ IhI 2Igl- Za e-'" dm :::: -CX-llf,2Ig,-2(a+l)e-'" dm, cx - 1
in
n
With Ihl 2 = L'>llhj Ih j l2 I2 • In the same way as above, one obtains that L'>l J_ J_ Converges uniformly over compact subsets of Q.
148
2. Interpolation and the Algebras A p
Let us now show that f = Lj;::1 gjhj . Since, for every n» I, f = LI:':j:,:n gjhj,n, it is enough to show that the sequence (Lj:~! gjhj,n,h;::1 converges locally unifonnly in n, We have already observed that L};::I Ihj,nk 12 is uniformly bounded over any compact subset of n and, moreover, Lj;::1 Igjl2 converges locally uniformly in n by hypothesis, hence
which converges uniformly to zero on every compact subset of n, Let us now assume that 1{1 is defined in a neighborhood of Q, but it is not necessarily of class C 2• One can find a sequence (1{In)n;::1 of subharrnonic functions, Coo in a neighborhood U of Q, such that 'ifrn ~ 1/In+l, 'ifrn(z) ~ 1{I(z) for any Z E U (see, e.g., [BG, §4.4,16]), For every n we can find a sequence of holomorphic functions hj,n such that: (i)
f =
L gjhJ,n; j;::1
(ii)
:::: _Ci_ Ci -1
r
In
IfI 2 Igl- 2a - 2e-'" dm.
As before, for every n the corresponding sequence (hj,n)j;::! is locally uniformly bounded in n, hence we can find nk < nk+! ~ 00 such that hj,nk --+ h j as k ~ 00 in ~(n). A reasoning completely analogous to those above allows us to conclude that the sequence (hj)j;::1 satisfies the two conditions required by the theorem. Let us now assume that n is an arbitrary open set. We claim we can find a sequence of bounded open sets (Qn)n;::1 such that: (a) (b)
Qn £ nn+l;
n = Un>1 nn; (c) for every n, ann is a regular boundary of class Coo.
The idea of the construction of the nn is as follows. First, we construct a function p: n ~ [0, +oo[ of class Coo such that p-I([O, r]) cc n for every r > 0. For this purpose, just take any compact exhaustion (Kj )j;::1 of nand choose rpj E D(n) such that rpj :::: I, rpj == 1, on a neighborhood of Kj, and supp(rpj) £ Kj+l. Let
°: :
p := rpl
+L
j (rpJ+1 - rpj-l).
j;::2
This function is Coo and proper. Moreover, by Sard's lemma the set of critical values of p is of zero measure. Therefore, there is an increasing sequence of
2.3. Ideal Theory in Ap
149
values rn -+ 00 such that On = p-I([O, rnD fonns an exhaustion of 0 with the desired properties. Now, by what we have already proved, for every n ~ 1 there exists a sequence (hj,n)j?!1 of holomorphic functions in nn such that
1= Egjhj,n
(i)
in On;
r~1
~
_0/_ 0/ - I
r 1/12Igl-2a-2e-Vr dm.
In
For m fixed and n > m, the sequence (h j •n ) is bounded in Om. It follows that we can find an increasing sequence nk -+ 00 such that for every j ~ I the sequence hj,nk has a limit hj in Jt"(0) when k -+ 00. It is easy to see now that the sequence (hj)j?!1 has the required properties. To conclude the proof we only have to eliminate the supplementary hypothesis that Ig I > 0 on O. Let 2 be the discrete set of common zeros in 0 of all the gj' and apply the theorem to 0' = 0\2. Since 1/1 and Igl are locally bounded above, the functions hj are locally square integrable in n, hence their possible singularities at the points of 2 are removable, as shown by the following elementary lemma:
2.3.11. Lemma. Let I be holomorphic in 0 < Izl < r, ~zI
1
r
I/(z)1 2 dm = 21l'1a_11 2 1og -
£:5121:5r
Letting
B -+
+ 1l'
2k + r + E lakI2-"""k--+ 2k 2 -
*7"-1
B
0 we reach a contradiction unless ak
Let us now present two simple corollaries of Theorem 2.3.10.
2.3.12. Corollary. II there exist constants C I, C2 such that ~
Igl
~
eCzVr ,
then,for every holomorphic function f in 0 such that
L
IfI 2 e- C3 tJ! dm
<
2
1
= 0 for every k < O.
This lemma concludes the proof of Theorem 2.3.10.
e-C,Vr
B
00,
0 0
ISO
2. Interpolation and the Algebras A I'
for some C3 such that C3 + C 1 (2a such that f = Lj~l gjh j and
+ 2)
~
0, there are holomorphicfunctions hi
rIhI 2e- c.", dm ::: _a_ inrIfI 2e- C,'" dm,
in where C4 = C3
a-I
+ C 1(2a + 2) + 2aC2.
Proof. Let C; := C3 + C 1(2a
+ 2)
~
0, hence the function C;1/I is subhannonic
and l'f,2Ig,-2u-2e-c;", dm ::: l'f,2e-C'''' dm. Therefore, there are (hj)j~l holomorphic in Q such that f
= Lj~l gjh j and
rIhI 2Igl- 2U e-c;", dm ::: a_a_ rIfI2e-CJ"'. - in
in Let C4 :=
I
q + 2aC2, then 1,h,2e-C'''' dm :::
In
IhI 2Igl- 2u e- C;", dm,
o
which concludes the proof of the corollary. 2.3.13. Corollary. If there are constants c and M such that
o<
c ::: Igl ::: M
< 00,
then,for every function f E L 2(Q) n Jf'(Q), there are holomorphic functions hj such that:
= Lj~l gjhj ; and In Ih 12 dm < 00.
(i) f (ii)
Proof. This is the previous corollary for the case 1/1
= constant.
o
2.3.14. Remark. Let us recall from 1.1 that the Corona Problem consists in finding bounded functions hj such that f = Lj~l gJh j , when f is bounded and 0 < C ::: Ig I ::: M < 00. The proof of the Corona Theorem for L 2 - functions which we have just given is much simpler than the Corona Theorem for bounded functions. As another application of Theorem 2.3.10 we shall see what are some of its consequences for the algebra Ap(Q), Q an open set in C. We assume the weight p satisfies the conditions (i)' and (ii)' of §2.1. That is, there are positive constants KJ, K2, K 3 , K4 such that zEQ,~EC, an
2.3. Ideal Theory in AI'
151
We remind the reader that C ::: 0 such that
I
E Ap(Q)
In
if and only if I
1/1 2e- cp dm
<
E J!'(Q)
and there is
00.
From the second assumption above, it follows that Ap(Q) contains the restrictions to Q of the polynomials. More generally, consider a family P of subharmonic functions such that if PI, P2 E P, then PI + P2 E P and max(PI, P2) E P. We associate to P the algebra Ap(Q) of those holomorphic functions in Q such that III ~ Ce P for some pEP (Le., Ap(Q) = UpEP Ap(Q». It follows that Ap(Q) contains the restrictions of the polynomials and that I E Ap(Q) if and only if there is PEP and C > 0 such that
In
1/1 2e- cp dm
< 00.
That is, one can define the growth conditions in A p (Q) either by pc estimates or by L 2 estimates. As an immediate corollary Theorem 2.3.10 we have the following theorem due to Horrnander [H03]:
2.3.15. Corollary. Let !?I, are equivalent:
... ,
gn E Ap(Q), then thelollowing three conditions
(I) /(gl, ... ,gn) = Ap(Q). (2) There is pEP and A > 0 such that
Igl ::: (3) There is
Ae- p •
a > I and PEP such that
1
-p
e2a+2 dm <
(2lgl
00.
2.3.16. Remarks. (1) Horrnander considered the case P =
{cp: c > OJ, P a fixed subharmonic function. (2) If there is a sequence (t:j)j'!l of holomorphic functions in Q such that
Ale-PI ~
Igl
~
A2eP'
for some constants A" A2 > 0, and PI, P2 E P, then for every IE Ap(Q), there is a sequence (hj)j~1 ~ J!'(Q) such that: (i) (ii)
I
=
Lj>1
e
Ih I ~ A 3
gjh j in Q; and for some P3 E P, A.~ > 0,
P'
which is Horrnander's result extended to the case of infinitely many generators.
Let us now give a situation where the need to consider an infinite sequence arises naturally. Let P be a subharmonic function in an open set Q. Assume that for every j E N*, the set OJ := [z E 0; p(z) < j} is relatively compact in O. We recall
2. Interpolation and the Algebras A p
152
that Ap(O) carnes an inductive limit topology, which is not metrizable. Nevertheless, the spaces Ap,B (0) are Banach spaces and thus, from Proposition 1.4.13, the closure of any subspace F of Ap(O) coincides with its sequential closure. Therefore, an ideal I in Ap(Q) is dense if and only if it is sequentially dense in Ap(O), i.e., there is a sequence (lkk?1 in I that converges to 1 uniformly over every compact subset of 0 and besides, there are two constants C), C2 :::: 0 such that (z E 0, k:::: 1). In this case, for every j :::: 1 there is kj such that ~
:s Ilk (z)1 2 :s ~,
for all
j
Z E
OJ.
Let
then
Igl'
~ ~ Igil':o (~e
-';) ci
exp(2C,p).
and, moreover, for every z E 0 there is a j such that j - 1 :s p(z) < j, hence -2
Te-ZP(Z)
:s ~e-2j :s e-2jlfk/z)12 =
Igj(z)1 2
:s Ig(z)1 2 •
This shows that the existence of four positive constants, A, B, A', and B' so that A'e-B'p
:s Igl :s Ae BP .
Conversely, Remark 2.3.16(2) shows that if we can find a sequence gj E I, and positive constants A, B, A', and B' such that the previous inequality holds, then I is sequentially dense. Therefore, we have proved the following proposition:
2.3.17. Proposition. Let p be a nonnegative subharmonic function in 0 such that for every c > 0 the set Oc = {z E 0: p(z) < c) is relatively compact in O. Assume p satisfies the other properties stated in the definition of the spaces Ap(O). An ideal I of the algebra Ap (0) is dense if and only if there is a sequence (gj)j?:.) in I and positive constants A, B, A', and B', such that A'e- B' p
:s
(L: Igj 12) 1/2 :s Ae BP .
As an example of application of this proposition, let p(z) = I Imzl
+ log(l + IzI 2 ),
the weight defining the algebra F(e'(JR». For nEZ, let
inez) := - 1
21r
1
2"
0
..
e- IIZ e-· nt dt
=1.- e-
21Ciz
21rI(z+n)
.
153
2.3. Ideal Theory in Ap
Since for each Z E C, the fn (z) are the Fourier coefficients of the function t ~ e- itz , we can apply Parseval's fonnula and obtain 1
"lfn(z)1 2 = L; 2rr nE ..
12Jr le-
itz l2dt
0
1 e 4Jr (lrnz)
= -4rr
1
- . Irnz
It is easy to verify the inequalities of Proposition 2.3.17 and conclude that the
ideal generated by the In, nEZ, is dense in F(£'(JR». Let us consider a finite family g" ... , gn E Ap(r.l). Denote h = I(g" ... , gn), the ideal generated by them in Ap(r.l) and let Z(h) be the set of their common zeros in r.l. A function f E ./Ti (the radical of I,) if there are k E W, h j E Ap(r.l), such that
fk =
L
gjhj .
We conclude that for some PEP and A > o. Conversely, suppose f E "e(r.l) is such that for some k E N*, A > 0, and PEP, one has Iflk ::s Algle P • In this situation we can apply Theorem 2.3.10 to the function f3k, choose 01 E ]1, 3[ and let 1{1 = Bp for B a sufficiently large integer, and obtain that f3k E I" i.e., f E ./Ti. We have therefore proved the following version of Hilbert's Nullstellensatz:
2.3.18. Corollary (Global Nullstellensatz). In order for a function f in the algebra Ap(r.l) to belong to the radical of the ideal h, generated by g" ... , gn, it is necessary and sufficient that there are k E N*, A > 0, and PEP, such that Iflk
::s Algle P
in r.l. Let now /z be the family of f such that
E
Ap(r.l) for which there are A > 0, PEP
If I ::s Algle P • It is easy to see that lz is an ideal in Ap(r.l) and that I, s; /z. If f E Ii, then there are A > 0, PEP, such that
If I ::s Algl 3e P , which implies, via Theorem 2.3.10, that f statement:
E
I,. Therefore, we have the following
2.3.19. Corollary. The ideals h, lz verify
Ii £; h In particular,
.Jli = .../12.
£;
/z.
2. Interpolation and the Algebras Ap
154
2.3.20. Proposition. If either II or 12 is a principal ideal, then II
= /z.
Proof. We need to show that in either case h 5; II. Let II (h) and let f E h Hence, there are A > 0, PEP, such that
=
If I ~ Alhle P •
=
It is immediate that fI h E )fen) and, if q; f / h, then q; E Ap(Q). Therefore, fEq;hE/I. If h is principal, h (h). Since h S;; h we have that for each generator gj of II there is h E Ap(n) such that
=
gj = ,{jh
(j = I, ... , n)
and, moreover, by the definition of h, there are A > 0, PEP, such that Ihl ~ Algle P • It follows that If I =
(2:I:::;j:::;n
IhI2)1/2 '#
If I =
°
in Q and that
!!l > ~e-p. Ihl - A
Therefore, i.e., there are h j
E
Ap(n) such that
1
=
L
hj,{j,
I:::;j:::;n
and hence
h
L
=
hj/jh
=
L
hjgj
E
h.
I:::;j:::;n
I:::;j:::;n
o
This concludes the proof of the proposition.
= 1. In that case we have
Let us return now to Proposition 2.3.9 and let ex
II Tt(u) +
T2*(v)II~ ~
(
021/1_lvI 2e-rpt dm,
in oz oz
for u E G 2 , V E dom(Tz*), q;1 = 1/1 + log(lgI 2), q;2 = q;1 + log(lgI 2) = 1/f + 2Iog(lgI 2). Let us replace 1/1 by l/f + 210g(l + IzI2). Then, for this new weight, the meaning of G2 changes but, since
02
oz oz (1/f + 210g(1 + Iz12» ~
2
(1
+ IzI2)2'
one obtains the inequality
I Tt(u) + T2*(V) liT ~ 21n (l1 i:1 v
2 )2 e -rpt
dm,
2.3. Ideal Theory in Ap
155
for u E G 2 , V E dom(TD, ({ll = 1/1 + log(lgI 2) + 2 log (I + IzI2), ({l2 = 'PI + log(lgI 2 ). We apply this observation to prove the following proposition:
2.3.21. Proposition. Let Q be a bounded open set with regular boundary of class Coo, let 1/1 be a subharmonic function of class C 2 in a neighborhood of and let gl, ... , gn be holomorphic functions in a neighborhood of such that Igl 2 = L: l:5j:5n Igj 12 > 0 on Let w = (WI, ... , wn ) be a system of measurable functions such that
n.
Let p := 1/1 + 210g(lgI 2 ) + 2 log (1 + IzI 2 ), then there are functions h) such that: (a)
n,
n
E
Lloc(Q)
ah·
a{ = Wj in the sense of distributions;
(b) In(Ll:5j:5nhjgj)iie-Pdm = Ofor everyholomorphicfunctionU E LZ(Q,p); Ihl 2 e->J! 1 Iw12_ (c) In jgj2 (1 + Iz12)2 dm ~ "2 In Igl2 e >J! dm.
Proof. The hypotheses show that for
'PI
:=
1/1 + log(lgI2) + 2 log (1 + Iz12)
k1W12(l + IzI 2le-'I'1 dm <
00.
We have also that l(wlv)d 2 =
1L 0.
~
2
w/Uje-'I'I dm
l:5j:5n
(inrIw12(1 + IZI2)2e-'I'1
dm) (
r
Ivl2 e-'I'l dm) . in (1 + IzIZ?
Therefore, if we let ({l2 = p, to conform with the notation of Proposition 2.3.9, we have that for u E G2 and v E dom(Tn l(wlv)11 2 S
t (fo. Iw12(1 + IZI2)2e-'I'1 dm) II TtCu) + T2*(V) IIi·
Let E be the linear subspace of (L 2 CQ, ({ll»n of elements of the form Tt(u) + T2*Cv), for some u E G 2 , V E dom(Tz*). Define a linear form (): E --+ C by the relation Then: (i) () is well defined because if Tt(u)
r
+ T2*(v) = 0 then
IvI2 -'1'1 d 0 inCI+lzI2)2e m=, hence v
= 0;
and
2. Interpolation and the Algebras A p
156
(ii) () is continuous, with nonn ~
-vff72(JnlwI2(l + IZI2)2e-q11 dm)I/2.
The Hahn - Banach theorem allows us to conclude that there is h
=
(hi, ... , h n) E (L2(Q,
(a' ) (wlv)1 = (hITt(u)
(b') IIhllr =
+ T2*(v»J,
whenever u E G2 and v E dom(TD; and
1Inlwl2(l + IZI2)2e-q11 dm.
The identity (a/) is equivalent to the pair of identities
In other words,
a: =
8h (a)
(b)
Wj,
1 ( 2: n
1 ~ j ~ n.
gjhj ) iie-'P2 dm = 0 (u E G 2)·
I:5:J:5:n
The inequality (b /) is simply (c)
Ih 12 e- 1fJ Ih 12 r Ih1 2e-'I'2 dm = r dm < 1 r - e - 1fJ dm In JnlgI2(l+lzI2)2 -2Jnlgl2 '
because of the choice of
o
2.3.22. Theorem. Let Q be an open subset of IC, and let 1/1 be a subharmonic function in Q. Let (g1. ... ,gn) (resp. (gj )j~ I) be a system of n holomorphic functions in Q (resp. a sequence of such functions). Let Z be the set of common zeros of the gj in Q. Then,for every function f E -'f(Q) such that
1
Ifl2
- 4 (1
n\z Igi
there are hi, ... , hn
E
f =
+ ~(Iog Igl)e- 1fJ dm
<
00,
-'f(Q) (resp. a sequence (hj)j~1 in -'f(Q» such that
L
gjhj
l~j~n
and
[lfl2 e- 1fJ [ Ifl2 Igl2 (1 + Iz12)2 dm ~ 2 in\Z Igl 4 (1 + ~(Iog Igl)e- 1fJ dm.
in
Proof. One starts by assuming that Q is a bounded open set with regular boundary of class Coo, 1/1 is C 2 and subhannonic in a neighborhood of Q, the n functions gj are holomorphic in a fixed neighborhood of and Igl > O. Afterward, one eliminates these restrictions one by one as done in the proof of Theorem 2.3.10. We shall just indicate how to prove the first step. One can write f = 2:1:5:j:5: n gjhj with
n,
hj:= f &2' Igi
2.3. Ideal Theory in Ap
157
Let Wj := ohj/oz. An easy computation shows that
Wi
= L 4 '" Igl
L.J
g. (g. Ogi _ gi Ogj) .
J:::,:::n
'oz
,
8z
Applying the Cauchy-Schwartz inequality one obtains
Iwl 2 =
L
Hence.
i.
Iwl2 -e-l/Idm< Igl 2
Q
-
2
L
-1/12 Igl8 J:::i:::n
Ogj) g. (Ogi g'--g; " 8z OZ J:;:j:;:n
i. Q
-1/12
Igl 4
L -Igl1
J:::i.j:;:n
4
1gOgi ogj '--gi - 12 e-l/Idm. 'OZ
OZ
On the other hand.
Therefore.
From Proposition 2.3.21 we see there are functions h'j. 1 :::: j :::: n. such that:
oh"
( 1) oz'
(2)
1( Q
(3)
= W·· ,.
L
gjh'j) ue-'1'2 dm
=0
(u E G 2 ); and
J:;:j:::n
Ih"1 2 e-l/I Iwl2 r dm < r -e-l/I dm. JQ Igl 2 (l + Iz12)2 - JQ Igl 2
Let hj := hj - h'j. Then (o/8z)hj = O. for j = 1•...• n. We also have that TJ(h") is square integrable in n, and condition (2) can be written as for Hence TJ (h)
=/ -
TJ (hI!)
u
E G2-
is also square integrable in (TJ (h) -
fluh = 0
n. TI (h) -
f
E G2.
and
158
2. Interpolation and the Algebras Ap
=
I,
for all u E G2. Since we can take u T\ (h) conclude the proof we only need to estimate h. We have e-Y, Ih' - h"1 2 e-l/I _Ihl2 dm -
1.
o Igl 2 (1
1.
+ Iz12)2
-
Igl 2
0
(1
lh l12
<2 1 - 0 Igl 2 (1 But
e-l/I
+ Iz12)2
Ih'1 = 111 /181 hence lhl2 e-l/I dm <2 11112 2
2
1
o 181 2 (1
+ Iz12)2
we see that T\ (h) =
dm+2
+ Iz12)2
-
:s 2
e-l/I
(1
1.o 1112 181
- 4 (1
+ Iz12)2
To
dm 2 11h"1 -0
2,
0 Igl 4
I.
dm+2
e-l/I
Igl 2 (l + Iz12)2
1lw 0
dm
.
l2
-e-"'dm Igl 2
+ ~(log 181)e-Y, dm.
As pointed out earlier, the remainder of the proof can be carried out as in the case of Theorem 2.3.10. 0 2.3.23. Corollary.
If g\. ... ,gn
r
}0\2
E Ap(n)
and there is apE P such that
~(log Igj)e- P dm
< 00,
then
EXERCISES
g)
2.3.
1. Let gl, ...• gn. 11 •...• In, and h be hololTlophic functions in a disk B satisfying :::: j :::: n) and for some A > 0,
= /jh (1
Ih(z)1 :::: A
(~n Igj(Z)1
2 ) 1/2
Show that the functions /j cannot have a common zero in B (see the proof of Proposition 2.3.20). 2. (The objective of this exercise is to prove a very simple case of a theorem of Bombieri [Ho3. Theorem 4.4.4].) Let p be a subharmonic function in an open set n. (a) Let U E Jf'(n) satisfy In lul 2 e- p dm < 00. Show that if ~o E n is such that e- P is not integrable on any neighborhood of ~o, then u(~o) = O. (b) Let 8(zo, r) ~ r > O. be such that e- P E LI (B(zo, r), dm). Show that there is U E Jf'cn), such that u(zo) = 1 and
n.
r lu(z)1 e2
p (Z)
Jn '--(-'-1-'-+'--1z-I-:--2)7"3 dm (z)
< 00.
(Hint: Try to find u in the fonn u(z) := I/J(z) - (z - zo)v(z).
where I/J E 1'(B(zo. r».1/J
==
1 in B(zo. r/2).)
159
2.3. Ideal Theory in Ap
Conclude that there is a nonzero U E Je(n) such that U satisfies the condition (*) and vanishes at every point/; En such that e- P ¢ LI(N{) for any neighborhood N{ of /;.
=
=
*3. Let 1/1 be a subhannonic function in B B(O, 1) such that 1/1(0) 0 and 1/I(z) < 1 for every z E B. Let J,L = /l1/l. (a) Show there is a Radon measure da on aB(O, 1), da(/;) =::; Id/;I such that 1/I(z)
= 2:rr1 18r log I1z _- z~/; IdJ,L(n +
1 a8
1 - Iz 12 Iz _ /; 12 da(/;).
(Hint: Use the Riesz decomposition theorem for 8(0, r) and let r -+ 1-. [BG. §4.4.24J.) (b) Show that
r log ~dJ,L(/;) + 1(Id/; I - da) = h. I/; I
18
a8
Conclude that
(c) Show that
da (/;) I =::; 6 1 .!....=J.:.f I~ 2:rr Iz - /;1
if
2
aB
Izl =::; ~.
(d) For R < e- 1/2 , let a = a(R) :=
r
~
1BW.R)
2:rr
dJ,L(/;).
Show that O:::;a=::;(-logR)-J <2. (e) Fix R E has
14,
e- J/2 [.
Show there is a constant C > 0 such that for Z E B(O, 1/2) one
11
I-
Iz-/;I log - - djJ.(r;) < C II-z/;I
R
and exp
(_~
flog 18W.R)
r ( Iz - /;J )
Iz - /;_1 dJ,L(/;») < II - z/;I -
JBW.R)
11- z/;l
-0
dJ,L(/;) < C. 2:rra-
(f) Conclude that
~
e-1/r(:)dm(z) < 00.
18(0.1/2)
(g) Use the previous parts of this exercise. and preceding Exercise 2.3.2, to prove that
if rp is a subhannonic function in a domain n. there is an open set G, G dense in n, and G 2 (z E n: rp(z) > -00) such that e-
r
JQ
U
E
Je(n) such that
lu(z)1 2 e-
(I
+ Iz12)3
dm(z) < 00.
'4. Let rp be a subharmonic function in C and H
for some N ~ O.
U
such
2. Interpolation and the Algebras Ap
160
(a) Let U E ;f(C) be such that Iufe-Ip E L!oc(C)' Fix R > 0 and let X on B(O, R + 2). Let v := XU, f := av/az. For t > 0 define !Pr(Z) := cp(z)
Show that
lim 1-+00
+ max
Jcr IfI
HOO
C
(I
(iii) lim
r_oo
Ur
+ Iz12)2 IUr
12 dm
==
I
(0, t log (RI~ 1) ) .
2 exp(-cpr)dm
(b) Show there exists U r E Coo such that 8u r/az (i) lim]; IUr 12 exp( -l{Jr) dm = 0; (ii) r~ kl
E 1)(C), X
=0.
=f
and satisfies
= 0; and
= 0 uniformly in
Izl < R.
(c) Show that
and
J
(l
Iv -
U r 12
+ IzI2)2+(r/2) e
-Ip d
m<
00.
(d) Conclude that the closure of Hlp in ;f(C) contains the set of all that lul 2e-1p E L!oc(C).
U E
;f(C) such
2.4. Dense Ideals in Ap(Q) In this section we will consider in more detail the question of when an ideal 1 of the algebra Ap(O) is sequentially dense (see [KTl], [KT2] , [Tal)). From Ap(Q), and clearly Proposition 1.4.13 we obtain that this is equivalent to j
=
the necessary and sufficient condition is that the function 1 must be a limit of a sequence in I. Another obvious necessary condition is that the elements of 1 do not have a common zero. This condition is not sufficient for arbitrary weights p, as will be shown in the exercises below. We now introduce a new concept, very useful in this kind of problem. 2.4.1. Definition. A subset S S; Ap (0) is called exhaustive if there is a sequence (Uj)j~I of subharmonic functions in Q, a sequence (Kj)j~l of compact subsets of Q, and two constants C" C2 ~ 0 such that: (1) Q
= Uj~l Kj;
(2) Uj(z) ~ 0 for every z E K j ; (3) Uj(z) :::: C 1 + C2P(Z) for every Z E Q; and
(4) for every j ~ I, there are constants subset Sj of S such that, for every Z Uj(z) :::: Cl,j
C1,j,
C2,j ~ 0
and a finite, nonempty
E Q,
+ C 2,jp(z) + ~ log
(L: feS,
If(Z)1 2)
•
2.4. Dense Ideals in Ap(Q)
161
2.4.2. Remarks. (I) It is clear that we can assume that the sequences (Uj)j:=l, (Kj)j:=l, and (Sj)j:=l are increasing. Under these conditions we shaH denote Sj = (Fl .... , FN), where (Nj)j:=] is an increasing sequence of integers. (2) A set S is exhaustive if and only if the ideal I (S) generated by S in Ap (Q) is exhaustive. Let us show, for instance, that if I (S) is exhaustive, then one can find subsets Sj ~ S, verifying (4) for the same sequence (Uj)j:=1 associated to I (S). In fact, there is a finite, nonempty set 6 j = {iI, ... , INj } ~ I (S) such that
Let us write lk = Li ak,iFk,i, for some finite collections Fk.i E S, Qk.i E Ap(Q), both collections depending on k Let Sj be the (finite) collection of all Fk.i E S appearing in one of the above representations of Jk E 6 j • Then
1~t=Nj 1!k(z)l2 ~ 1~t:Ni (~Iak.i (Z)1 2) (~IFk.i (Z)1 2) for some Aj , Bj > O. This gives a choice of Sj and constants C;.j' q.j for the set S. (3) If S is exhaustive, its elements cannot have a common zero Zo E Q. Otherwise, log
(2:
11(ZO)1 2)
=
-00
jES)
for every j, and this contradicts the fact that Uj ::: 0 on K j • (4) If each Uj is of the form log Igjl, gj E Jf(Q), then gj E Ap(Q) and, moreover, g] E I (S) by Theorem 2.3.10. (5) If S is an ideal without common zeros, the conditions to be exhaustive are easily seen to be equivalent to the following "pointwise conditions": There are constants C I , C 2 ::: 0 such that for every l; E Q there is a function Us which is subharmonic in Q, a compact subset KI; of Q, a nonempty finite subset SI; of S, and constants C I (l;), C2(l;) ::: 0 such that: (I) ul;(z) SCI + C 2 P(z) fOT Z E Q; (2) ul; (z) ::: 0 in a neighborhood of ~; and
(3)
fOT
every
Z E
Q\K, one has ul;(z)
~ CI(~) + C2(~)p(Z) +!
(E jESj
The main result of this section is the following:
11(Z)1 2)
2. Interpolation and the Algebras A"
162
if and only if it is dense.
2.4.3. Proposition. An ideal J in Ap(Q) is exhaustive
Proof. Let us assume first that J is dense. It follows that there is a sequence (hi)j~1 of elements in J which converges to 1 uniformly on compact subsets of Q and such that, for some constants C), C2 ~ 0, one has
We set
Uj
~
(j
Ihj(z)1 :5 CI exp(C2P(Z»
1, z
Q).
E
:= 1 + log Ih j I, and
K j :=
{z
E
Q: Uj(z)
~ O} n {z E Q:
Izl :5 j and d(z,
QC)
~
7}'
It is easy to verify that the conditions of Definition 2.4.1 are satisfied for this choice. Conversely, let us assume that I is exhaustive. Let (Uj)j~), (Kj)j~I' and (Sj )j~1 be the corresponding sequences assumed increasing, satisfying the conditions of Definition 2.4.1. Let us choose a sequence (l(Jj)j~1 5; 1J(Q) such that: (i) 0:5 I(Jj :5 1, j ~ 1; (ii) I(Jj(z) 1 for every z E Kj; (iii) there is a constant Co > 0 such that
=
(Z E Q, j
For j
~
~
1).
1, we can define new subharrnonic functions in Q by Vj(z) := 4
+ IZI2»)) ( Uj(z) + log ( 0d(z, QC)
.
These auxiliary functions have the following property: 2.4.4. Lemma.
lim (lim sup In{ 1alP}az 12 e-
J-+OO
v,
dm) = O.
k-+oo
Proof of Lemma 2.4.4. Choose k(j) ~ 1 such that Uk ~ 0 on sUPP(l(Jj) for all k :::: k(j). Let K; = supp(l(Jj) n (Q\Kj ), then for k :::: k(j) we have
1 n
1
aCP'I2 - ' e- V ' dm <
ai
-
c02
1 ,
K'
e- 4Uk (z)
d(z QC)4 ' Iz1 2)4
d(z, QC)2 (I
+
dm
.
Since Mj = sUPzeO\K, d(z, QC)2/[(1 + IzI2)2]:5 l/j2 and Uk:::: Oon sUPPCPj, we have { 1acpj 12 -Vk < C6 { dm Jo az e dm - j2 In (1 + IzI2)2' This inequality proves the lemma.
o
2.4. Dense Ideals in Ap(S"l)
163
Let Cj := If/. lacpj /aiI 2e- v'(I) dm, with the strictly increasing sequence k(j) chosen as in the proof of the preceding lemma, so that limj .... oo Cj 0 and Vk(j) ::: 0 on the support of CPj. From Theorem 2.4.3 it follows that there are functions l/Ij such that 8l/1j/oi 8cpj/ai and
=
=
r
if/. Il/Ijl
e- VkU )
2
(1
+ Iz12)2 dm
I
~ 'iCj.
The functions hj := CPj - l/Ij are clearly holomorphic in Q. We claim that all h j E I and that the sequence (hj)j~1 converges to I in Ap(Q). Let us prove first
the last item. We need to show first that there exist constants C > 0, B > 0, such that, for every k ::: 1 and every Z E Q, e-V'CJ)
e-Cp(z) < B -----=-"""" -
In fact, letting
(1
+ IzI2)2·
fJ = log B, this inequality is equivalent to -Cp(z) ~ fJ - Vk(Z) - 2 log (1 + Iz12),
i.e., Vk
But, Vk = 4(Uk + log(l B 1 , A2, B2 such that
~
fJ + Cp(z)
- 2log(1
+ IzI 2 )/d(z, QC)) log(1
+ Iz12)
and there are positive constants A"
~ Al
1
d(z, QC) ~
+ IzI 2 ).
+ BIP(z),
A2 exp(B2P(Z)),
by the defining properties of the weight p. Since Uk satisfies the estimate of Definition 2.4.1, it is then clear that for C » 1 and fJ > 0 convenient, the desired inequality holds. Hence,
!oll/ljl2 e-CP dm
~ B !oll/ljl2 (1 :~:jl)2)2 dm ~ BCj,
and, as a consequence
On the other hand,
r11 - cpjl2e-
if/. and, since if C
»
1 we have e- cp
CP
dm
~
r
if/.\Kj
ELI (Q),
e- cp dm,
then
.lim [11 - cp_1 2 e- cp dm = O. )-+00
Therefore,
Jf/.
J
2. Interpolation and the Algebras A p
164
It follows that hj ~ 1 in Jt"(Q), by a mean-value argument that we used several times. Moreover, from the fact that In 11/1jle- cp dm ::: M < 00 for all j ::: 1, and the estimates of d(z, QC) in terms of p, we conclude that there exist constants A 3 , B3 > 0 such that Ih j I ::: A e e 113P •
This shows that hj ~ I in Ap(Q). To complete the proof of the Proposition 2.4.3 it suffices to show that hj E I. Choosing 01 = 2 in Theorem 2.3.10 allows us to prove that if we can show there are constants Cj > 0 such that
Jr.f n
Ihl 2
3
J
(I~Nj IFk1
2
e- CjP dm <
00,
)
then there are functions hj,I,"" hj,Nj E Ap(Q) such that hj =
L
hj.kFb
l:".k:".Nj
and, moreover,
1 n
L
Ihj.k 12
l:".k5 N)
2
e-C;P dm <
C~N) IFkI2)
-
21 n
2
Ihjl
3
e-Cjp dm.
C5~Nj IFkI 2)
The existence of the constants Cj > 0 follows from property (4) of Definition 2.4.1 and a reasoning entirely similar to those leading to the estimates for 1/Ij. 0
2.4.5. Corollary. Let 1 be an ideal in Ap(Q), which has no common zeros, and for which there exist finitely many F I , •.• , FN Eland E > 0, C > 0, such that every connected component of the set
is relatively compact in
Q.
Then 1 is dense in Ap(Q).
Proof. Let (Dj)j?,1 be the connected components of the open set D, then each
Vj
is a compact subset of Q. Consider the subharmonic function in Q, u(z):=
4log (
L
IFj (Z)1 2) -loge + Cp(z).
i:".j:".N We have D
= {z E Q: u(z) < OJ.
2.4. Dense Ideals in
165
Ap(n)
Define
u.(z).- {U(Z) J .max(O, u(z»
if z ¢ K j , if Z E Kj,
and Sj being equal to the set {FI, ... , F N) to which we have adjoined a finite subset 6 j of J with no common zeros in Kj • It is easy to verify that (Uj)j~), (Kj)j~i, (Sj )j~J, satisfy the condition of Definition 2.4.1. Hence I is exhaustive and the corollary has been proved. 0
2.4.6. Remark. The proof of Lemma 2.4.4 also shows that if satisfy the conditions (I), (2), and (4) of Definition 2.4.1, and if u(z) :=
UI ::: U2 ::: •••
lim Uj(z), J-+OO
then j contains h 3 for every h
E Ap(Q)
such that
Ih(z)1 ::: C) exp(C2 P(z) EXERCISES
+ u(z».
2.4.
I. Prove Remark 2.4.5. 2. Let n ::::: {z E C: Re z > OJ, and recall that C({l) denotes the space of continuous functions in (l. Let Ao(n) :::::: {f E Jf"(n):
lim I(z)::::: OJ
l'l~oo
n C(Q).
tEll
(a) For
1E
Ao(n) define
11/11 = sup I/(z)l. tEO
Show that Ao(n) is a Banach algebra under multiplication, without a unit. (b) Show that the function (z) := e- Z does not belong to Ao(n), but that (·Ao(n) is a proper ideal of Ao(n). (c) Prove that L I : I ~ (·1 is an isometry of Ao(n) into itself. Conclude that ( . Ao(n) is a closed ideal of Ao(n). (d) Let 'P E Ao(Q)\{O} and let I be the ideal (·'PAo(n). Show that the hypothesis 'P E j implies the existence of a sequence (f.)n~1 in Ao(n) such that ('Pln).~1 converges in Ao(n) to an element !PI E Ao(n) verifying 'P ::::: (·'PI. Show that it follows that 'P ::::: lim.-+ oo e2 q;.ln in Ao(n), and infer that ('Pd.)."?1 has a limit ({Jz in Ao(Q), and that rp = (2({J2. Derive by induction the existence of a sequence ('Pkh"?1 <; Ao(n) such that rp t k ·rpk for all k 2: 1. Conclude that, in fact, 'P ¢ i. (e) Let rp be as in part (d), and let J = rpAo(n). Show that t·rp is the limit in Ao(n) of the sequence (e.'P)n~I' en(z) := (n/(z + n»n. Infer that j = LI (J) <; J.
=
3. Let pCz) := log(l + Iz12) and n = (z: Rez > OJ as in the previous exercise. Also let e.(z) = (n/(z +n»n,n 2: 1. (a) Show that en E Ap(n) and lim._ oo en 1 in Ap(Q).
=
2. Interpolation and the Algebras Ar
166
=
(b) Find explicitly the function lfJo E Ap(Q) such that lfJo lim n _ oc 0:. (c) Prove that the ideal I = lfJoAp(Q) is different from Ap(Q). (d) Is lfJo invertible in Ap(Q)?
2.5. Local Ideals and Conductor Ideals in Ap Let us recall that in Definition 2.2.7 we defined the local ideal associated to a finitely generated ideal I of ApCQ). In fact, that definition makes sense for any ideal I. 2.S.1. Definition. The local ideal I loc associated to an ideal I in ApCQ) is the collection of all I E ApCQ) such that for every Z E Q there is a neighborhood U of z and there exist functions 11, ... , In E I, gl, ... ,gn E JlfCU), verifying
1=
L
/jgj
in U.
l~j~n
=
It is clear that if V CZk, mkh~l denotes the multiplicity variety of the ideal I, and ICV) (resp. J(V» is the ideal in Ap(Q) Crespo in JlfCQ», corresponding to V, i.e., I (V) = (f E ApCQ):
I
vanishes at every Zk with multiplicity ::: mk},
then I loc = I CV). Moreover, if i: ApCQ) -+ JlfCQ) is the canonical injection, J is the closed ideal generated by I in JlfCQ) (i.e., by i(l», then J = ICV) and Iloc = i-I (I). Hence, I loc is closed in Ap(Q). Furthermore, the identities Iloc
= (lloc)loc = (i)loc,
are immediate consequences that all these ideals have the same multiplicity variety. In particular, if V = 0, then lIoc = ApCQ). Clearly if I = to} then I loc = {OJ. We now introduce another ideal associated to I. 2.5.2. Definition. Given an ideal I in Ap(Q) the conductor ideal (or quotient ideal) of I is the ideal 1* defined by I* := (f E ApCQ): Ig E I for every g E floc}. It is clear that [
~
1*.
2.5.3. Proposition. if [ is a closed ideal in Ap(Q), then 1* is also a closed ideal. Proof. Let (fn)n~l be a sequence in 1* converging to I in Ap(Q). For any g E floe, the sequence (fng)n~l lies in I and converges to Ig. Since I is closed,
IgeI.Hence/EI*.
0
2.5. Local Ideals and Conductor Ideals in Ap
167
2.5.4. Proposition. We have the inclusion
F r;;; (i)*. Proof. We already know that (1)* is closed and that floc = (1)loc' Hence for any f E 1* and g E (i)loc, then f gEl r;;; 1. Therefore, f E (i)*. It follows that
F r;;;
0
(i)*.
The following simple proposition is crucial to the study of localizability of ideals (see Definition 2.5.7). The idea to use it in this context is apparently due to Ehrenpreis [Eh 1). 2.5.5. Proposition. Let [ be an ideal, I
= IIeJC if and only if 1* = Ap(Q).
Proof. If [ = I loc , then for every f E Ap(Q) and g E I loc , we have fg E I, hence f E 1*. Conversely, if 1* = Ap(Q), then 1 E 1*, hence for every g E I loc we have g = 1· gEl. 0
2.5.6. Proposition. If /* is dense in Ap(Q), then Proof. Namely, by Proposition 2.5.4, ifF
i =
i
=
Iloc-
= Ap(Q), then (1)* = Ap(Q). Hence,
(i)loc = Il oc •
0
2.5.7. Definition. We say that an ideal/of Ap(Q) localizes if
i
= I loc •
This concept will play an important role in the study of convolution equations later on. We see that for I to localize it suffices to prove that I * is dense in Ap(Q), but this requires at least that /* has no common zeros. In fact, we can show that this is always the case. 2.5.8. Proposition. For every ideal I of Ap(Q), the ideal 1* has no common zeros.
# {OJ (Why?). Let us argue by contradiction and assume there is Zo E Z(I*). We claim that if g E 1*, then the function G(z) := g(z)/(z - zo) belongs to /*. To continue the proof we need the following very simple lemma:
Proof. We can assume that I
2.5.9. Lemma. /f f
E Ap(Q)
and f(zo) = 0, then F(z) := f(z)/(z - zo) is in
Ap(Q).
Proof of Lemma 2.5.9. Let fJ(zo, r) C; Q, then for z E B(zo, r) we have IF(z)1 ~
max
I/;-zol=r
IF(~)I
1
= -r
max
II;-zol=r
If(~)I,
2. Interpolation and the Algebras A"
168
and for z
E
0\8(zo, r) we have clearly
IF(z)1
~
I
-If(z)l· r
The defining properties of the weight p allow us to conclude that F
E Ap(O).
o We return to the proof of the proposition. Since V(l*) S; V(l), the multiplicity m of I at zo verifies that m :::: 1. Let h E I be such that the multiplicity of h at Zo is exactly m. Then we can write
h(z) = (z - zo)m H(z), with H(zo)
'# 0,
HE Ap(O). Similarly, any f E Iloc can be written as fez) = (z - zo)m F(z),
F E Ap(Q). It follows that the identity
f(z)H(z)
holds. We want to show that Gf we can write
E [,
= h(z)F(z)
for any f
E
G(z)f(z) = g(z)f(z) [H(Zo) - H(Z)]
H (zo)(z
-
zo)
= g(z)f(z) [H(Zo) - H(Z)] H (Zo)(z - zo)
Il oc • Using the previous notation
+(
g(z) ) f(z)H(z) z - Zo
+ h(z)
H (zo)
[G(Z)F(Z)] . H (zo)
Since gf E I and hE I, we obtain Gf E I. Therefore, G E 1*. This shows that the multiplicity variety V (1*) must be 0 empty. In other words, 1* is an ideal without zeros. This proposition allows us to extend Corollary 2.4.5 to an arbitrary ideal in Ap(Q).
2.5.10. Corollary. Let [ be an ideal in Ap(Q) for which there are functions F I , ••• , FN E I and constants s > 0, C > 0, such that every connected componentofthe set D = {z E Q: (Lt:Sj:SN IFj(z)I)I/2 < sexp(-Cp(z))} is relatively compact in Q. Then i = floc' Proof. The preceding Proposition 2.5.8 shows that /* has no common zeros and, since F t , ••• , FN E 1*, we can apply Corollary 2.4.5. We conclude that 1* is sequentially dense in Ap(Q). Hence j = [Ioc by Proposition 2.5.6. 0
In general, it is not true that j = Iloc. We show this with an example due to Gurevich ([Gu2]. see also [KT2] and Exercise 2.4.2).
2.5. Local Ideals and Conductor Ideals in Ap
2.5.11. Example. Let let
n=
169
-n/2 <
{z E C: Rez > OJ. For z = re i9 ,
p(z) := exp( -ar" cos 0
e < n/2,
+ r P) + r Y ,
with 0 < (3 < 01, 0 ~ Y < 1, a > O. Since fJ < 01, the quantity -ar" cos e + rfJ -+ -00 whenever r -+ 00 for 0 fixed in 1- n /2, n /2[. Therefore, every function f E Ap(n) is of order y along every ray starting at the origin, but the type of this function could increase to 00 as this ray approaches the vertical axis. We shall show that the principal ideal I = e- z Ap (n) is not localizable in Apcn). It is clear that I ¥- Apcn) since eZ ¢ Ap(n). Since I has no common zeros, Iloe = Ap(n). We want to show that j =1= Ap(Q). Assume that the opposite is true. We would then have a sequence (gnk::1 in Ap(n) such that: (i) gn (z )e- Z -+ 1, unifonnly on every compact subset of n; and (ii) Ign(z)e-ZI ~ C l eC2P (z) for all n ~ 1 and some C I , C2 > O.
Consider the open set
nl
:= {z
E
n:
Re z > rY}. On the set Q I we have
-ar" cosO + r P < -ar y +a -
I
+ r P.
z
E
Assume that (3
then there is a constant Co > 0 such that p(z) ~ Co
+ rY
for all
nl.
Therefore, we have Ign(z)e-ZI ~ C I exp(COC2 + C 2 r Y )
and for
z
E
ani,
since Rez = r Y on ani' In order to apply the Phragmen-LindelOf principle consider the auxiliary function fez) := exp( -c~zY),
c' _
C2 + 1
2 - cos(ny /2)
We have Ifgnl ~
c;
and the order of fgn is at most y < 1 in asserts that Hence, for
Z E
on
an)
n l . The Phragmen-LindelOf principle
nJ, we have Ign(z)e-ZI ~ C;
e- Rez
If(z)1 ::: C) exp(C~zY - Re z).
2. Interpolation and the Algebras A p
170
In particular, for z real, gn(z)e- Z ~ 0 as z dicts the condition (i) mentioned above.
~ 00,
uniformly in n, which contra-
The preceding example raises the natural question of whether there are weights p such that every ideal in Ap(C) localizes. In the next section we prove this for p(z) = IzIP,O < P < 00. The very important case p(z) = IImzl + log (1 + Iz12) is left to Chapter 6.
2.6. The Algebra Ap of Entire Functions of Order at Most p We denote here Ap the algebra Ap(C), p(z) = IzI P, 0 < P < 00. Note that all the functions of this algebra are either of order < p or order p and finite type. (In the literature sometimes a function of order p and infinite type is also said to be of order at most p.) For this algebra we want to give more precision to the results of this chapter about interpolation and ideals. Many reasonings extend to the case where p is a radial weight satisfying the doubling condition p(2z) = O(p(z)). We start with a result of Lindelof which we take from [Lin]. and we also suggest [Lev]. [Bo] for a complete study of entire functions and their zeros. Part of this section is based on the doctoral thesis of Squires [Sql], [Sq2]. Let f be a nonconstant entire function and recall that n (r) denotes the number of zeros of f in B(O, r), counted with multiplicities. We assume, for simplicity, that f (0) = I. Let then denote, as usual, M(r)
= max I/(z)1 Izl='
N(r) =
and
l'
net)
-dt. o t
Hence Jensen's formula (see [BO, Chapter 4]) is simply N(r)
= -211'1 12Jr log I/(re i8 )1 dO. 0
2.6.1. Lemma. If I has finite order p > 0 and finite type., then . nCr) hmsup - -
'_00
rP
~
ep •.
In particular, nCr) = O(r P ). Proof. In fact, for e > 0 given there is r(e) > 0 such that
logM(r)
~
~r(8),
(.+e)r P ,
r
.+8,
r ~ r(e).
hence N(r)
-- <
rP
-
171
2.6. The Algebra Ap of Entire Functions of Order at Most p
Now let
~
> I and r ~ reel, then nCr) being increasing implies that
n(r)log~:5
1
ft, net) dt:5 ,t
loP' -n(t)d t = t
0
N(~r)
:5
~P(7:
+e)r P.
Therefore
nCr) :5
(7:
rP
+e) inf ( P>I
~PR.) = ep(7: +e),
r
log,...
~
r(e).
o
Since e > 0 was arbitrary, the lemma has been proved.
2.6.2. Lemma. If p > 0, but is not an integer, an entire function f of order at most p (i.e., f E Ap) is of type zero f E Ap if and only if nCr) = O(r P).
if and only if nCr) = o(r P ).
Note that if the order of f is strictly less than p, then of type 7: = 0 with respect to p.
In any case,
f is considered to
be
Proof. We write the Hadamard factorization of f in the form fez) = zmeQ(z) P(z),
Q a polynomial of degree deg Q = q < p, and p the canonical product of genus p, p < p < p + l. Since log leQ(z)IO(r q ) = o(r P) with Izl = r, we only need concern ourselves with the canonical product P(z) = TIn>' E(z/zn, p), where E(z, p) = (l - z) exp(z + (z2/2) + ... + (zP / p» is the Weierstrass primary factor. We know from [Lev, p.12], [BG, §2.6.1O] that for some constant K = K(p) one has log IP(z)1 = '" ~ log IE
(z )I zn' p
r
:5 Kr P { io tn(t) P+ 1 dt
+r
1
00
T
tnet) p+2 dt } .
n;::1
Let nCr) .:5 ArP, nCr)
== 0 for 0:5 r
log IP(z)1 .:5 KArP Since p < p < p
{1:
< ro, then t P- P- I dt
+r
1
00
t p- p - 2 dt }.
+ 1, the second integral is convergent and
p p logIP(z)l:5 KArP {r p-p
+
r P- P }.:5 K,Ar P, p+l-p
so that the type of P is at most K I A. If nCr) = o(r P), for e > 0 given, let rl be such that nCr) .::: er P whenever r ~ rl. Then 10gIP(z)1< Kr P -
--dt + Ker P {IT tp-p-Idt+r L ~n(t) t + T,
:5 KorP
0
P 1
+ Klu P
so that P is of type zero.
.:5 K 2 u
1"" ,
t p- p - 2 dt
}
P,
o
172
2. Interpolation and the Algebras A p
If the order p of the function f is an integer, one cannot expect such a simple result. The reason is that either the term e Q can dominate or the infinite product P can have infinite order while nCr) O(r P ). These possibilities occur in the O(r) but the function has examples e: and II r(z). In the latter case, nCr) infinite order. There is a classical result of Lindelof that deals with this case.
=
=
2.6.3. Theorem. Let p be a positive integer and let f be an entire function of order ::5 p, f(O) = I, and let (Zn)n~1 be the sequence of its zeros repeated according to multiplicity and with increasing absolute values. Denote by p the genus of the canonical product P and by ao the coefficient of zP in the polynomial Q, where fez) e QC :) P(z). Then:
=
(I) the function f is offinite type if and only if (i) nCr) O(r P ); and (ii) S(r) L:lznl~,(l/z~) is a bounded function.
= =
(2) The function f is of zero type if and only if either (i) p = p, nCr) = o(r P ), and L:n>I(l/z~) = -aop; or (ii) p = p - I and ao = O. -
Note that in case 2(i) we are also assuming that the series is convergent. For the proof we need some preliminary results.
2.6.4. Lemma. Let f be a nonconstant entire function such that f(O) = 1. Let a > 0 and let 27rrf(a, r) denote the linear measure of the set A(a, r) := {z then for r
»
E
aB(O, r): If(z)1 :::: M(r)-I1),
1 one has 1
f(a, r) < - - . -I+a
Proof. Let a(a, r) := (O and _I
r
E
[0, 27r[: re i8 E A(a, r»), then its measure is lea, r)
10glf(reili)ldO:::: -al(a,r)logM(r),
271" JaCI1,r) while on the rest of the circle aB(O, r), one has loglf(reili)1 :::: logM(r).
Hence, _I 27r
r
log If(re i8 )1 dO ::5 (I - lea, r» log M(r).
JaBC(7,r)\aCl1,r)
Therefore, 0= log If(O)1 ::::
2~
fo
21f
log If(reili)1 dO :::: (I - (a
+ l)l(a, r» log M(r).
As a consequence, as soon as log M(r) > 0, we have 1 - (a
+ I)l(a, r)
~
O.
[
173
2.6. The Algebra Ap of Entire Functions of Order at Most p
2.6.5. Lemma. Let P E N*, converging to product
00,
(Znk~1
IZnl S IZn+tI. If
IT E ( : ,p -
=
P(Z)
be a sequence of nonzero complex numbers Ln~I(l/lznIP) < 00, then the canonical
n~1
I)
n
has order at most p and type zero.
Proof. The convergence of the series is equivalent to
1
00
o
Therefore, n(r) -
1
00
n(t) --I
t P+
dt <
00.
n(t) - d t=o (I) t P+ 1
r
r -+
as
00.
We also have the inequality (see [BO, §2.6.1O], [Lev, p. 12]) 10gIP(z)1 = LlOgjE ( : ,P-I)j n~l
n
< Krp-I
-
f' n(t) dt + KrP Jo t P
1
00
r
n(t) dt. t P+ 1
Let C > 0 be such that n(t) S CtP, t > O. Oiven E > 0, let R > 0 be such that n(t)/t P S E, then
1R
1 r
-n(t) dt S C o tP 0
dt
1
+E
r
dt = (C - E)R
+ er.
R
Hence, log IP(z)1
s
(C - E)K Rr p- I + Ker P + o(1)r P = o(r P),
o
as we wanted to prove.
2.6.6. Lemma. Let n (t) be the counting function of entire function f of integral order p, f(O)
= I, then for Izi = r we have
L
log jE ( : ,p - I) j +
I::;n::;n(r)
n
< C (r P- 1 -
P
L
log
n>n(r)
r
Jo
n(t) dt tP
+ rP+11OO r
IE (~ , p ) I n(t) dt) . tP+2
Proof. For fixed r > 0 consider two auxiliary sequences: the first, (un)n, is finite, Un = Zn for I S n S n(r); the second, (vn)n, is the cofinal part of (zn)n, for n > n(r). Let J-Lr' Vr be their respective counting functions. Then J-Lr(t)
={
n(t) n(r)
if 0 :::: t :::: r, if t > r.
174
2. Interpolation and the Algebras A p
o
IJ,(t)
= { n(t) -
if 0 < t < r, if t ;: r.-
nCr)
Then the first tenn we want to estimate corresponds to the canonical product of the sequence (un)n, the second to the sequence (vn)n. Let p = 1, then from [BG, §2.6.lO] (or [Lev, p. 12]) we have
'"'
I!':;;-(,)
I (z 0) I ~ ior -JL,(t) 1 JL,(t) t - dt + r , ~dt 00
log E
zn'
=
'"' L log IE
(z zn' I )
l' {l o
I~ 4
net) -dt+n(r), t
OO
r
v,(t) dt + 2r2 f2
r
n>n~)
1T 00
v,(t) dt }
,
= 4 { 2r21OO n;t) dt - nCr) } . So, certainly the sum can be estimated by 8
{1'
n;t) dt
+ r2100 n~t) dt} .
In case p > I, we have similarly
~ 2P { (p +2
P
~C
P
+1
l)r P - 1
l' nt~) l
{(p + l)r + p
{r P- 1
dt
OO
1
r
+ nCr) }
net) dt - ncr)}
t P+2
dt + rP+1lOO net) dt}. ior net) tP , tP+2
o
Proof of Theorem 2.6.3. Let us first prove that the conditions (i) and (ii) in part (1) are enough to conclude that f is of finite type in the case when p p. We denote P + alz p - I + ... + p =
=
Q(z) aoz
a
and note that E(u, p) = eUP / p E(u, p - 1).
Consider the quantity D := log
If(z)1 - Re [zp
{ao + ~p E z~}] I:;:n:;:n(r)
2.6. The Algebra Ap of Entire Functions of Order at Most p
=Re{alz P -
1
L
+ ... + a p + m logz} +
175
log
IE (:n' p -
I) I
l:5n:5n(r)
The first tenn in D is bounded by 0 (r P - I ) when p > I, and by 0 (log r) when p = 1, we shall write this as O(r P - I ). Using now the preceding Lemma 2.6.6, we have D
= O(r P - 1) + C -
(r P-
1
dt + rP+lj"" net) dt) . ior net) t t + P
P
r
P 2
We have net) ~ At P for some A > O. It is clear that the two integrals together are dominated by 2C p Ar P, so that D = O(r P ). Since S(r) is bounded, it follows that: log Ifez)1
= D + Re (zp
{ao+ ~s(r)}) =
O(r P ).
In other words, f has finite type dominated by 2Cp A + S, where S is an upper bound for lao + (\/ p)S(r)l. A similar argument can now be made for part (2), also when p = p. For £ > 0 given, we can find ro such that n (r) = uP for all r ~ roo Then the integral tenns in the earlier estimate of D can be bound by
C r PP
1 ro
1
0
net)
-
tP
dt
+ 2CP up.
Since the series L:n~1 z;P is convergent to -aop, for r ~ rl we have ao
1
+ - '""' ~ p
z-P < £' n -
l:5n:5n(r)
whence, 10glf(z)1 = O(r P -
I)
+ (2C p + i)uP ,
so that f is of type zero. Consider now the case where p = p - 1. Then the series L:n>1 IZnrp < and the infinite product P(z) has type zero by Lemma 2.6.5. Wehave
D = log If(z)l- Re[Q(z)
+ m logzJ
00,
= log IP(z)1 = o(r P ).
Therefore, in part (I) we immediately obtain that log If(z)1 = O(rP), the type of f is exactly laol in this case. In part (2), ao = 0 means that deg Q < p, then log If(z)1 = o(r P ). This last argument concludes the proof of the sufficiency of the given conditions. We now want to show these conditions are also necessary. In part (I), We already know that f being of finite type implies nCr) = O(r P ). The boundedness of S(r) is only a problem when p = p. Returning to the proof of the
2. Interpolation and the Algebras Ap
176
sufficiency, we see that we have shown that - Re
{zp
lao + ~ L z~l} = -log If(z)1 + p I~n~n(r)
O(r P ).
For r ::: I we can estimate log If (z) I :'S log M (r) :'S B r P , but we need to estimate I log If (z) II. We can apply Lemma 2.6.4, with a = 3, and obtain that the measure of the set A(3, r) ~ aB(O, r), where log If(z)1 :'S -3Br P is 2rrrl(3, r), with l(3e, r) :'S Ur>r _0 A(3, r) we have Re
!. Therefore,
{zp [aD +.!.
in the complementary of the set
L z~l} = O(r P).
p I~n~n(r)
We want to conclude from this that S(r) remains bounded. If S(r) is not bounded, then the expression between square brackets is not bounded. This means that for some sequence of values rj ~ 00, we have for some I{Jj E [0,2rr[ and Rj > 0
4}
The measure of the set {O E [0, 2rr [: cos(O + I{Jj) ::: is 2rr /3, hence the set {z: Izl = rj, Rjrj cos(O + I{Jj) ::: iRjrfJ cannot be contained in A(3, rj), and this leads to a contradiction. We conclude that S(r) is bounded, as we wanted to show. (The reader will find an alternate proof of this fact in Exercise 2.6.1.) The necessity of the given conditions in part (2) is proved as follows: In the case p p - 1, we have already seen that the type of f is laol. Hence, 0 when f is of type zero. Let us now consider the case p = p. Since f is of type zero, then nCr) = o(r P ), and as proved earlier this leads to
ao = (*)
=
- Re {zp
lao +.!. L z~l} p I~n~n(r)
= -log If(z)1
+ o(r P ).
Let e > 0 be given, then we have ro > 0 such that the term o(r P ) is bounded by er P for r ::: ro, and we can also assume log M(r) :::: er P for r :::: rD. As above, the measure of the set A(3, r) £ aB(O, r) where
I log If(z)II ::: 3er P ,
Izl = r :::: ro,
2.6. The Algebra Ap of Entire Functions of Order at Most p
177
is at most (rr /2)r. We claim that lim sup
(lo + -1 L
p
P l:sn:Sn(r)
r->oo
::: 8e.
Zn
If not, there are rj ~ 00, Rj > 8e, ({Jj E [0, 2rr [ such that for
Re
{Zp [(l0 + ~
L z~l} =
Rjr; cos(8
+ ((Jj)
::::
Z
= rje iB ,
4Rjr;
P l:sn:Sn(r)
>
4er;
on a subset of aB(O, rj) of linear measure (2rr/3h. But, any point of this set belongs to the exceptional set A(3, rj). Using the expression (*) we obtain a contradiction, so that the inequality (**) is true. We need to show that (**) implies that the series l:n>l z;;P is convergent and has the sum -(loP. Consider a finite sum
L
1 p'
l::on::oN Zn
and let r =
IZNI,
then
L
l:Sn:sN
1
zP n
L
l:Sn:sn(r)
if r :::: ro. Hence, lim sup N--+oo
~
z~
::: L
1',I=r Zn
1(l0 + ~ P
It is now clear that
L
1
n~l
p
I
liP:::
L
nCr)
-P- ::: e
r
41::: ge.
l:sn,,;N Zn
= -(lop·
Zn
This concludes the proof of Lindelof's theorem.
o
From the above proof we can extract the following corollary of Theorem 2.6.3:
2.6.7. Proposition. Let the canonical product P(z) = TIn>l E(z/zn, p) represent a/unction in the space Ap, and PN(z) = TIl:sn:SN E(zjzn, p). Then PN
N--+do
P
in the space Ap.
Proof. It is well known that the partial products PN converge to P uniformly Over compact sets. Let VN(t) be the counting function of the sequence (Znh,,;n:sN.
2. Interpolation and the Algebras Ap
178
and let n(t) be the counting function of the whole sequence. First, consider the case p fJ N. Then we have p < p < p + I and some constant K = K(p) > 0, log IPN(z)1 ~ Kr P
-<
K rP
{i
r
VN(t) dt o t P+
{Jo
--I
r
n(t)
( - - dt t p 4-1
1 oo} + J -00
+r
VN(t) } dt t P+
--2
r
n(t)
r
t P+2
dt
r ~
KArP,
where n(r) ~ Ar P, since P in A p , therefore
E
Ap. So that the sequence PN is a bounded sequence
lim PN
N-'>oo
=P
in Ap.
In the case p E N*, p = p, we recall that the function S(r) = I:I:-on:;:n(r) z;;P is bounded, say IS(r)1 ~ C. It follows that the functions ISN(r)1 =I I:1:;:n:;:vN(r) z':;-PI are bounded independently of N, say by a constant C 1 > 0. (Here we use n(r) = O(r P ) as in the proof of the necessity of conditions (i), part (2) of Theorem 2.6.3.) As in Theorem 2.6.3 we rewrite
logIPN(')1
~R{: ".E" ,;,} + "I", log IE (~, I) I + L IE (:n ' I' p-
log
p)
n>VN(r)
and we conclude that
for some constant K = K(p), C 1 > 0, A > 0, as defined earlier. The case p E N*, p = p - I, is treated the same as the case p concludes the proof of the corollary.
fJ N. This 0
In the last chapter (Definition 1.3.24) we found that the space Expo(C) of entire functions of infraexponential type, i.e., order 1 and type 0, is isomorphic to the space of infinite order differential operators. For that reason it is interesting to consider the spaces of functions of minimal type. 2.6.8. Definition. Let Ap,o denote the Frechet space of entire functions of order p, p > 0, and type zero. The norms are given by
Ilfllm
W1m }, = max{lf(z)lezee
It is easy to see that Ap,o is an FS space.
mEN·.
2.6. The Algebra Ap of Entire Functions of Order at Most
179
p
2.6.9. Proposition. For p rt N, if the canonical product P(Z) = TIn>1 E(z/zn, p) belongs to Ap.o, and we let P N denote its partial products, then -
PN N~60P, in Ap.o. The same statement if true if p E N* and the genus
p
=p-
1.
Proof. It is the same as that of the first part of the preceding proposition. Note that in the case pEN', P = p - I, the products PN E Ap.o. 0
2.6.10. Remarks. (1) In the case p
E N*, we have a difficulty when the genus p p. In this case, the canonical factor E«z/zn), p) does not belong to Ap.o. Hence, in general, we have the same problem for the partial products. On the other hand,let us consider the case of an even function I E Expo(C), 1(0) = l. Then the zeros appear in pairs ±ako and if we order them so that
=
then
1 L-=O, n~1
so that if Ln~l (l/Izn J)
I(z) =
g
E
Zn
= 00, the canonical product has genus p =
(~, 1) =
D{(1 - :J (1 + :J ez/a,
1 and
e- Zja, }
This time, the partial products
IN(z)
= J~L
(1 - (:J 2)
do belong to Expo(C), they converge uniformly over compact sets, and for every (1IINllm)N~l is a bounded sequence. This implies that
mE N*,
IN N-+601 in Expo(C). Note that if we denote by iN (z) = TIk>N(1 - (z/ak)2), then iNez) E Expo(C) and iN -+- 1 in Expo(C). (2) Similarly to the last part of the previous remark we have that if P(z) = 1L.~1 E(z/zn' p) is a canonical product of a function in Ap, p > 0 (resp. Ap.o, P ¢ N*), then PN(Z)
=
II E ( : ,p) -+- I
n>N
in Ap (resp. Ap,o).
n
2. Interpolation and the Algebras A p
180
(3) A number of the preceding properties, as well as the problems about localizability of ideals in Ap and Expo(IC), and algebraic properties of closed ideals, that shall be considered below, can be generalized to the spaces Ap(IC), where p is radial and satisfies the doubling condition (see [RuT], [KT2], [Bra], [Nil], [KrD. We now return to the question of localizable ideals in Ap. 2.6.11. Proposition. Every ideal in Ap is localizable.
Proof. From the previous section we know that it is enough to prove that if I is a closed ideal without zeros, then 1 = A p. Let f be a nonzero function in I. If I(a) = 0, then there is gEl such that g(a) "# O. As we have done before, we can write
[ g(Z) - g(a)] fez) z-a
= g(z) [/(Z)] _g(a) I(z) . z-a
z-a
The two functions in square brackets belong to Ap, hence g(a)[/(z)/(z - a)] E I. Since g(a) "# 0, I(z)/(z - a) E I. If a "# 0, it is also true that for any pOlynomial Q of degree::::: p, [/(z)/(l - (z/a»]e-Q(z) E I. Therefore, if I has finitely many zeros, we conclude that 1 E I. If I has infinitely many zeros we have
I(z)
= zmeR(z) IT E k~l
(!.., p) , Zn
p ::::: p, deg R ::::: p. We can eliminate zm and eR(z) by the previous remark and assume I coincides with the infinite product. The Remark 2.6.10(2) shows that the sequence III
converges to the function 1 in Ap. Hence, 1 E I.
o
Let us prove that the same result holds for the space Expo(IC). 2.6.12. Proposition. Every ideal in Expo(1C) is localizable.
Proof. Let us assume first that 1 is a closed ideal in Expo(lC) such that V (1) == 0. Let IE I\{O}. Multiplying I by j (j(z) = I(-z», if necessary, we can assume I is an even function. The same reasoning as that of the previous proposition, shows that if Zo is a zero of multiplicity m, then I(z)/[(z - zo)m], as well as f(z)/[(z2 - z~)m], belong to I. Therefore we can assume that 1(0) = 1, that
fez)
= II k~l
(1 -(: r) k
2.6. The Algebra Ap of Entire Functions of Order at Most p
181
(why is there no exponential factor?), and that for any n ::: 1,
Since fn ~ 1 in Expo(C), we conclude that I = Expo(C). We now observe that the proofs of Propositions 2.5.3 through 2.5.6 and 2.5.8 can be carried on verbatim for the space Expo(C). It follows that every ideal in Expo(C) is localizable. 0 As a consequence of the localizability of the ideals in Ap one can easily prove that if I is closed ideal in Ap there are two functions fJ, 12 E I such that I = i(fl, h) (see Exercise 2.6.1). In fact, one can do much better, but at the cost of substantial amount of work. Squires [Sq I] proved that one can even find a pair g" g2 E I such that I = I (gl, g2) i.e., they are jointly invertible in Ap. As expected. this has interesting consequences about interpolation. We shall explain this work in the rest of this section. Recently Braun [Bra] has extended the result of Squires to the algebras Ap(C), p a radial weight satisfying the doubling condition. The reader can safely skip the proof of the following proposition in a first reading:
*2.6.13. Proposition (Squires). Given a multiplicity variety V there is a pair invertible.
S; V (g), g E Ap,
ft, hEAp such that V = V(fl. h) and ft, h are jointly
Recall that fl. h being jointly invertible means that if hEAp, V(h) ;2 V, then there are gl, g2 E Ap such that h = gIll + g2/z'
Proof of Proposition 2.6.13. Let V =
(Zk. mdk::I' then its counting function n v is given by nv(r) = Llz'I~" mk. The condition V S; V(g) implies
nv(r) .:'S ng(r) = O(r P ). Hence, if p ¢ N*, we have V = V(f), for a single function f E Ap, as a consequence of Lemma 2.6.2. Moreover, every function in Ap is invertible (see Proposition 2.2.14). Henceforth, we shall assume p E N*. The main part of the proof consists in finding fl E Ap, and 8, C > O. such that S(ft, 8, C) = {z E C: Ifl(z)1 < Ce- W }
has the following properties. Let Z counting multiplicities), then:
= (Zk Jk::I, Z (fl) = set of zeros of fl
(i) there are two disjoint open sets SI, S2. such that
(without
2. Interpolation and the Algebras A"
182
(ii) S2 is the union of disks B(zo, r) with the property that
d(B(zo, r), SI) ~ Kdzol-P for some convenient constant K I > 0; and (iii) for every Zk E Z, the multiplicity of Zk as a zero of fl is exactly mAo Once fl has been found we conclude the proof as follows: Let cp be a COO function which is equal to I on S2, 0 on SI, 0 :::: cp :::: I, and satisfies everywhere
I~~(Z)I:::: K2IzI
P,
for some K2 > O. This is possible due to condition (ii). The function Bcpjaz vanishes on SI U S2, hence the function (l!fI)(BcpjBz), which is well defined outside Z(h), can be extended to be Coo everywhere. Therefore, there is a Coo solution u of the equation BujBz = (l/fd(Brp/Bz), and the function h := cp - ufl is entire. We note that
hl(Z(fI)\Z) = 1
and
hlV
= O.
Moreover, the multiplicity of any ZI E Z(h)\Z as a root of the equation h(zl) = I is at least equal to the multiplicity of ZI as zero of fl. These properties of f2 follow from the fact that u is holomorphic on SI U S2. Hence
V(f ..
h) =
V.
What we need to know is that hEAp, for that we need to choose u satisfying the correct growth conditions, and, furthennore, we would like fl, h to be jointly invertible. Let us estimate 1(l/fd(ocp/Bz)l. Since supp(Bcp/Bz) n S(fl, e, C) = 0, then IfI (z)1 ~ ee- CW on supp lacp/Bzl, therefore
_1_1
acp Ifl(z)1 az
(Z)I <- Ae BW
(z
E
C),
for some constants A, B > O. From here we conclude there is a choice of u satisfying the estimate .
l
lu,2e-MW
C (1
+ Iz12)2
dm <
00
for some M > 0, as well as the equation au/az = (l/fl)(acp/az). It is possible to obtain a pointwise estimate for u from these two properties. Recall the following lemma from [BO, Exercise 3.2.8].
2.6.14. Lemma. Let n be an open set in C and let K be a compact subset ofn, there is a constant C > 0 such that for every function v of class C I in n
s~Plvl5 (s~pl:~I+ llv'dm).
2.6. The Algebra Ap of Entire Functions of Order at Most p
183
Proof. Let X E D(Q), X = 1, in a neighborhood of K, 0 ~ X ~ 1. Apply Pompeiu's formula [BO, §2.IAJ to the function XU for a E K, then v(a)
= x(a)v(a) =
1 -. 2m
1.
aX v(z) -::-(z)--dz 1\ dz az z- a
Q
1.
I av dz 1\ dz + -. X(z)-=-(z) . 2m!2 az z- a
Since Iz - al :::: .5 > 0 on supp(aX/az), we can estimate the first term by const. J!2 Ivl dm. Using that 1/(z - a) is locally integrable, we can estimate the second 0 integrand by sUP!2lav/azl. We apply this lemma to Q = B(O, 1), K =
to},
v(z) = u(zo
+ z), for a fixed
Zo E C, and obtain
lu(zo)1
~C(
sup Izo-zl:" 1
IfI a~1 + r 1
az
} B(zo.1)
IUldm).
On the other hand, for some constants A', A" > 0, we have sup Izo-xl:"1
I~ a~ I ~ It
az
sup
AeBlzlP
=
A' eBlzolP
Izo-zl:,,1
and
X
(1
I 12 -MlziP U e
B(zo.l) (1
+ Iz12)2
dm
) 1/2
< A" eM'lzolP
-
with M' = M /2. It follows that lu(z)1 ~ C1ec,W and 12 E Ap. Let hEAp be such that V(h) 2 V. The function h(l - h) E Ap and V(h(l- 12» 2 V(/t). Therefore, there is gl E Ap such that h(l- h) gtfl. Hence h = gtfl +hh,
=
which shows that fl, 12 are jointly invertible. We shall now construct fl. The idea is to add zeros to Z, with convenient multiplicities so that these new zeros are sufficiently distant from Z and the corresponding multiplicity variety V' satisfies V' 2 V and the conditions of Lindelofs Theorem 2.6.3. We call these new zeros {aj )j~1 (not necessarily distinct). There are three conditions to be satisfied: there is a constant K :::: 0 such that: (i)
(ii)
IRe { IZkl:"r L m: + lajlY E a~} I~ K; zk j
IIm{ L m: + L IZklY Zk
(iii) nv,(r) = O(rP).
~}I ~ K; and
lajlyaj
2. Interpolation and the Algebras A p
184
The reason to separate the real and imaginary part of LindelOf's condition is that for p odd we shall need to place the aj on the real and imaginary axes respectively to satisfy those two conditions, while for p even we need to place them on the rays (J = 0, (J = 7r / p, for the condition (i), and on the rays (J = 7r /2p and (J = 37r 12p, for the condition (ii). We shall show in detail for the case of p odd how to place zeros on the real axis to achieve the boundedness condition. The other cases, being entirely analogous, are left to the reader. The proof is a bit delicate; it takes six steps and follows the blueprint of the proof of the Cartan-Boutroux lemma [Lev], [BG, §4.5.13]. For n ~ I, let An = {z E C: 2n- 1 ::: Izl ::: 2n }, Bn = B(O, 2n ), and denote the points of Z n An by Vn = (b7h~j~An' repeated according to their multiplicities. Therefore, for AO = nv(l), we have Ao
+ AI + ... + An =
nv(2n).
For a disk C we denote by R(C) its radius. First Step. Let 0 < "f/ ::: 1/10 be a fixed constant that will be chosen later. Following Cartan- Boutroux one can find disjoint disks Cf, i = 1, ... , in, such that 2H:= R(Cn = "f/2n+ 1
L
l~i~in
and the number of points of Vn interior to q is exactly (HIAn)R(Cf). Set lln := HI An = "f/2n 1An. It is shown in [BG, §4.S .13] that the q have the property that if z 1- UI~i~ln q, then any disk B(z, kiln), kEN", contains at most k - I points of Vn • We replace the disks q by concentric disks Cf of radii R(Cf) = R(Cn + 311n. For z rt U1
L
R(Cn::: SH
= S"f/ 2n .
l~;~in
The disks
Cf
will be called the "omitted disks."
Second Step. We shall estimate the measure of the intersection of the real axis with the omitted circles. The points in the set
can only appear as points from disks starting from points of Vn_ l , Vn, or Vn+l· The length is at most 2 x 571(2n- 1 + 2" + 2"+1) = 351]2n. This indicates that eventually we shall need 1] < 1/70 to have enough room to maneuver.
2.6. The Algebra Ap of Entire Functions of Order at Most
185
p
Third Step. Let us estimate the maximum value of IRe El!:j9. 1/(bj)PI. Since Ibjl ~ 2n - l , we have
Re
L
A"
1!:j9,.
( b" )P < - -(--1)-' 2 n- p j
Fourth Step. We are now trying to find where to locate the new zeros. One of the conditions is that the disk of radius 28" centered at a new zero does not intersect any of the omitted circles or any of the other previously added disks ofradius 28". We shall show that the new zeros can be chosen as a subsequence of the sequence «~j)l/p)jEZ' for a conveniently chosen constant 0 < ~ < 1. We shall choose ~ so that d", the smallest distance among the points in «~j)l/P)jEZ n A", is smaller than 8", and this holds for every n EN. For that reason we need to estimate d". The distance between successive points is
(~(j + I))I/p
-
(~j)I/p
= (~d/p
I)l/P ] [( 1 + -: -I
(C')llp
~ _"_1_._
1
Pl
clip
= '~-(l/P)' Pl
Since this distance is decreasing as a function of j, the value dn is as can be estimated taking j = [2 np / Hence,
n
d< "
~l/p
On the other hand, since nv(r) 8
"
~
<---..,.,.
p(2"P /~)(p-l)/p -
= rJAn2n
~ CorP,
> -
r
~
~ = np Co2
p2n(p-l)
1, we have An rJ
C 0 2,,(p-l)
~
Co2 np , and
.
If we take ~ = rJP/Co, we obtain dn < 8n . Note that by choosing the new zeros among the points of {(~j)I/P: j E Z} we automatically satisfy the requirement n v' (r) 0 (r P ). To choose the subsequence (aJ)I!:j!:j. of «~j)l/p)jEZ n An that we shall use as new zeros we impose two extra conditions:
=
(iv) The distance between two of the points (aj)l~j~j, is at least 48n ; and (v)
-I-}I
2: _1_ +Re { 2: II!:j!:}. (aJ)P l!:j!:)". (bJ)P
< _1_. 2(n-1)
We note that (v) implies (i) almost immediately. In fact, if r E [2n , 2n+ 1 [ and, by abuse of language, S(r) denotes the sum in (i), then IS(2") - S(r)1 ~ C2np /2(n-l)p = C2 P , while IS(2")1
~ IS(I)I +
L -(' ~ 1 ~ 2 + ISO)I· )
l!:j!:n2 1
186
2. Interpolation and the Algebras Ap
Since dn < On. we can choose a positive integer e such that 40 n < edn < 8on • By choosing the (aj)j as a subset of the sequence «~ej)llp)jEZ n An. we guarantee (iv). To simplify the notation. let ~ = e;. The idea about how to satisfy (v) consists in showing we can do it in the worst possible case. e.g .• I:1~j~).n l/(bJ)P = -A n/2(n-l)p. (The case An/2(n-l)p is dealt with similarly. since An n «~j)llp)jEZ is symmetric with respect to the origin. and p is odd.) Let us start by proving an auxiliary lemma. 2.6.15. Lemma. Let In = lan• bn[ £; I = la. b[. 0 < a < b. I ~ n ~ N. I:l
- > -10 I~M~A - 2~ g aP+(~/W+l)T
1
1
[
]
.
Proof. The largest measure of a subset of I P = laP, b P[ that can be covered by the union of the intervals = laf, bf [ is obtained when they cover the interval ](b - L)P, bP[, this value is precisely T = b P - (b - L)p.
I:
The minimum value of the sum is obtained when the union of the intervals covers consecutive points of the sequence (~j)j~O starting with a P • The shortest possible length of the intervals I/: is precisely W = (a + a)P - a P. On the other hand. the length of the intervals is less than ~ , since no consecutive pair ~j, i;(j + 1) lies in the same interval. Therefore, the maximum interval covered by the I: in case the sum is minimized. is J = laP, aP(i;/W + 1)T[. Let J' = IP\]. Hence
I:
I:
I 2: -log I {bL -I 2: ~jEJ' L -: l~k~M ~Ok ~J 21; I;
P
[ (a P
+ ( -i; + I ) T)/~ ] W
-I}
I { bP } - -10 - 21; g aP+(I;/W+l)T .
The last inequality is a consequence of the following one:
2: -kI 2: I
ms k ~n
+ 1 -dt = log m+1 t n
(n-m ++-I1 )
(n ) .
2: ~ log m
o
We apply the lemma to the case the omitted circles intersect the interval = 2n - 1, b = 2n ,
[2 n - 1 ,2n ] in a set of the largest possible measure. We have a a 2: 4on , T = 2np (1 - (1 - 3517)P) (from the second step),
W
= (2n-1 + 4o
n
)P - 2(n-l)p 2: p4on2(n-l)(p-I) ,
2.6. The Algebra Ap of Entire Functions of Order at Most p
and ~ :::: 8on P2n (p-I), hence ~/W able points (an l::;k ::;kn' we obtain "I
I I
L....- (an)p:::: 2>- og
l::;k::;kn
+I
{2 P
~
k
187
:::: 2P + 1. Therefore, using all the availlog({2P 1(1 + I'}')j 168 2n(p-l) , nP
'}
1(1 + I'})::::
with 1)' = (p2P + 1)2P (I - (1 - 351)P). We need to obtain that this lower bound exceeds AnI2(n-l)p = 1)2nlo n2(n-l)p, i.e., we need log({2 P 1(1
+ I'}')}2n 16p2np
1
I'}2n ----- 2(n-l)p'
~-------------->
It is clear that, after dividing out the common factors, letting .,., ....... 0 makes the left-hand side approach (1/32) log 2, while the right-hand side tends to zero. Therefore, for a convenient choice of I'} > 0 we can achieve 1
L ( n)p +Re L
I::;k::;k;
ak
I (bn)p:::: 0,
l::;j::;An
j
ar
in the case the second term is negative, choosing all the > 0 (as we pointed out, the situation is symmetrical and we only need to consider this case). By eliminating points we decrease this sum and, since each term (ar)-P :::: 2-(n-l)p, we can stop just before the sum becomes negative. In this case we obtain
ar
0< "
-
I --
L....- (n)p I::;j::;jn aj
+ Re
I
"
- - < 2-(n-l)p < 2-(n-1)
L....- (bn)P
l::;j::;An
-,
j
which is the estimate (v).
Fifth Step. Let Wn := UI::;k::;n Vk U {aJ: I :::: k :::: n, 1 :::: j :::: jn}, and Pn(z):=
II (z -
Wk)·
W!EWn
We want to show that for z E A n- I U An and outside any of the omitted disks, corresponding to k :::: n, and any of the disks B(a}, 20k), I :::: k :::: n, I :::: j :::: jn, the polynomial Pn satisfies the inequality IPn(z)l::::
(s: 2n f
n ,
Pn := L:I
aJ
m
188
2. Interpolation and the Algebras Ap
points of Vi. Now, if for some e, we have Iz - yfl < (f.j2)8k. let m = (ej2) or (e + 1)/2, according to e being even or odd, then B(z, m8k) contains at least the e points yt, ... , yf, while 2m - I ~ e. Since (5mj4) - I < 2m - I, this is a contradiction. Hence, the claim is correct. Consider now the polynomial
n
Qk(Z) =
(z - yl)·
I::;i::;qk
As a consequence of the previous claim, outside the excluded disks it satisfies IQk(Z)1 :;:
q, (/8 ) D -f
:;:qk!
(8 )q,
= qk!;
Y/2k- 1 )q, (> qk
:;: qk!
(Y/2k)q, 2Ak
(Y/-2k-I)q, . e
For z E An-I U An, the only values of interest for excluded circles are k = n - 2, n - I, n. For the other values of We we have Iz - wei:;: d(z, Bn - 3) :;: 2n - 3 • Hence
as claimed.
Sixth Step. The estimation of Pn will help us to show that the entire function defined as the canonical Hadamard product with zeros in W = UI
Ii
for every z E An-I U An outside the omitted disks and all the B(aJ, 28 k ), H is a strictly positive constant independent of n. The proof is essentially the same as that of the Minimum Modulus Theorem. We can assume Ii (0) = 1. We have W = {Wk: k :;: I}, IWkl ::::: IWk+d, II(z)
= II E (~, p) = g(z)h(z), Wk
k:::1
g(z) =
IT
O
11 - ~I· Wk
As before, let Pn be the number of zeros of /1 in Bn. For outside the omitted disks and B(aj, 28k), then Ig(z)1 =
IWI' "Wp,I-1
IT
Iz -
wkl > IWp,I-P'
Z
E
An_I U An and
(;2 f'· n- 3
I ::;k::;p,
We have Pn
1 = n(2n)::::: log 2
1
2"+1 n(t)
2"
--dt ~ t
log M(2 n + l ) log 2
,
2.6. The Algebra Ap of Entire Functions of Order at Most
189
p
YJ) logM(2n+l) log Ig(z)l::: ( log 8e log 2 ' outside the omitted disks and the B(a}, 20k). If we choose 17 > 0 smaller if necessary, we can assume that the sum of the radii of the omitted disks and the disks B(aj, 20k ) which intersect An-I U An is less than 2n-2. Therefore, there is an R, 3 x 2n - 2 < R < 2n such that the last inequality holds on the circle Izl = R. Let us now choose () such that Ih(Re i9 )1 = M(h, R) = sUPlzl=R Ih(z)l. Since M(R, f) = M(R) ::: M(2 n +I ), we have log M(h, R) ::: log M(R) -log Ig(Re i9 )1
(!L)} .
::: log M(2 n+ l ) {I - _I-log log 2 Be
On the other hand, h has no zeros in Bn and h(O) = 1, hence the Borel-Caratheodory lemma [BG, §4.5.9l allows us to conclude that 2r
log Ih(z)1 ::: ---log M(h, R),
Izl ::: r < R.
R-r
Let r = 2n-l, then loglh(z)1 ::: -4IogM(h, R),
Izl ::: 2n- l .
We conclude that for Z E A n- Io outside the omitted circles and B(aj,28k ) we have log If(z)1 ::: -Cllog M(2 n +l ) ::: -C2IzI P - e3, for some el, e2, e3 > O. This shows that if e = e- c.., the set SUI, e, e) has the three required properties, namely, we let S2 = Up B(aj, 20k ) and SI its complement. This concludes the proof of Proposition 2.6.13. 0 2.6.16. Corollary. Given V = (Zk, mkh~1 ~ V(g), g E Ap. The necessary and sufficient condition for V to be an interpolation variety for Ap is the existence of fl' hEAp, and constants e, e > 0 such that V = V(fl, h) and Ifl(m,l(zk)1 mk!
+ I fi m,l (zt> I > mk!-
(CliP)
eexp -
Zk
•
In the context of this corollary, let us introduce another kind of interpolation problem, to distinguish it from those of Section 2.2. We shall call it the universal interpolation problem. The remainder of this section is based on [SqIl, [Sq2J. 2.6.17. Definition. A multiplicity variety V = (zt. mk)k>1 in Q is called a universal interpolation variety for Ap(Q) if for every sequence
2. Interpolation and the Algebras Ap
190
Yk.j(k ::: I, 0
~
j
~
mk - 1) for which there are constants A, B > 0 so that
We note that this concept only coincides with that of interpolation variety if mk = O(exp(Bp(Zk») for some B > 0, for instance, if p(z) = IzlP or p(z) = I Imzl + log(l + Izl). If V = (Zb mkh~l = V(j[, ... , fn), the functions ft, " ., fn are jointly invertible in Ap(C) and V is a universal interpolation variety for Ap(C), then there is F E Ap(C) such that
2.6.18. Proposition.
F(mkl(Zk) ---=1 md
(k ::: 1)
and V(F) 2 V.
We start with two preliminary lemmas:
2.6.19. Lemma. If V is a universal interpolation variety for Ap(C), then for every C > 0 there are constants A, B > 0 and functions fk,j E Ap(C) such that: (i) fi:j(zd = 0 except when k (ii) fk(,jj(Zk) = j!; AeBp(z) (iii) IA,(z)1 ~ (C ( exp P Zk
= I and i = j;
»
Proof. Let Ap,u(V):= {(Yk.j)k,j: 3A, B > 0, IYk.jl ~ Aexp(Bp(zd)} and p: Ap(C) ~ Ap,u(V), p(!) = f(j) (Zk)/j !, which is surjective by hypothesis. For a given C > 0, let D:= ((Yk,j) E Ap,u(V): IYk,jl ~ exp(Cp(Zk»}, topologized by the distance induced by the norm II(Yk,j)11 = sup{lYk.jle-Cp(z.J}. k,j It is easy to see that D is a complete metric space. Let Un := If E .*'(C): If(z)1 ~ n exp(np(z»)}. One shows, as in Lemma 2.2.6, that p(Un ) D is closed in D. By hypothesis, D Un>! (P(Un) D), hence one of the terms has a nonempty interior. Therefore, for some nand e > 0
n
=
n
we have It follows that p(Un/e)
n D = D. Hence there are entire functions .h,j such that
l.frc.j(z)1 ~ ~enP(Z), e
2.6. The Algebra Ap of Entire Functions of Order at Most p
if I = k and i otherwise.
h.j(Zt) = {exP(Cp(Zd)
j !
191
0
= j,
The functions fk,j := exp( -Cp(Zk».h,j satisfy the conditions (i), (ii), and (iii) of the lemma. 0
= V (fl, ... , fn). with fl, ... , fn jointly invertible in and V = (Zb mkh::: I is a universal interpolation variety for Ap(IC), then there is a constant Co > 0 such that 2.6.20. Lemma. If V Ap(IC),
L
e-COP(Zk)
<
00.
k:::1
Proof. Since I
E Ap(IC),
it follows that for some
1
e-C1P(Z)
Let dj := infllzk - Zjl: k =1= disks are disjoint, hence
L
j~1
r
lB)
dm <
CI >
0 we have
00.
n, dj := min{l, dj/2}, and Bj = B(zj, dj ). These
e-C1P(z)
dm :::::
1
e-C1P(Z)
dm <
00.
IC
Now, since a universal interpolation variety is an interpolation variety, we know (see proof of Theorem 2.2.10) that there exists 8 > 0, C2 > 0, such that
On the other hand, there are Ao, Bo > 0 such that for any Z E Bj, p(Z) ::::: AOP(Zj) + Bo. Therefore, for some, 8J, 82 > 0,
r
lB
e-C1P(z)
dm 2: "[(dJ81 exp(-C3P(Zj» 2: 82exP(-Cop(Zj».
j
Hence
o We can go back to the proof of Proposition 2.6.18.
Proof of Proposition 2.6.18. Let us consider the series F(z) = Lmk(Z - Zk)!k.mk-1(Z), j:::l
with the functions obtained in Lemma 2.6.12 for a value C > 0 to be chosen below. If this series were uniformly convergent over compact sets and FE Ap(IC), then we could differentiate it term by term and evaluate it at Zk. We would then
2. Interpolation and the Algebras A p
192
obtain
F(md (zd = Am, -I (Zk) = l. mk! (mk - I)! Let us prove the convergence of the series and estimate F. IF(z)1
:s Lmk(lzl + IZkDIAm,-I(Z)1 k:::1
:s IzlA exp(Bp(z»
mk exp( -Cp(Zk»
L k:::1
+ Aexp(Bp(z»
Lmklzklexp(-Cp(Zk». k:::1
Since we are in the conditions of Theorem 2.2.10, it follows that the property (*) given there holds. In the course of the proof that the property (*) just mentioned implies that V is an interpolation variety, we proved that there are constants E, F > such that
°
for every
k::: 1.
Using that estimate and the fact that from the properties of p we have for some L, K >0,
Izl
:s Lexp(Kp(z»,
we obtain IF(z)!
:s 2A exp«B + K)p(z»
Lexp(-(C - E - K)p(Zk». k:::1
°
Choose C > so that C - E - K ::: Co, Co the constant of Lemma 2.6.20, then the series converges uniformly and IF(z)1
:s A' exp«B + K)p(z».
This concludes the proof of the proposition.
o
2.6.21. Remark. Let us remark that we have also shown that if V:::; V (fl, ... , In), with II, ... , In jointly invertible, then V is a universal interpolation variety if and only if V is an interpolation variety. In order to find necessary and sufficient "geometric" conditions for a multiplicity variety to be an interpolation variety, we need to introduce a little extra notation. Given two positive constants K I , K2 and a point zo, we denote r(zo) the largest positive number such that if z E 8(zo, r(zo» then p(z) :s KIP(zo) + K2· The hypotheses on p guarantee that we can find Kl> K2 > 0 so that r(zo) ::: 2 for every zoo We fix these values henceforth. If V = (Zk. mk)k:::l, we denote rk := r(zk). We also denote n(zt, r, V) the number of points of V, counted with multiplicity, in B(z*, r)\{zd. If V :;: V(h). we write n(z*, r, h) instead of n(zl, r, V).
2.6. The Algebra Ap of Entire Functions of Order at Most p
193
2.6.22. Proposition. Let V = (Zk> mkk~1 = V(fl, ... , fn), where fl, ... , fn are jointly invertible in Ap(C). Assume V is an interpolation variety for Ap (
1 r,
(1)
o
n(z
t V)
/." t
(2)
dt ~ Cp(Zk) mk < -
+ D,
Cp(zd + D . logrk
Proof. Let F be the function obtained in Proposition 2.6.18. For every k ::: 1 we have _1_
2;ri
r
J2-:.I=rk U; -
d~
F(O
=
F(mk)(zd
Zk)m,+1
I
Since F E Ap(
+ K2
= 1.
mk!
for
Z
E B(zk, rk),
we obtain
1 ~ A exp(8p(zd)/r;"
for some A > 0, 8 > o. Taking logarithms we obtain (2). Let us now apply Jensen's formula to the function fez) = F(z)/(z - Zk)m,. We have for f (0) = 1, hence
1'; o
n(z t F) 1 k> , dt = t 2;r
12JT 10g1F(Zk +rke,6)lde . -mklogrk. 0
Recalling we have assumed rk ::: 2, we obtain
1
n(zk> t, F) --'---'-- dt o t r,
By construction we have n(zk> t, V)
~
~ Cp(Zk)
+ D.
n(zko t, F), hence (I) also holds.
0
In the case of the algebras Ap, one can obtain now clear geometric conditions to decide when a multiplicity variety is an interpolation variety. Note that these are the conditions that exclude the examples of the type occurring in Exercises 2.2.12 and 2.2.13. 2.6.23. Proposition. Given a multiplicity variety V = (Zk. mkh:::1 ~ V (g) for some g E Ap \{O}, a necessary condition for V to be an interpolation variety for Ap is that (1')
(2')
for some C, D > 0 and alllzkl ::: 2. If V = V(g), these conditions are sufficient.
2. Interpolation and the Algebras A"
194
Proof. From Proposition 2.6.13 we know that V = V (II, h), II, h jointly invertible in Ap. The proof of the necessity is a verbatim copy of the proof of Proposition 2.6.22. To prove the sufficiency of the conditions (I'), (2') when V = V(g), we need to show that for some £ > 0, A > 0, one has
I: : Ig(md(Zk) mk!
(k :::: 1).
ee-A1z!lp
It is enough to prove it for IZkl :::: 2, and also to assume that g(O) = 1 (Why?). We can apply the minimum modulus theorem ([BG, §4.5.l4], [Lev, p. 21]) to obtain constants £1 > 0, A I > 0 (independent of k) such that for Z E B(O, 21zk I) and outside a finite collection of exceptional disks, Bk • l , ••• , Bk,jk the sum of whose radii does not exceed ~IZkl, one has
Ig(z)1 ::::
£le-A11z,IP.
It follows that there must exist a value r, !IZk I < r < IZk!. such that CJB(zt, r)
n Bk,j
= 0,
1 ::: j :::
ik.
Apply now Jensen's formula, [BG, §4.4.30] to g in the disk B(zt, r). One obtains log I
g<md(Zk) mk!
1+ mk logr +
1
r n(V,
Zk,
t
0
t)
1
dl = -2 7r
1
2".
log Ig(Zk
0
.
+ re '8 )1 de.
From this identity, (1'), (2'), and the choice of r, it is clear that log I
g(md(zd mk!
I :::: -AlzklP - B
o
for some A, B > O.
2.6.24. Remark. [Sq2J. For the weight p(z) = I Imzl + 10g(1 + Izl), and a multiplicity variety of the form V = V (h), given by h slowly decreasing for Ap(C), one has that V is a (universal) interpolation variety if and only if
1
P(Z') n(V Zk t) , , dt::: Cp(Zk)
o
and
+D
t
Cp(Zk) + D mk < - - ' - - - - log P(Zk)
for some C, D > 0, independent of k. One can find necessary and sufficient geometric conditions interpolation, generalizing Proposition 2.6.2 for arbitrary radial weights satisfying the doubling condition [BL]. The correct expressions to consider are just the integrated Nevanlinna functions N (V, Zk> r) = (n(V, Zb 1)/1) dt. In this form the criteria obtained extend also for meromorphic functions and to several complex variables. This can be generalized to other radial weights, but the necessary and
J;
2.6. The Algebra Ap of Entire Functions of Order at Most p
195
sufficient conditions do not coincide, though they are close to each other. This is a phenomenon characteristic of functions of infinite order of growth [BL]. In [GR] similar conditions for spaces of functions with prescribed indicator of growth have been found. Interpolation in the space of functions of infraexponential type has very interesting applications to the theory of Dirichlet series, as we shall see in Chapter 6. A classical theorem of Polya- Levinson [Levs, Theorem XXXI] states the following: Let V = (Zk}n~1 be a sequence of nonzero complex numbers such that Re Zn > 0,
. -n = 0, lIm
n~oo Zn
I1m Zn I =
ll.m
0
IZn I
n-'oo
and there exists c > 0 such that
Then there is an even function
f
of infraexponential type such that V S; V (f),
-log If'(zn)1 = o(lznl) and for
Z
as h --+
00
such that Re z > 0, dist(z, V) ::: c 18, we have -log If(z)1 = o(lzl)
(tt)
asizi --+
00.
Note that (t) implies that points Zn are distinct. One can prove that the main point (t) of the statement is exactly the condition that V is an interpolation variety for functions of infraexponential type. In [Vi2] it has been proved that the conditions on Rezn > 0 and I Imznl/lznl --+ 0 are not necessary. In fact, given any sequence V = (Zn In:::1 , n(r) = o(r), and {znln:::1 satisfies V, then there is a function f of infraexponential type such that (i) V 5; V (J); (ii) all zeros of f are simple; and (iii) V (J) is an interpolating variety in the space of functions of infraexponential type. (The condition (tt) also holds when dist(z, V(J» ::: c/8.) We provide a proof of this result in Chapter 6.
It is possible also to find necessary and sufficient geometric conditions for a variety to be interpolating in space of type zero for any radial weight satisfying the doubling condition [BLV]. EXERCISES
2.6.
I. Let j be an entire function, j(O) = 1. For r > 0, let ai, ... ,an denote the zeros of f in B(O, r) repeated according to multiplicity. (a) Show that
1121r e- i8 2" 7r
0
log If(re i9 )1 de
= ir/,(O) + i L"( ~ _ -) k=1
ak
ak r
•
Use this formula to show that if f is an entire function of exponential type then SI (r) Lld tl:5r ak' is bounded as r ~ 00.
=
2. Interpolation and the Algebras A p
196 (b) Let P E N*, find an explicit expression for -
I
27T
121< e-'p& . 10gl/(re,e)ldO . 0
=
involving Sp(r) L:1a,IY a;p. Use this expression to show that Sp remains bounded as r ~ 00, whenever lEAp. (Hint: Divide out the zeros of I to obtain a nonvanishing holomorphic function g in B(O, r). All this exercise requires is to compute the Fourier coefficients of the hannonic function Re(log g) (see also [BG, Exercise 4.6.9)).
2. Let E p , Ex be respectively the spaces of entire functions of order at most p (resp. finite order). For 0 < p < 00, nEW consider the Banach spaces Ep .• of all entire functions such that 11/11. = sup(l/(z)1 exp(-lzIP+'/.» <
00
and F., the Banach space of all entire functions such that 111/111.
= sup(l/(z)le- I: I') <
00.
(a) Show that Ep = nn>' Ep.n , Eoc = Un>1 F•. Ep is an FS space and is a DFS space. (Hint: Use Proposition 1.4.8.) (b) Show that every closed ideal in Ep (resp. E",,) is principal. (Hint: Find g such that V(g) = V(I).)
3. For 0 < p < such that
00,
n
E
II/lIn
N*, let Ap.n be the Banach space of all entire functions
= sup (I/(Z)I exp ( -~IZIP))
<
00.
(a) Show that Ap.o = nn,,' Ap,o is a FS space. (b) If p ¢ N, show every closed ideal is principal. (c) If p E N*, let w = ei21f / P , and assume that I E Ap,o satisfies I(wz) = I(z) for all z. Show that in the Hadamard canonical expansion of I, I(z) == z"'e Q(:) IT'>I E(z/zn, p), the polynomial Q has degree::: p - I, also that by grouping conveniently the Weierstrass factors E(z/zn, p) one can write the infinite product as ITk,,1 f{)" each f{)k E A p.o, and the infinite product converges in Ap.o. (d) Show that every ideal in Ap.o is localizable.
4. Use Proposition 2.6.11 to show that every closed ideal in Ap is principal if p ¢ N. 5. The object of this exercise is to prove that if [ is a closed ideal in Ap (resp. Ap.o), 12 E Ap (resp. Ap.o) such that [ = [(/1, h)· (a) Use Proposition 2.6.11 to show that if V = V(I) then [ = I(V). (b) Let 0 < IZII ::: IZ21 ::: "', IZn I ~ 00, (nkh,,1 a strictly increasing sequence ofpositive integers, €k > 0, E,>, €k < 00. Show that one can find a sequence of distinct Wk E C, 0 < IWk I /' 00, su~h that the set {Wk: k ~ I} is disjoint from {z.: n ~ I}, Iw;P z,;tl < €" and if the series En:::k z;;P converges to the value 01, and (Un)n"l is the sequence given by Un = Zn if n 1: nk(k ~ 1), Un, = Wko then En,,1 u;;P = S. (c) Let II E [\{O}. Use parts (a) and (b) to find 12 E Ap (resp. Ap.o) such that V(/I) n V (h) = V. Conclude that I = 1(11. h)· PEW, then there are two functions II,
6. Let the canonical product of a function I of finite order be I (z) ITn"l E(z/zn, p). Let PR(z) := zm Il1',I:::R E(z/zn, p), R > O.
zme Q
=
2.6. The Algebra Ap of Entire Functions of Order at Most p (a) Assume
f
E
A p, show that e- Q fI PR e- Q f
.
lIm - - g PR
R-co
E
A p, and that for every g
=g
.
In
197
E
Ap one has
Ap •
(b) Let [ be a closed proper ideal in A p , V = V(l), f E [\{Ol, g E l(V). Write P R as a product PH = Pk P; such that V(P;) = V n B(O, R). Show that fI Pk E I. Use this to prove that gEl. (c) Prove parts (a) and (b) in the case Ap.o, P f/. N, and also when p E N* but P = p-1. (d) Let f E Ap.o, p = p, Q(z) = aoz P + QI (z), deg QI ::: P - I. Show that
pf
exp (zp - ~
z;;P )
E
Ap.o
p l'nl"R
R
and
holds in Ap.o. Use (c) and (d) to prove that all ideals in Ap.o are localizable.
7. Let V = V(jl, ... .jn), fI, ... ,fn jointly invertible in Ap(lC). Show that V is a universal interpolating variety if and only if V is an interpolating variety.
=
8. Let p(z) Izl P , r > 0, kl > \, k2 > 0, compute the function r(z) > 0 that has the property: p(t;)::: kIP(z) +k2 if and only if t; E B(z, r(z».
9. Suppose V C, D>O
= (Zk> mdk~1
1 rl
o
satisfies condition (I) of Proposition 2.6.22, i.e., for some
n(zk> t, V)
---'-''---'-dt ::: C(P(Zk)
t
+ I)
(k ::: 1).
Show that there are e > 0, A > 0 such that for any j =1= k IZk - Zj I ~ u-AP('
10. (a) Use Proposition 2.6.23 to show that the lattice IE + iZ is interpolating for Ap if p = 2. (b) Give a direct proof of the same result by using the Weierstrass a function [Mark].
CHAPTER 3
Exponential Polynomials
3.1. The Ring of Exponential Polynomials An exponential polynomial is an entire function
L
I(z) =
1 of the form
Pi (z)e"jt,
I~j~m
where {Xi E 1[, Pi E lC[z J. We assume that the {Xi are distinct and the polynomials Pi not zero. The Pi are called the coefficients of 1 and {Xi the frequencies. (Sometimes the {Xi are called the exponents, especially in the Russian literature. In some contexts {Xi = i)\,i, Ai E JR, and the Ai are called the frequencies and ri = 2rr/Ai (when Ai t= 0) the periods; clearly eiA)t is periodic of period ri') It is immediate that there is a unique analytic functional T whose Fourier-Borel transform J(T) coincides with I, i.e., J(T)(z) = (TI;' e:l;) = I(z). Namely, T is representable as a distribution in I[ of the form T
= '"' a·).v 8(v) a, ~
1
j,v
8t)
where 8a ) is the Dirac measure at the point (Xi' is a "holomorphic" derivative of order v, 8~~) = (a/az)"8"j' and the ai.v are complex constants. In the case where the frequencies {Xi are purely imaginary, i.e., (Xi = iAi' AJ E JR, then 1 is the Fourier transform of a distribution fJ- E £'(lR). That is, we let d L. ai.v iv -d v 81..) X V
fJ- :=
J.v
with fhj the Dirac mass at the point Ai
I(z)
=
E
JR (acting on Coo functions in JR) and
(fJ-x,
e- iXz ).
These simple observations correspond to the fact that if p(z) = Izl, then every exponential polynomial 1 E Ap(C), while if the frequencies are purely imaginary, f E F(£'(lR», i.e., one can use the smaller weight function p(z) = I 1m z I + log(l + Iz 12 ). We shall see in this chapter that, among other properties, a nonzero exponential polynomial is always slowly decreasing in Ap(C). 198
3. I. The Ring of Exponential Polynomials
199
We denote by E the ring of exponential polynomials (including the zero function) and by Eo the subring of E consisting of exponential polynomials with constant coefficients (also called exponential sums). Let £*, denote the sets E\{O) and Eo\{O), and U(E) and U(£o) the groups of invertible elements of E and Eo, respectively. It is clear that U(E) = U(Eo) = {ae b:: a E C', bE q. In the set of complex numbers we consider the lexicographical order ~ defined by a ~ fJ if and only if either Rea < RefJ or Rea = RefJ and Ima:5 ImfJ. If a ~ fJ and a =1= fJ, we shall write a -< fJ. From now on, in the above representation of the exponential polynomial f we shall always assume aj -< aj+l, l:5j:5 n -l. We start by studying divisibility in the rings E and Eo. Since eYz is invertible and a -< fJ is equivalent to (a - y) -< (fJ - y), we can assume f has the form
Eo
fez)
=
L
Pj(z)e ajZ ,
0= ao -<
al
-< ... -< am,
O~j~m
for some m
~
1.
3.1.1. Proposition. Let fez) = LO~j~m Pj(z)e ajZ , g(z) = LO~k~n Qk(z)e flkZ be two exponential polynomials of the form (* ).If g divides f in E, then the frequencies fJI, ... ,fJn of g are linear combinations with rational coefficients of the frequencies al,"" am of f·
Proof. Since g divides f in E there is h E E*, h (z) = Lo~/~p R,e Y' z , such that f = hg. Since we are assuming Yo -< YI -< ... -< yp, it is immediate that Yo = O. If P = 0, then the frequencies of f and g are the same, and the coefficients are related by Pj = RoQj. We can assume therefore that p ~ I. Let 0 = Wo -< WI -< ... -< Wr represent all the frequencies of the form fJk + y" 0:5 k :5 n, 0 :51:5 p. Clearly (aD, ... , am) £ (Wo, ... , Wr I. In order to proceed we need an elementary, albeit important, lemma. 3.1.2. Lemma. Let F(z) = LI~j~n cpj(z)e AJZ be a nonzero exponential polynomial with frequencies AI -< A2 -< ... -< An, and coefficients f{Jj(z) = ajzmJ + ... , polynomials of degree mj ~ 0 (aj E C*). For every j, 1 :5 j :5 n, we have
II (~ - Ak)m.+l] fez) = cjeAjZ , [( ~dz - Aj)mJ k#j dz Cj = mj! aj II(Aj - Ak)mk+1 o. =1=
k'f-j Proof of Lemma 3.1.2. One proves by induction on the degree m of P E C[z], P(z) = az m +"', a E Co, that
(:z -
A)
m
(P(z)e AZ ) = m! ae AZ .
200
3. Exponential Polynomials
Hence
)m+1
d ( dz - A Therefore, if i
i=
(P(z)e AZ ) = O.
j, we have
and
~ _ A)ml II (~ - Ak)mdl(p.(z)e AjZ ) = II (~ ~ Ak)m'+I(m'! a·e A1Z ) ( dz dz dz J
J
J
k1'J
J
kh
=mj! aj
IICAj ~Ak)m,+leAJZ.
D
k#j
3.1.3. Corollary. The family CeAzhEIC is a basis of the C[z]-module E.
Proof of Corollary 3.1.3. The spanning property is clear. If an exponential polynomial F with frequencies AI, ... , An is the identically zero function, then Cj = 0 for i :::: j :::: n, which is impossible unless all the coefficients are zero.D Let us now return to the proof of Proposition 3.1.1. We have
hence, from Corollary 3.1.3 we have Qk(z)R/(z) = {
P(Z)
oj
ifwj=aj, . { } If Wi t/. aO,···, an .
Let us assume there is a tho, 1 :::: ko :::: n, which is not a linear combination with rational coefficients of ao, ... , am. Let V be the Q-vector space spanned by ao, ... ,am, and let cl, ... ,Cq be a basis for V. Then fJko,Cl, ... ,Cq is a Q-linearly independent family. Therefore, we can find Cq+l, ... , Cf such that tho, .0 I, ... , Cq, Cq+ I, ... , Cf forms a basis of the Q-vector space W spanned by ao, , .. , am, fJl, ... , fJn, YI, ... , Yp' Let us denote co := fJko' Every fJk, 1 :::: k :::: n, can be written in a unique way as fJk =
L
rkiCi,
O::::i~t
Let Uo := max{rk.O, 1 :::: k :::: n}, Uo 2: I. Consider the nonempty subset of fJk such that the coefficient rkO of co coincides with uo. Among these choose one whose coefficient rj.l of CI is as large as possible, say Ul, UI
:= max{rk,l: for k such that rk,O
= uo}.
3.1. The Ring of Exponential Polynomials
201
Continuing this way we find a unique {hi denoted f3*, such that f3* =
2:= Uiei, 0:5i:51
whose successive coefficients Ui are maximal. Proceeding in the same way with YI, 0 ::: 1 ::: p, we find y* y* =
= Yk"
2:= Viei, O:5i :5t
whose successive coefficients are maximal. Here Vo ~ 0 since Yo = o. Let w* := f3* + y* = L:O
2:=
Pk R, = Pk, Ril
"# 0,
fh+YI=w*
°
since Pkl "# 0, RI, "# 0. It follows that w* E lao, ... , an}. This means that Wi = for i fj. {I, ... , q}. In particular, Wu = O. This is a contradiction. We conclude 0 that every f3k is in V, and the statement of Proposition 3.1.1 is correct.
We are going to consider now the problem of the factorization of exponential polynomials. 3.1.4. Definition. An element
f
f
E
Eo is normalized if it is of the form
(z) =
2:=
aje"'Z,
O:5j:5m
°
with = ao -< al -< ... -< am and ao = 1. A normalized element f E Eo is said to be simple if all the aj, 1 ::: j ::: m, are commeasurable, i.e., there is A E
= q;A,
E Q, qi > O. Let N be the common denominator of all the qi, then we can write qi = Pi/N, Pi E N*, and
qi
lXi
= Pif.l,
f.l=)"/N,
0< PI < ... < Pm.
202
3. Exponential Polynomials
Hence. if we let l; = eJ1.:. we have f(z) =
1+ a,l;P' + ... + am l; Pm •
which is a polynomial of degree Pm in the variable l;. Therefore. it can be factorized into Pm factors of the form 1 + cl; . In other words. f can be factorized into Pm simple factors of the form 1 + ceJ1.z. The same argument can now be repeated by writing
1 + ceJ1.Z
= 1 + c(eJ1.z j r)'.
for any r E N*. Hence. each of them factorizes into r normalized exponential sums of two terms. Therefore. not only is a normalized simple exponential sum not irreducible. but it admits infinitely many distinct factors (see Theorem 3.1.9 below).
3.1.6. Lemma. Let 0 -< a, -< ... -< am and P := dimQ V. with V =
=
L'~j~mQaj.
the Q-vector space spanned by a, ..... a m. Let Q+ Q n ]0.00[. One can find a basis /-L, ..... /-Lp of V such that aj E L'~k~pQ+ /-Lk. 1 :::: j :::: m.
Proof. For any j we have either Reaj > 0 or Reaj = 0 and Imaj > O. Accordingly. we can find 8 > 0 sufficiently small so that the numbers Aj = e- i8 aj satisfy ReAj > 0.1 :::: j :::: m. It is clear that e- i8 V = L'~j~m QAj also has a Q-dimension equal to p. Let m, •...• m P be a Q-basis of this space. Assume Aj =
L
aj/m/.
'~I~p
with ajl E Q. We shall first find another basis Mi •...• Mp of e- i8 V so that mk E L'~19 Q+ M/. for 1 :::: k :::: p. For that purpose. let us choose rational numbers tl,k so that for every I they approximate Reml for 1 :::: k :::: p. Assume further that det(t/k)/,k =J:. O. Then we can detennine M, . ...• Mp from the system of equations
m, = L
tlkMko
1 :::: I :::: p.
'~k~p
This indicates that the Aj can be written as
Since we want
tlk
approximately equal to Reml. then the rational numbers
bjk := L'~I~p aj/tlk are approximately equal to Re(L'~I~p aj/ml) = Re Aj > O. In other words. if we choose tlk sufficiently close to Reml. 1:::: k :::: p ••we can guarantee that bjk E Q+ and. clearly. this can be done so that simultaneously
the detenninant det(t/k}J,k =J:. O.
3.1. The Ring of Exponential Polynomials We now let /-Lk LI!'Ok!'OP Q+/-Lk.
= ei8 Mb
203
I ~ k ~ p. They form a basis for V and Cij
E
0
Let us now return to the situation of Proposition 3.1.1.
3.1.7. Proposition. Let fez) = LO::;j!'Om Pj(z)e Uj ; and g(z) = LO!'Oj::;n Q/z)e PjZ be exponential polynomials of the form (*), with more than one term. Assume that g divides f in E and that Cil, ... , Ci m are linear combinations with nonnegative rational coefficients of p Q-linearly independent complex numbers /-LJ, ••• , /-Lp. In this case, the frequencies /3t, ... , 13m are also Q-linear combinations of /-LI, .•• , /-Lp with nonnegative coefficients. Proof. From Proposition 3.1.1 we conclude that
We want to show that rjk ::: 0 for every j, k, 1 ~ j ~ n, I ~ k ~ p. To fix ideas let us assume rj I < O. We choose among the f:3j, those for which rj I is minimal; among these, those for which rj2 is minimal, etc. One finds in this way, a certain f:3io, denoted 13*, whose coefficients with respect to /-LI, ••• , /-Lp are all successively minimal. Let us write
13* =
UI/-LI
+ ... + Up/-Lp,
Uj
E
Q,
1
~
k
~
p.
We have UI ~ rjo, 1 < O. Let 0 = Yo -< YI -< ' .. -< Yr, and Rk be polynomials such that
We augment the collection /-LI, ••• , /-Lp to a basis /-LI, ., . , /-Lq of the Q-vector space generated by Cil, ... , Ci m, 131, ... , f:3n, YI, ... , Yr' Among Yo, ... , Yr we find Y * whose successive coefficients with respect to /-L I, ••• , /-Lq are minimal. Let y* = VI/-LI + ... + vq/-Lq, with VI ~ 0 since Yo = O. As in Proposition 3.1.1, we can see that 13* for some i, 1 ~ i ~ m. Then
+ y* = Cii
+ VI )JLI + ... + (up + vp)JLp + Vp+1 JLp+1 + ... + Vq/-Lq. = ... = Vq = 0 and UI + VI < 0, which is a contradiction with the
Cii = (u I
Clearly Vp+1 hypothesis that every oti has nonnegative coefficients with respect to JLl, .•. , JLp. Therefore, it must be true that 'ik ::: 0, 1 ~ j ~ n, 1 ~ k ~ p. 0
3.1.8. Remark. It follows from Proposition 3.1.7 that the Yj are also Q-Iinear combinations of the JLk with nonnegative coefficients. We shall accept the following result of Ritt about factorization in the ring Eo. The proof is entirely algebraic and rather long (see [Ril], [Go], [Schi]).
204
3. Exponential Polynomials
3.1.9. Theorem. Every normalized exponential sum f E Eo which is ¥= 0, 1 can be written in one and only one way (up to reordering) as a finite product of normalized exponential sums
where the ai are simple, any two of them have no nonzero common frequency, and the lrj are irreducible in Eo.
3.1.10. Remarks. (1) Let f(z) = 1 + ale"": + ... + ame"'m z , 0 -< al -< •.. -< am. If all the aj, 1 ::s j ::s m, are Q-linear combinations of J-tl, •.. , J-tp with nonnegative coefficients, and f = It ... ft is the decomposition of the previous theorem, then the frequencies of all the jj are Q-linear combinations of J-t I, ... , J-tp with nonnegative coefficients. (2) If P = rank of the additive group generated by al, ... , am, then it coincides with the dimension of the Q-vector space V defined in Lemma 3.1.6. Changing, if necessary, the J-tk obtained in the lemma by a submultiple, we can assume l::sj::sm. aj E N* J-tk>
2:
I~k~p
(3) Every nonzero exponential sum form
f
can be written in a unique way in the
with ae"'z E U(Eo), ai, ... , as normalized, simple, sharing no common nonzero frequencies, and lrl, ... , lrr, normalized irreducible exponential sums. We recall some classical results of commutative algebra that we will use in the sequel, see, e.g., [AM]. Let A be a commutative integral domain with identity and let Jt be the field of quotients of A. Denote A* := A\{O}, U(A) := the group of invertible elements of A, Jt* := the multiplicative group of the nonzero elements of Jt, and A[X], resp. Jt[X], the rings of polynomials in the variable X, with coefficients in A and Jt, respectively. We assume A satisfies the condition (~)
Any two elements of A * admit a g.c.d.
3.1.11. Definitions. (1) The content of a nonzero polynomial f E A[X] is a g.c.d. of the coefficients of f. It is denoted c(f). (2) A polynomial f E A[X] will be said to be primitive if c(f) is invertible. For every f E A[X] we shall write f = c(f)j, with j E A[X] primitive. The product of two primitive polynomials is primitive. If f, g E A[X]\{O}, then c(fg) = ac(f)c(g), a E U(A). Let us recall finally that there is a family P S; A[X] such that every element of P is primitive in A [X] and irreducible in Jt[X] and, moreover, any f E Jt[X]
3.1. The Ring of Exponential Polynomials
205
can be written in a unique way in the fonn
I
= a II pn(p), peP
with a E .It, n(p) E N, and only a finite number of n(p) are different from zero. It follows that if I E A[X], then
I
cU) II pn(p),
=
peP
and the decomposition is unique. We shall now show that Eo verifies the property (Ll) and that E is isomorphic (as a ring) to Eo[X].
3.1.12. Lemma. Every exponential polynomial lEE can be written in a unique way in the lorm I(z) = lo(z)
+ II (z)z + ... + II(z)/,
where 10, ... , II E Eo. In other words, E is isomorphic to Eo[X].
Proof. Let I(z) = LI
L
(X2
-< ... -< am.
bijzj,
05cj 5cd;
so that if 1:= max{d l , I(z)
... ,
=
dm },
L (L bije";z) zj = L 05cj 5c1
d; ~j
jj(z)zj,
05cj 5c1
with jj(z) := Ld;~j bije<:t;z E Eo. This decomposition is unique. If L05cj5c1 jj(z)zj == 0, then Pi(z) == 0 by Corollary 3.1.3, I :5 i :5 m. Hence, bi • j = 0 for all i, j. Therefore, 1==0 for O:5j:51. Finally, it is now clear that Eo[X]
L
~
jj(z)xj t-+
05cj 5c1
E,
L
jj(z)zj
05cj5c1
o
is a ring isomorphism.
3.1.13. Lemma. Let
I, g E
Eo, then there is a g.c.d. in Eo-
Proof. As a consequence of Theorem 3.1.9 we have to consider the following three cases: (i) (ii) (iii)
I and g are simple; I and g are irreducible; I is simple and g is irreducible.
206
3. Exponential Polynomials
Case (i). Let 1(::.) = I + ale"lo + ... + ame"m o, 0 -< 0'1 -< ... -< am, ai = PiA, Pi E N*, and g(z) = 1 + b1e PIz + ... + bneP"z, 0 -< b l -< ... -< b n, bj = qj/1, qi E N*. We have two possibilities: either 0'1 I {31 E Q or not. The first case is equivalent to ai I (3j E Q and to A//1 E Q. Thus, in this case, there is v E C* such that ai
= riV,
ri,
Sj E
I ::::
N*,
Moreover, we have rl < r2 < ... < r m and consider the two polynomials in l;
SI
j ::::
< ... <
I:::: j :::: n.
m,
Sn.
Let l;
=e
Vz ,
and
F(l;) := 1+ all;r l + ... + aml;'m,
G(l;) := 1 + bll;sl + ... + bnl;Sn.
Let H be the g.c.d. of F and G in IC[l;]. It is clear that h(z) = H(e VZ ) divides I and g in Eo. Furthennore, h is nonnalized and simple. The Bezout identity H=FU+GV
for some U, V E IC[ l;] shows that h = uf
+ vg,
where u(z) := U(e":) and v(z) := V(e VZ ) belong to Eo. Hence if ii E Eo divides I and g, it follows ii divides h. Therefore, in this case, h is the g.c.d. of I and g. We claim that if A//1 ¢ Q, then I is the g.c.d. of I and g. In fact, if h divides both I and g, h is nonnalized and not I, then by Proposition 3.1.7 all the frequencies of h are integral multiples of A, hence h is simple. By the same proposition, all the frequencies of h must also be multiples of /1. Hence A//1 E Q, which is impossible. Case (ii). If I and g are irreducible, then either 1= ag, a E U(Eo), in which case 1= g (if they are nonnalized), or 1 is the g.c.d. of I and g. Case (iii). If I is simple and g is irreducible, the g.c.d. must be g or l. If g divides I, then it is simple by Proposition 3.1.7. Accordingly, the g.c.d. is 1. This concludes the verification that Eo has the property (~). 0
jj
Given lEE, we can write it in a unique way as I(z) = I:o:oj:o/ /j(z)zj, E Eo. We denote by c(f) E Eo the g.c.d. of the coefficients 10, ... , It-
3.1.14. Lemma. Let lEE, g divides c(f) in Eo.
E
Eo. Then g divides I in E if and only if g
Proof. Assume g divides I in E. Therefore, there is h (z) = I:O:oj:or h j (z )zi , hj E Eo, such that g(,)
(,E,
hJ(,),J)
~ /(,) ~ ,E, fJ(,),J,
3.1. The Ring of Exponential Polynomials
207
By the uniqueness part of Lemma 3.1.12 we have 1= rand g(z)hj(z) = jj(z), I ~ j ~ I. This shows that g divides all the coefficients of f in Eo, hence g divides c(f) in Eo. The converse is clear. D
3.1.15. Lemma. Let f,g
E
E. There is a
E
U(Eo) such that c(fg) = ac(f)c(g).
Proof. This follows from E ;;:: Eo[X] and Eo verifies
(L~).
D
3.1.16. Proposition. Every exponential polynomial fEE can be factorized in a unique way (up to invertible elements) f
= C(f)TC~' ... TC;' ,
where c(f) E Eo, TC!, ... ,TCr are exponential polynomials in E\Eo, which are irreducible in E, mutually coprime, and n!, ... , nr E N*. Proof. Let f(z) = LO::):051 Ii (z)zj, Ii E Eo, and j(X) = LO:Oj:051 fj(z)xj, the corresponding polynomial in Eo[X]. Then, there are unique polynomials if!, ... , ifr , primitive and irreducible in EolX], and n!, ... , nr E N* such that j
= c(f)if~'
... if;' .
Let TCj be the exponential polynomials in E that correspond to ifj via the isomorphism Eo[X] ::::: E. Since this is a ring isomorphism, it follows that TCj are irreducible in E. This proves the factorization. The uniqueness follows the D same way.
3.1.17. Theorem. Every exponential polynomial can be factorized, in a unique way up to invertible factors, as the product of a finite number of simple exponential sums with frequencies pairwise incommeasurable, a finite number of irreducible normalized exponential sums, and a finite number of irreducible exponential sums. Proof. We apply Ritl's Theorem 3.1.9 to c(f) and Lemma 3.1.13. The isomor0 phism E ;;:: EolX] allows us to conclude the factorization is unique.
3.1.1S. Theorem. The ring E also satisfies the condition
(~):
any two exponen-
tial polynomials have a g.c.d. in E. Proof. Given f, g E E*, it is enough, by the preceding theorem, to consider the following two cases:
(i) f E Eo and it is simple. Then f and c(g) admit a g.c.d. in Eo. By Lemma 3.1.14, this is also the g.c.d. of f and gin E. (ii) f is irreducible. If f divides g in E, then f is the g.c.d. If not, is the g.c.d. D
208
3. Exponential Polynomials
We now use Lemma 3.1.2 to obtain some inequalities about exponential polynomials. We first note that if /(z) = LI;Sj;SN Pj(z)e AjZ , AI ~ A2 ~ ... ~ AN, Pj a nonzero polynomial of degree mj ~ 0, then these coefficients are uniquely determined \::y the function / and, in particular, we can define the degree of /, dO/ := mI + ... + mN + N - 1. We can also consider the supporting function H of the convex hull K of the points Xl, ... , XN , conjugates of the frequencies, namely, H(z) := m~xRe(Ajz) = max(Xjlz), J
J
where (,1,) denotes the Euclidean inner product in ]R2. This function is also uniquely determined by /. Recall that M(j, z, r) max{l/(~)I: I~ - zl :s r}.
=
3.1.19. Proposition. Let /(z) = Ll::oj::oN Pj(z)e AjZ E E*. For any ro > 0, there is a constant C > 0 such that/or every z E C and every 0 < r :s ro, the inequality
rdol exp(H(z))
:s CM(j, z, r)
holds.
Proof. Let us denote
Cj,l
E
C. Then
(:z)
Qj
/(z)
= Cje
AjZ
,
=
with Cj mj! aj ITk;l)(Aj - Ak)mk+l as in Lemma 3.1.2. Therefore, we can apply Cauchy's formula and obtain
e AjZ
=~
L
Cj.d(/)(z)
L
r
[!cI· t (~ O:s/:sdOI 2;r I Cj .I1~-zl=r
=
Cj O;st::odOf
This shows that given ro > 0 there is a constant
K
- z)do/-I
(~
/(~;ol Id~. - z)
+
> 0 such that
o Let us remark that the size of K in the last estimate depends on min{lAk - Ajl: k =f:. min{lajl: 1 :s j :s N}, max {!AjI: I :s j :s N} and dot· Let us give here a corollary of the above proof.
n,
3.1.20. Proposition. For any
f
E
E* there is a constant A > 0 such that
AeH(z):s
L
1/(/)(z)l.
O:s/:sdO I In particular, the multiplicity of any zero of f does not exceed dOf.
3.1. The Ring of Exponential Polynomials
209
Proof. We have already shown that
e AjZ = ~ C. J
hence
leAjZI
~ c-J,/ f(/)(z) , ~
05.i5.dO f
.:s m~x ICj,//Cjl
L
If(/)(z)l,
05.i5.dO f
J.
which implies the required estimate. Moreover, if fez) = 0 and f.L = multiplicity of this zero, then f(l)(z) = 0, o .:s I .:s f.L - 1, which shows immediately that f.L .:s dOf. 0 Let us recall that, as a corollary of the Minimum Modulus Theorem [BG, Theorem 4.5.14], we obtained in Lemma 2.2.11 the following inequality: Let F be an entire function, F(n =j:. O. For any r > 0 there is p E ]r/4, r/2[ such that log (
IF(Z)I) inf - > -8logM ( -F- , l;, 2er ) . IF(l;)1 FU;)
Iz-~I=p
3.1.21. Proposition. Let fez) = L:15.j5. N Pj(z)e AjZ , be a nonzero exponential polynomial, mj = deg Pj, M = maxl5.j5.n mj' Then: (i) there is a constant A > 0 such that for any r E [0, 1] and z Ere we have
M(j, z, 2er)
.:s A(1 + Izl)M eH(z);
(ii) there are constants AI, MI > 0 such that for any r E (0, 1], z Ere, and any function g holomorphic in B(z, r), we have
Ig(z)leH(Z)
.:s
Al (l
+ Izl)Ml M(jg~ z, r), r
with v = 9dof. Proof. The first part is immediate. For the second part, let us choose any l;, Il; - zl = r/4, such that If(nl
= M(j, z, r/4) > O.
From the previous observation, there is apE ]r/4,r/2[ such that minlw-~I=plf(w)1 can be estimated from below. Let us write g = f g / f, then max
Iw-~I=p
If(w)g(w)1
M(jg, z, r) Ig(z)l.:s -'--m-i-n-If-(-w-)-I-.:s -m-in'--"-I-f-(w--)I· Iw-~I=p
Iw-~I=p
From the above estimates, if Iw -Z; I = P. then log If (w) I - log If (z;) I ::: -8 [log M (f,z;, 2er) - log If (z;)I]· Hence log If(w)1 ::: 9 log M(t, z, r/4) - 8 log M(f, l;, 2er).
3. Exponential Polynomials
210
From Proposition 3.1.19, we have log M(f, z, r/4) ::: H(z)
+ (do/) log r + C',
for some C' E IR. (C' = -logC - (do/) log 4, ro = 1, in the notation of that proposition). On the other hand, by part (i) we have
+ M logO + Is I) + A' H(z) + M log(l + Izl) + A",
logM(f, S, 2er) .:::: H(s) .::::
since Is - zl .:::: r/4 .:::: 1/4. It follows that log [ min If(W)I]::: H(z) - MJlog(l Iw-{I=p
+ Izl) + vlogr -
M, = 8M, v = 9dof. This estimate implies that part (ii) is correct.
A",
0
We remark that if f E E~, then M = MJ = 0 in the previous proposition, but dOf = N - 1 > 0 (unless f is just an exponential), hence v > 0 also. The following consequence of Proposition 3.1.21 is an explicit form of the Lojasiewicz inequality for functions in E:
3.1.22. Theorem (Global Lojasiewicz Inequality). Let f E E*, and let V = {z E C: fez) = O} =1= 0. There exists a positive constant C such that
f
I (z)l:::
C d(z, V)" H(') (l + Izl)1' e -,
where v=9dof, /-L=v+8max'~J:"'Nm). Moreover, C can be explicitly estimated in terms of the frequencies and coefficients of f. Proof. Let z ¢ V and consider g = l/f in B(z, d(z, V». Let us choose r = inf(l, d (z, V». We can therefore apply the previous proposition and obtain eH(z)
---- <
If(z)1 -
with MJ = 8 max'.::;).::;N mj, v d (z, V) and we obtain
(1
+ Izl)MI
A1 - - - - - - - -
= 9d of,
rV
and A, > O. If d(z, V) .:::: 1, then r =
1 H(-) d(z, V)V If(z)l::: Ate • (1 + Izi)MI . If d(z, V) > 1, then r = 1, but d(z, V) .:::: Izl we still obtain the required inequality.
+ const, hence letting /-L = M, + v, 0
3.1.23. Remark. If we replace d(z, V) by d(z, V) = inf(1, d(z, V», then one can take /-L = 0 in Theorem 3.1.22 when fEED, at least when V =f 0. When V = 0, fez) = ae AZ , a E C*, then d(z, V) = 00 but d(z, V) = 1, v = 0, /-L = 0, H(z) = Re(Az) and the inequality is trivially satisfied.
3.1. The Ring of Exponential Polynomials
211
From previous remarks we see that we have a rather precise knowledge of all the constants C, f.1, v in the Lojasiewicz inequality. It is rather annoying that f.1 could be positive. We shall show in what follows that one can take f.1 = 0 (after replacing d(z, V) by d(z, V», but at the cost of losing track of the nature of C. This is an argument due to Grudzinski [Grudl]. Let us consider a vector a = (ai, ... , aN) E eN such that al -< a2 -< ... -< aN, m = (ml, ... , mN) E NN. Set Iml = ml + ... + mN, f.1 = Iml + N1, lIall = maxI:::;j:::;N lajl. Let E(a,m) be the vector space of all exponentialpolynomials of the fonn I(z) = Ll:::;j:::;N Pj(z)e"jZ, deg Pj ~ mj, 1 ~ j ~ N, where we allow some of the Pj to be zero. Denote Pj(z) = LO
3.1.24. Lemma. There is a constant c = c(a, m) > 0 such that lor any IE E(a, m) and any r, 0 < r ~ 1, M(f, 0, r) ~ cr/LIIIII.
Proof. We have (M(f, 0, r)l ~ -
1
1
21l'
2".
I/(re ii1 )1 2 dO =
0
L 1/(1)(0) --, - 12 r21 l. 12:0
L I(~!(O) ~ (L
~
1
12 r21
0:::;1:::;f.L
1I(I:!(O) 12)
r2f.L.
O:::;I:::;/L
On the other hand, the map I 1-+ (Lo:::;I:::;/L 1/(I)(0)/1!12)1/2 clearly defines a seminonn in E(a, m), but the previous observation about the multiplicity of zeros for functions in E(a, m), shows that it is really a nonn. Since any two nonns in E(a, m) are equivalent, there is a constant c = c(a, m) > 0 such that (
0~f.L 1I(~!(O) 12
'/2 )
~ clllll· o
This proves the lemma.
3.1.25. Corollary. For any I
E E(a, m),
0< r
~
1, we have
M(f, 0, r) ~ cr/L max laj I, 15}:::;N with the same positive constant c
Proof. From the definition of
lajl::: II!II.
01 the previous lemma. aj = aj(!) = leading term
of Pj, we have 0
3. Exponential Polynomials
212
3.1.26. Remark. If none of the aj is zero, an inequality of the fonn M(f, 0, r)
~
crl"ll/lI
is still valid for r ~ 1, with c = c(a, m) > 0, independent of I. In fact, in this case I E E(a, m)\{O} cannot be a polynomial, hence
is also a nonn in £(a, m). Moreover, the previous proof yields for r
~
1
which shows the existence of the constant c. The main point of the argument of Grudzinski, to improve the previous Lojasiewicz inequality of Theorem 3.1.22, is the following observation. Let L z (f) be the exponential polynomial defined by Lz(f)(W) := I(w
+ z),
which also lies in E(a, m). Moreover, the leading coefficients of are related by identities
I
and L:(f)
As an immediate corollary of these identities, we obtain: 3.1.27. Corollary. For any r
E
]0, 1],
I
E
E(a, m), and z
E
C, we have
M(f, z, r) ~ erl" max {laj(f)leReajZ} ~ e(f)rl"eH(z), l~j~N
=
where c c(a, m) > 0 is independent of f, and the constant e(f) depends on f, but c(f) > 0 if f =1= O. Let I E £(a, m) be a nonzero exponential polynomial such that f(O) = O. Let v be the multiplicity of the origin as zero of f. Let 0 < r S 1 be given and fix p E ]0. r] such that f has no zeros in 0 < Izl s p. The monic polynomial Qj(z) = ZV has exactly the same zeros as f in 0 S Izi s p, hence f/Qj is a holomorphic nonvanishing function in B(O, p). Therefore, we have inf1z l9 If(z)/Qj(z)1 > O. Since maxlzl=p Ig(z)1 is also a nonn in E(a. m), if IIf - gil is sufficiently small. the function g E E(a. m) does not vanish on Izi = p and has the same number of zeros v as f in B(O, p). Moreover, if Qg is the monic polynomial of degree v whose roots coincide with those of g in B(O. p). we have that Qg is continuous as a function of g, for g close to f. Namely, recall that if ZI •••• , Zv
3.1. The Ring of Exponential Polynomials
are the roots of g, then for k
E
213
N* the Newton sums of the roots are given by
L zf = -1-1 2:n:i
O:'Olsv
zkg'(z) dz g(z) ,
Izl=p
and the coefficients of Qg are fixed polynomials in the Newton sums. Therefore, one can find a fixed number K > 0 such that inf Izl:'Op
IQg(z) g(z) I >
K
> 0
for any g in E(a, m), which is sufficiently close to f. Now let h be an arbitrary function holomorphic in B(O, r). We have M(gh, 0, r)
~ M(gh, 0, p) = M
( Qg . h·
~g ,0, p) ~ KM(Qg . h, 0, p).
In order to estimate this last quantity, let us prove the following simple lemma: 3.1.28. Lemma. There is a constant C v > 0 such that for any monic polynomial Q of degree v and any p > 0, there exists pi E [p /2, p] for which
IQ(z)1
if lzl
~ Cvpv
= p'.
Proof. (This lemma is also a consequence of the Cartan-Boutroux lemma [BO, §4.5.13].) Decompose the annulus p/2:s lzl :s p into v + I concentric annuli using circles such that the difference of the successive radii is exactly p/2(v + 1). One of these annuli contains no roots of Q. Let pi be the radius of the median circle of one such annuli. Let Q(z) = (z - ZI)" • (z - zv). For lzl = pi we have
Iz - zjl ~
p
4(v
+ 1)
for every j. Hence pV
IQ(z)1 ~
(4(v
+
1))"'
o
Izl = p'.
Returning to the previous inequalities we have M(gh, 0, r)
~
KM(Qg . h, 0, pi)
~
KCvpv M(h, 0, pi)
~
KCvp"lh(O)I.
Since K and p depend on f, all one can conclude is that there is a constant c(f) > 0 such that for every gin E(a, m) sufficiently close to f. one has M(gh, 0, r)
~
c(f)rVlh(O)I,
where c(1) has been chosen so that c(f)r V = KCvpv. Moreover, since r and v :s J.t, we also have M(gh, 0, r)
~
:s
1
c(l)rillh(O)I.
The unit sphere of E(a, m), IIfll = I, is compact, hence there is a constant Co = coCa, m) > 0 such that for any g. IIgll = I, we have M(gh, 0, r)
~
corlLlh(O)I.
3. Exponential Polynomials
214
f
Finally, we can conclude that for any M(fh, 0, r)
~ collfllr
l1
lh(O)1
~
E(a, m),
E
Co
C~~XN laj(f)I) r
l1
lh(O)1
for any h holomorphic in B(O, r). This inequality leads almost immediately to the following:
3.1.29. Proposition ([Grudl]). Let f minl-oj-oN laj (f)I. Then, for any z a neighborhood of B(z, r), we have Co
E(a, m), 0< r ::; I, and let K(f) = C and any holomorphic function g in
E E
M(fg, z, r) ~ K(f)r l1 e H(::)jg(z)l.
Proof. Let h = L:(g), then h(O) = g(z), and M(fg, z, r) = M(L:(f)h, 0, r)
Now, aj(L::(f»
~ Co (max la;(L:(f)I) r l1 lh(O)I. [-o'-oN
= aj(f)ecx}: #- 0 if and only if aj
is a frequency of f. But
max{Reajz: a, is a frequency of f} = H(z).
Hence,
and we obtain the stated inequality.
o
One should compare this statement with Proposition 3.1.21(ii). Here one can take JL = dOf and, moreover, there is no factor (1 + Izl)M,. The trade off is the very unpredictable behavior of the constant Co (see Remark 3.1.21). As a further corollary we obtain a global Lojasiewicz inequality without denominator.
3.1.30. Proposition ([Grudl]). Let f be a nonzero exponential polynomial with frequencies a[ -< 0i2 -< ... -< aN, V = {z E C: f (z) = OJ, and let d(z, V) = inf{l, d(z, V)}. There is a constant c = c(f) > 0 such that
If(z)1
~ c(d(z, V»dO j eH(z).
Proof. It is enough to consider fez) = L[-oi-oN Pi (z)eCX,Z ,mj = deg Pj , Pj #- O. Then JL = Iml + N - 1 = dOf, and the same proof of Theorem 3.1.22 applies, 0 replacing Proposition 3.1.21 by Proposition 3.1.28. We shall now show that the exponential polynomials are not only slowly decreasing in the sense of the previous chapter, but that one can give a very precise description of the size of the components of S(f, e, C) (see Definition 2.2.13).
3.1.31. Proposition. Let f be a nonzero exponential polynomial. For every s > 0, C > 0, one can find s[, C[ > Osuch thatfor every connected component 0 of
3.1. The Ring of Exponential Polynomials
215
S(f, cI, C1) = {Z E C: If(z)1 < cle- Cilz1 } one has IZI - z21 < ee- Cizil for every Zl, Z2 E
O.
°
Proof. We can assume c :s 1. Fix Z E 0 and let < r :s 1, to be chosen later (depending on z). Consider l; E aB(z, r) such that If(nl = M(f, z, r). As pointed out after the proof of Lemma 4.2.11, the Minimum Modulus Theorem yields P E ]~r, 2r[ such that for every w, Iw - l;1 = p, one has If(w)1
log If(nl > -810gM
(
f ) f(l;),l;,4er .
Hence, inf Iw-,I=p
If(w)1 > If(l;W - M(f, C;, 4er)8
We know that for some C > 0, J1 > 0,
If(l;)1 = M(f, z, r):::
Crf.LeH(z),
and, for some D > 0, M(f, l;, 4er) :S M(f, z, 5er):s D(l
+ IzI)MeH(zl.
Therefore,
'w~~f=p If(w)1
Er"e H (:) ::: (l
+ Izl)N
for some E > 0, N ::: 0, v > O. Let us now choose r = r(z) = ce-Clzlj6ec. We are going to choose CI > 0, C 1 > 0 so that for any z E C one has C
_ Er(z)"eH(z) e- C11 - 1 < ---....,-;I (1 + Izl)N .
To verify that this choice is possible, let us observe that there is a > 0 such that for every z E C H(z) :::
Since r(z)
-alzl.
= ee-Clzlj6eC, we have to choose £1,
CI so that
£" E e(C,-a-"C)lzl cI<-----....,.,.- (6e C )" (l Iz I)N .
+
This shows that once we choose C 1 > a + vC, e.g., C 1 = a easily find £1 > 0 satisfying the inequality (t). Hence, inf
Iw-;I=p
+ vC + 1, we can
If(w)l::: cle-C,lzl,
which implies that
o Therefore, if
Zl, Z2 E
IZI - z21
~ B(l;, p) ~ B(z, 3r).
0 we have
:s IZI -
zl
+ Iz -
The last inequality follows from
:s ee- Cizl - C :s ce- C1zil . the fact that Iz - zll < 3r :s 3£/6 .::: z21 < 6r
~.
D
3. Exponential Polynomials
216
3.1.32. Remark. In the case where all the frequencies of I are purely imaginary, then H (z) ::: -O! I 1m z I for some O! > 0 and we can replace Iz I throughout the proof by p(z) = logO Iz12) I Imzl. In this case
+
+
SU, B" C,) = {z
C: I/(z)l <
E
B,e- C1P (:)}
and the diameter condition is IZI - z21 < Be-CP(z,) , EXERCISES
Zl, Z2 E O.
for
3.1.
1. Let f be an exponential polynomial with frequencies AI"", AN, and K = cv{):,J, ... , ):,N J, and let H = HK = the supporting function of K. Show there is a set E ~ [0, oo[ of relative zero measure (see §3.5.9) such that '0
· log If(re' I1m r
r-oo
• 'IE
)1 -_
H(
e
iO)
.
Conclude that f is of completely regular growth and that if /-i E JIf' (C) is such that ;'H/-i) = f. then /-i*: JIf(n + K) -4 .}f(n) is surjective for any open and convex subset n of C.
2. Let f be an exponential polynomial with purely imaginary frequencies. Let g be an entire function such that for some m E Z;, A, B > 0, the product fg satisfies the estimate (z E C).
(a) Show that g satisfies an estimate of the form
Ig(z)1
~ B'(l
+ IzD""e A'11m:l,
and quantify the dependency of B', m' in terms of B, A, m. and constants depending only on f. What can you say about h g? (b) Conclude that if g is an entire function such that fg E .r(e'(~» (resp . .r('D(~») then g belongs to .r(e'(~» (resp. .r('D(~»). 3. Let f be an exponential polynomial, V = V(f) = (Zk> mkh?I' p(z) = Izi. Show that V is an interpolating variety for Ap(C) if and only if there exist E > 0, C > 0, such that for every pair (k, j), k 1= j.
IZk -
Zj
I ~ ee- CI : kl •
What is the corresponding statement for p(z) = Iz IP, P > I? Could V be an interpolation variety when 0 < p < I if f has more than one frequency? 4. Let f be an exponential polynomial, V = V(f) = (z .. md.(':I' p ::c: I. Show that the following two statements are equivalent: (i) There is A > 0 such that IZk - Zj I ~ e-A
°
5. Let f be an exponential polynomial with purely imaginary frequencies, and p(z) = Ilmzl +Iog(l + Izl), V = V(n = (Zk>mdk?l. Show that V is an interpolation variety
3.2. Distributions of Zeros of an Exponential Polynomial
217
for Ap(C) if and only if there are e > 0, A > 0, such that for every pair (k, j), k #- j. Equivalently, show this condition can be replaced by the existence of 0 > 0, B > 0, such that for every k I/(md(Zk)1 ~ oe-Bp(zd. 6. Let 1 be an exponential polynomial with purely imaginary frequencies, icxj' al < a2 < ... < a., and ml, ... , m., the degrees of the corresponding coefficients, and let J.L > max{mk!(ak - al): 2:5 k :5 n}. Show that if Imz > J.Lloglzl,
Izl» 1,
then I(z) #- 0. (Hint: Consider e- ia1z I(z).) Conclude that there is A > V
= V(j) <; {z:
I Imzl :5 A(I
°such that
+ log(l + Izl)))·
Show that V is an interpolation variety for :F(t:'(~» if and only if there exists e, B > such that IZk - zjl ~ e(IZkl + IZjJ)-B for any pair of distinct zeros of
°
I.
3.2. Distributions of Zeros of an Exponential Polynomial Given ep E [-n /2, 3n /2[. f-L E 1R, and 1/ E]O, +00[. we denote by V (ep, J.L, H) the half-strip:
Veep, /-L, H) := {z
E
+ /-Llog Izll ~ HI· angle q; + n /2 with the
C: Im(ze- i ",) ~ 0 and I Re(ze- i ",)
Let 8 denote the ray starting at 0 and making an positive real axis. By T(e) we denote the angular sector of width 2e and having 8 as a bisectrix. (See Figures 3.1 and 3.2.) If /-L = 0, V(cp, 0, H) is a half-strip limited by two half-lines parallel to 8 and equidistant from 8 at a distance H. In this case, it is a straight half-strip. If /-L ::/= 0, the two unbounded portions of the boundary are curves asymptotic to lines paralled to 8. We then say that V(ep, /-L, H) is a logarithmic half-strip. Let us choose argz E [0, 2n[, then lim
Izl-oo
argz
n
= ep + -, 2
ZEV("'.I1. H j
which means that, for every
e>
0, there is an r > 0 such that
Veep, /-L, H) n B(O, r)C
~
T(e).
3.2.1. Lemma. The intersection o/two half-strips V(cp) , f-L), H) is an unbounded set if and only if ep) = cpz and f-L) = f-L2·
n V (ep2, J.L2,
H2)
218
3. Exponential Polynomials
Figure 3.1
T(e)
Figure 3.2
3.2. Distributions of Zeros of an Exponential Polynomial
219
Proof. From the previous observation, if V = V(CPI, /11, Ht} n V(cpz, /12, H2 ) is unbounded, then
lim arg z ZEV
Izl .... oo
1r
1r
= CPI + -2 = CP2 + -. 2
Moreover, /11
=
.
Re(ze-i\O\)
lIm
log Izl
ZEV
Iz I"" 00
.
=
hm ZEV
Izl .... oo
Re(ze-\02)
log Izl
o
= /12.
We shall now prove a very famous result of P6lya concerning the distribution of zeros of exponential polynomials. Following the work of Dickson [Dill we shall give simultaneously the result for a larger class of functions, those that are asymptotically exponential polynomials (AEP) (see also [RoJ). 3.2.2. Definition. We say that f f E .Tf(C\K) and has the form fez)
=
E
L
AEP if there is a compact set K such that Ajzmj(l +Cj(z»e WjZ ,
l:sj:sn
for a finite collection of distinct frequencies WI, •.. , Wn E C, 2 and the functions Cj are holomorphic in K C and satisfy
~ n, Aj E
C*,
mj EN,
lim Cj(z) = O.
Izl .... oo
Note that any exponential polynomial with more than one frequency is in AEP. It is enough to write the polynomial coefficients Pj as Pj(z)
where
mj =
deg P, Cj(z) =
= ajZmj (l
Pj(z)/ajZmj -
+ Cj(z», 1, which is holomorphic for Izi
»
1.
3.2.3. Definition. For a function f E APE represented by (*), we denote by P := P (f), the P61ya polygon of f, the convex hull of the set WI, ... , Wn , and p := {z E C: Z E PCf)}' the set of conjugates. In what follows we shall assume that WI, ••• ,Wn are indexed in such a way that the following geometric condition holds: the points WI, ... ,wa are the vertices of P and, moreover, the oriented segments LI := [WI, ci}z], ... , L a- I := [Wa-I, w,,], L" := [w", WI], are precisely the successive segments of the posititively oriented boundary ap. The remaining points are ordered arbitrarily. Note that some may still be in aP. Let us denote by hk the ray that starts at the origin and has the direction of the outer normal to aP at any point in the open segment ]Wb Wk+1 [, 1 ~ k ~ a-I, h" is the one that corresponds to [wa , wd. We make a further assumption on the choice of WI. if '!/fl, ..• , '!/fa are the arguments of the points in hI •... , h",
220
3. Exponential Polynomials
respectively, then 0 ::: 1/11 < 1/12 < ... < 1/1" < 2rr. Note that for z E hk one has Re(wkz) = Re(wk+lz) for 1 ::: k ::: a - I (with the obvious extension to the case k = a). Let 51 := {z E c: 1/1" ::: arg z ::: 1/11 + 2rr} and, for k = 2, ... , a,
5k := {z
E
C: 1/Ik-l ::: argz ::: 1/Id.
Fora fixed 0 verifying 0 < 20 < min{1/I1 let us define SI
-1/1" + 2rr, 1/Ik -1/Ik-1 (k =
+ 0 ::: arg z ::: 1/11 + 2rr E C: 1/Ik-1 + 0::: argz ::: 1/Ik - O}
:= {z E C: 1/1"
Sk := {z
2, ... , a)),
O},
(2::: k ::: a).
Note that because n :::: 2 one has 0 < 0 < rr /2. Finally, let us denote
Tk := {z E C: 1/Ik -
(J :::
argz ::: 1/Ik
(See Figure 3.3.)
Figure 3.3
+ O}
(1 ::: k ::: a).
3.2. Distributions of Zeros of an Exponential Polynomial
221
3.2.4. Lemma. (a) If z E Sk. ~ E P, then Re«wk - ~)z) ~ o. (b) Ifz E Sk. ~ E P,then Re«wk -~)z) ~ Izllwk -~lsinO. (c) If z E Sk n Tk and Wj ¢ Lb then there is a value eo > 0 such that Re«wk - Wj )z) ~ Iz Ilwk - Wj I sin 00 . (d) If z E Sk+1 n Tk and Wj ¢ L k . then there is a value 01 > 0 such that Re«wk+1 - Wj)z) ~ Izllwk+1 - wjl sine l .
Proof. Recall that (·1·) denotes the Euclidean scalar product on ]R2 and that for any a, b E C, Reab = (alb) = (alb). We shall give the proof of the lemma for 2 :::: k :::: a - I, leaving it to the reader to adapt it to the cases k = I, k = a.
Part (a). We observe that if h-I denotes the ray perpendicular to hk-h chosen so that the angle from h-I to h k- I is n/2, and h, the ray perpendicular to hk. so that the angle from hk to h is n /2, then the sector Sk is complementary to the two quadrants determined by h-I and hk-I and by hk and [k. The point ~ E P and, therefore, ~ - Wk belongs to the closed sector determined by h-I and h. This is just the sector opposite to Sk. Hence, the angle between any z E Sk and ~ - Wk is at least n /2. In other words, Re(z(~ - Wk» = (zl~ - wd ::::
o.
This is equivalent to statement (a).
Part (b). Here the angle between z E Sk and ~ - Wk is ~ nl2 + 0, hence (zl~ - Wk) :::: -lzll~ -
wkl sinO.
Part (c). Since Wj ¢ Lko one can immediately see there is a Of > 0 such that
n, _ _ 3n 1/Jk + "2 + 0 :::: arg(wj - wd :::: 1/Jk-1 + T· On the other hand, z E Sk n Tb so that 1/Jk-1
+ e < 1/Jk -
0 :::: arg z :::: 1/Jk.
Hence, the angle between z and Wj - Wk lies between n/2 + 0' and 3nl2 - O. Letting 00 = min{O, O'}, we obtain (c).
Part (d). This time there is Oil > 0 such that ./,
'f'k+1
n + "2 :::: arg (Wj_
_
)
- Wk+1 ::::
./,
'f'k
n + "2 -
nil (7
,
and
1/Jk:::: argz:::: 1/Jk +0 < 1/Jk+1
-e.
Choosing 01 = mintO, Oil) we can proceed as before.
3.2.5. Lemma. (a) Ifz E Sk and j =1= k, then zmjeWjZ = e(z)zm'e w*,. (b) liz E Tk n Sk and Wj If Lb then zmjeWjZ = e(z)zm'eWkZ .
D
222
3. Exponential Polynomials
n Sk+l and Wj f/ Lb then zmjeWjZ = e(z)zmt+JeWk+Jz, where e(z) denotes a quantity, different in each case, such that
(c) If z E Tk
lim e(z) z~oo
= 0.
whenever z is restricted to the corresponding region. Proof. (a) By Lemma 3.2.4(b), we have
which vanishes at 00. (b) It is the same as part (a) using Lemma 3.2.4(c), and (c) Just use Lemma 3.2.4(d).
3.2.6. Lemma. For f
E
APE represented by (*) and any z fez) = Akzmk(l + e(z»e WkZ ,
where lim e(z)
z_oo
eo
instead of e. 0
E Sb
= 0.
ZESt
Proof. This is an immediate corollary of Lemma 3.2.5(a), and the definition (*) ~f.
0
3.2.7. Corollary. Under the same conditions of the previous lemma, there is r > such that
°
for all z
E
Sk
n B(O, r)c.
if
Proof. From Corollary 3.2.7 we have If(z)1 > (IA k I/2)lz mk e""zl whenever z Sk n B(O, ro)c. From Lemma 3.2.4(a), we have
E
T
> 0, such that
1e~~Zz~mk
~
le~ zz~mk
f(z)1 :::
if z E
n B(O, ro)'
and
3.2.8. Lemma. There are ro > 0, ~ E ft, then
Sk
T.
IAkl f(z)1 > -2- exp[Re«wk -
~)z)l :::
IAkl 2'
0
We remark that this lemma shows that all the zeros of f, of large absolute value, must lie within some sector Tk. The condition on T is just 0 < T < infk IAd/2. On the other hand, the definition of Sk depends on e, hence the value ro of this lemma depends on e. We proceed to study what happens in the sector Tk •
3.2.9. Lemma. Let A be the set of indices j such that Wj ELk. For have fez) = LAjzmj(1 + Bj(z»eWjZ , jElk
Z E
Tko we
3.2. Distributions of Zeros of an Exponential Polynomial
223
with lim £j(z) =0.
zET,
z~oo
Proof. This is a corollary of Lemma 3.2.5(b) and (c), the functions £j are the same as in (*) for j =I k, k + 1, those with these two indices will change. D In order to continue our study of zeros, it is convenient to introduce, for j E i k , the following quantities: 1:j
= Wj + mjei1/l"
which both depend on the frequencies and the degrees of the corresponding coefficients. To simplify the notation we rotate the coordinates and assume 1/Ik = 11/2. Let Qk denote the convex hull of the points Wb Wk+!, and 1:j, j E ik' We choose a new indexing of the vertices of Qb while traversing 8Qk in the positive sense. Start at Wk. We denote Wk,1 = Wb 1:k,l = Wk + imk (it could coincide with Wk,1 if mk 0), and denote the successive vertices 1:k,2, ... , 1:k,(71-1, 1:k.al = Wk+l + imk+l' Denote Wk.al := Wk+l' If ak > 2, all the mj that appear in 1:k,/ for 2 .::; I .::; ak - 1, are positive. We denote Wk,/ the corresponding projection onto the horizontal segment Lk = [Wb Wk+l]. The other Wj, j E iko are indexed arbitrarily as 1:k,/ = Wk.l + imk,l, with ak < I .::; nk. For 1 .::; j .::; ak-I. the segments [1:k,/, 1:k,/+)] are denoted Lk,I and the slopes f.Lk,l. Then
=
mk,l - mk,/+1
f.lk.l
= _
Wk,l - Wk,l+1
=
mk,l -
mk,l+1
Cik.1 - Cik,/+1
= Rewk.l = Rewk,l.
with
Cik.1
... >
Cik,a,.
Note that with our conventions Due to the convexity of Qko these slopes satisfy
ak,l
> CiU >
-cx:; < f.Lk,1 < f.Lk,2 < ... < J-Lk,a-l < 00.
Let z = x +iy, for H > 0 we define Uk•1 := {z E C: y 2: 0 and x + J-Lk.llog Izl > H}, Uk.l
:= {z E C: Y 2: 0, x + J-Lk,l-llog Izl < -H and x + f.Lk,llog Izl > H},
for I = 2, ... , ak
-
1, and
Uk.a, :=
{z E C: y 2: 0, x + J-Lk.a, log Izl < -H}.
Let us also introduce V k./
:=
V(O, f.Lk,/,
H) = {x E C: y 2: 0, Ix +J-Lk,I!oglzll.::; H}.
Figure 3.4 depicts these regions for 1 .::; I .::; ak - 1. The Uk,1 together with the V k .1 cover Tk (see Exercise 3.2.1). For R » 1, the different sets Vk,1 n B(O, Rr, Uk,! n B(O, r)', are connected and disjoint. Remark further that the "median" of Vk,1 is the curve x + J-Lk,/!og Izl = 0, which is asymptotic to the curve x + J-Lk./!Og Y = 0. We keep from now until Remark 3.2.22 the assumption that 1/Ik = 11/2 and the preceding notation.
3. Exponential Polynomials
224
Vk.3
Vk2
"tk.o,_1
"tk•O,
Figure 3.4
3.2.10. Lemma. There is v > 0 such that for every
-v,
mk,p - mk,l+l - I1k,/(ak,p - ak,/+l) < -v.
Proof. Because of our choice of coordinates it is natural to consider variables a, m, so that the a-axis is the real axis. In these variables, the line Lk.l can be represented by either of the two equations
In these coordinates, the point
O.
Since there are only finitely many possible
o
3.2.11. Lemma. Let Z
E Uk,j' then:
(a) If j = l,assume
with
lim e(z) = O.
z.... oo
zeUk,j
3.2. Distributions of Zeros of an Exponential Polynomial
(b) If I .::: j .:::
Uk -
1 and
Lk.p E L k. j ,
225
or if2 .::: j .:::
Uk
and
Lk,p E Lk,j-h
then
Proof. (a) If j = 1, ... , Uk - 1, there is a iJ- < iJ-k,j (fl- depends on z) such that x + JL log Izi = O. If j = 2, ... ,Uk< there is a iJ- > iJ-k,j-I with the same property. (Just use the definition of Uk . j .) Since Im(wk.p - Wk,j) = 0 and Re(wk,p - Wk,j) = ak,p - ak,j' Iz m , p-m, j e(Wk,p-W, j)Z
I = Izl m! p-mC}-I1-(ak,p-a, j) log Izl
= Izlm,p-mkj-l1-(akP-a,j). Consider the case when ak,p - ak,j < O. Then definitely j Lk. i, and iJ- < iJ-k,j' The previous lemma implies that mk,p - mk.j -
fl-(ak,p -
ak,j)
<
mk,p - mk,j -
i- Uk, hence
iJ-k.) (ak,p - ak,j)
mk,j -
fl-(ak.p - ak,j)
<
mk,p -
mk,j -
iJ-k,j-l(ak,p - ak,j)
¢
< -v
for some v > O. In case ak,p - ak,j > 0, we have j > 1 and use the hypothesis that Lk,j-l when j = 2, . , , ,(Tk. We obtain mk.p -
Lk,p
Lk,p
¢
< -v.
Finally, if ak,p - ak,j = 0, we must have that mk,p < mk,j' Choosing v smaller if necessary, we have in any case, 1£(z)1 < Izl-v. (b) If j = 1, ... ,Uk - 1 and rk,p E Lk,j, then mk,p - mk,j = iJ-k,j (ak,p - ak,j)' On the other hand, we have x + iJ-10g Izi = 0 (with the same choice of iJas in (a» and x + iJ-k,j log Izl :::: H. Hence (iJ-k.j - fl-) log Izi :::: H and, since iJ-k,j - iJ- > 0, Iz I :::: exp(H /(iJ-k,j - iJ-», Because rk,p E Lk,j, we have ak,p .::: ak,j and therefore,
= Iz 111-' j (a, p-ct, j )-I1-(ct k p-ctk j) .::: eH(ct,p-akj)
=
e-HiaLP-akjl
so that
o
The other case is verified in exactly the same way.
3.2.12. Lemma. There exists r > 0 such that if z (a) Izm ,
(b)
jeW! jZ
I ::::
Izmk" ew"lz I;
E
Uk,j
n B(O, rY
then:
and
Izmkjew'.jZI :::: Izmk'"kewk,u,zl.
Proof. (a) If j = 1, the result is obvious. If j = 2, Lk, I E Lk,l, and we can apply (b) of the previous lemma with p = 1, j = 2. Hence
3. Exponential Polynomials
226
If 3 S j S ab then Lk,I ¢ Lk.j-I, and by Lemma 3.2.1l(a), we have zm, I e"'l.I z = £(z)zm"le""jZ, hence the inequality (a) holds if Izl» 1. The proof of (b) is similar, with Lk,rI, replacing Lk,l. 0
3.2.13. Lemma. If z
E
Uk,j n Tko then
f(z) = LAk,pzm,p(l +£k,p(z»e""p Z , where the sum is taken over the indices p such that: (i) Lk,p E Lk,I when j = 1; (ii) Lk,p E Lk,j n Lk,)-I when 2 S j S ak - 1; (iii) Lk,p E Lk,rI,-1 when j = ak; and, as always, lim £k,p(Z) = 0, Izl-H)O
ZEU, jnT,
Proof. It is an immediate consequence of Lemmas 3.2.9 and 3.2,Il(a).
0
3.2.14. Lemma. There are constants ro > 0, Ho > 0, such that if r ::: ro, and H ::: Ho, then for any z E Uk,j n Tk n B(O, r)f one has If(z)l> !IAk,jllzm,jeW,jZI.
Proof. We use the asymptotic development of f given in the previous lemma, and then use Lemma 3.2.11(b). We obtain If(z) I ::: IAk,jZm'j eW'j (1 = IAk,jZm, jeW, j x
+ £k,j (z»1 -
L IAk,pZm"p (l p,;,j
+ £k,p (z»e W, I pZ
I
(11 +£k,;(z)l- L
II~k'Pllll +£k,p(z)lzm"p-m'le(W,p-W,j)zJ) k,J
::: IAk,jZm'le"",jl {II +£k,j(z)l-
L II~k'Pllll +£k,p(Z)le-HIWl.p-Wkll} , ph
k,J
It is clear that by choosing r and H sufficiently big, the required estimate holds.
o
From now on we will assume r ::: ro ::: 1 and H ::: Ho, so that the conclusion of Lemma 3.2.14 holds. Note that the regions Tko and correspondingly the value ro, depend on the opening 9. On the other hand, we can take Ho independent of e. Namely, it is enough to choose it so that
"lax ,J
(E
IAk,pl e-HOlw"P-W"jl) S p,;,j IAk.j I
~,
3:2. Distributions of Zeros of an Exponential Polynomial
227
since we can take ro so that for Izl :::: ro, z E Tko we have
ISk,)(z)1 :::: ~ forallk,j.
3.2.15. Lemma. Let 0 < P, then
i"
< infk ,) IAk,) 1/2. If z E Uk,) n Tk, Iz I :::: ro, and ~ E le-~z f(z)1 > .,
Proof. If z E Uk,) Lemma 3,2.12(a) le-~z f(z)1
n Tk n Sk we have from the previous lemma and from :::: !IAk,jzmk,je"'k jZe-~ZI :::: !IAk,)zmk.le"'k,'Ze-~zl IAk,) II Imk 'I > . > -z -
2
= IA;,)IIZlmk.leRe«wk-~)Z)
'
where the previous to last inequality uses Lemma 3.2.4(a). If z E Uk,) n Tk n SHI, the reasoning is the same, with mk,! replaced by mk'''k'
o
3.2.16. Proposition. For r » 1,0 < i" < infk ,) IAk,)1/2, by switching the conditions ~ E P and z E (Uk,) Vk,jY n B(O, r)C, then le-~z
f(z)1 > .,
Hence, If(z)1 > uH(z), where H (z) is the supporting function of P. Proof. We fix some () > 0, as we have done above and let Tk = Tk «(}), Then, by Lemma 3.2.8, the inequality le-~z f(z)1 > i" holds if z E Sk, r :::: ro » 1. This value ro depends on the choice of (), but if we make it sufficiently large the conclusion of the previous Lemma 3.2.16 also holds for z E Uk,) n Tk for some k, j. On the other hand, if Z E Vk,j and Izl > rio then z E Tk; if z E Uk,) Vk,) either z does not belong to any Tko i.e., it belongs to some Si, or it belongs for Izl :::: r[, to some Uk,) n Tk. In both cases the inequality le-{Z f(z)1 whenever z rf. Uk,) Vk ,). Recalling that
>.
ma~Re(~z) ~EP
= max Re(wz) = max(wlz) = H(z), WEP WEP
concludes the proof of Proposition 3.2.16.
o
A consequence of the proposition is that all the zeros of f with large absolute value lie in Uk,j Vk,j' Note that because the Vk,j do not depend on e, the rl can be considered independent of e. The reader should verify that for those
228
3. Exponential Polynomials
segments Lb which contain no Wj in their interior, everything we did above works, it is even simpler. We also suggest comparing Proposition 3.2.16 with Corollary 3.1.27, we get more precision on the constants, and not only about local maxima of f, albeit only for r ~ rl. We shall presently study the behavior of f in one of the half-strips Vk.j.
3.2.17. Lemma. Let z (a)
If 'k,p f/
L~.j,
E
Vk,j' Then
then
for some function 8(Z) --. 0 as z --. rt Lk,j, then we also have
00
(z
E
Vk,j)'
(b) If'k,p
same condition on £ as that in the previous item, (c) If,,,p E Lk,j' then
Proof. Since z E Vk,j we have x + /lk,j log Izl = K, for some K such that IK I :s H. By Lemma 3,2.10, if 'k,p f/ Lk,j, we have Izm, p-m'.j e(w, p-OJ, J): I = Iz 1m, p-m, j eX (a, p-a, J)
= Izlm, p-m, j-J.l., j(a, p-ex, j)e(a, p-OI., j)K
This proves part (a). Part (b) is entirely analogous. If 'k,p E Lk,j then mk,p - mk,j = /lk,j(CXk,p - CXk,j), so that the previous computation shows that
o 3.2.18. Lemma. For r
»
1, one has:
(a) Izm"ew"zl:s Izm'iew'Jzl,ijl:sj :sak-l;and (b) Izm'u'ew,u"1 :s Izm, ieWkjZI, ij2:s j :s (J'k,
for every z E Vk,j n B(O, r)C, Proof. (a) If j = 1, the estimate is clear. If2:s j :s ak - 1, then we can use Lemma 3.2.17(a). (b) Similar proof, using Lemma 3.2,17(b).
3.2.19. Corollary. If z
E
fez) =
Vk.j
n Tb then
L fk,pELIt. j
A k • p(1
+ sk,p(z»zmk·peWk pZ,
'k,1
rt L k ,; and 0
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.20. Lemma. For r
229
» 1, if t; E P and z E Vk,j n B(O, rY, then Izmk,je(wkJ-nzl:::: 1.
Proof If z
E Vk,j
Similarly, if
Z E
n Sk use Lemmas 3.2.18 and 3.2.4 to conclude that Izm!,j errol j-I;)z I :::: Izl m !" :::: 1. Vk,j
n Sk+I, the same argument holds with
mk,l
by mk,lJk'
0
3.2.21. Lemma. Let z f(z)
replaced
E Vk,j
= zmk"eWk.j
n Tb
L
then
Ak,p(l +£k,p(z»e(WkP-Wkj)(Z+/LkjLogz),
fJ..,pELk,j
where hp -+ 0 as z -+ 00 within Yj,k value of the logarithm.
n Tb
and
Log z represents the principal
Proof When
Lk,p E Lk,j, we have mk,p - mk,j = JLk,j(Wk,p - Wk,j)' We then 0 apply Corollary 3.2.19, and we are done.
3.2.22. Remark. If we do not make the assumption that 1{!k = 71: /2, and hence that Im(wk,p - Wk,j) = 0, then we can introduce CPk = 1{!k - 71:/2 in the expansion of f as follows:
2::
f(z)=zm'jeWk,jZ
A k,p(I+£k,p(z»
f1c,p EL k,j
x exp[ei'l" (Wk,p - Wk,j)(ze-i'l"
+ JLk,j logz)].
In this case, ei'l'k(wk.p - Wk,j) E JR, and f becomes, up to a multiplicative factor, a sum of exponentials in the variable ze-i'l'k + /-Lk.j log z, with coefficients almost constant. It also follows that the Vk,j coincide with V(cpt, JLk,j, H). In fact, the main relation to remember is that 1{!k = Arg(wk+l - wd - 71: /2, so that one has CPk = Arg(wk+1 - Wk) - 71:, hence CPk = - Arg(wk,p - Wk.j) (modulo 71:).
=
= +
3.2.23. Proposition. Let g(z) 2:1::;j::;n AjeWjZ , z x iy, Aj "# 0, Wj E JR, WI < W2 < ... < Wn • For H > 0, h > 0, and YI E R consider the rectangles R := {z E C: Ixl :::: H, Iy - YII :::: hI. There is a constant Ho :::: 0, independent of YI and h, such that every zero of g lies in the strip Ixl :::: Ho. Moreover. if H :::: Ho. the number N (R) of zeros of g in the rectangle R satisfies
/N(R) -
~(wn - WI)/ :::: n -
1.
G(z) = (l/AI)e-W1Zg(z) = 1 + 2:2::;j::;n Bje yjz • Bj = AdAI. and Yj = Wj - WI. 0 < Y2 < ... < Yn' Both functions G and g have the same zeros. For Re z = x > 0 we have
Proof Let
3. Exponential Polynomials
230
with lim.l->oo !31 (z) = O. Hence, if x » 1 we have II + !31 (z) I ::: ~, and G cannot have any zeros. Similarly, G(z) = (1 + !32(Z», where !32(Z) -')0 0 as x -')0 -00. Hence, there is Ho > 0 such that if IRe zl > Ho, G(z) #- O. Clearly, if H ::: Ho the number NCR) is independent of H. Let us fix £ > O. Choose H ::: Ho such that ne and I Arg(l + !3j(z))l < 4 on the lines x = ±H, j = 1,2. For a given h > 0, YI E R let us choose 8> 0, 0 < 8 < min{h, ne/2Yn}, with property that G has no zeros on the boundaries of the rectangles RI := [- H, H] X [YI - h - 8, YI + h + 8] and R2 := [-H, H] x [YI - h + 8, YI + h - 8]. This is possible because the zeros of G are isolated. Let us investigate now the variation of the argument of G along aRI. In the vertical portion x = - H, it is at most n e /2. On x = H, it is at most ne/2 + (2h + 28)Yn. On the horizontal sides, the function Re G(z) has the form Re G(z) = 1+
L
EjeYjX,
2sjSn
On the other hand, by Rolle's theorem, a function of the form K (x) = LI:::J:om Fje aJ '\, Fj E R*, al < a2 < ... < am, can have at most m - 1 real zeros (counting multiplicities). We show this by induction on the number of frequencies. It is clear for m = 1. If it is true for m - 1 frequencies, then we consider e- atX K(x) = L(x), L has the same number of zeros as K does. Its derivative L' is an exponential sum of the above type with m - 1 terms, hence L' has at most m - 2 zeros. Since between any two zeros of L there must be one of L', the claim is correct. Return to the function Re G(z) on a horizontal segment of aRlo We have that either Re G(z) == 0 on that segment or it has at most n - 1 zeros. In the first case, the variation of arg G along the segment is zero since G does not vanish on this segment. In the second case, the variation of arg G does not exceed ~ (n - 1)2n. Therefore, the total variation of arg G along aR I does not exceed ne
+ 2hYn + 28Yn + (n
-
1)2n.
Hence, by the argument principle N(R) ~ N(R I ) ~ n - 1 +
-hYn + e. n
By a similar reasoning hYn N(R)::: N(R2) ::: - n
n
+1-
e.
o
Since e > 0 was arbitrary, we obtain the proposition. The reader can easily verify that the example g(z) mate in Proposition 3.2.23 is optimal.
= 1+e
Z
shows the esti-
3.2. Distributions of Zeros of an Exponential Polynomial
231
We continue toward our goal of understanding the localization of the zeros of f E AEP. We are now trying to prove a sort of Lojasiewicz's inequality (see Proposition 3.1.30). We are still keeping the previous notation, i.e., 1/Ik = If /2. Let us prove a few technical observations about the regions Vk • j , 1 S j S ako which were stated at the beginning of this section. (I) If z = x + iy E Vk,j, then liml:l-+oo Iy/xl = 00. In fact, if J1.k.j = 0 the result is immediate. If J1.k,j :f= 0, let J1. = J1.k,j and let K (z) = x + 4J1.log(x 2 + yZ) E [-H. HI. Therefore, when Izl -+ 00 we also have Ix I -+ 00 and the sign of x must be the opposite to that of J1.. We can solve for y / x and obtain
(~r = x- 2 exp [2KJ1.(Z)
-
~]
The term 2K(z)/J1. is bounded, while -2x/J1. -+ Iy/xl -+ 00 as Izl -+ 00, Z E Vk,j' (2) If Z E Vk,j, then liml:l-+oo Arg Z = If /2. Namely, for large Izl we have 0 < Argz < be that Argz -+ If/2 when Izl-+ 00. (3) The curves x
+ J1.k.j log Izl = ±H
If
-1.
+00 as Izl -+ 00, hence
and, since Iy/xl -+
00,
it must
are a symptotic to the curves x
+
J1.k.j logy = ±H.
Let us show this for x + J1.log Izl = H, J1. = J1.k.j :f= 0, the case J1. = 0 being trivial. Let + (J1./2) log(x; + y2) = H and X2 + (J1./2) log y = H, then we have
Xl
1J1.1 ( 1+ IXI-x21=Tlog since XI/Y -+ 0 as Izi -+
00,
(Xly )2)
-+0
by the previous item (2).
(4) For 2 E Vk.j denote W = z + J1. Logz, J1. = J1.k.j, Logz the principal branch of the logarithm. For 8 > 0, there is rl ::: I such that if IZI - 221 ::: 8, and also zj, z2 E Vk. j n B(O, rr)c, then IWI - w21 ::: 8/2. Let Yj = Imzj. Vj = 1m Wj. If IYI - Y21 ::: 8, then we can use that by (2) VI - V2
= YI
- Y2
+ J1.(ArgzI
- Arg z2) -+ YI - Y2,
as Izl -+ 00, to conclude that IWI - w21 ::: IVI - v21 ::: 8/2 if rl If IYI - Y21 < 8, let q = YI - Y2. We have
:!. = Z2
YI Y2
(Xl + YI
i)
(X2 Y2
+
i) -I = (I + !L) (Xl + i) Y2
YI
In this case, we have XI/YI -+ 0, X2/Y2 -+ 0 and Y2 -+ ZI/22 -+ 1. Therefore, WI -
W2
= 2\
- 22
+ J1.(LogzI
- Logz2) = ZI - Z2
»
(X2 Y2
1.
+ i)-I
00 as rr -+ 00, hence
+ J1. Log (;~)
232
3. Exponential Polynomials
and Log(zl/z2) --+ 0 as r1 --+ IWI - w21 ;::: 8/2 if rl » 1.
00.
Hence, in this case, we can also conclude that
E AEP, Z = Z(n = (z E IC: Izl > R, fez) = O}, and H the supporting function of the P6lya polygon P of f. For any 8 > 0 there exist T > 0, r > 0, such that for Izl > r if d(z, Z) ;::: 8, then
3.2.24. Proposition. Let f
If(z)1 ;:::
reH(z).
Proof. From Proposition 3.2.16 we know the existence of Ho > 0 and TO > 0 such that if we fix H ;::: Ho there is ro > 0, so that Iz I > ro, z if Uk,j VA,j, implies that If(z)1 ;::: ToeH(z). Therefore, it is enough to consider z E Vk,j' Izl --+ 00. Following Remark 3,2.22, we can expand f as follows: fez)
L
= zm, jeW' jZ
Ak,p(1
+ Bk.p(Z»
T/.. pEL/.. j
x exp[ei'l" (Wk,p - Wk,j)(ze-i'l"
+ f.1k.j log z)],
and recall that the "frequencies" ei'l" (Wk,p - Wk.j) are real. Since, by Lemma 3,2.20 we know that for I'; E P, z E Vk,j, and Izl has Iz m , j e(Wk j-~)Z I ;::: I, we can just consider
»
1, one
Then we can change variables and study the sum gl(W)
=
L
Ak,p(l
+ 11k,p(w»exp({J/..,pw),
fl.,pELk,)
where W = ze-i'l'k + f.1k.j log Z, {Jk,p = ei'l" (Wk,p - Wk,j), and 11k,p (w) = £k,p (z), Clearly, 11k,p(W) --+ 0 as Iwl--+ 00, and if d(z, Z(f» ;::: 8, we have dew, Z(gl» ;::: 8/2 as long as Izl » 1. Here Z(n = (z: Izl > R, fez) = OJ, Z(gd = (w: gl (W) = OJ, the variables z, W being restricted to z E Vk,j' Izl > R, and W to the corresponding region. Note that as a consequence, I Re wi ~ Hand 1m W --+ 00, as z --+ 00, We are thus led to introduce the following exponential sum, with real frequencies:
where the summation takes place over the same set of indices as above. From what we have just said, if z --+ 00 in Vk,j, then gl (w) - go(w) --+ 0, Using Remark 3,1.23, we conclude that if Z(go) = (w E C: go(w) 0< 80 ~ 1, then dew, Z(go» ;::: 80 and I Rewl ~ H implies Ig(w) I
;::: c80
= OJ,
and
3.2. Distributions of Zeros of an Exponential Polynomial
233
for some c > 0, v :::: I. Therefore. there is r :::: 1 such that Iwl :::: r, I Re wi ::: H, and dew, Z(go» :::: 80 • then C
Ig) (w)1 ::::
280'
In fact, it is enough to choose r so that IgI(w) - go(w)1 ::: (c/2)8 0 for Iwl :::: r. As a consequence, we see that if WI E Z(gl), IRe wII ::: H, IWII :::: r, we must have d(w), Z(go» ::: 80 . Hence, if Iwl:::: r, I Rewl :::: r, and dew, Z(gl» :::: 8/2 we will have dew, Z(go» :::: 8/2 - 80 , Take 80 = 8/4, then 8/2 - 80 = 80, so that dew, Z(go» :::: 80, and thus 1!t(z)1 = IgI(w)1 ::::
~ (~r.
for z E Vk • j • Izl» 1, and d(z, Zen) :::: 8 (since this would imply that dew, Z(gl» :::: 8/2). It follows that:
If(z)e-~;;I
=
Izm'je(wkj-~)zfl(z)l:::: ~ (~r
for z E Vk,j. Izi » 1, d(z, Z(f» :::: 8, and ~ E Proposition 3.2.24.
P.
This concludes the proof of 0
Let us define. with the preceding notations for a function
f
E
AEP, ex > 0.
s > 0, H > 0, R.,j (ex, S, H) :=(z E C : Im(ze-i
+ I-ik,j arg z E
+ I-ik.j log Izil
[ex, ex + s],
::: H},
with argz E [qib qik + Jl'], and let N(Rk,j(ex, s, H» be the number of zeros of f in Rk,j(ex, s, H). We are now ready to prove the P6lya-Dickson theorem on the distribution of zeros of functions that are asymptotically exponential polynomials.
3.2.25. Theorem. Let fez) = L:I:::j:::n Ajzmj (l + ej(z»eWjZ , Aj E C*, mj EN, Wj E C, and ej are holomorphic functions in the set Izl > R, such that lim z_ co ej(z) = 0, There are constants H > 0, r :::: R such that: (a) all the zeros of f in B(O, r)' lie in Uk.j Vk,j; and (b) for every e > 0 there is ex (e) > 0 such that ifex :::: ex(e). s > 0,
IN(Rk,j(ex. s. H» - 2:
IWk,j+1 -
Wk,jll ::: nk,j - 1 + e,
where nk,j is the number of frequencies lying in the segment Lk,j [Wk,j, Wk,j+!l.
=
Proof. The statement (a) is a consequence of Proposition 3.2.16. The proof of (b) is a refinement of the proof of Proposition 3.2,24. Using the notation introduced
3. Exponential Polynomials
234
there, let f3k.p = e irp , (Wk,p - Wk,j) gk,j (w)
E
JR, and
2:
=
Ak,pe P, pW.
lk,pELk,j
°
Assume H > so that the conclusion of Proposition 3.2.23 holds for every gk,j' Note that the quantity If3k,HI - f3k,j I = IWk,HI - Wk,j I is the distance between the furthermost frequencies of gk,j' Choosing rI » I we can guarantee that for different choices of (k, j) the sets V (cpt, J-Lk,j, H + 1) n B(O, r])C are disjoint. (Recall Vk,j = V (cpt, J-Lk,j, H).) Note that rI can also be chosen so that for z E V(cpt, J-Lk,j, H + I) n B(O, rI)C we can take arg Z E [CPk, CPk + rr] and that for a convenient choice of ao Im(ze- iCP ,)
+ J-Lk,j argz :::: ao >
0.
Therefore, for a :::: ao, the regions Rk,j(a, s, H) and Rk,j(a, s, H + 1) cover Vk,j n B(O, rIY and V (cpt, J-Lk,j, H + 1) n B(O, r])C, respectively. Once we fix 8 E ]0, rr/2[ we can also choose aI » 1 so that for a :::: aI, S > 0, Rk,j(a, s, H
+ 1) £
Q := V(cpt, J-Lk,j, H
+
1)
n Tk (8) n B(O, rd c .
From Remark 3.2.22 we have fez)
2:
= zm, jeW,,) 'Ct
Ak,p(1
+ ek,p(Z» exp[f3k,p(ze- irp, + J-Lk.j logz)]
pEL" j
=
in the region Q (arg z E [cpt. CPk + rr]). The transformation Fk,j: z 1--+ W + iv := ze- irp, + J-Lk,j logz maps Q conformally into the half-strip lui:::::: H + 1, v :::: ao. The regions Rk,j(a, s, H + 1) are mapped conform ally onto the rectangles U
R(a,s, H
+ 8) = [-H -8, H +8]
+i[a,a +s],
for a :::: ab 8 > 0, 0:::::: 8 :::::: 1. Let fk,j(Z):= f(z)z-m'je-w'jZ and hk,j be defined by hk,joFk.j = fk,j' Hence hk,j(w) = gk,/W) + l1k,j(W), with l1k,j(W) -+ as lui:::::: H + 1 and v -+ 00. We clearly have for a :::: ab s > 0, 8 :::::: 1, that
°: :
Nj(Rk,j(a, s, H
+ 8»
°
= Nj,/Rk,j(a, s, H + 8» = N h, j(R(a, s, H + 8»,
where the index of N indicates the function whose zeros we are counting. Moreover, these values are independent of 8, since all the zeros of f lie in Vk,j' It follows from Proposition 3,2.23, applied to the gk,j, that there is a constant M > 0 such that for any a E JR, < 8 :::::: 1,
°
Ngkj(R(a,
8, H
+ 8)) :::::: M.
Hence, for any a, there is a horizontal line segment A, in R(a, s, H that dew, Z(gk,j» :::: 2(M
+ 1)
+ 8)
such
3.2. Distributions of Zeros of an Exponential Polynomial
235
for any point W EA. (The line segment A depends clearly on 8, ar, and (k, j).) It is enough to divide the rectangle in parallel horizontal strips of equal width 1/(M + 1), and take as A the bisectrix of a strip void of zeros of gk.j' Fix (k, j), s > 0, and ar ~ arl + 1, let Rex,s be the rectangle whose vertical sides lie in u = ±(H + 8) and whose horizontal sides are given
by the line segments Al and A2 corresponding to 'R(ar - 8,8, H + 8) and + s, 8, H + 8), respectively. We have that ROI,s 2 'R(ar, s, H) and that, for every point W E a ROI,s one has
'R(a
8
dew, Z(gk.j» ~ 2(M From 3.1.3 we conclude that there is r
+ 1)
= r(8)
Igk.j(w)1 ~ r
> 0 such that
on aROI,s.
If v ~ vo» 1, we have IYJk,j(w)1 :5 r/2, hence, for ar ~ ar2 ~ arl
+ 1, we have
Ng,/ROI,s) = Nh, ,(ROI,s)'
We remark that the choice of ar2 depends on 8, and eventually on E. The height of ROI .s lies between sand s + 28, and now from Proposition 3.2.23 we infer that
lN g,
J
SI<
(R OI.S ) - -IWk 21T· j+1 - Wk" .j
81 wk '+1 - Wk .j·1 . - nk .j. - 1 + . j 1T
Given B > 0, we can chose 0 < 8 :5 1 so that the left hand side does not exceed nk.j - I + E. for any (k, j). Hence Nj ('Rk,j (a, s,
H» = N h( /'R(ar, s, H»
:5 N g, j (ROI,s)'
Repeating this construction with 'R(a, 8, H + 8) and 'R(ar + s - 8, 8, H + 8) we obtain a rectangle R~,s ~ 'R(ar, s, H), such that all the previous reasoning applies, so that N g" (R~.s) = Nh, j (R~.s) and the height of R~,s lies between s - 28, s. Therefore
and
,
s
N g, j (ROI,s) ~ 21T IWk,j+1 - Wk,j 1 - nk,j
+ 1-
E.
This concludes the proof of the P6lya-Dickson Theorem.
o
3.2.26. Remarks. (1) We have already used in the proof of Theorem 3.2.25 the factthat for any HI > H, and as soon as ar » I, the number Nj('Rk,j(ar, s, H» = NrC'Rk.j(ar,s, HI»' (2) Taking E = 1 in the statement of Theorem 3.2.25, one gets the bound INjC'Rk,j(ar, s, H» -
for every s > 0, once a
»
1.
2:
IWk,j+1 - Wk,j1i :5 nk,j
236
3. Exponential Polynomials
=!.
Taking £ for those s such that (s/2rr)lwk.j+l -wk,jl sharper bound nk.) - 1 for the above quantity. In any case. we get
E
N* one gets the
We shall give below some consequences of the P6lya-Dickson theorem. 3.2.27. Lemma. Let fez) = Ll::<:}::<:n A}z} (1 + £j(z»e Wj : be an AEP. Assume f has infinitely many zeros going to 00 in a half-strip of the form V ('P. J,L, K), Then there exists (k. j) such that 'Pk = 'P. J,Lk.j = JL
Proof. One starts by remarking that V('P. J,L, K) n (Uk.) Vk,}) is unbounded since it contains a sequence converging to 00. Hence, for some (k, j), the region V ('P. J,L, K) n Vk.} is unbounded. We have shown in Lemma 3.2.1 that this implies that 'P = 'Pb J.L = J.Lk.j' 0 Let f, g be two exponential polynomials, fez)
=
L
Ai(z)ea,Z,
l~i:5m
g(z) =
L
B}(z)e fl,=.
l:5J~n
Let L be a side of the P6lya polygon peg) of g, and denote by l(L) its length. Let us denote by 1/fL the direction of the exterior normal of peg) on this side L. Set 'PL = 1/fL + rr /2. Let 0 < 8 < rr /2 be fixed and denote h(8) := {z E C
I argz -1/fLi
< 8}.
3.2.28. Proposition. If there is ro > 0 such that the function fig is holomorphic in h(8) n (B(O, ro»', then there is a side L * of the Polya polygon P(f) of f such that L * is parallel to Land l(L) ::: l(L *).
Proof. Let i3h, 1 ::: h :::: a', be the vertices of peg), L = Lh = [i3h, i3h+d, and 1/fL = 1/fh' Let Qh(g) be the corresponding polygonal region. constructed as in Lemma 3.2.9, resting on the side L, !'~,p' I ::: p ::: a~, the vertices of Qh(g) lying outside L h• we also denote by L h.p = [!'~,p' !'';,p+l]' Vh,p = V('P~, J.Lh,p' H), the corresponding half-strips, and Th(8) = {z E C: largz -1/fhl < 8} = h(8), We are going to show that if fig is holomorphic in Th(8) n (B(O, r)Y, then there is a side L * = Lk of P (f) such that Lk is parallel to Lh and the corresponding vertices ak, ak+l satisfy 1i3h+l - i3hl :::: lak+l - akl· We assume that H » I and r » I so that we can use the conclusion of Theorem 3.2.26, if necessary (for both f and g). For r» 1 we can also assume that V('P~, J.Lh,p' H) ~ Th(B) n (B(O, r)r. Since fig is holomorphic in this region, then all the zeros of gin V('P;', J.Lh,p' H) are also zeros of f. Therefore, f has infinitely many zeros, in V('P~, J.Lh,pH).
3.2. Distributions of Zeros of an Exponential Polynomial
237
=
It follows from the previous lemma that there are CPt, /-Lk,) such that cp;' CPk and /-L~,p = /-Lk,). Therefore, there is a side Lk of P(f) directed by CPk, hence parallel to L~. By the same reason, for every p, I :::: p :::: uj, - I, there is j E {I, ... , Uk - I} such that /-L~.P = ILk,). For a » I, s > 0, let
'R~.p(a, s, H)
= (z E
V~,p: Im(ze-i
+ /-L~.P argz
E
[a, a
It coincides with the region 'R k .) (a, s, H), previously defined for every zero of g is a zero of t (counting multiplicities) then
+ sn.
t, and since
Ng('R~.p(a, s, H» :::: NfCRk.) (a, s, H».
By Theorem 3.2.26 we have s
I
-
-
Nil ('Rh,p(a, s, H» = 27T l.Bh,p+J - .Bh,pl Nf('Rk,j(a, s, H»
Since we can let s
~
=
2:
lak,j+J - ak,)1
+ 0(1), + 0(1).
+00, we conclude l.Bh,p+J - .Bh,pl :::: lak,j+J - ak,jl·
Recall that the .Bh,p lie in the segment L'" and are ordered in such a way that
L
e(L~) =
l.Bh,p+J - fJh,pl·
l~p~l1~-l
So that
e(L~)::::
L
lak,)+J - ak,) I::::
L
lak,j+l - ak,)1 = e(Ld,
o
which is what we wanted to prove.
We are now going to give a rather general division theorem in the class of exponential polynomials due to Hajj Diab [Hajj], The first such result was proved by Ritt [Ril], [Ri2], it states the following: Let t, g E Eo be such that fig = h E )fCC), then h E Eo. There have been several successive ameliorations due to [Shi], [BD2], [HRS], among others. We start by describing a division algorithm, a variation of the Euclidean division algorithm for polynomials.
3.2.29. Definition. If fez) = Ll
min (Re(aj)},
l~l~m
i.e., the length of the projection of P(f) onto JR., 8'R,(f) ::: O. By p'R,[aj, ai+d we denote the length of the projection onto JR. of the segment [ai, ai+l]'
238
3. Exponential Polynomials
3.2.30. Lemma. Let f, g be two exponential polynomials, g #- O. There is an exponential polynomial X #- 0, with purely imaginary frequencies, and two other exponential polynomials q and r such that
xf =
qg
+r,
with either r = 0 or 8JR(r) < 8JR(g).
Proof. Let u I, ... , u" be real numbers such that u I < ... < up (if p ~ 2) and let {UI,"" up} = {Re(ai): I :::: i :::: mI. Let us define Ij := {i: Re(ai) = Uj} and
=
LAk(z)exp[i(lmadz]. I.EI]
Then f(z)
= L I::;i::;p
Note that the frequencies of the
Eh(z)eY'Z,O:::: YI < ... < Ys, 8 k E (5\{O}) U {OJ.
1::;I.::;s
Let us write g(z) = LI::;j::;n Wj(z)eWiZ, WI < ... < Wn , Wj E 6\{0}. Assume first that UI = W, = O. Then f, g E F, 8iR(f) = up, and 8JR(g) = W n • If up < W n , then we can choose X = 1, q = 0, f = r, and we are done. If up ~ Wn and n = 1, we can take X = w" q = f, r = 0, and we are also done. Hence, we assume up ~ WIl , n ~ 2. Let up, ... , Up-k be all the Uj ::: Wn . Consider
+ ... +
ro(z)
= Wn(z)f(z)
- (p(z)e UpZ
qo(z)
=
+ ... +
(p(z)e UpZ
Then we have wnf
= qog + ro,
=
F, ro E F, and either ro = 0, or 8JR(ro) :::: up - (w n - Wn-I). If ro 0 or if 8JR (ro) < W Il , we are done. If not, repeat the procedure for ro and thus obtain q" E F, such that WnrO = q,g + r"
qo E
r,
where either r, = 0 or 81R (rl) :::: 81R(ro) - (w n - w n-,). Since Wn - Wn-I > 0 we can continue to the point we find qo, ... , qs-' E F, rs E F, and either rs = 0 or 81R (rs ) < W n , and one can write W~f
= (W~-lqO + w~-2ql + ... + qs-,)g + r s,
which proves the lemma in the case UI = W, = O.
3.2. Distributions of Zeros of an Exponential Polynomial
239
Let us now consider the case of two arbitrary values UI, WI. The functions F(z) := f(z)e-U\Z, G(z) := g(z)e-W\Z, are now in the previous situation. Hence, there are 1/1 E 6, q, rEF, such that either r = 0 or 8~(r) < 8IR(G), and
1/IF
= qG + r.
It follows that
1/I(z)f(z)
= [q(z)e(U\-w,)Z]g(z) + eU\Zr(z).
Clearly, either eUtZr = 0 or 81R(eU\Zr) = 8IR(r) < 81R(G) = 8IR(g).
0
3.2.31. Theorem. Let f, gEE, g =1= 0, and assume the coefficients of g are relatively prime. If there is R :::: 0, and an angular sector S of opening strictly bigger than T( such that fig is holomorphic in S n (B(O, R)Y, then there is h E E such that f = gh. Proof. Rotating coordinates if necessary, we can assume that the sector is S = {z E C: - 17 < Arg z < T(}, with 0 < 17 < T(. The proof proceeds in three stages. First, we shall show that if f / g is holomorphic in {1m z > 0, Izi > R}, then fig = q /1/1, with q, 1/1 E E, and 1/1 has only purely imaginary frequencies (i.e., 1/1 E 6* in the notation of the previous lemma.) We show later that q /1/1 being holomorphic in {z E
3.2.32. Lemma. If fig is holomorphic when Izl > R, Imz > 0, then there are q E E, X E 6* such that xf = gq. Proof of Lemma 3.2.32. Replacing z by z + i R we can assume f / g is holomorphic in 1m z > O. This transformation is compatible with the condition 1/1 E 6*. We show first the following:
£g is holomorphic in Imz > 0, then 81R(g) .:::: 81R(/)' In fact, let 8R(g) = Wn - W], 8R(f) = up - UI as in Lemma 3.2.30. The convexity of peg) and P(f) tells us that they have vertices PI, ... , Pn (resp. aI, ... , ap ), such that Re Pn = Wn > ... > Re PI = WI (resp. Re ap = up > If
... > Real = UI), hence these vertices span the upper part of peg) and P(f), respectively. For 1 .:::: k .:::: n, let Ok be the line through 0, with the direction of the outer normal to peg) along [Pb Pk+d. Let 1/Ik be the angle it makes with the positive real axis, then 0 < 1/Ik < T(. Choose (J > 0 such that 0 < 1/Ik - (J < 1/Ik + < T(. Let Tk «(J) be the sector of bisectrix Ok and opening 2(). Since fig is holomorphic in Tk«(}), we can apply Proposition 3.2.28 to conclude there is a side [aj" ajl+t] of pC!) which is parallel to LBko .Bk+JJ and whose length laA+t - ah I :::: IPk+l Pkl. Clearly, the length of the respective prOjections on the real axis preserves
e
3. Exponential Polynomials
240
Figure 3.5 the same size relation, let us say PIII.[t3k.t3HIJ ~ pIII.[aj"ajk+,J. Therefore, 8I11.(g) =
L
PIII.[t3b
PHd ~ ~
L
pIII.[aj" aj,+,]
L
pIII.[aj, aj+d
= 8R(f).
I~j~p
Appealing now to the preceding Lemma 3.2.30, we infer there are X E 6*, and exponential polynomials q and r such that
xl
= qg+r,
and, either r = 0, or 8I11.(r) < 8I11.(g). We can rewrite the previous equation as r I -=X--q, g g
which shows that r / g is holomorphic in {1m z > OJ. From what we have just shown we must have We conclude that r = 0. Hence,
xl =
qg,
o
as we claimed.
3.2.33. Lemma. Let h E E, 1/f E 6*, and assume there are 1], 0 < 1] < ']'(, and R > 0, such that h/1/f is holomorphic in {z E C: Izl > R, I Argzl < 1]}. Then there are b E C[z]*, q E E, such that bh
= q1/f.
3.2. Distributions of Zeros of an Exponential Polynomial
241
Proof of Lemma 3.2.33. Let l/f(z) = b l (z)eii'\z + ... + bn(z)eiA,Z, Al < A2 < ... < An, bj E (:[z]. The P61ya polygon P(l/f) is a segment of the imaginary axis, [-iAn, -iAIl. Proposition 3.2.28 states that there is a side [ab ak+Il of the P6lya polygon P(h) which is parallel to the imaginary axis and lak+1 - akl ::: IAn Ad. Let us change to new variables, ~ = iz, h'(~) = h(z), l/f'(~) = l/f(z). In this case, IAn - Ad = 8JR(l/f*) and lak+1 ~ akl :s 8JR(h*), hence 8JR(l/f*) :s 8R(h*).
Copying tbe proof of Lemma 3.2.30, we can show there is an s E N* and two exponential polynomials q*(n, r*(~), such that b~(-inh'(~)
= q*(~)l/f*(n
+ r*(n
and, either r* = 0, or 8JR(r*) < 8R(l/f*). Let r(z) = r*(iz). Then r(z)/l/f(z) is holomorphic in {z: Izl > R, I Argzl < 17}. From the first part of the proof we can conclude that 8JR(l/f*) :s 8JR(r*). This shows that r* O. Hence,
=
b~(z)h(z)
3.2.34. Lemma. Let f, g, h
E and c(f), c(g), and c(h), respectively, the
E
g.c.d. aftheir coefficients. If fg
o
= q*(iz)l/f(z).
= h, then c(f)c(g) = c(h).
=
=
Proofof Lemma 3.2.34. Let fez) LI~i~m Ai(z)e"'; g(z) L1:5j~n Bj(z)e PJ ; and h(z) = LI~k~p Ck(z)e YkZ • We can assume that al -< ... -< am, fiI -< ... -< f3n, and YI -< ... -< Yp· Consider first the case where c(f) = c(g) = I. We want to show that then the Ck must be coprime, i.e., c(h) = I. Let WI, .•. , Wq be the different values taken by ai + f3j, 1 :s i :s m, 1 :s j :s
n. Assume
WI
-<
W2
-< ... -<
Wq .
We have
From the independence of {eAZhEc over C[z) (Corollary 3.1.3) we obtain that p :s q, that for each frequency Yk = WI.. for some h, and that Ck(z) =
L
Ai(z)Bj(z),
1 :s k
:s p,
",+pj=Wl,
0=
L
Ai(z)Bj(z),
(X,+P,=Wl
Let us suppose that c(h) i= 1. This means there is a value z = a which is a common zero of all polynomials Ck. Since c(f) = 1, z = a is not a common zero of all Ai. Let io be such that Ai (a) = 0 for 1 :5 i < io and Aio (a) =P O. Let WI = exio + f3I. whether I E {h, ... , lp} or not, the corresponding sum vanishes at z = a. On the other hand, if IXi + f3j = IXio + f3I. it is not possible that i > io
242
3. Exponential Polynomials
because of the ordering of the frequencies. Moreover, if i = ;0, then j = I. Then ;<;0 u;+{3j=w/
Since the left-hand side vanishes at z = a, and the second sum also does, we have Aio(a)BI(a) = O. In other words, BI(a) = O. Let us now assume Bj(a) = 0 for 1 ~ j < k ~ n. We want to show that Bk(a) = O. We consider ;<;0
i=;o
i>;o
The total sum vanishes at z = a, as well as the sum for those indices (i, j) with; < io. If i > io, then aio -< ai and, since ai + fJj = aio + fJk we conclude fJj -< fJko i.e., j < k. Hence the sum with indices; > ;0 also vanishes at z = a. For; = io we have only one term AioBk' Therefore, Bk(a) = O. We have arrived at the conclusion that Bda) = 0, 1 ~ k ~ n, thus e(g) i= I, contrary to the hypotheses. Therefore, is must be the case that e(h) = 1. We consider now the case where e(f), e(g) are arbitrary nonzero polynomials. Then F = fle(f) E E, G = g/e(g) E E, e(F) = e(G) = l. Hence, H = FG also has e(H) = l. H = h/(e(f) e(g» is an exponential polynomial, and by Corollary 3.1.3 (i.e., the independence of (eA:helC over C[z]), we conclude that e(f) e(g) divides all the coefficients of h. That is, e(f) e(g) divides c(h). On the other hand, since e(H) = 1, we conclude that we have c(f)e(g) = c(h). 0 Let us now return to the proof of Theorem 3.2.31. By Lemma 3.2.32 we have that there is X E 6*, h E E, such that xf = gh. It follows from the hypotheses that h / X is holomorphic in {Iz I > R, I Arg zI < 1]}. We can apply Lemma 3.2.33 and conclude there are bE C[z]*, q E E
such that bh
=qx.
That is, xbf = Xgq·
Because X i= 0 we have bf
= gq.
From the assumption c(g) = I and Lemma 3.2.34, we conclude that b divides c (q ). Hence, q / bEE and, a fortiori
£g EE, as asserted by the theorem.
o
3.2. Distributions of Zeros of an Exponential Polynomial
243
3.2.35. Remarks. (1) The condition c(g) = 1 is necessary, as shown by the entire functions (sinz)/z and (1 - eZ)/z that are not exponential polynomials (use Corollary 3.1.3). (2) On the other hand, the proof shows that if f, gEE, and fig is holomorphic in a sector of opening bigger than 7r, then there is h E E, b E e[z] such that fig = hlb. (3) The condition on the opening of the angle is necessary. The function eZ I (1 - eZ+ I) is holomorphic in the half-plane Re z > -I, but it is not an exponential polynomial. (4) There is an open conjecture of H.S. Shapiro that is related to Theorem 3.2.31. Assume f, g E Eo have infinitely many zeros in common, does there exist h E Eo, not a unit, such that h divides both f and g? (See [Sha], [VT].) Let us consider now the question of estimating the number of zeros of an exponential polynomial in a disk. We first note for future reference the following corollary of Proposition 3.2.24: 3.2.36. Lemma. If f E AEP is an entire function, then f is a function of completely regular growth.
Proof. It is immediate that
f satisfies an inequality of the form
If(z)1 < A(1
+ IzI)NeH(Z),
where H is the supporting function of the P61ya polygon P of f. On the other hand, from Proposition 3.2.24 we have that for any k E N* there are rk > 0, Tk > 0, such that if Izl > rk and d(z, Z(f)) ~ 11k, then ~
log If(z)1
log Tk
+ H(z).
It is then clear that hf (e i8) = l'1m sup T--'*'OO
log If(re i8 )1 = H( ei8) . r
We can assume the values rk satisfy rk+l > 2rk and
I log Tkl
1 - k
- - - <-.
rk
Let Z = {zn: n ~ I} be the sequence of distinct zeros of f, IZnl'::: IZn+d, and define Pn = 11k if rk < IZnl :s rk+l' Let E = Un",:,[lznl- Pn' IZnl + Pn], then
." )1 = H(ei6).
lim log If(re' ';:;r r This clearly implies that
L
r, !Slz.l!Sr
f is of completely regular growth. In fact,
Pn = (n(Z, r2) - n(Z, rl»
1
+ ... + k(n(Z, r) -
n(Z, rk»,
244 when rk < r
3. Exponential Polynomials ~
rk+ 1. Hence, given s > 0, let j so that 1/j < I r
""' ~
Pn
neZ, rj)
~
then
neZ, r)
+S---,
r
rl:5lz.l:9
8,
r
o
which shows that E has zero relative measure.
3.2.37. Corollary. Let n(V, Zo, r) denote the number of zeros (counted with multiplicities) of an entire function f E AEP in B(zo, r) and e(p) the length of the boundary of the P6lya polygon of f. Then . n(V, Zo, r) e(p) hm =--.
r
r-->oo
2]I"
Proof. This follows from properties of functions of completely regular growth. See [Lev, p. 288]. 0 Another way to express this corollary is that
n(V, zo, r) =
e(p)
~r
+ 0(1),
so that, in particular, for every Zo there is a constant C :::: 0 such that
n(V, zo, r)
e(p)
~ ~r
+ C,
which is due originally to P61ya. In [Tu] it is shown that if f is an exponential polynomial one can prove that the inequality holds with a constant independent of zoo Tunln also showed that such an inequality has many applications in number theory and elsewhere. The original estimate of Tunin has been improved by Tijdeman [Ti], Waldschmidt [Wa 1], and others. By a very ingenious argument, generalizing Rolle's theorem to holomorphic functions, Voorhoeve [Voor] obtained the following sharpening of Turan's estimate for the number of zeros of an exponential polynomial f: n(V, Zo,
r) ~ inr + 2(JOf, ]I"
where n = max{IO!jl: obtained
O!j
frequency of
n(V, Zo, r)
fl.
Using a geometric argument, he also
e(p)
~ ~r
+ vdof,
where v denotes the number of vertices of the P61ya polygon. Let us start by recalling some notation. If f is a meromorphic function in an open set U of the complex plane, f ¢. 0, then for Zo E U, v(zo) = v(zo, f) is the index of the first nonvanishing coefficients of the Laurent development of f about Zo, i.e., if V(zo) < 0 then Zo is a pole of f and -v(zo) is the order of this pole, if v(zo) > 0, then it is the order of Zo as a zero of f, finally, if V(Zo) = 0, then Zo is neither a zero nor a pole of f. Following [Voor] we introduce a generalization of the total variation of the argument of f along a closed interval [a, b) of the real axis.
3.2. Distributions of Zeros of an Exponential Polynomial
245
3.2.38. Definition. Let f 1= 0 be a merom orphic function in neighborhood of [a, b] and Xl < X2 < ... < Xm the collection of zeros and poles of f in la, b[. We set A(a, b, f):=
lb Im(f'(~)lf(~»1 d~ I
L
+JT{
IV(Xbf)I+4Iv(a,f)1+4IV(b,f)I}.
l:,:k:,:m
Let V be a complex neighborhood of [a, b], f E M(V), and [a, b) £ ]C, d[ c c V, then if we denote by ~ I, ... the collection of zeros and poles of f in ]C, they are the only poles of 1'(z)lf(z) on ]e, and there is a domain V, ]e, d[ £ V £ V and a function g E Je(V) such that
,;n
dr.
1'~)
L
=
f()
V;~~'
l:,:j:,:n
f) ~J
dr.
+ g(z)
(z E V).
On the real axis the sum is real valued, hence, Im(f'(~)lf(~»
=
Img(~),
and we conclude that Im(f' If) can be considered as a continuous function in ]C, dr. In particular, A(a, b, f) is well defined. It is also clear that A is additive with respect to intervals, i.e., if e < r < s < t < d, then A Cr. t, f)
=
ACr, s, f)
+ A(s, t,
f).
Another useful property is the following. For YJ real, IYJI« 1, the function fry(z) := fez + iYJ) is meromorphic in a fixed complex neighborhood, still denoted V, of ]e, dr. Moreover, we can assume that for 0"# YJ, fry has no zeros or poles on the real axis, so that ACa. b, fry)
=
lb Im(f~(~)1fry(~»1 d~. I
3.2.39. Lemma. lim A(a, b, fry) = A(a, b, f).
ry--+O
Proof. By the additivity of the functional A we can assume that or poles in ]a, b[, so that for some YJo > 0 we have
1'(z) ex f( z) = z-a
ex
= v(a, f), f3 = v(b, f),
and g holomorphic in
a-Y/o::::Rez::::b+y/o,
Then, for
~ E
f3
+ --b +g(z), z-
JR, exYJ
IImzl::::YJo.
f has no zeros
246
3. Exponential Polynomials
For 0 < c « 1, a < ~ < a + c, the last two tenns are bounded for all values of rJ, 0 < IrJl < rJo· So that
+ c, fry) =
A(a, a
l
aH
I Im(f~(~)lfry(~»1 d~
1
!J~
a +o
=O(c)+larJl
rJ2+(~-a)2
a
= O(c) + lal arctan(c/lrJl). Similarly,
+ 1.81 arctan(c/ll7l).
A(b - c, b, fry) = O(c)
Since
= Img(o, we also have
Im(f'(~)lf(~»
1 IIm(f'(~)lf(~»ld~ 1 b
b e -
=
a
a+e
IImg(~)ld~ + O(c).
Therefore, we are led to consider
1 1 b-o'lm(f~(nlf~(~))! -lb-& IImg(~)ld~1 a+&
a+e
:: 1 b
8
-
I
a+8
Im(j~(~)1f~(~»
1
+
1 b
-
a+8
8
rJ2 +
(~ -
g(~)1 d~
1
d~
b- e
::::: larJl a+e
- 1m
d~
h- 8
a)2 + l.8rJl aH
I Im(g(~ + irJ)
-
rJ2 +
(~ -
b)2
g(~»1 d~
We infer that IA(a, b, f~) - A(a, b, j)1 :::::O(c) + O(lrJD
+ (lal + If we let IrJl
+ O(lrJl/c 2 )
I.8DI arctan(e/l rJ I) - n,/21.
= O(e 3 ), then all the tenns above are O(c), which proves that lim A(a, b,
~->o
f~)
o
= A(a, b, j).
3.2.40. Corollary. Let f, g be two nontrivial meromorphic functions on [a, b], and let k E Z, then:
(i) IA(a, b, j) - A(a, b, g)1 ::::: A(a, b, fg) ::::: A(a, b, j) + A(a, b, g); (ii) A(a, b, fk) = IkIA(a, b, j).
Proof. For 0 < rJ « 1 we have that neither in [a, b]. Moreover,
f~
nor
g~
have any zeros or poles
3.2. Distributions of Zeros of an Exponential Polynomial
247
and
(f;}'fl; = k(f~/I~), hence the identity (ii) is clear for I~ instead of I, and the estimates (i) are a consequence of the triangle inequality for I~, g~. Using the continuity at 1'/ = 0 proved in the previous lemma, we conclude the corollary holds. 0
3.2.41. Proposition. Let I be a meromorphiclunction on [a, b) such that neither I nor f' have zeros or poles at the endpoints 01 the interval. Then A(a, b, f) ::: A(a, b, I') where e(z):= I Arg(f'(z)/I(z»1 argument of ~ ).
(Arg~
+ B(a) -
B(b),
is, as usual, the principal value
01 the
Proof. Note first that I Arg ~ I is a continuous function in C* with values in [0, rr]. Hence, 19 is a continuous function in a neighborhood of a and of b. Moreover, for small T} real, e~(z) := I Arg(f'(z)/I~(z»1 = e(z
+ i1'/),
so that e~(a)
e~(b)
--+ e(a),
--+ e(b)
as
1'/ --+ O.
Therefore, by Lemma 3.2.39, if we can prove the inequality for I~, 0 < 1'/ « 1, it follows for I. Therefore, we assume that I and f' have neither zeros nor poles on [a. b). In other words, we can assume that f' /1 is a holomorphic function on [a, b]. and thus Im(f'/f) has only finitely many zeros on [a, b]. Let a < XI < ... < xn < b be the zeros ofIm(f' If) in la, b[, denote a = Xo, b = Xn+l, which could be zeros of Im(f'/f) or could be not. Whenever Im(f'(~)/I(~» :::: 0 we have Arg(f'(O/I(~»::::O and conversely, so that in any interval]xt,Xk+l[ we have with
ak
= ±1, and
for the sarne value
ak.
Consequently, for
~ E ]Xk, Xk+I[,
~ ee~) = ak :~ Im(Log(f'(~)/I(~)))
(:~ LOg(f'(~)/I(;»)
=ak1m =
ak
Im(f"en/I'(;) - I'(~)/f(~».
Using this identity we obtain,
l
X, + I
x"
I Im(f'(~)/f(~»1 d~ = ak
l
x' + 1 Im(f'(~)/f(~»d~
Xk
3. Exponential Polynomials
248
1 Xk
=ak
+J
~
Im(j"(~)/f'(~»d~-
lx'+J q
d
-8(~)d~ d~
so that A(a, b, f) =
=
~ 1:k+1 I Im(j'(~)/f(~»1 d~
ta 1:'+1 adm(j"(~)/f'(~» d~
~ A(a, b,
1') + 8(a) -
- (8(b) - 8(a»
o
8(b).
3.2.42. Corollary. Let f be meromorphic and not constant on [a, b], then
°
A(a, b, f) ~ A(a. b, f')
Proof. For < T} « 1, the functions From the previous proposition A(a, b, f~) ~ A(a, b, f~)
since
°
~ 8~(~) ~
n. Letting
T} ~
f~, f~
+ 8~(a) -
°
+ n.
have no zeros or poles at a and h. 8~(b) ~ A(a, b, f~)
we obtain the corollary.
+ n,
o
Corollary 3.2.42 is the key to good estimates. It allows us to estimate the variation of the argument of solutions of differential equations, as shown by the following result:
3.2.43. Proposition. Let f he a nontrivial meromorphic function on [a, b] satisfying a nontrivial differential equation of the form
!!... [1/In!!... dz dz
[...
!!...
dz
[1/I1!!... f] ... ]] dz
= 0,
where 1/110"" 1/In are meromorphicfunctions on [a, b]. (For simplicity we shall drop the brackets in the future applications of this proposition.) Then A(a, b, f) ~
L
A(a, b, 1/Id
+ nn.
l:::k::5n
Proof. If n = 0, i.e., df/dz = 0, then f == c =1= 0, and it is easy to see that A(a, b, c) = 0 (see Exercise 3.2.3). The right-hand side is also zero in this case. For n ::: 1, let g = 1/11 f'. If g == 0 then we are in the previous case, so we can assume g is nontrivial. We can apply induction to obtain A(a, h, g) ~
L 2skSn
A(a, b, 1/Ik)
+ (n -
l)n.
3.2. Distributions of Zeros of an Exponential Polynomial
249
On the other hand, the previous corollary and Corollary 3.2.40(i) and (ii) yield
+ n = A(a, b, g1/f j l) + n A(a, b, g) + A(a, b, 1/fjl) + n A(a, b, g) + A(a, b, 1/f1) + n
A(a, b, f) ~ A(a, b, 1') ~
=
~
L
o
A(a, b, 1/fk) + nn.
One can prove, using this proposition, that if rpl, ... ,rpm are linearly independent meromorphic functions on [a, b], and CI, ... ,Cm are arbitrary constants such that f = 2:: c)rp) ¢:. 0, then an upper bound for A(a, b, f) is independent of the values of the constants. The same holds for an upper bound of the total number of zeros and poles of f on [a, b] (see Exercise 3.2.8). We are now ready to go back to exponential polynomials. We recall from Section 3.1 that we denote
L
fez) =
p)(z)eC(jZ,
1:S):Sn
3.2.44. Proposition. Let a < b, N (a, b, f) be the number of zeros (counted with multiplicity) of the exponential polynomial f in the segment ]a, b[. Set J:= maxIma).
/ := minIma), J
J
Then (b - a)/ = n(dOf - N(a, b, f)
~
lb
~
(b - a)J
Im(f'(x)/f(x» dx
+ n(dOf -
N(a, b, f).
Proof. We can assume the indices have been chosen so that
It is easy to prove that
where D = d/dz (see Exercise 3.2.6 and the proof of Lemma 3.1.2). We are using the notation from the statement of Proposition 3.2.43. Further, one can write D2 = DID, D3 = DIDID, ... , where 1 is the constant function 1 acting as a multiplication operator. Hence (*), has the form described in Proposition 3.2.43, applied to ! (x )e-a,X, where we have a total dO! functions 1/fko some
250
3. Exponential Polynomials
1/Ik == 1, and the others 1/Ik(X) = exp«ak - ak+')x). From Proposition 3.2.43 we conclude that
A(a, b, !(x)e- Ci1X
):5 L
A(a, b, e(Ci.-Cik+il x )
+ ndO!,
l:sk:Sn-l
L
=
(b - a)1 Im(ak - ak+t>1
+ 77:dO!,
l:sk:Sn-'
since A(a, b, 1) = 0 and A(a, b, eWX ) = (b - a)1 Imwl (Exercise 3.2.3). Since 1m ak is increasing, we have
L
1Im(ak - ak+,)1 =
L
(Imak+l - Imak)
= 1m an and hence,
A(a, b, !(x)e- Ci1X ) :5 ndO!
Imal
+ (b -
=J -
/,
a)(J -I).
On the other hand,
lb
Im(f'(x)/!(x» dx
:51
b
1Im«f'(x)/!(x» - a,)1 dx
+ (b -
:5 A(a, b, !(x)e-a1X ) - n N(a, b, f) :5 ndO!
+ (b -
a)J -
77: N(a,
a) Imal
+ (b -
a)/
b, f),
where in the third inequality we used the definition of the functional A and the fact that N(a, b, f) = N(a, b, !(x)e- a1X ). This is half of the inequalities we wanted to prove. To prove the other half, we observe that it is also true that so that As earlier, we have
lb
Im(f'(x)/J(x» dx ::: (b - a) Iman ::: (b - a)J
-lb
1Im«f'(x)/!(x» - an)1 dx
+ 77: N(a, b, f) -
A(a, b, l(x)e-a,X),
which proves (b - a)/
+ nN(a, b, f) -
77:dO!
:51 Im(f'(x)/!(x»dx, b
o
as desired.
3.2.45. Corollary. The number N(a, b, f) o! zeros o! J in a closed interval [a, b] oJthe real line satisfies the inequality -
N(a, b, f) :5 dOl
+ 2n1 (b -
a)(J - I).
3.2. Distributions of Zeros of an Exponential Polynomial
°
Proof. For s > we have N(a, b, f) :'S N(a - s, b proposition we have (b - a + 2£)1
+ rrN(a -
s, b
+ s, f)
251
+ s,
f). From the previous
-rrdof
:'S (b - a
+ 2£)J + rrdof -
rr N(a - s, b
+ £,
f),
so that 2rr N(a, b, f) :'S 2rr N(a -
Letting
£ -+
£,
b
:'S 2rrdof + (b - a
+ £, f)
+ 2s)(J -I).
0+ we obtain the required inequality.
D
We can obtain from Proposition 3.2.44 inequalities holding in any line segment of the complex plane.
3.2.46. Corollary. Let a, bEe and denote by [a, b] the line segment joining these two distinct points. If f has no zeros on la, b[, then
1m
(1
f'(z) dZ) :'S rrdOf
[a.b)
f
(z)
+ max Im«b -
a)aj)
)
::::: rrdOf
+ max{lm(i(b -
a»: Z E P},
where P is the P6lya polygon of f. Proof. Let g(n = f (a
+ (b -
ag) =
L
p) (1;) exp«b -
a)aj1;),
'~j~n
for some new polynomials
h(n =
eaaj pj(a
+ (b -
a)l;),
of degrees mj as before. The exponential polynomial g satisfies dOf N(O, 1, g) = 0, and
r' g'(~)
Jo
g(n
d~ =
=
r'
Jo
= dOg,
!'(a + (b - a)~) d (b - a) f(a + (b - a)~) ~
r J[a,b]
!'(z) dz.
fez)
Hence, from the second inequality of Proposition 3.2.44 we infer that 1m
(1
[a,b]
Since P
ff'(Z) dZ) ::::: rrdOf
+ max{lm«b -
(z)
= cv{a, ... , an}, we have maxj{Im«b -
a)aj)}.
)
a)aj)} ::::: maxZEP 1m (i(b - a». D
As a consequence, we obtain an estimate of the number of zeros of f in any square with sides parallel to the axes.
252
3. Exponential Polynomials
3.2.47. Theorem. Let Q be the square
Q := (z
E
C: a
~
~
Re z
b, c
~
1m z
~
d)
and N Q the number of zeros of the exponential polynomial f in Q. Further, let t:.y:= max (Re(a, -aj)). IS.j:::n
Then, NQ ~ 2d °f
+ 2rr1 {(b -
a)t:.x
+ (c -
d)t:.y}.
Proof. For 0 < e« 1 we have that if Qe:= {z E C: a -e ~ Rez ~ b+e, c - e ~ 1m z ~ d + e}, then BQe contains no zeros of f. (If BQ contains no zeros of f we can let e = 0.) Rouche's theorem states that N
r
< N = _1_ f'(z) dz = _11m QQ, 2rri laQ, fez) 2rr
(r
/,(z) dZ) . laQ, fez)
We divide BQe into four segments and apply Corollary 3.2.46 to each of them. We obtain 1m
(1
/,(z) dZ) aQ, fez)
~ 4rrdOf + max (lm«b - a + 2e)aj» I:::j:::n
+ I:::j:::n max (Im«a -
b - 2e)aj»
+ l:::J:::n max (lm(i(d -
c - 2e)aj»
+ max (lm(i(c -
d - 2e)aj»
l:::J:::n
= 4rrdOf
+ (b
- a
+ 2e)Llx + (d -
c + 2e)t:.y.
o
It is clear that this inequality proves the theorem.
We now obtain the promised Tunin type inequality. Let, as earlier, denote by V = V(n = the multiplicity variety of f, let n(V, Zo, r) be the number of points of V in 8(zo, r), and let Q = maxj lajl. First, we state a result that the reader should compare with Proposition 3.2.23.
3.2.48. Corollary. If f has real frequencies, then the number Ns of zeros of f in the strip S := (z E C: c ~ 1m z ~ d) can be bounded by Ns
~
2dof
(c - d)
+ -2rr - - l:::i.j:::n max (a,
- aj).
Proof. It is easy to see that there is R > 0 such that f has no zeros in the region I Rezl > R. Let Q = (z E C, -R :5 Rez :5 R, c :5 Imz :5 d), and apply the previous theorem. 0
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.49. Corollary. For any
Zo E
253
C, r > 0, we have
n(V, zo, r) ::::: 2dof
4Q
+ -r. 7r
Proof. Let Q = {z E C: Xo - r ::::: Re Z B(zo, r) ~ Q, so that
:::::
Xo
+ r, Yo -
r ::::: 1m Z
Yo
:::::
+ r}. Then
Clearly, Llx ::::: 2Q,
Lly ::::: 2Q,
hence, n(V, zo, r) ::::: 2dof
4Qr
o
+ -. 7r
Finally, we need a little geometric argument to obtain Voorhoeve's strengthening of the P6lya estimate involving the perimeter R(P) of the P6lya polygon of f.
3.2.50. Lemma. Let P and Q be convex polygons with exactly v vertices aI, ... , a v and f31, ... , f3v, respectively, where their indexing corresponds to the positive orientation of the respective boundaries. Let av+1 := aJ, f3v+1 := f31. Assume that 1
~
k
~
v.
Then,
L I~k~v
max{lm(z(ak+1 - ad)} = zeQ
L I~k~v
=
L
max{lm(z(f3k+1 zeP
.Bk»}
Im(.Bk(ak+1 - ad)
I~j~u
=L
Im(akCf3k+1 - f3k».
I~k~v
Proof. Let HQ be the supporting function of Q and evaluate it at the exterior unit normal nk to the side [f3k. f3k+d of Q, i.e., consider HQ(nd
= max Re(Znk) zeQ
as, in fact, the maximum is attained at any point of [f3k, f3k+I]. The condition Arg(ak+1 - ad = Arg(f3k+1 - f3k) indicates that
nk
=-
i(ak+1 - ak) . lak+1 - akl
254
3. Exponential Polynomials
Since H Q is a homogeneous function. we have HQ(-i(ak+l - ak»
= Re(-i(ak+l
= Re(-i(ak+l
- adPd
- ak)PHl)
= Im(Pk(ak+l - ak» = Im(Pk+l(ak+l - ak», and also Similarly, max Im(z(.Bk+l - f3d) = Im(ak(f3k+l - f3d) = Im(ak+l (f3kH - f3d). ZEP
Moreover, because of the periodicity of the indices,
L
L
Pk+lak+l =
Pkak,
and for a, bEe, Imiib = - 1m ba.
Consider now the expression we want to identify 'L....J " maxlm(z(aHl - ak» ZEQ
l~k~v
= '" L....J
Im(Pk(ak+l - ak»
l~k~v
L
=
Im(Pk+l (ak+1 - ak»
l~k~v
= 1m
(L
Pk+lak+1 -
l~k~v
= 1m
(L
Pkak -
I~k~v
L
= 1m (
L
Pk+lak)
l~k~v
L
PkHak)
I~k~v
(Pk - Pk+I)ak)
I~k~v
L
=
Im(ak(f3k+l - f3d)
l~k9
= '" L....J
max Im(z(f3k+1 - f3d).
l~k~v
ZEQ
This long chain of identities proves the lemma.
o
We remark that each sum in the previous lemma is invariant under translation of P or Q. e.g., for any Zo, Zl, Z2 E C we have
L l~k~v
max Im«z - zO)(ak+l - ak» ZEQ
=
L l~k~v
Im«ak - ZI)(f3k+l - 13k»
255
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.51. Theorem. Let v be the number of vertices of the P61ya polygon P of the exponential polynomial f and V = V(f). Then, for any Zo E C l(P) n(V, Zo, r):::: 2rr r + v dOf·
Proof. If v = 1, fez) = e a ,! PI (z), dOf = ml = deg PI and l(P) = 0, so that n(V, zo, f) :::: dOf is best possible. If v = 2, then all the frequencies lie in a straight line in the complex plane. Moreover, considering f(z)e- Cl1Z we can assume that al = O. Rotate the variables, z = (an/lanl)~. Then, in the variable ~ we have an exponential polynomial with real frequencies, say 0 = WI < W2 < ... < W n • Then l(P) = 2wn and we can apply Corollary 3.2.48, which states that in any strip S of the form S = {~ E C: c :::: 1m ~ :::: d) we have ° (c - d) N~::::2df+~wn.
Clearly, B(zo, r) ~ S when c
= Imzo -
r, d
n(V, zo, r) :::: 2dof
= Imzo + r. Thus, l(P)
+ 2rr
r.
Assume now v ~ 3. Let us index the frequencies of f in such a way that ai, .... cl u are the vertices of P in the positive direction, we set av+1 = al. We can find a polygon Q, which circumscribes B(zo. r) and such that if PI, ... , Pv are the vertices of Q, then Arg(uk+1 - ad = Arg(pk+1 - Pk), with Pv+1 = PI. (See Figure 3.6.) If CJQ contains any zeros of f, we replace B(zo. r) by B(zo, p) for p > r and let Q still be the corresponding polygon. We can do this with p so close to r so that n(V, zo, r) = n(V, zo, p), and if we prove the estimate for p we will be done by letting p -+ r. Hence, we may as well assume that CJ Q contains no zeros of f. Therefore, if N Q denotes the number of zeros of fin Q, we have n(V, Zo, r) < N Q
-
1 = -2Jri
1 aQ
1'(z) -dz fez)
av-I
Figure 3.6
= -2rr1 1m
1 aQ
1'(z) -dz fez)
3. Exponential Polynomials
256
::::
L
~dol + _I 2
2n
I:<=k:<=v
max{Im«z - zo)(ak+1 - ak))}. ZEQ
From the proof of Lemma 3.2.50 we know the maxima are achieved at any point of the segment [~k> jjk+tl. Since Q circumscribes B(zo, r) we can choose a point z that belongs to aQ so that Iz - =01 = rand max Im«z - zo)(ak+1 - ad) :::: rlak+1 - akl· ZEQ
As t(P) = L:1:<=k:<=v lak+1 - akl, we obtain v t(P) + --r, r 2n when v ::: 3. It is slightly better than the stated inequality. n(V, Zo, r) :::: -dol
o
Comparing the estimates from Theorem 3.2.51 or Corollary 3.2.49 with the conditions from Squires' Theorem 2.6.23, we see once more that if I is an exponential polynomial and p(z) ::: Izl (or p(Z) ::: I Imzl + 10g(1 + Izl) if the frequencies are purely imaginary), then the only obstruction for the variety V (f) to be an interpolation variety for Ap (iC) is the separation condition of the distinct zeros. That is, we would like to guarantee that if Zj 1= Zk on any two zeros of I, then IZj -
zkl :::
eexp(-Cp(Zk»
for some e, C > 0 independent of j, k. In which case we say that the zeros are well separated for the weight p. The example I(z) = sinz sinAz, A > 0, shows that given p we can choose A so that this condition is not satisfied. Namely, we let A = L:~I IO- n " for a sequence of integers nk t 00 sufficiently fast (depending on p). In fact, the zeros of I are nn, nEZ, and mn lA, m E Z. If A rt Ql then all zeros are simple, with the exception of Z = O. The distance between two zeros is bounded below in each sequence, the difficulty arises between the sequences, so that
Inn - mnlAI = ilnllA -
~I,
and the problem is to see how fast A is approximated by rational numbers. If A is a quadratic irrational one can easily see that [HW] > IA- mj n -
for some e > O.
~ n2 '
n
E
Z*,
mE
Z
3.2. Distributions of Zeros of an Exponential Polynomial
More generally, if A E there is 8 > 0 such that
Q, i.e., A is an algebraic number, then for any fJ
I A-~I n
257 > 0,
>_8 - n2+~'
see [Ba]. Note that in this case the frequencies of fare ±i ± iA, which are all algebraic numbers if A E Q. Moreover, the coefficients of f written as an exponential sum are also algebraic numbers. So that for f(z) = sinzsinAz, A E Q, guarantees that the zeros are well separated for any weight. If we consider f(z) = sin(z - a) - sin(J2z - a), a > 0, then the frequencies are again algebraic, but the coefficients are combinations of sin a and cos a with algebraic coefficients, which in general are transcendental numbers. On the other hand the zeros form the sequences 2krr 2a + (2k + l)rr - - - and kEZ. I-J2 1+J2 Again, depending on the choice of a we can make the zeros well separated or not. Moreover, note that from the P6lya-Dickson Theorem 3.2.25 it follows that when the frequencies of an exponential polynomial f are purely imaginary, then there is a positive constant A such that the zeros of f lie in the logarithmic strip
IImzl:::: A(l +log(l + Izl)) so that, if the zeros of f are well separated for the weight p(z) = 11m zl log(l + Izl), it follows that there are 8 > 0, N > 0, such that IZj - zkl
~
+
8
(Izkl
+ IZjl)N
for any two distinct zeros Zj, Zk. These considerations led to the following conjecture of Ehrenpreis [Eh5], [BY2]: 3.2.52. Conjecture. Let f be an exponential polynomial with algebraic frequencies and with polynomial coefficients in Q[z], then the zeros are well separated for the weight p(z) = Izl. That is, there are constants 8 > 0, A > 0, such that for any pair of distinct zeros Zj and Zk we have IZj - zkl ~ 8e-Alzkl.
Moreover, if the frequencies are purely imaginary, then the zeros are well separated with respect to the weight p(z) = 11m zi + log(l + Izl). That is, there are constants 8 > 0, N > 0, such that
for any pair of distinct zeros Zj and Zk. It is very little we know about this conjecture. In trying to prove it, it is natural to consider the Q-vector space r generated by the frequencies ai, ... , an
3. Exponential Polynomials
258
of I. The dimension of r as a Q-vector space is called the rank of r. The Conjecture 3.2.52 has been shown to the true when rank r = I, and also when rank r = 2 under the further assumption that I is an exponential sum (see [BY2]). This conjecture is equivalent to the fact that if the frequencies and the coefficients of f are algebraic, then V (f) is an interpolation variety for the space Exp(C), or for the space F(£'(IR». Another consequence of the conjecture is that if II, h are exponential polynomials with algebraic coefficients and frequencies, without any common zeros, then I E /(fI, 12) in Ap(C) (for p(z) = Izl or p(z) = lIm zl + log(l + Izl». We shall mention some related questions about mean periodicity in Chapter 6. It would seem strange that one could relate the separation properties of the zeros of f to the arithmetic nature of its frequencies on coefficients, but, in fact, following the work of Siegel and Shidlovsky one has quite a bit of knowledge about the zeros of f. For instance, they are not only transcendental numbers but one can even measure how closely can they be approximated by algebraic numbers. The reader is referred to [Va] and the references therein for some results of this type. Nevertheless, some simple looking arithmetic conjectures are still open and may be related to Conjecture 3.2.52. For instance, the conjecture of Schanuel stating that if (XI, ••• ,(Xn E C are linearly independent over Q, then the degree of transcendency of the field k generated by (X I, ... , (Xn, e'" , ... , e'" , is at least n, which has not yet been solved for n ::: 2 (see [Wa2]). EXERCISES
3.2.
1. Using the notation from the proof of the P6lya-Dickson theorem show that: (a) If z = x + iy E V,,)' then lim:~oo Iy/xl = 00. (b) For z E V. j, Iim:~oc Arg z = 7T /2. (c) Show that the curves of equation y + /-Lkj Arg z c in VA,j are asymptotic to y C - /-Lk,j7T/2 as Izl --+ 00. Similarly, the curve x + /-LA,j log Izl H is asymptotic to x + /-LA,j logy = H. (d) Let w Z + /-Lj,A logz, z E VA,j' Show that if Iz, - z21 ~ 8> 0, then for z), Z2 large one has IWI - w21 ~ 8/2. (e) For some R» 1, (U Uu) U (U Vk,I) 2 (B(O, R)' n (y ~ OJ).
=
=
=
=
2. Assuming the Conjecture 3.2.52 is correct, show that if fl' 12 are exponential polynomials with algebraic frequencies and algebraic coefficients, which have no common zeros, then I E l(f" 12), the ideal generated by flo 12 in Ap(lC)(p(z) = Izl or I Imzl + 10g(1 + Izl). 3. Let A(a, b, f) be the functional defined in Definition 3.2.38. Evaluate it in the following cases: (i) fez) == C E IC; (ii) fez) = e'''; (iii) fez) = e'P<=), cp holomorphic; and (iv) If f is a polynomial of degree m, show that A(a, b, f) :'57rm. 4. (a) Show that Iime~o+ A(a -
E,
b
+ E,
f)
= A(a, b, f);
3.2. Distributions of Zeros of an Exponential Polynomial (b) Let
I
259
be a meromorphic real valued function in [a, bl, show that A(a, b, I)
= 11: t~m IV(Xko 1)1 + ~Iv(a, 1)1 + !Iv(b, /)I} ,
where XI, .•• , Xm are the distinct zeros and poles of I in la, br. (c) Let I be holomorphic, real valued function on [a, bl, with simple zeros at a and b. For any e > 0 show that 27r ~ A(a - e, b
Conclude that theorem).
+ e, I)
~ A(a -
e, b + e, J')
+ 11:.
I' must have a zero in [a - e, b + fl. Thus, I' has a zero in la, b[ (Rolle's
5. Let I be a meromorphic function on JR, periodic of period p > O. (a) Show that O(s) = Arg(l'(s)/I(s»1 = O(s + p) if s E JR is such that neither I nor I' have a zero or a pole at s. (b) Show that A(O, p, I) :::: A(O, p, 1'). (Hint: Use Proposition 3.2.41, for a convenient interval.) 1
6. Prove that if I(z) Then
= L:I:U:;" p/z)e
ajz
is an exponential polynomial, deg Pj
= mj.
where D = d/dz. 7. Use the example I(z) is sharp.
= (2cOSZ)"-1
to show that the estimate in Corollary 3.2.45
8. Let rplo .•. , rpm be meromorphic functions in the disk B(O, R), and let Wk = W(rpl,.'" rpk) be their Wronskian, i.e.,
Wk
= det
(a) Show that for any other meromorphic function h, W(hrpl, ... hrpk) = hi Wk. (b) Show that !PI, ... , rpk are linearly dependent if and only if Wk == O. (c) Show that for any k and I E N* W(W(rplo ... , rpko rpHd, W (rpl, ... , rpko rpH2), ... , W(rpl, •.. , rpk, rpk+/»
= WI-I Wk+l •
(Hint: Prove it by induction on k.) (d) Let f = LI:;k:;1 Ckrpko Ck E IC and rpl, ...• rpm linearly independent. Show that D
W~_I Wm Wm- 2
D
W~_2 Wm_1 Wm- 3
Wr
D ... D Wi D DL. W3 WI W2 WI
= o.
(Hint: Consider the auxiliary functions hk = W(!PI, ... , fPk-1o /), for 2 ~ k ~ m + 1.) (e) Let I be as in (d) and let N(a. b, I) be the number of zeros and poles of f (counting multiplicities) in the closed interval [a, b), when -R < a < b < R. Give an upper bound of N(a, b, f) independent of the choice of constants CI, ... , em.
CHAPTER 4
Integral Valued Entire Functions
4.1. The G-Transfonn In this section we study a transform of analytic functionals akin to the Cauchy transform considered in Chapter 1. This transform will allow us to obtain rather easily those properties of entire functions of exponential type that can be derived from their behavior on sequences of the form n :::: no, n E Z. It also provides an elementary method to study the analytic continuation of power series of the form Ln>o I (n )tn, where I is an entire function of exponential type. The main references for this section are [Bo], [Av I], [Av2], [AG I], [AG2], [AG3]. If K is a convex compact subset of the strip Q := {z E C: I 1m zl < 1l'},
we denote by
e- K
:=
{e- z :
z E K} and by Q(K) the open neighborhood of zero Q(K) := C\e-K.
Let .Jt"o(Q(K» denote the space of those holomorphic functions in Q(K) which vanish at 00, considered with the usual topology of uniform convergence over compact subsets of Q(K). Note, that since every function in .Jt"o(Q(K» is holomorphic at infinity, this topology coincides with that of uniform convergence on compact subsets of S2\e- K • If T is an analytic functional carried by such a compact set K we denote by G(T) its G-transform:
G(T)(z) :=
(T~'
_1_,.). 1- ze'
4.1.1. Proposition. liT is an analyticlunctional carried by the compact convex set K cc Q, then G(T) belongs to .Jt"o(Q(K». Moreover, its Taylor series development about z = 0 is given by
G(T)(z) =
L ~(T)(n)zn, n:,:O
and its Laurent development about z
= 00 is
G(T)(z) = _ ""' ~(T)(-n). L zn n:;:O
260
4.1. The G-Transfonn
261
Proof. The proof depends on the relations between the Cauchy and Fourier-Borel transfonns of an analytic functional which were established in Section 1.3. To see that G(T) is holomorphic in Q (K) we can appeal to the representation (T, h)
= ~1 21
1
~ T(w)h(w) dw,
y
valid for every h E .J'f(U), U a convex open neighborhood of K, y a loop in U \ K of index 1 with respect to every point of K, and T the Cauchy transfonn of T. Namely, if B(zo, e) <; Q(K), one can find an open bounded convex set U such that K <; U and e- u n B(zo, e) = 0, Hence, for z ¢ e- u we have 1 G(T)(z) = ~
21
1 y
~ T(w)
dw
1 - zew
,
which immediately implies that G(T) is holomorphic in B(zo, s). Let rl = rl(K) := d(O, e- K ), and r2 = r2(K) := supun ~ E e- K }. For any fixed z with Izl < r1 we have the development
which is unifonnly and absolutely convergent compact for w in a convex open neighborhood Uz of K. Moreover, if Izl ~ PI < rl, we can take this neighborhood to be independent of z. Hence, G(T)(z) =
2: (;i 1T(w)e nW dW) zn n2:0
y
n2:0
n2:0
is the Taylor development of G(T) about z = 0, convergent at least for Izi < rl. Similarly, for Izi > r2 we have 1 1 - zew = -
e- nw
2: --;;;-' n2:1
so that G(T)(z) = _ ' " J(T)( -n) ~ zn n2:i
is the Laurent development of G(T) about z =
00,
convergent for Izl > r2. 0
4.1.2. Proposition. Let K be a convex compact subset of n, U a convex open set, K <; U <; Q, ct a loop in U\K of index 1 with respect to the points in K,
262
4. Integral Valued Entire Functions
and y the loop y = e- a lying in e-U\e-K. Then,Jor every analytic functional T carried by K and every h E .JIf(U) we have (T,h) = __1_. ( G(T)(z)h(-Logz/ z = -1-.1G(T)(e-S)h(s)ds. 2:n:l
Proof. Let z
}y
z
2:n:1
ct
= e- s , for s E U, hence, s = - Log z. Therefore, ( G(T)(z)h( - Log z/z .
1 ct G(T)(e-S)h(s) ds = -
}y
Z
On the other hand, I I - - = - - +rp(u), 1 - eU u
where rp is a function holomorphic in 2Q = {u G(T)(e- S )
= / Tw, \
E
C: I 1m u I < 2:n:}. Hence,
I _.) = _ / _1_) +
1- e W
Tw,
,
\
w - s
e(s),
and 8(s) := (Tw, rp(w - s» is well defined, since w - s E 2Q and It follows from Cauchy's theorem that
e E .JIf(Q).
1 8(s)h(s) ds = O.
Since
_1_)
/ Tw , \ s- w
we have - -I . 2:n:l
1 y
dz G(T)(z)h(- Logz)-
z
= :n:T(s),
= --;I 21
1
T(s)h(s) ds A
= (T, h).
o
ct
The previous propositions have the following corollary:
4.1.3. Corollary. Under the same hypotheses as the preceding proposition we have J(T)(~) = - -I .
2:n:l
i
y
dz = -I. G(T)(z)e-' Logz_
z
2:n:l
1
G(T)(e-S)e'S ds.
ct
4.1.4. Corollary. Let K be a compact convex subset of G: .JIf'(K) -+ .JIfo(Q(K», given by T t-+ G(T). is bijective.
Q,
then the map
Proof. It is clear from Proposition 4.1.'1. that G is injective. To prove the surjectivity, let g E .JIfo(Q(K» and define (T, h)
=
--1-.1 2:n:l
g(z)h( - Log z) dz,
y
z
C\e- K ,
which is independent of the choice of loop y in as long as it has index 1 with respect to every point of e- K • We can thus assume it is a Jordan curve. It
4.1. The G-Transfonn
263
is an easy consequence of this assumption that this fonnula defines an analytic functional carried by K. In order to verify that G (T) = g, we use the definition of G(T), for l; E Ext(y) G(T)(l;) = (T:, (l-l;e z )-I)
1 --I-·l
= - -1. 2m
=
2lfl
g(z)(1 -l;exp(-Logz»- l -dz
y
y
Z
g (Z)-I- dZ
z-l;
= g(l;).
The last identity holds because Indy(1;) = -I and g(oo) = O.
o
4.1.5. Corollary. A necessary and sufficient condition for two analytic functionals T l , T2 E off' (Q) to coincide is that there is no E Z such that the identity J(Tl )(n) J(T2)(n) for every n ~ no, n E Z.
=
Proof. Evidently there is a convex compact set K S; Q such that both Tl and T2 are carried by K. The analytic functionals Sj = enoz'0 (i.e., (Sj' h) = ('0, enOZh) are also carried by K. Moreover, J(SI(n» = J(S2)(n) for all n ~ O. By Proposition 4.1.1, G(SI)(1;) = G(S2)(1;) for l; in a neighborhood of O. Since Q (K) is connected, G(SI) == G(S2). The previous corollary implies now that SI = S2, hence Tl = T2. 0 4.1.6. Theorem (Wigert, Leau, P6Iya-Carlson). Let f be an entire function of exponential type such that there is a convex compact subset K of Q with the property that for every e > 0 there exists a constant C. > 0 verifying for all ZEe
If(z)1
~
Ce exp(HK(z)
+ elzl)·
(a) If there is no E Z such that f (n) = 0 for every nEZ, n ~ no, then f = O. (b) The Z-transform Z(j)(z) := Ln>D f(n)zn of the sequence (j(n»n~o is holomorphic near z = 0, it has an analytic continuation to the whole open set Q (K), and its Laurent development about z = 00 is
_ ' " f(-n). ~
zn
n::>:O (c) The discrete Laplace transform of the sequence (f(n»n::>:o is defined by the Dirichlet series F(s) := L.m>D f(n)e- nS , and it converges in the halfplane Re s > -log rl (K). Moreover, it can be analytically continued to the open set C\ U(K + 2ilfj), je'Z.
admitting the expansions: (i)
F(s) = -
E f(-n)e n~1
nS
for
Res < -)ogr2(K),
4. Integral Valued Entire Functions
264
and (ii)
F(s) = f(O) 2
+ N-+oo lim
+ 2imr)
' " B(s L...
in IC\ U(K JET.
Inl:::oN
+ 2inj) ,
where B is the Borel transform of f, rl(K) =d(O,e- K ), and r2(K) = max{I~I: ~ E e- K }.
Proof. Since f = J(T) for a uniquely determined T E .Yf1(K), (a) and (b) are immediate consequences of the previous two propositions and their corollaries. Furthermore, Z(f) = G(T). It is also clear that F(s) = G(T)(e- S ) and, hence, F admits an analytic to continuation to the connected open set {s
E
IC: e- s
E Q(K)} =
IC\ U(K + 2inj). JET.
The statement (i) of part (c) is now obvious from Proposition 4.1.1. To prove (ii) we need to prove first a little lemma.
4.1. 7. Lemma. The following expansion holds uniformly on compact subsets of 1C\2niZ: 1 . 1 - = -2I + N-+oo lIm . 1 - e- W W - 2nin
L
Inl:::oN
Furthermore, for every s ¢ UjET.(K + 2inj), there is a neighborhood U of K such that 1. + lim ' " ___1_ __ 1 1 - e!;-s 2 N-+oo L... ~ + 2nin - s
=
Inl~N
and this series converges uniformly for
~ E
U.
Proof of Lemma 4.1.7. The function k(w) := 1/(1 - e- W ) - ~ is meromorphic in IC, is 2ni-periodic, its poles are simple, located at the points 2nij, j E Z, and of residue exactly 1, since --- -
1-
e- W
I
-2
= -w1 - -w6 + O(w 2).
On the other hand, the auxiliary function defined by
l(w):= lim
N-+oo
L Inl~N
W -
Il = -L +2 2nin
w
n~1
w
w2
+ 4n 2 n 2
is meromorphic in IC with poles exactly in 2niZ, since the convergence is uniform in compact sets of 1C\2niZ. This function is 2ni-periodic, its poles are simple and with residue 1. Therefore, the function k - I is entire and 2n iperiodic. Moreover, when 11m wi ::: n we have clearly, lim Re w-++oo
1 = 1, 1 - e- W
4.1. The G-Transform
265
. hm
1
Rew->-oo
=0.
1 - e- W
Fix some value 0 < a < rrj2. If WE Q and Iwl» 1, we have I Argw 21 ~ a. Then A := 2rr I w will also satisfy I Arg A2 I ~ a and
-+2 1 1
w
Lw
2
n:::1
+w4n 2 rr2
I
ILl I
-IAI 1+2 2rr
-
n:::l
< -IAI ( 1+2 - 2rr
~
< - 2rr
(1
IAI (
~ 2rr
+
1 + n 2 A2
jn
~
dt ) 1 + (.J2"j2)t 2 1A[2
n-l
2~ roo ~) IAI Jo 1 + u2 rr~) ,
1 + 1):1
which is bounded when IAI « 1, i.e., when Iwl » 1. These two observations imply that k - I is bounded in I1m w I ~ rr, hence everywhere. Therefore, the difference k(w) -/(w) = C = constant, which can be evaluated at w = O. It turns out that C = O. We conclude that k == I, which proves the first part of the lemma. The second part follows from the first by letting w = s - ~. We simply take s > 0 sufficiently small, so that the sets Ve(K) + 2rrij remain pairwise disjoint (recall that Ve(K) = (z E C: d(z, K) ~ then ~ - s remains in a 0 compact set disjoint from 2rriZ when ~ E Ve(K).
sn,
Returning now to the proof of Theorem 4.1.6. We have already seen that F(s)
= \/ T,
1~) = !J'(T)(O) + N->oo lim -s
1- e
L / T, Inl:::N
\
S
1.
+ 2rrln -
~
).
On the other hand, from the diagram following Definition 3.3.12 we see that / T,
\
s
1.
+ 2rrm -
~
) = rrT(s
+ 2rrin)
= C(T)(s + 2rrin) = B(f)(s Thus, F(s)
as we wanted to prove.
=
f(O) -2-
. + N->oo hm
+ 2rrin)
L
B(s
= B(s
+ 2rrin).
. + 2rrtn),
Inl::::N
o
266
4. Integral Valued Entire Functions
4.1.8. Example. Let T = [(-I)k/k!}8 k , so that K = {O}. Then 3'(T)(n = ~k / k!, C(z) = 1fT(z) = l/zk+l and 1 ) 1 dk G(T)(z) = ( - - - k! dl;k 1 - ze~ ~=o For k = 0 we have G(T)(z) = 1/(1 - z) and I
G(T)(e- S ) = - - = I - e- S
4+ N-+oo lim" ~
Inl:::N
S -
1 2.
(s ¢ 21fiZ).
1f1n
For k 2: 1 we have G(T)(z) = Pk(Z)/(l - z)k+l, where Pk is a polynomial of degree k. Moreover,
1
-S
G(T)(e
)
(d
k
= k! d~k 1 -
=L (s nEZ
1)~=o = ~ d k
(-ll
dsk
el;-s
1 21fin)k+l
(
1 1- r
) S
'
(s ¢ 21fiZ),
which can be developed as for
Res> 0, Res < O.
More generally, the G-transform of T = 8~k) is of a rational function of the form Pk(z)/(l - zea )k+l, with Pk a polynomial of degree k.
4.1.9. Remark (On the Inversion Formula for the G-Transform). Let y be a loop in e-fl\e- K of index 1 with respect to the points of e- K , we denote by OO1(g), the Mellin transform of g E Jr o(lC\e- K ), the entire function
1
g(w) 1 OO1(g)(z) := - - . dz, 21f1 y w z+ 1
where W Z = eZ Log w. The inversion formula of G can be obtained from the commutativity of the following diagram (see Proposition 4.1.2): ;j
4.1. The G-Transfonn
267
where Exp(K) :=
If
E
-*'(C) : "Is > 0, 3C. 2:: 0
such that If(z)1 ~ C.eHdz)+'lzl, z E IC}. We also consider Exp(Q) := UKccn Exp(K). We would like to find now some explicit formulas for the inverse 9)1-1 of the bijective map 9)1: -*'o(Q(K» ~ Exp(K). Let f = ;reT). For 0 < ({J+ < 7r/2 and -7r/2 < ({J- < 0 we denote by y+ = y+(a, ({J+) (resp. y_ = y_(a, rp_» the half-lines of origin a E] - 1, O[ making an angle ({J+ (resp. rp_) with the positive real axis. Let y be the broken line y_ y+. As we shall now see, for Izi sufficiently small we have the integral representation G(T)(z) =
~ 21
1 y
!(w) e- i7CW z W dw
Slll7rW
(0 <
Izi
«
1).
For n 2:: 0 let Yn be the loop of base point a suggested by Figure 4.1. Let 0 < z « 1, the residue theorem yields 1 ---:
1
21
v
. z W dw _.few) _ _ e- I1fW
OJ
= 7r
'"" ~ Res (f(W)' _._ _ e- I7rW z W , 0 k
Slll7rW
~
=
L
:5n
Slll7rW
f(k)zk.
O:":k:":n
(n
+ 1/2 -
a) tan !p+
- - - - - - -
a -1
n + 1/2
(n
+ 1/2 - a) tan !p_
_ _ _ _ _ _ _
Figure 4.1
W
= k)
268
z
4. Integral Valued Entire Functions
Let us now verify that the integrand is integrable along y+. In fact, we let and w = pei'P+ + a. The denominator is bounded below, since
= reil}
I sin rr w I 2: C exp(rr p sin qi+) for some
C
= c(qi+, a)
> 0. Hence, when w E y+ we have for arbitrary
£
> 0,
If(w)e-iJrwzWI
I sin rrwl On the other hand, () sinqi+ - Hk(ei'P+) for all 0 < r
«
(cosqi+)logr >
£ -
°
1, since cos qi+ > 0. Therefore, the integrand satisfies If(w)e- i 1!"W z WI ,- ' - - - - - - < C e po
I sinrrwl
-
,
for some c' > 0 and I) > 0 that can be chosen independent of r, (), as long as 1. This confirms that the integral over y+ converges uniformly and absolutely for all z E 8(0, ro)\{O}. The same is true for y_. On the other hand, consider the vertical segment r n of Yn, w = n + + i u, (n + 4- a)tancp_ .:::: u .:::: (n + 4- a) tanqi+. We have
o< r «
4
If(w)1 .:::: C,e A (n+(1/2)+lul) .:::: C;e Bn
for some A, 8 > O. Furthermore,
and
I sin rrwl Isin(rr(n + 4) + iu)1 = I cosh ul 2:
e1!"lul
2'
so that
for some C, 8' > O. Clearly, the upper bound tends to zero when n 0< r < ro « 1. As a consequence, we obtain that for 0 < Izl < ro « 1 1 -:
1
21
y
, z Wdw = "L....., f(n)zn _.f(w) _ _ e-I1!"W
sm rrw
n~O
~ 00
if
= G(T)(z),
since the series also converges for small Izl. Note that the value of the integral we obtained does not depend on the choice of arg z. On the other hand. for z = 0, the value of the integral obtained setting Ow = 0 does not necessarily coincide with G (T) (0) unless f (0) = O. We are going to use these remarks to show that the integral converges and represents G(T) for all z ¢ (e- L U CO}). L := K +] - 00,0].
269
4.1. The G-Transfonn
For that purpose we consider separately the convergence of the integral over f/ (e- L U (O}), z = re ilJ , then the integral over y+ is absolutely convergent if there is an £ > 0 such that y+ and over y_. Let z
On the other hand, the definition of HK(z) = sup{Rezt: t E K} implies that a + i/3 E K if and only if for every rp with -rr ~ rp ~ rr one has
Thus, a + i/3 E L if and only if the inequalities (**) hold for -rr /2 ~ rp ~ rr /2_ Since z f/ (e- L U (O}), then the points -log Izl - i argz f/ L for any choice of argz. Clearly, if a + i/3 f/ L, then a + i/3 + t f/ L for any t :::: O. Therefore, for the given rp+(O < rp+ < rr/2), since the inequality (**), with rp = rp+, determines the supporting line of L slope tan(rr/2 - rp+), we can choose a determination of arg z such that the point - log Iz I - i arg z lies on the open side of this line and the set L on the other. (Namely, if argz E ] - rr, rr] does not work, we can choose argz E] - (2n + l)rr, -(2n - l)rr] for n» 1.) (See Figure 4.2.) It is clear that we can find a half-strip R+ = {a + i/3: a > aD > -00, /31 < /3 < /32, /32 - /31 < 2rr} such that -log Iz I - i arg z E R+ and there is £ > 0 such that for every a + i/3 E R+
a cosrp+ - /3 sinrp+ :::: Hde iifJ +) + 2£. The image of R+ by the map a + i/3 1-+ e-(a+i{3) will be the open angular sector S+ = (t E C, 0 < Izl < e-ao , -/32 < argt < -/3d and z E S+. Since (**) holds now for every point t E S we see that the integral over y+ defines a holomorphic function of t in the sector S+. In exactly the same way we construct a sector S_ such that the integral over y_ defines a holomorphic function of t. Now Z E S = S+ n S_ and we have a holomorphic function defined in S, which for 0 < Izl < ro coincides with G(T)(t) since, as we said before,
7t
-7t
Figure 4.2
4. Integral Valued Entire Functions
270
for those values the integral representation does not depend on the choice of arg l;. We conclude that the integral representation is now valid everywhere in C\(e- L U {O}) because G(T) is holomorphic and single valued there. In the same fashion, if we take ifJ+ E lrr, rr/2[, ifJ- E l- rr, -rr/2[ and consider the path y (ifJ+, ifJ-, a) = y defined as before, then the integral
_~ { !(w) 21
i
(_z)W
dw
smrrw
y
+ [0, oo[ (see Figure 4.3).
represents G(T)(z) outside e- A , A = K
1t
1t/2 - (1\
L
Figure 4.3
4.1.10. Remark. If carrier of T. Then,
f =
J(T), T E Jt"(n), let K denote the minimal convex
hj(O) = lim
log If(r)1 r
r----)-oo
= supRez ZEK
and the radius of convergence of the series Ln>o f(n)zn is exactly exp( -hj(O». Hence, lim sup yilf(n)1 = exp(hj(O» n.... oo
and · sup log If(n)1 I1m n----+oo
n
= h j (0) .
This is a theorem originally due to P61ya which can be easily derived from the previous Remark 4.1.9. We are going to use the G-transform to study some classical results about holomorphic functions. The first question to study is that of the convergence of
4.1. The G-Transform
271
the Newton series of an entire function. (An excellent reference for this subject is [Ge].) We recall from [BG, p. 237] that if f is an entire function, its nth divided difference An = An (f) is given by An =
(~) (_l)n- j f(})·
L: O~J~n
The Newton series
f is the series " z(z - l) ... (z - n ~An , n~O
+ 1)
n.
.
Clearly, if f is a polynomial, the series terminates and its sum coincides with f. The empty product has the value one, as always. Let U = {z E Q: le z - 11 < I}. This set is an unbounded open convex subset of Q whose boundary is the curve
x
= log cos y + log 2,
Iyl <
Jr.
Note that B(O, log 2) S; U. (See Figure 4.4.)
1t
Log 2
-1t
Figure 4.4
4.1.11. Proposition. Let f be an entire function in Exp(K) for some convex compact set K ~ U. The Newton series of f converges uniformly over any
4. Integral Valued Entire Functions
272
compact subset of C and fez)
" z(z = '~ I:!..n(f)
1) ... (z - n
,
Proof. For z
E
C fixed, s > 0, and f; eZl;
+ 1)
(z E C).
n.
n":O
E Ve (K)
= [eel; -
1)
cc
U, we can write
+ 1]2
and use the binomial series to obtain "l; e" =
E (e l; -
1)
11,,:0
n
z(z - 1) ... (z - n n!
+ 1) .
This series converges unifonnly on Ve(K). Hence, if T E .Tf1(K) is such that f = J'(T) we have '" z(z - 1) ... (z - n + 1) fez) = J'(T)(z) = (Tl;. e"l;) = ~(T, (el; - l)n) , .
n.
n,,:O
It is easy to see that (T, (el; -I)") =
L (;) (-l)n- (T.el;j) = I:!..n(f). j
O:o}:on
Therefore, the Newton series converges pointwise to K, there is an Me > 0 such that Il:!..n(f)1
= I(T. (el; -I)n}l:'S Me
f. Since
T is carried by
sup lel; -Iln:'S Me[)'(K,s)t, l;EV,(K)
with )'(K, €) := sup{lel; - 11: f; E Ve(K)} < 1. We leave it as an exercise to the reader to show that for any R > 0 the power series __________________ R(R + 1) ... (R + n - 1) r n
L
n~l
n!
has radius of convergence 1. Letting Izl :'S R. r = )'(K. e), this ensures that the Newton series converges unifonnly over compact subsets of C. 0
4.1.12. Corollary. If the function f in the preceding proposition satisfies fen) 7l..for every integer n ::: 0, then f is a polynomial with rational coefficients.
E
Proof. The divided differences I:!..n (f) E 7l.. for every n ::: 0 but I:!..n (f) ~ 0 as n ~ 00, since the Newton series converges. Hence. I:!..n (f) 0 for n ::: no. 0
=
We are now going to study the influence on functions nomial behavior on the integers, i.e.,
If(n)1 :'S Alnl P + B
(n
E
f
E
Exp(Q) of a poly"
7l..).
For that purpose, let us consider the space A of analytic functionals carried by a compact subset of {O} + i] - 71:, 71:[. and by B the space of those carried by a compact subset of 1f'\{-1}, 'Jr = aB(O. 1). There are two special subspaces in A and B. In B we have the elements of V'('Jr) with compact support in
273
4.1. The G-Transfonn
1l'\ {-I}, let us call B this subspace. In A we have the subspace A of those analytic functionals for which there is a distribution R E £' (] - rr, rr [) such that (T, h) = (R y , h(iy»)
for every
h E £(Q).
The relations between these spaces are given by the following statement:
4.1.13. Proposition. The map which to every TEA associates the element S E B defined by C(S)(z) = -G(T)(z),
ZEIC\1l',
is a linear isomorphism of A onto B that maps
A onto B.
Proof. Let us recall (Definition 1.6.9) that the Fantappie transform of S is the pair (Sl, S2) given by Sl(Z) = / SI,
\
S2(Z)
= / SI, \
_1_)z _1_) zt-
t
E £'(1')
1,
for
Izl <
for
Izl> 1,
so that Sl(Z) = -rrS(z) = -C(S)(z) for Izl < 1 and S2(Z) = rrS(z) = C(S)(z) for Iz I > 1. Moreover, the map S ~ (S l, S2) is an isomorphism of £' (1') onto £(B(O, x £0(1C\8(0, If
1»
1».
=
(T,h) = __1_. (
2rrl ly z By Corollary 4.1.4, T is carried by any compact convex set K such that L S; e- K , in particular, TEA. Furthermore,
=
for some C, D > O. Since S = -(1/n)G(T), we can apply Propositions 1.6.10, 1.6.14, and 4.1.1 to conclude that the Fourier coefficients cn(S) satisfy Icn(S)1 ~ C(1
+ Inl)D
and S E V'(1l'). Hence (A) S; B. Now let S E 13, then Sl and S2 have slow growth on a neighborhood of 1l', i.e.,
274
4. Integral Valued Entire Functions
for some M ~ 0, k ~ O. As a consequence, if G(T) = -C(S), we have that G(T)(e- S ) satisfies a similar inequality in a neighborhood of iJ - Jr, Jr[. In the proof of Proposition 4.1.2 we showed that G(T)(e- S ) = JrT(s)
+ 8(s),
I Imsl
< Jr,
with 8 holomorphic in 11m sl < Jr. Therefore, T has slow growth near i] - Jr, Jr[. Since T is holomorphic inside a compact subset of i] - n:, Jr [, the Edge-of-theWedge Theorem [BG, Theorem 3.6.23] ensures the existence of distribution R with compact support in ] - Jr, Jr [ such that (T, h) = (R y , h(iy»)
for every h E Je(Q). We conclude that <1>(.4) =
B.
o
4.1.14. Proposition. Let T E Je'(Q). A necessary and sufficient condition for the existence of a distribution R E £' (] - Jr, n: D such that h
(T, h) = (R y , h(iy»),
is that
~(T)(n)
E Je(n),
= O(lnI P) for some pEN.
Proof. To show that the condition is necessary we observe that as a consequence of the Paley-Wiener-Schwartz Theorem 1.4.15
for some 0
~
a < n:. Hence 1~(T)(n)1 ~ C(1
+ Inj)p.
=
Conversely, if ~(T)(n) O(lnl)P and S = <1>(T), then cn(S) = O(lnj)P, so we conclude that S E V'(T). The conclusion follows from the preceding propo-
0
~~
4.1.15. Corollary (Cartwright's Theorem). Let f be an entire function of exponential type for which there is a convex compact set K £ Q such that for every 6 > 0 there exists Co ~ 0 such that (z E C).
If for some pEN, fen) = O(lnIP),for all nEZ, then there is a E [0, Jr[ and C > 0 such that If(z)1 ~ C(1
+ Izj)Peollrnz l
(z E C).
4.1.16. Corollary. If f is an entire function of exponential type zero such that for some pEN, fen) O(jnI P ) (n E Z), then f is a polynomial of degree at most p.
=
Proof. The distribution R from the preceding Proposition 4.1.14 will have
support at the origin.
0
275
4.1. The G-Transfonn
There are elements of Exp(Q) such that their zero sets contain the complement of a set of the fonn Z\pZ, for some p E N*. For instance, if w == e21ri / 3 is the cubic root of unity, consider the function fez) := (l
with g
E
Exp(t Q ). Then
f
+e +e WZ
E
Z. Let g =
f(3n
~(T),
f
)g(z)
Exp(Q) and it verifies
E
f(3n) = 3g(3n),
for every n
W2Z
+ 1) =
f(3n
+ 2) =
0
then
= ~«8
+ 8w + Owl) * T)
and the function G(z)
= G«o + 8w + 8wl) * T)(z) = Z(f)(z)
satisfies the functional equation G(wz) = G(z).
In fact, for Iz I « 1 we have G(z) =
L
f(n)zn =
n~O
and G(wz) =
L
f(3k)z3k
k~O
L f(3k)(wz)3k = L f(3k)z3k = G(z). k~O
k~O
Let .ifo := UK .ifo(Q(K» and let K be a convex compact subset of Q. We would like to detennine all the G E .if0 such that for some w E C" one has G(wz) = G(z),
Z
E Q(K).
4.1.17. Lemma. If G
E .ifo\{O} and WE C* satisfy G(wz) = G(z) for all Q (K), then w is a root of unity.
Proof. We have G = G(T), T G(wz)
=L n~O
E .if'(Q)\{O},
~(T)(n)(wzt
=L
z
E
and
~(T)(n)zn
for
Izl« 1.
n~O
If J(T)(n) = 0 for n ~ 1, then G is a constant, hence G == O. Therefore, there is n ~ 1 such that J(T)(n) t= O. We conclude that wn = 1. 0
For G, w as in Lemma 4.1.17, let p = inf{n E N*: w n = I}. Then we can assume w = e 21ri / p in the equation G(wz) = G(z). Let us denote by Sk the angular sector 21T 21T} , Sk:= { ZE
1 ~ k ~ p.
276 Let
4. Integral Valued Entire Functions
Dk
be the ray Dk
:= {z
= -J...wk , J...:::: O},
1~k
~
p.
4.1.18. Lemma. There are compacts Lk ~ Sk of the form Lk = e- Kk , where Kk are convex compact subsets of Q, pairwise disjoint, such that G is holomorphic outside the set Ul:ok:op L k •
Proof. We know G is holomorphic in a neighborhood of Dp =] - 00,0]. The functional relation G(wz) = G(z) implies G is holomorphic in a neighborhood of each of the rays Db 1 ~ I ~ p. Therefore, there is a compact K c Q, K carrier of T, G = G(T), such that L = e- K is disjoint from all the rays Dk • We can hence assume L = Ul:ok:op Lko Lk = e- Kk , with Kk convex compact, 0 pairwise disjoint. (See Figure 4.5.)
For every k choose Jordan curves Yk in Sk, with Lk ~ Int(Yk). We have from an obvious extension of Proposition 4.1.2 and Corollary 4.1.3 that J(T)(n = __1_.
L lv.r G(T)(z)e-~Logzdz.
2rrl l:o:op k ,.
Z
Since any point in Yk is of the form z = al-1u, where u + [2(k - l)rri]/p, we have
E
Ylo and Logz =
Logu
r G(T)(z)e-I;LogZ dz = e-[2(k-l) rri lpR r G(T)(u)e-I;IOgU du .
~
hi
z
u
On the other hand, there is an analytic functional S carried by (l/p)Q such that (S,h) = __1_. G(T)(e-i(l-IIP)rrz)h(_Logz)dz, 2m ly Z
r
Tt/3 Tt
Tt/3 0
Do
-Tt/3 -Tt
-Tt/3 (p = 3)
Figure 4.5
4.1. The G-Transfonn
277
Y = ei(l-I/p)JT YI· Hence,
~(S)(n = __l_.ei(l-I/P)JT~ and
1
G(T)(z)e-~Log=dz
n
2nl
~(T)(n = ei(l-I/p)JT~ ( L
z
e- 12 (k-IJ1Ti/ PR )
~(S)(~).
I ""Ck""CP
We can now state the following proposition:
4.1.19. Proposition. Let p E N*, w = e 2:n:i/p. For any T E
.}f"
(Q) the following
three conditions are equivalent: (1) The function G = G(T) verifies the functional equation
G(wz) = G(z) (2)
~(T)(~) = ei(I-I/p):n:~
(
2:=
(z E Q(K».
e-[2(k-I)JTi/ PR )
~(S)(n
l""Ck""CP
for some S E .}f"«(l/p)Q). (3) The function ~(T) satisfies ~(T)(n) = 0 for every n E Z\pZ.
Proof We already know that (1) implies (2). The fact that (2) implies (3) follows from the fact that for n E Z\pZ one has 1 - w- np 1 + w- n + w- 2n + ... + w-(p-l)n = = O. I-w- n
Finally, if
~(T)(n)
= 0 for all n ¢ pZ, one clearly has
L J(T)(n)zn = L n:,::O
J(T)(n)(wz)n
n,,:O
o
and thus, G(wz) = G(z).
= ~(T), where T is an analytic functional carried by a convex compact set K ~ Q, be such that f (Z) = O. Then f has the form
4.1.20. Corollary. Let f
fen = (sinnn~(S)(~), where S is an analytic functional carried by a compact subset of IR.
4K
Proof Let R be the analytic functional carried by such that J(R)(S) = f(~/2). Then J(R)(2n) = 0 for all n E Z. Hence, G(R)(-z) = -G(R)(z). Define RI by zG(RI)(z) = G(R)(z). Then G(RI)(-z) = G(RI)(z) and J(Rd(~) = ~(R)(~ + 1). This means that RJ = e WR, i.e., (R I , hew») = (R, eWh(w»). Clearly R) is also carried by ~K ~ Q, so that G(RI) is holomorphic in the half-plane Rez < O. Therefore G(Rd is holomorphic in CVIR. From the preceding Proposition 4.1.19,
~(RJ)(O
= (1 + e-i7r~)ei{7r/2~(SI)(O = (2 cos i~) ~(SJ)(~),
278
4. Integral Valued Entire Functions
with SI E Jt"'(~n), and G(S,)(z) = 2G(R,)(iz). Hence G(Sd is holomorphic in C\R as a consequence S, is carried by a compact subset of R Moreover,
~(R)(t;)
=
~(R,)(t;
=
(2 sin ~t; ) ~(e-'" Sl)(n
-1) = (2 sin
~t;) ~(Sd(t;
-1)
and f(t;) = (sinrrt;)~(2e-WS1)(2t;).
From the properties of the Fourier-Borel transform we conclude there exists S, carried by a compact subset of JR, such that ~(S)(t;) = ~(2e-W S1)(2t;). Hence, f(t;) = (sinrrnJ(S)(t;),
with S carried by a compact subset of R as we wanted to show.
o
4.1.21. Example. Consider a special case of the previous Corollary 4.1.20 when K is the minimal compact convex carrier of T, K ~ n, and on n K = {a irr, a + irr}, for some a E R always with the assumption feZ) = o. We have fen = (sinrrt;)h(t;), with hen = ~(S)(t;), S carried by a compact subset of R The function e-al; f(t;) corresponds to the minimal carrier K - a. We have (K - a) nan = {±irr}, Since, e-al; fU;) = (sinrrOe-al;h(~),
we have that get;) = e-al; hen must be an entire function of exponential type zero and g = ~(R), with R carried by {OJ. If not, let the minimal convex carrier of R be [a, .8] and T be such that f = ~(T). Then 1 T = 2i (8 in
-
Lin)
* R * 8a
would have its minimal convex carrier of the form [a + a, .8 + a] = i[ -rr, rr] which contradicts the hypothesis on ann K = {a ± i rr}. Therefore, we conclude that f(t;) = (sinrrt;)eal; g(t;),
with g a function of exponential type zero.
4.2. Integral Valued Entire Functions In this section we shall consider entire functions of exponential type that are integral valued, i.e., feN) ~ Z. In signal processing it is standard to call Ztransform of a sequence (un)n~O the series "" un.
L
n~O
zn
279
4.2. Integral Valued Entire Functions
If Un = I(n), IE Exp(Q) and z = l/s we obtain the Z-transfonn Z(f)(s) introduced in the last section. What we are be trying to find out is when Z(n is a rational function in case I is also integral valued. In this case it will tum out that I must be an exponential polynomial. Let us start by considering a rational function G, which is zero at 00 and has no pole at z = 0,
N(z) ao + alz + ... + as_Iz s- 1 = = D(z) bo + biz + ... + bs ZS
G(z) = -
2: -
Un
n2:1
zn
with bobs i:. 0, although as-I may be zero. We write the decomposition of G in partial fractions in the convenient form
where the values Zj i:. 0 are the poles of G of respective mj ::: 1, 1.::: j .::: M. The advantage of writing G in this form depends on the fact that the Laurent expansion at 00 of the different terms is very simple. In fact, for v ::: 0, a i:. 0,
a v+1
2:
-(--)V-+-:-I = z- a
n",v+1
an Pv(n)-;;, z
where the polynomials Pv are given by Po(W)
= 1,
~(w)=
(w - l)(w - 2) ... (w - v) I
'
v.
v ::: 1.
Since Pv (1) = ... = Pv(v) = 0 for v ::: 1, we can rewrite the previous expression as
Applying this formula to the Laurent expansion of G at
00
we obtain
where Qj is a polynomial of degree at most mj - 1. Conversely, assume that the Laurent expansion at 00 of a function G defined for Iz I » 1 has the fonn Un G(z) = -
2: zn n~l
and the coefficients
Un
can be written as Un
=
L l""j""M
Qj(n)z'j
4. Integral Valued Entire Functions
280
for some collection of nonzero polynomials Qj, deg Qj :::: mj - 1, and complex numbers Zj =1= 0, 1 :::: j :::: M. We can appeal to the fact that {Pv }v2:0 is a basis of the vector space q w 1 to express each Qj as
L
Qj =
Aj.vPv,
O::sv~mj-l
for some uniquely determined complex numbers Aj.v. Then, for
G(Z)
=L
:: = L ( L
n2:1
=
1 we have
Qj(n)z;) z-n
I-:IsM
n2:1
L (L L n2:1
Izl »
Aj.vpv(n)z;) z-n
I:Sj::;MO::;v::;m,-1
which shows that G is a rational function. In other words, a necessary and sufficient condition for the Z -transform G(z) = Ln>1 unz- n to be a rational function is that the Un are the values taken at the natural numbers n by an exponential polynomial of the form LI::;j::;M Qj(w)zt· One can obtain a different characterization by comparing coefficients in the identity N(z)
=
D(z)
(L ::) . n2:1
That is, s I s (U-,;-+-,;-+"" I U2 ) ao+alz+···+as-Iz=(bo+···+bsz)
which implies
boul
+
blul blU2
boun
+
blun+l
+ + +
b 2uI b2U2
bs-IUI
+ ... +
+ +
+ ... + ... + +
b3U2
bSUI = as-I, bsU2 = as-2, bsUs-1 = aI, bsu s = ao, bsUs+1 = 0,
+ bsus+n
= O.
Hence, the hypothesis b s =1= 0 implies that the sequence (U n )n2:1 satisfies the linear recurrence relation
n
~
1,
4.2. Integral Valued Entire Functions
281
(which is said to be of order s when bobs =1= 0), with the initial conditions U2
=as_l/b.. = (a s-2 - bs-IUI)/bs ,
Us-I
= (al - b2UI - b3U2 - ... - bs-Ius-z)/bs .
UI
Conversely, let us show that if a sequence (Un)n,,:l satisfies the linear recurrence relation
n
~
1,
with b s =1= 0 and Ul, ... , Us-I given, then the Z-transform En>1 unz-n is the Laurent development at 00 of a rational function. To see this, we define ao, •.. ,as-I by the formulas as-l := bsUl, a s-2 := b s U2
+ bs-IU1,
which, together with the recurrence relation (*), show that ~ Un
ao +alz
~zn = n~1
+ ... +asz s - I
bo+blZ+.··+bz s s
Note this reasoning did not use anywhere that bo #- O. There is a third way to conclude that En> I Un z-n represents a rational function near z = 00. Consider the relations k=n,n+l, ... ,n+s.
They can be interpreted to say that the nonzero vector (bo, ... , bs ) belongs to the kernel of the linear transformation defined by the (s + 1) x (s + I) symmetric matrix Mn S
=
C'
Un+1 . Un+s
Un+l Un+2
Un+s+1
Un+s+1
Un+2s
",+, )
.
Therefore the sequence of their determinants must be zero, i.e., H; := det M~ = 0,
n;:::1.
These determinants are called Hankel determinants of order s. We remark that the nonzero vector (b o, ... , bs , 0) is in the kernel of M~+I, hence H:+ I = 0 for all n ~ 1.
4.2.1. Proposition (Kronecker). A necessary and sufficient condition for the series L:n:::1 unz- n , convergent for all large Izl, to be the Laurent development
4. Integral Valued Entire Functions
282
at z = 00 of a rational function is that for some s determinants H: are zero for all n ~ N.
~
0 and N
~
1 the Hankel
Proof. It only remains to show that the condition is sufficient. For that purpose we need a preliminary lemma. 4.2.2. Lemma. Let A = (aij)l~j.j~n be a square matrix with complex coefficients and let Aij be the cofactor of aij' Denote D = detA and d = det(aijh~j.j~n-I' then we have the relation
AllA"n - AlnAnl = Dd. Proof of Lemma 4.2.2. One can easily verify the following matrix identity:
all
ani
al2
an2
aln
ann
A I•I A2.1 A 3. 1
0 1 0
An-I.I An.l
0 0
(I
1· ..
0 0 0
A I.n A2.n A3.n
1 0
An- I.n An.n
a 1.2 a2.2
al.n-l a2.n-1
an-I.2 an.2
an-l.n-I an.n-I
and conclude that
D(AIIAnn - AlnAnl) If D#-O we obtain,
AIlAnn - AlnAnl
:)
= D2d.
= Dd
as we wanted. Since the invertible matrices are dense in the space of all n x n matrices, this formula remains valid by continuity. 0 Let us now return to the proof of Kronecker's Proposition 4.2.1. Applying to the Hankel determinants the last lemma, for k ~ 2 we obtain the relations
If for some fixed n one has H j = 0 for all j ~ jo then one obtains from these identities that H!+I = 0 for aU j ~ jo, hence H~ = 0 for all j ~ jo and all m ~ n. Let s be the smallest integer such that = 0 for all n sufficiently big. By hypothesis such an s exists. If s = 0, then Un = 0 for all sufficiently big n, and then the conclusion of the proposition is true. If s = 1, then we conclude from Lemma 4.2.2 that u~ - Un-IUn+1 = 0 for all n sufficiently big. Hence, either uno = 0 for some no » 1 and then Un = 0 for
H:
, 4.2. Integral Valued Entire Functions
283
all n ~ no, in which case we are done, or Un i= 0 for all n case the quadratic equations leads to the relations
~ =
Un+1 Un+2
Un+1
=
Un+2 Un+3
= ... = /l.
~
no
»
1. In this
i= 0,
and letting bo = 1, b I = - /l. we obtain the linear recurrence for
n
~
no,
which implies the desired conclusion. If s ~ 2, it must be that H ;-1 i= 0 for all n ~ N. If not, let no ~ N such that H;;I = O. Since we have that H; 0 for n ~ N by hypotheses, from (t) we conclude that also H;o;'1 = 0, this clearly contradicts the minimality of s. Hence, for n ~ N we can consider the system of s equations and s + I unknowns
=
+ +
boun
(tt)
{ bOUn+1
~oUn+s-1 +
+ bs - l un+2s-2 + bsUn+2s-1 = o. s x s determinant H;-I i= O. All the
It has rank exactly s since the (b o, ... , bs ) of the system are of the form Un+1
bO=Adet
(
U
: U n+ s
n +s
Un
b l = (-l)Adet
n s
U + Un+s-l
)
: Un+s-I
Un+s+1
Un
b s = (-l)SAdet
s- I = I\.'Hn+I'
Un+2s-1
Un+1
(
)
solutions
(
Un+1
Un+2s-1 Un+s-I
: Un+s-I
)
= (_l)SAH~-l.
Un+2s-2
Thus, the solution is nontrivial if and only if A i= O. For such a solution we also have since Un
o = H~
= det
(
: Un+s -! Un+s
4. Integral Valued Entire Functions
284
= (-I) s+1 Un+s (hO) T
+ (-1) s+2 Un+s+1
+ ... + (-1) 2s-1 Un +2s (_1)S+1
=
')..
[bou n +s + ...
( _').. hi )
(b
- -s - ) (-1)'')..
+ bsUn +2sl
.
Therefore, exactly the same (b o, ... , bs ) chosen above will be a solution of the system bOun+l + ... + bsUn+S+1 = 0, { bOU n +2 + ... + b s Un + s +2 = 0, bou n +s +
... + bsUn +2s
= O.
This means that if we make a particular choice of (bo, ...• bs ) forthe system (tt) when n = N, for instance, taking').. = 1, then, for this particular choice, all the systems (tt) with n 2: N are satisfied and, in particular, the sequence (Un)n'?N satisfies a linear recurrence equation. This shows that 2:::n'?O unz- n represents a D rational function in a neighborhood of 00. 4.2.3. Corollary. Let the series G(z) = 2:::n>1 unz- n he convergent for large Izi. A necessary and sufficient condition for G to-be the Laurent expansion at z = 00 of a rational function, which vanishes at 00, is the existence of an integer s 2: 1 such that the Hankel determinants Ht vanish for all k ::: s.
+ ... + bsu n +s = 0 can be rewritten as bou n + ... + bsu n +s + OUn+s+1 + ... + OUn+k = 0
Proof. Namely, the linear recurrence bou n
for k ::: s. Thus, as we already know that if G is a rational function then H~ = 0 for n ::: 1 and some s ::: 1, we conclude that H: = 0 for all k ::: s and all n. In particular Ht = 0 for all k ::: s. Conversely, if Ht = 0 for all k ::: s ::: 1, then from (t) we will have (H{)2 = Ht Ht - Ht+ 1H;-I = 0 for all k 2: s. etc. We conclude that H: = 0 for all n ::: 1 and all k ::: s. The preceding proposition can now be used to finish the proof. D 4.2.4. Theorem (Patou). Let G(z) = 2:::n>O unz- n he the expansion about z = 00 of a rational function. Let s ::: 1 be the smallest of the integers k ::: 1 for which there is a polynomial Nko deg Nk .:::: k - 1, and a polynomial Dk, deg Dk = k, such that G = N k / Db and let N(z) ao+alz+ ... +as_lzs-1 G(z) = - - = - - - - - - - , - - - - D(z) bo + biz + ... +bsz s be such a minimal degree representation of G. Then, if all the Un are integers we can choose the denominator D(z) = bo + biz + ... + bsz s to be a polynomial
285
4.2. Integral Valued Entire Functions
with integral coefficients and b s have integral coefficients.
= 1.
In this case, the numerator N will also
Proof. We want to show that bs = 1. For that purpose, choose integers CI, ... , Cs such that if Wn := CIU n + ... + CsUn+s-l, then WI = W2 = ... = Ws oF O. From the factthatthere is a representation G = Nsf D s , with deg CDs) = s, we know that it follows that H~ = 0 for n ::: 1, and hence that there are nontrivial solutions bo, ... , bs of the system boul {
+ ... + bsUs+1 = 0,
~OUS+I + ... + b,U2>+1 = 0,
and, for each of them, the recurrence equations
will be satisfied. Since the Un E Z we can choose bj E Z and further assume that gcd(bo, ... , bs ) = 1. Choosing then aj by the rule as-I := bsUI, as-2 := bsU2 + bs-IUI,
we obtain two polynomials N(z) = ao + alz + ... + as_lz s- 1 and D(z) = b o + + ... + bsz s in Z[zj and such that N f D = G. Moreover bs oF 0, otherwise we would contradict the minimality of the degree s. For the same reason Nand D are coprime. We can therefore assume bs > O. All that remains to do now is to prove that bs = 1. We can make a further simplification of the problem. We can assume there is no common factor to all the Un. If dlu n for n ::: 1, then from the definition of aj we see dlaj, 0 :::: j :::: s - 1. Hence we can divide all the aj and all the Un by d and preserve the identities. To continue the proof we need a slight variation on the well-known Gauss' lemma for primitive polynomials. biZ
4.2.5. Lemma. Let P(z) = I:k<1 Pkzk, Q(z) = I:k<m qkzk, R(z) = I:k
Proof of Lemma 4.2.5. Since PI oF 0 and qm oF 0, then n = 1 + m and rn oF O. Assume there is a prime d which divides all the rko k :::: n. If d does not divide PI, then we have rn = Plqm and this implies that dlqm. From the next identity, rn-I = Plqm-l + PI-lqm, we conclude that dlqm-I. Continuing in this fashion we see dlqk for all k :::: n. Therefore, we can assume dlpl, and hence there is
286
4. Integral Valued Entire Functions
j 2: 0 such that dlpl, ... , dlpl_j and dlpl-j-l' Consider now the identity
qmPI-j-l
+ qm-lPI-j + ... + qm-j-lPI
= rm+l-j-l,
we can conclude as before that dlqm. Using the corresponding identities for rk, k :::: m + 1- j - 2, we obtain dlqi for all i ~ m. This contradiction proves the lemma. [] Let us return to the proof of Theorem 4.2.4. Since Nand D are coprime, there are polynomials A, B E Z[z] and an integer m E N* such that AN+BD=m. We can rewrite this as m = (A
~ + B) D = (AG + B)D =
CD,
where C is a Laurent series with integral coefficients (which is in fact convergent for Izi » 1), e(z) = Lk<1 CkZ k , CI =1= 0 (l E Z). If m =1= 1 then is must be that mlcb for all k. This is a consequence of Lemma 4.2.5, because the coefficients bo, ... , bs of D are coprime. Therefore we can assume that m = 1. Then the coefficient of the larger power of z in the product CD is precisely c1bs =1= O. It must be that 1+ s = 0 and c1b s = 1. Hence b s = ±1, therefore bs = 1 by the assumption b s > O. [] 4.2.6. Corollary. If Ln>l unz- n is the Laurent series expansion at 00 of the rational function G an£all the Un are integers, then all the poles of G are algebraic integers. In order to continue our study of the z-transform G(z) = Ln>o unz- n we need to remind the reader of the notion of transfinite diameter and-some of its properties (see, e.g., [BG, Chapter 4]). Let E be a compact nonempty subset of C, then one considers the quantities Mn (E) = inf{max IP(z)l: P(z) = zEE
Zn
+ alz n- 1 + ... + an}.
One knows that this infimum is achieved for some monic polynomial of degree n whose roots line in cv(E). Any polynomial that achieves this minimum is called a Chebyschev polynomial for E. The transfinite diameter of E is the quantity i(E):= lim (Mn(E»I/n. n~OO
If Y is a union of Jordan curves YI. ... , YI, with disjoint interiors, we will denote Int(y) = Ul:: j :9 Int(Yj). We shall need the following fact about the transfinite diameter. For every e > 0 there is a rectifiable curve YE, which is the union of finite many Jordan
287
4.2. Integral Valued Entire Functions
curves with disjoint interiors, such that E S; Int(y) and 7:(E) < 7:(Int(y,» :::: 7:(E)
+ E.
The first inequality is just a consequence of the monotonicity of the transfinite diameter. To find the curve Ye we proceed as follows. Let m ~ 1 be such that Mm(E):::: (7:(E) +sj2)m. Choose a Chebyschev polynomial Tm of degree m, and let Ye be the curve defined by the equation ITm(z)1 = (7:(E)
+ s)m.
This curve Y is a rectifiable union of finitely many Jordan curves with disjoint interiors and Int(y,) = {z E
=
For z E E we have (Tm(z» Mm(E) :::: (7:(E) + ej2)2, hence E S; Int(y,). Consider now the monic polynomials of degree mn Pn(z)
For z
E
= (Tm(z»n.
Int(y,) we have IPn(z)1 :::: (7:(E)
+ e)mn, hence + e)mn
Mmn(Int(Ye» :::: (7:(E)
and therefore, 7:(Int(y,» :::: 7:(E)
+ s.
Now let E be a compact subset of
1
Z n-I G(z)
d Z.
y
Extend the domain of definition of the sequence Un to Z by letting Un = 0 if n < O. Let Pk(Z) = zk + pk,lZk-1 + ... + Pu be a family of monic polynomials, k ~ l. We set Pk«U n
»
:=
Un
+ Pk.IUn-1 + ... + Pk,kUn-k.
Then it is easy to see that we have G(z)Pk(z) =
L
Pkz~~;»'
n?,O
Note that if we let Po(z) == 1, Po«u n» = Un, the last formula remains valid for k = O. Consider the Hankel determinants UO
Ho := det
( UI
:
Un
Un ) Un+l
U2n
,
n~l.
288
4. Integral Valued Entire Functions
It is clear that for any choice of polynomials Pk we have
Po«uo» PI «UI» PO«UI» PI «U2» ( H~ = d e t . .
Pn«u n » ) Pn«Un+I» .
Po«'un» PI«~n+j» Pn«~2n» since the determinant is a multilinear alternating function in the columns. Applying the same argument to the rows, we obtain Po«uo» PO«UI» n_ Pj«Uj» PI «U2» [ Ho - d e t : :
... ...
j
PO«u n» Pj«un+j» :'
.
..
Pn«un» Pn«Un+l» Pn «U2n» Observe that if for fixed k ::::: 0 we consider the new sequence u~ := Pk«U n», nEZ, then PI«U~» = (PkP/)«U n» = PkP/«Un», that is, the composition of the operators corresponds to the product of the polynomials. Therefore, we also have PO«UO» Pj«UI» Pn«U n» ) ( H~ = det PO«Uj» Pn«~n+t»
Po(~Un» =
PJ«UO)) PjPo«ud) ( det
PnPo~(Un))
Pj
«~n+j»
Pn«U2n»
POP1«Ut» Pt«U2»
... ...
p,,«un») PjPn«Un+I» .
PnPI«Un+l»
P;«U2n»
-1-.1
z
Moreover,
Vj,k = Pj Pk«U n» =
2~1
y
Pj(Z)Pk(Z)G(Z) dz.
After these preliminaries we can state and prove the following theorem due to P6Iya-Carlson:
4.2.7. Theorem. Let L:n>O unz- n be the Laurent series expansion at 00 of a function G hoiomorphic D = C\E, where E is a compact set with 7:(E) < 1. If the coefficients Un are integers, then G is a rational function.
in
Proof. Let £ > 0 be such that 7:' = 7:(E) + 2£ < 1 and let y be the previously found rectifiable curve such that E ~ Int(y), 7:(Int(y» :::; 7:(E) + £. Let (Tkk~.o (To == 1) be a sequence of Chebyschev polynomials for Int(y).
4.2. Integral Valued Entire Functions
289
Ho
We are going to transform the sequence of Hankel determinants of (un)n?,:O with the help of the sequence (Tkh?':o. As we have seen above, if Vj,k
= -I . 27rl
1 y
dz G(z)1j(Z)Tk(Z)-, z
then H~ = det(vp)O:::j,k:::n'
Let M = max zEy IG (z) / z I and L be the length of y. Since the Tk are Chebyschev polynomials for Int(y), there is a constant A > 0 (independent of k) such that for z E Int(y) ITk(z)1 .::::
+ e/
A(r(Int(y»
.::::
A(rY.
Therefore,
and
From the Hadamard inequality for determinants we have
IHnl o -<
II. (Iv·
j,O
1+'" + Iv·J,n I)
An +1 < (l _ 1r,)n+1 (r,)n(n+l)/2.
-
O""J:::n
Since r' < 1 there is an no such that for all n A 1(r,)n/2
---=----=-1 - r' Hence, for n
~
~
no we have
< 1.
no we have IH~I < 1.
On the other hand, the Un E Z for all n, whence H(j E Z for all n. We conclude = 0 for n ~ no. This implies that G is a rational function. 0 that
Ho
It is now immediate from the discussion at the beginning of this section that the following result holds:
4.2.8. Corollary. Under the same hypotheses as the P6lya-Carlson theorem we have Un
=
L
Qj(n)zj,
l:::j:::N
where the Zj are algebraic integers lying in E, together with all their conjugates, and the Qj are polynomials with algebraic coefficients in the field generated by Zl, .•• , ZN and their conjugates.
4. Integral Valued Entire Functions
290
In order to be able to apply the preceding theorem and its corollary we need to find an elementary way to estimate 'fee) for rather simple compact sets E. In particular, we assume that E has no holes, i.e., S2\E is simply connected. It is known (see [BG, Chapter 4]) that there is a conformal map ({J from EC onto B(O, rY, r = 'fee), ({J(OO) = 00, of the form W = ((J(Z) =
Z
+ !Yo + -!Yl + ... Z
hence, its inverse 1/1 has the form Z
= 1/I(w) = w
+ ao + -alZ + ...
(w E B(O,
rn
(see, e.g., [BG, Exercise 4.9.6]). Let us assume for simplicity that E is connected and its boundary BE is regular of class Coo. We know that 1/1, ({J admit a Coo extension up to the boundary of B(O, r) and E, respectively (see, [BG, Theorem 4.8.17]). For w = re iiJ we have dz = 1/I'(w)dw, and the arc-length parameter s in BE is given by ds
= rI1/l'(re i9 )ldl:l.
On the other hand, one can compute by residues
-1. 27Tl
1. ' Iwl=r
1/1
dw (w)w
=1
or, equivalently,
127r 1/I'(rei8 )dl:l = 2n. Hence, for the length L of aE one has L =
and we conclude that 2n
127r ds(l:I) = r 127r 11/I'(rei8 )ldl:l, ~
Llr, i.e.,
'fee)
r =
L ~ 2n'
Moreover, there is equality if and only if
127r 1/I'(reiiJ)d(} = 127r 11/I'(reiIJ)ldl:l, which can only occur if 1/I'(reiIJ ) = 1, hence when E is a disk of radius r. Let A = meA) be the area of E, then A
=! JaE r xdy -
ydx
= n (r2 _ lad 2 r2
_ ... _
n Ian 12 r 2n
... )
< -
nr2
4.2. Integral Valued Entire Functions
291
as an elementary computation shows. (Parametrize aE, as earlier, using 1/1'.) Equality holds if and only if an = 0 for all n ::: I, i.e., if and only if aE is a circle. It follows that
~ ~ r =T(E). Finally, we have obtained
L - < T(E) < ~ 7r - 27r' and observed that any of the two equalities implies that E is a disk of radius T(E) and one has equality throughout. One can relax the hypothesis of COO regularity of aE to C l regularity ([Porn]). 4.2.9. Corollary. Let En>o unz- n be the Laurent expansion of a function f holomorphic in the exterior of the unit disk. Assume further that the radius of convergence of the series in liz is exactly 1 and that Un E Z for all n ::: O. Then, either 8(0, IY is the domain of holomorphy for f or f is a rational function whose poles are algebraic integers lying, together with their conjugates, in 8(0, 1). Proof. If there is a regular point Zo of aB(O, I), it means that f admits an analytic continuation across an arc of the unit circle, one can therefore construct a compact set E, with Coo boundary, EC simply connected and L < 27r, such that f is holomorphic in E C • Namely, just replace a convenient arc of aB(O, 1) by its secant and smooth up the endpoints. By the previous inequality r(E) < 1 0 and we can thus apply Corollary 4.2.8.
4.2.10. Corollary. Let T be an analytic functional carried by 8(0, 1) such that F(T)( -n) E Z for all n E N* then, either the minimal convex carrier of T is 8(0, 1) or T has the form of a finite combination of derivatives of Dirac masses T
= '"' ak -o(k) ~ .J
Zj'
k.j
where the values log Zj are algebraic integers lying together with their conjugates in 8(0, 1). Proof. The last part is just a consequence of Theorem 4.2.4 and computation of G-transforms in Example 4.2.8. 0
4.2.11. Lemma (Fekete). Let E be a compact subset ofe with T(E) < 1, then there can be at most a finite number of algebraic integers with the property that they and all their conjugates lie in E.
4. Integral Valued Entire Functions
292
Proof. We recall that the quantities lin(E) = max {
II
IZj -
zkI 2/ n(n-I):
ZI,""
Zn
E
E}
I~j#~n
have the property that r(E) = limn--+oolin(E). If we had infinitely many algebraic integers lying together with their conjugates in E, there is an increasing sequence of integers nk ~ 00 that for any nk there is a family ZI, ... , zn, of distinct algebraic integers that contains all its conjugates. The product IlI~j
II
IZj -
zhl 2
:::
1.
I~j#h~nk
It follows that and thus r(E) ::: 1.
o
4.2.12. Lemma. Let P be a monic polynomial with integral coefficients and S := {z E IC: IP(z)1 < I}. The only algebraic integers lying together with all their conjugates in the closure S are the zeros of P and the solutions of any equation of the form (p(z»m = 1, m E N*.
Proof. Let
VI E S be an algebraic integer such that its conjugates V2, ... , Vn E S also. Their product a = P(YI)' P(Y2)'" P(Yn) must be an integer, since it is a symmetric function of the Yj, but lal < 1 by the definition of S, hence a = 0 and so P(Yj) = 0 for at least one j. It follows that P(YI) = ... = P(Yn) = O. If YI E S is an algebraic integer, Y2, ... , Yn its conjugates, Yj E Sand P(Yj) to, then P(YI), ... , P(Yn) are all algebraic integers of absolute value 1, and this set contains all its conjugates. It follows from a theorem due to Kronecker (see [PS, Part VIII, Exercise 200]) that they are roots of unity. This concludes the proof of the lemma. 0
4.2.13. Remark. The previous two lemmas are due to Fekete [Fe], [Bu], who also showed that if r(E) < 1 there is a monic polynomial P E Z[z] such that E ~ S = {z E IC: IP(z)1 < I} (see the construction of Y. after Corollary 4.2.6). Let us now return to the consideration of integral-valued entire functions of exponential type.
4.2.14. Proposition. Let K be a convex compact subset ofn = {z: 11m zi < 1T} such that reeK) < 1. Then, every entire function f satisfying f(N*) ~ Z and such that for every e > 0 there is C. > 0 such that If(z)1 :::; C.eHdz)+elzl (z E IC),
4.2. Integral Valued Entire Functions
293
has the form
L
fez) =
Pj(z)cJ.
I:5j:5N
where Pj E Q[z] and Cj are algebraic integers belonging together with their e Z Iogcj). conjugates to e K , 1 ~ j ~ N. (cJ
=
Proof. Let T be the analytic functional carried by K such that f = ~(T). Recall that t is such that (t. h) = (T. h). where h(z) = h(-z). Then is carried by -K and ~(T) = j. The function G(T) is holomorphic in (C\e K and has the Laurent expansion
t
about z = 00. From the previous results of this section we conclude that G(T) is a rational function, all whose poles ZI •... , ZN are algebraic integers. which belong to e K together with all their conjugates, and G(T)(z) = ao
+ alz + ... + as_lz s- I bo + bIZ + ... + bsz s
ao •...• as-I. bo.···. bs E Z. bs = 1. Applying the inversion formula 6.1.3. we have v f(l;) =
v
~(T)(l;)
= - -1.
1
v {L dz G(T)(z)eogz_
27r1
=
L
Z
y
P/l;)e-{LOgzj
•
I:5j:5N
When we compute the integral by residues we see that the polynomials Pj have algebraic coefficients. Namely. this follows the usual rule of computation of the residues of a quotient. In fact the coefficients are in the field generated over IQ by Zl •..•• ZN and their conjugates. 0
4.2.15. Corollary. Let K and f be as in the preceding proposition. If there is an integer m
~
2 such that K £ {z E Q: le z
-
ml
< I},
then fez) = P(z)m Z for some polynomial P
E
Q[z].
Proof. The compact e K cc B(m, 1), hence r:(e K ) < 1. Moreover, by Lemma 4.2.12, m is the only algebraic integer that together with its conjugates belongs to B(m. 1) = {z: Iz - ml < I}. The remark at the end of the proof of Proposition 0 4.2.14 implies that the polynomial P has rational coefficients.
294
4. Integral Valued Entire Functions
4.2.16. Corollary. If K and f satisfy the hypothesis of the Proposition 4.2.14 and if, moreover, K S; {z E Q: le'l < I},
then fez)
=
L
Pk(z)cZ,
l:::,k:::,N
there the Ck are roots of unity, Pk E Q[z].
Proof. There is an £ > 0 such that e K S; B(O, 1) n {z: Rez > -1 + £}, so that reeK) < I. We can apply Lemma 4.2.12 to P(z) = z and obtain that the Ck are roots of unity and moreover the Pk have coefficients in the field Q(w), where w m = 1 for some mEN'. 0
4.2.17. Corollary. Let f, K be as in Proposition 4.2.14, and K S; {z 1I :::: I} then
fez) = Po(z)
+
L
E Q:
Ie: -
Pk(Z)(I +e2rrir,),
l:::,k:::,N I
I
-
rk E] - 2' 2[nQ, Pk E Q[z].
Proof. There is £ > 0 such that e K S; B(l, I) n {z: Rez > £}, which ensures reeK) < I. We can now apply Lemma 4.2.12 to P(z) = z - I. The nonzero roots of (p(Z»m = 1 have the form 1 + e2rrik/n, -n/2 < k < n/2. 0 One can prove that if ao > 0 is such that r(exp(B(O, ao») = 1, then ao E ]0.834,0.845[. (See [Pi].) With this notation, we can prove the following result:
4.2.18. Corollary. Let f be an entire function of exponential type such that
If(z)1 < Me alzl for some M > 0 and 0:::: a < ao. Assume feN') S; Z. Then fez) =
L
Pk(z)c",
l:::,k:::,N
where Ck are algebraic integers in B(O, a) (together with their conjugates) and Pk E Q[z]. Moreover,
. .
(l)ifa<0.8thenckE (ii) ifa < log
I
{ 1,2, 3+i.J3 2 '
( 3 +2i.J3) I then Ck E
3-i.J3} 2
;
{l,2};
(iii) if a < log 2 then Ck = 1 (and f is a polynomial).
4.2. Integral Valued Entire Functions
295
Proof. See [Pi].
The reader will find other interesting extensions of the results of this section in [Gra], [Grum], [Bez], [Bu]. EXERCISES
4.1 and 4.2.
1. Let T be an analytic functional carried by a convex compact subset K of Q. Show that the following two conditions are equivalent: (i) The analytic functional T has the form
="
T
"Ak 8(V) L Va.!:'
~
where a" ... , a p E Q, rnk E N*, AA.v E
II (I -
B(z) =
UK'CCQ
Jeo(Q(K'», of the
eU1 Z)J1.1
I"SJ~p
al •.. , a p E Q, fLk E
N*.
2. Let T, and T2 be two analytic functionals carried, respectively, by the compact convex subsets K, and K2 of Q. Assume that KI + K2 S; Q and K, - K, + K2 S; Q. Show that for z f/- Int(y,)· exp(-K 2 ), G(TI * T2 )(z) is given by G(T,
* T2)(z) =
__1_.
2m
r G(T)(t)G(T2) (~)t dtt
J
y,
where y, = exp( - r,). with r, a convex polygonal curve such that K, S; Int(r d S; Int(r,) S; Q. Use this expression to find the Laurent expansions of G(T, * T2 ) at the origin and at infinity. 3. Let us recall the generating functions of four classical families of orthogonal polynomials. (i) Chebyschev polynomials of the first kind, en: "
g,(t,Z)=~Cn(l)zn= n~O
1- lz 2' 1- 21Z + Z
(ii) Chebyschev polynomials of the second kind, Un:
(iii) Legendre polynomials, Ln:
(iv) Gegenbauer polynomials,
•
~ e'
e~(A >
g4(t,z)=~ n(t)Z ,,~O
n
0)
I = (1-2/Z+z2»)"
296
4. Integral Valued Entire Functions
(a) For ( E ] - 1, I [ let 0, E ]O,:rr [ be such that cos 0, = (, and define the convex sets K 1•1 = (O}, Kj,t = i[-O" 0,] (j = 2, 3, 4). Let r j ., be a positively oriented Jordan curve in C\] - 00,0] enclosing exp( - K j ,,). Show that the relations (for Iz I sufficiently small)
~(1},t)(z) = -2~i
1 gj(t,~)e-'Log,~~
(1 :::: j :::: 4)
i'
define analytic functions 1}., carried by Kj ". (b) Show that if n E N and Pn represents either of Cn , Un, Ln, C; and g is the corresponding generating function then
r
I
d~
Pn(t) = 2:rri }y g(t,~) ~n+1
'
where y is a Jordan curve enclosing the origin such that exp(-K2,r) is contained in the exterior of y. (c) Let K be a convex compact subset of Q and! E Jt"o(Q(K» be such that its Taylor series at the origin is 2::n>0 a.z'. Show that for every t E ] - 1, 1[, j = 1,2, such that K + Kj,t s::: Q and K - K + Kj., (or Kj" - Kj" + K s::: Q) there is an analytic functional Sj,' carried by K + K j ., with the property that: (i)
~(SI.r)(n)
= anC.Ct) for all n EN.
(ii)
:F(S2,)(n) = a.Un(t) for all n EN.
(iii)
G(SI.r)(z) =
!(ze i9,) + !(ze- iO,) 2
'
Z E
(iv)
C\exp(-(K Z E
+ KI.r))'
C\exp(-(K + K 2
,,».
(Hint: If K - K + Kj" s::: Q let c be a convex polygonal curve enclosing K, C exp( -c), and for z fj exp( -(Int(c) + Kj,t)) define Sj,t by
=
Use residue calculus to obtain (i)-(iv). If Kj" - Kj" + K s::: Q just replace K by Kj" in the previous construction.) (d) Let t,sEl-I,I[ be such that 2()s+(),<:rr (or 0,+20, <2:rr), and let Q(t, s, z) = 1 - 4tsz - 2(1 - 2t 2 - 2s 2 )Z2 - 4tsz 3 + Z4. Show that for Izi < I
L Un(t)U.(s)zn = • ,,:0
1-
Z2
Q(t,s,z)
.
1 - 3tsz + (2(2 + 2S2 - l)z2 - tsz L Cn (t )C. (s )zn = ------'---,,----,--'---Q(t,s,z) 3
4.2. Integral Valued Entire Functions
(c4)
297
"'U ()C • 1 -2lsz + (2S2 -1)z2 ~ n t • (s)z = ------::--,---'----'--
.,,0
Q(t,s,z)
==
(ds) For A E N*, U-2(1)
' " Un (t)C). (s )z· ~
n
n~O
(e) For every r
"'(n ~
E
N*, t
== 0,
-1, U-I (I)
~ Un -
2 (t)
rt
C
k
= Ck,
(~) (n ~ m) Cn-2m(s)] (-z)'
= -'-n-__=o_ _ _-=~"'_-__=o=_____ _ __ _ , _ - - - - - " - - Q(t • S , z»).
E)-
I, 1[, show that
+ 1) ... (n + r)Cn(t)zn
n~O
C I)
rl'f + Cn(t)(-z)n = _n_=_o__n_ _-::---:-__ (1 - 2tz + Z2),+1
'i:
'" ~(n
.,,0
1- Z2 (r ++ 21) Un(t)(-z)n .-0 n + 1)··· (n + r)Un(t)z n = rl--:.c.-::"--::2------,2:-:-)--:+-;-I-(l -
tz +z
r
(Hint: Use part (c) with I(z) = (1 - z)-r-I.) The reader will find other applications of this kind of the G-transform, as well as the akin G-transformation, in the recent work of Avanissian and his student, Supper [Av!], [Av2], [ASI), [AS2), [SuI), [Su2).
4. Let 1 be an entire function of exponential type such that for some convex compact K £ n and every e > 0 there exists a constant C, > 0 so that 1 satisfies the inequality I/(z)1 ::::
c, exp(HK(z) + elzl).
Recall that under these circumstances, if I(N) = {O}, then 1 == o. Assume now that I(N) is a finite set::j: to}. Let T be the analytic functional carried by K such that J(T) = I. (a) Show that G(T) is holomorphic in the unit disk. (b) Show further that there are m E N* and a polynomial P of degree < m such that G(T)(z)
P(z) =-. 1- zm
(c) Deduce that if y is a Jordan curve in C\] different from -1, then I(z)
(d) Conclude that
1
=
00,
0] enclosing the mth roots of unity
-1-1
P(l;) e-Zlogdl;. y l;m - I l;
2JCi
is m-periodic and can be written as I(z) =
L
I(n)km(z - n)
O:Sn<m
where sin JCZ km(z)
=.!. '" m ~
IiI<mj2
e21fij(z/m) =
{
. (JCZ) msm . ( m JCZ) sm (m - 1 ) m . (JCZ) msm -;;
if m is odd,
if m is even.
298 Thus. with m'
4. Integral Valued Entire Functions
= [mI2]-
1 if m is even and m' f(z)
=
"L
= [mI2]
if m is odd. we have
(2lfikZ) .
CkexP -~
Ikl::::m'
5. Let f be an entire function of exponential type that is periodic of period r and satisfies lim f(n) O. I"t-+oo
E
C\JR
=
nEZ
Conclude that f == O. (Hint: If If(z)! ::: Ae B1 : 1• let m = [Blri/lf] previous exercise to g(z) = f(rzlm).)
+ 1.
6. Let f be an entire function of exponential type and a. Im(a - fJ) =1= O. Assume that for some no E N we have Ref(n +a) = Ref(n
+ fJ) =0
and. moreover. lim f(n)
Inl--oo
for all
fJ
and apply the E
C such that
n:::: no.
= O.
nEZ
Show that f == O. (Hint: Consider IrAz) = !(f(z + a) + f(z + a» and fp(z) = 4(f(z + fJ) + fei + fJ»· Show that for x E JR. fa (x) = Ref(x +a). f{J(x) = Re f(x + fJ). and thus f(z + 2i Im(a - fJ» = f(z) for all z E rc.) Prove the same result if we assume Imf(n +a) = Imf(n + fJ) for all n:::: no. instead of (*).
CHAPTER 5
Summation Methods
5.1. Borel and Mittag-Leffler Summation Methods Given a power series f (z) = l:n>O anz n of radius of convergence R, R < 00, we are trying to find explicitly the analytic continuation of f to the largest domain, star-shaped with respect to the origin, to which f admits an analytic continuation. Let us denote by D(f) that domain. (Why is it well defined?) We shall obtain D(f) as the union of certain domains Bp(f), such that in each of them we shall be able to describe explicitly the analytic continuation of f, these domains are parametrized by p ~ 1. The domain D(f) is called the star of holomorphy of f. We start by explaining how to determine B(f) = B1 (f), usually called the Borel polygon of f. For z E C* let us denote by Kz the closed disk Kz := B(z!2, Izl!2). Kz can also be defined by
o<
Kz
= (<;
E
C*: 3. E C* such that ~
= n,
Since for. =1= 0, Re(1!.) :::: I if and only if. Let us denote M (f) = M I (f), the set M(f) := {z E C: Kz
ReO!.) :::: I} U {OJ.
E B(~, ~)
- {OJ.
£ D(f)}.
As an example, let Zk =1= 0, I :::: k :::: N, R := inf{lzkl; 1 :::: k :::: N}, and the series f (z) = l:n;::O an zn be the Taylor series about Z = 0 of the rational function N
L
Ak
k=1
z-
Zk·
U
{AZk:
Then one can easily verify that D(f)
=q (
A: : I})
l::;k::;N
and that M (f) is the intersection of the open half-places lrb which contain the origin and are bounded by the lines Lk perpendicular to Zk and passing through Zk (i.e., Lk = {z: Re«z - zk)zd = O}.) In this case M(f) is a convex polygon. On the other hand, if 8(0, R) is the domain of holomorphy of a function f then D(f) = M(f) = 8(0, R). 299
5. Summation Methods
300
One of the essential points of the theory is contained in the following simple lemma. (Compare with Theorem 1.3 .11.)
°
5.1.1. Lemma. Let < R < 00 be the radius of convergence of the series f (z) L:n:::O anz n. In the disk B(O, R) we have the integral representation
=
where 8 1 (f) (z) := L:n:::O anz n/ n!, which is an entire function of exponential type exactly 1/ R.
Proof. In fact, we have lim sUPn~oo ~ = 1/ R. Hence, for every is an integer no such that lanllzl n < n! -
(~ + R
8
>
°
there
8) l:r n!
This inequality shows that 8 1 (f) is an entire function and, moreover, it is of exponential type at most 1/ R. We leave to the reader the task of checking that the type is exactly 1/ R. From the obvious identity n! =
1
00
e-tt n dt,
n:::: 0,
we obtain
On the other hand, for z fixed, Izl < R, we can choose that JL:= Izl(l/R + 8) < 1. Hence, for n :::: no, t > 0,
8
>
e-I "'" lanllzlnt n < e-t "'" (tJL)n < e-(l L-; n! L-; n! n===.n£
°
with the property
f.1-)t
,
n?:.n E
which is an integrable function on [0,00[. This allows us to conclude that
o
and the proof is complete. A similar proof leads to the following lemma:
5.1.2. Lemma. With the same notations as in the previous lemma, assume that =0 E
C* is such that the function t
Then:
f-+
e- I8 1(f)(tzo) is integrable in [0, 00[.
5.1. Borel and Mittag-Leffler Summation Methods
301
(i) for every A, 0 < A :'S 1, z = AZo, the function t ~ e- t T3 1(f)(tz) is integrable on [0, oo[ and
1
00
e- t T3 1(f)(tz) dt = s
10
00
e-stT31(f)(tzo)dt,
with s = llA :::: 1; (ii) the function S t---+
s
10
00
e- st T3, (f)(tzo) dt
is holomorphic in the half-plane Res> 1; (iii) the function fzo defined by fzo(z) := Zo
z
roo e-tzo/zT3,(f)(tzo)dt
Jo
is holomorphic in K Zo and it coincides with f on the segment Z = AZo, 0 < A < min{1, R/lzoll.ln particular, f and fzo define a holomorphicfunction, still denoted fzo' in the open set B(O, R) U K zoo Proof. There is nothing to prove in (i), it is just a change of variables. On the other hand, if Res:::: 1 we have le-stT31(f)(tzo)1 :'S e-tlT3,(f)(tzo)l. This immediately guarantees the condition (ii) (see [BG, § 1.2.9]). To prove (iii), we use that the earlier description of Kzo tells us that Re(zolz) > 1 is equivalent to z E Kzo. Hence, fzo is holomorphic in Kzo by (ii) and coincides with f on the segment AZo, 0 < Amin{1, Rltzt} by (i). Therefore, fzo = f on B(O, R) n Kzo,
and the conclusion of the lemma is correct.
D
This lemma justifies the definition of the Borel polygon B(f) as the set of those of l; E C such that there is a compact neighborhood V~ of l; and a nonnegative function g~ E L' ([0, oo[) with the property that for every z E VI; the inequality a.e. t > 0, holds.
5.1.3. Proposition. The sets M(f) and B(f) coincide. Proof. Let us show that M(f) S; B(f). Let Zo E M(f), then Kzo Cc D(f), therefore we can find a positively oriented circle y in D(f) enclosing K zo ' For ~ E y, we have l; :f. 0 and l; t---+ Re(zog) is a continuous function strictly
smaller than 1 - 8, for some 0 < 8. Choosing y conveniently we can make 8 as small as necessary. Hence, we can write for;; E y __ 1_ =
1 - zoll;
1
00
0
e-Ue(zo/I;ludu.
5. Summation Methods
302
We denote still by Then
I the analytic continuation of the series
I(zo)
Let A := max y mate
-1-.1y
=
bn
I/(n/~1
I(t;) (
t;
2::n~o anz n
roo e-Ue(zog)u dU)
io
< 00, hence for
(S-, u)
E Y
to D(f).
d~.
x [0, oo[ one has the esti-
which allows us to interchange the order of integration so that I(zo)
=
ioroo e-
U
(-1-.1y
1(t;)e(zom U dt;) duo
2:rr1
t;
Let 0 < e < R be such that B(O, e) S; Int(y). In this case,
-1-.1 2m
I (S)e(zo/nu dt;
t;
y
= _1_.
(
2m il~l=e
I(t;)e(zom u dS-
S-
= 13 1(f)(uzo).
where the last identity is an elementary computation of residues. Moreover, this identity yields the estimate. 113, (f)(uzo) I :s A C~:) e(H)u.
Finally, we conclude that I(zo)
=
1
00
e- U 13, (f)(uzo) du,
and the integrand is bounded by the integrable function A[C(y)/2:rr]e- 8u • It is clear we can choose a compact neighborhood Vzo of Zo such that for all z E V:o' K: CC Int(y) and Re(zg) :s 1-8 for Z E Vzo ' S- E y. Under these conditions, all the previous reasonings are valid and, hence, I(z) =
1
00
e- t 13, (f)(tz) dt
for all z E V: o' and the estimate e- u 113, (f)(uz)1
:s A e~:) e-~u
holds for z E Vzo. u E [0,00[. This shows that Zo E B(f). We conclude that M(f)
~
B(f).
Let us now prove the other inclusion, B(f) S; M(f). Let Zo E B(f), and a > small enough so that B(zo, a) ~ B(f) and there is a nonnegative integrable function gzo such that e- t l13 1(f)(tz)1 ::: gzo(t), a.e. t > 0, z E B(zQ,a). Consider the open set
°
U zo := (
U KZ) zeB(zo,a)
\{O}.
5.1. Borel and Mittag-Leffler Summation Methods
303
From Lemma 5.1.2 we obtain a collection of analytic continuations fz of f, defined on B(O, R) U Kzo ' z E B(zo, a)\{O}. We claim that Vzo U B(O, R) S; D(f). In fact, the open set Vzo U B(O, R) is clearly star-shaped with respect to the origin, and the different fz define, together with f, a single valued holomorphic function in this set. Indeed, if KZI n KZ2 -I 0, then this intersection is fZ2 in the nonempty open set B(O, R) n k ZI n k Z2' so that connected anc! fZI fZI = fZ2 in K ZI n K z,. Due to the maximality of the set D (f), we conclude that Vzo U B(O, R) S; D(f). In particular, Kzo S; D(f), so that Zo E M(f). That is, B(f) S; M(f). 0
="
We can rephrase these results by means of the following proposition: 5.1.4. Proposition. Let f (z) = L:n>O anz n have radius of convergence < R < 00, 8 1(f)(z) = L:n>O anz nliz!. The function f has an analytic continuation still denoted f, -to the Borel polygon B(f). B(f) is an open set which can be characterized as B(f) = M(f) = (z E C: Kz S; D(f)}. This analytic continuation is given by
°
Z E
B(f).
This procedure to obtain the analytic continuation f in B(f) is called the Borel summation method. In order to obtain a similar explicit analytic continuation to the whole star of holomorphy D(f) we must replace the exponential function by the Mittag-Leffler function Ep. We proceed to give a succint description of Ep and its properties. (The reader will find a detailed study of this function and its applications in [Dr].) 5.1.5. Definition. For p > 0, the Mittag-Lerner function Ep of index p is the entire function Ep(z) =
L
Zk
f(l
+ k/ p) .
k":O
It is easy to see that E 1 (z) = e Z and that E p is a function of exponential type, for instance, from the decay rate of its coefficients (see [BO, Chapter 4]). We need somewhat more precise asymptotic information. For future use we shall restrict ourselves to p ~ 1. Let us recall that the entire function 1/ r can be expressed in terms of the Hankel integral I r(z)
1
= 2Jri
1
Ya.
e'
~ dt.
where Ya,a is the Hankel contour in Figure 5.1, the indices satisfy 0 < a, Jr/2 a < Jr, and t Z = eZ Logt (see [BO, Exercise 5.3.3]).
<
304
5. Summation Methods
Figure 5.1
Changing variables in the integral, t -1- = p-
2rr i
r(z)
= uP = ePLog u. we obtain
1
Y~a
euP u -pz+p-I d u,
where Y~,a is in fact Yb,fJ' with b = a l / p , fJ = (XI p. Since a > 0 is arbitrary and (XI p E lrr 12p, rr I p[, we can replace Y~,a by the Hankel contour Ya.O, with () E lrr 12p, rr I p[. In particular, 1
ro +nlp) =
p
2rri
5.1.6. Lemma. For Izl < a we have Ep(z) = -p.
where
2rrl
°
1
1
Yo .•
uP
e
du u n+ I '
euP -du -,
Yo.
U -
z
< a, rr/2p < () < rrlp.
Proof. Let u = rei'!', r :::: a,
upi 1 L I-un+! -e <------ (l r ' zn
n,,=O
erPcosp(l
f.J,)
5.1. Borel and Mittag-Leffler Summation Methods
305
and, on the circular part we have
L n,,:O
Therefore, p 2ni
r
Jv
YaO
e
IU::
l
eupi.:::: A <
00.
(rJv
uP du '" P u _ z = ~ 2ni n~O
YaO
uP du ) u +1
n
e
n Z
zn
o
--~r(l+n/p)'" - E p (z) . n":O
5.1.7. Lemma. Thefunction p
z~--
2ni
1Yae
uP due u-z
is holomorphic in C\Ya,e.
o
Proof It is left to the reader.
Let us say that z is to the left of Ya,e, if it belongs to the component of C\Ya,e which contains the origin, To the right, otherwise.
e<
5.1.S. Lemma. Let a > 0, n /2p < (i) For
Z
n / p.
to the left of Yu,o we have Ep(z) = -p.
2nl
1 Yd"
(ii) For z to the right of Ya,e we have -p
EpCz) = pe"
p + -.
eU p -duu-z
1
2nl
Yd"
eU
du
P
--
u -
Z
Proof The first formula is an immediate consequence of the preceding two lemmas. To prove the second identity, let a' > a and let Z be a point to the right of Ya,e and to the left of Ya',Ij. Let Y be the contour described by Figure 5.2, From Cauchy's theorem we obtain
r
1 uP du I 2ni JY~e e u -z - 2ni
r
Jya • e
uP du u -z
=
p 2ni
r
Jy e
uP du zP u -z =e ,
o
which yields (ii). 5.1.9. Corollary. Let p > 1, 0 < B < n / p, such that n /2p tion Ep has the asymptotic behavior: Ci)
lim
Izl"?oo I Argzl~"./p-e
IEp(z) - pezPI = 0,
+8
< n. The func-
5. Summation Methods
306
Figure 5.2
(ii)
lim
IEp(z)1 =
lzl->oo ".j2p+e:s:1 Argzl:s:".
o.
o
Proof. It is left to the reader.
These lemmas also show that Ep has order p and finite type. We are going to compute now its regularized Lindelof indicator function Lp with respect to the order p (see Definition 1.3.15. p = 1 in that case). Lp(z) := (lim sup 1->00
~P log IEp(tZ)I) * t
. log IE (Il;) I ) = lim sup ( hmsup p ~->z
1->00
•
tP
5.1.10. Lemma. For p > 1, we have
if I Argzl :s if
7f
7f
2p'
2p < I Argzl
:s 7f.
In particular, Lp 2: O. Moreover, jar every e > 0, there is a constant C e > 0 such that
5.1. Borel and Mittag-Leffler Summation Methods
307
Proof. Note that the result holds for p = 1 as E 1(z) = e'. Let 0 < 8 « 1 so that I Argzl :::: n/2p :::: nip - 8. Thus we have Ep(tz) = pe(IZ)' + cp(tz), with lim(-+oo cp(tz) = O. It follows immediately that
· log IEp(tz)1 I1m tP
/ ...... 00
= Re( zP).
and this limit is uniform for Izl ? 1. In particular, for z = Izleirp , we have that if e > 0 there is Re » I so that for Izl ? R.
+ e.
log IEp(z)1 < Re(eirp) IzlP For n 12p :::: I Arg z I :::: n we can write IIp E (z) = - +PrO-lip) z 2ni
since
1
and
-p 21ri
1 YoB
e U P du =
uP
e
Yu.B
u
1
--=--+ u - z z
1
z(u - z)
udu , z(u - z)
.
1
ro- lip) •
because p > 1. It follows that E
z --
p()-
1 I 0 (-1 ) r(l-l/p)z+ Iz1 2
'
It is now clear that 10gIEp(tz)1 = _Iogt tP tP
+0
(~). t 2p
where the "big oh" estimate is uniform if Izl ? 1. So that lim log IEp(tz)1 = O. tP
/--+00
The continuity, as a function of z. of the limits obtained, avoids the use of the regularization. Thus we obtain that Lp is the continuous function described in the statement of the lemma. We note that we have also shown that in the second angular region we have that for any e > 0 there is Re » I such that for Izl ~ Re one has log IEp(z)1 IzlP :::: e. The function Lp is continuous and homogeneous of degree p, whence the previous observations imply that for e > 0 there is R. » 1 with the property that
5. Summation Methods
308
This implies the existence of a constant C e > 0 such that
o
everywhere.
5.1.11. Lemma. In the open set (z -1- = 1- Z
C: Lp(z) < I} (p :::: 1) we have the relation
E
1
00
e- tP t p- 1Ep(tz) dt.
0
Proof. Let K CC (z: Lp(z) < I}, then we can find e > 0 such that
sup(Lp(z)
+ elzl P -
1)
= -k
< O.
zEK
On the other hand, there is a constant C e > 0 such that IEp(tz)1
s
Ceexp(tPLp(z)
+ etPz P),
so that e- tP tp-1IEp(tz)1 S Cet p- 1exp(tP{Lp(z) + elzl P
-
S Cet p- 1exp( -kt P) if z
E
K. This proves the uniform convergence of the integral zE K
~p
1
00
e- tP t p- 1EpCtz) dt.
Figure 5.3
I}),
5.1. Borel and Mittag-Leffler Summation Methods
309
This integral defines hence a holomorphic function in {z: L p (z) < I} and, furthermore, one can justify for Iz 1< 1 the following computation:
= -1- . 1- z
'" n =~z n~O
o
The lemma follows by analytic continuation.
We are going now to return to the Borel method, but this time with the sets
K; := {z for z
E
E
IC*: 3r
E
C* such that
C*. It is easy to show that Kf
Dp :=
Ki =
<=
= zDp,
rz, Lp{l/r)::: I} U to}, where
{r E IC*: Lp(l/r) ::: l} U
to},
which is a compact convex set with 0, I E aD p • Moreover, Dp is star-shaped with respect to the origin, and the equation of aD p in polar coordinates is ](
](
- - <8 < - . 2p - 2p
Figure 5.4
5. Summation
310
For -7r/2p < 0 < 7r/2p one has, with r' r2
+ 21"
-
1'1'''
= dr/dO
and r" = d 2r/d0 2 ,
+ P + (\ + p)(tan pO)2)
= (COSO)2/p(l
which shows that Dp is convex. For f (z) = LII'>o all z", with radius of convergence and p 0:: 1 we denote by Mp(f) the open set, Mp(f) := Iz E C*:
5.1.12. Lemma. D(n
Kf
Method~
°
< R <
> 0,
00
as above,
C; D(f)}.
= Up~1 Mp(f).
Proof. It is left to the reader. The main point is to observe that K r: lies in an 0 angular sector of opening 7r / p. As before, we denote by Bp(f) the family of those l; E C such that there is a compact neighborhood V~ of l; and a function & ELI ([0, oo[), gl; 0:: 0, such that for every ;; E VI: we have e- I " tP-1jBp(f)(tz)j ~ gl;(t)
a.e.
t > 0,
with
The function Bp(f) is sometimes called the Mittag-Leffler transform of f. One verifies without difficulty the following lemmas and propositions, whose proofs we leave to the reader:
order p of
5.1.13. Lemma. Bp (f) is an entire function of order p and finite type. 5.1.14. Lemma. Let Zo on [0,
00[.
E
C· be such that t
Then:
(i) For every z = AZo, 0<1. ~ I, t [0, oo! and
p
e-tl' t p- IBp(f)(tzo) is integrable
1--*
1--*
e-ll'tp-1Bp(f)(tz) is integrable on
r
{X e-t"tp-1Bp(f)(tz)dt = ~ )0 AP )0
c
e-u"/).,Pup-IBp(f)(uzo)du
= psP ('" e-
)0
with s = 1/1.. (ii) The function S 1--*
psP
1°C e-(ts)P t p- Bp(f)(tzo) dt, I
sP = exp(p Logs), is holomorphic in the region Is: Re(sP) > I}. (iii) The function f:o (z) := p(zo/z)P
1
00
e- tP (zo/z)Pt p- 1 Bp(f)(tzo) dt
5.1. Borel and Mittag-Leffler Summation Methods
311
is holomorphic in k~o and it coincides with / in the segment Z = AZo, 0< A. < min{R/lzol, I}, so that / and /:0 together dejine a single-valued holomorphic/unction, also denoted /:0' in B(O, R) U K~o'
5.1.15. Proposition. The sets BpC!) and Mp(f) coincide. There/ore, DC!) =
U Bp(f). p~l
5.1.16. Proposition. Let / (z) = Ln>o an zn have radius 0/ convergence 0 < R < 00. The analytic combination ol/ to the region Mp(f), p ::: I, is given by the formula
Up to now we have applied the summation methods of Borel and Mittag-Leffler to germs of holomorphic functions at z = O. It is a rather old idea (see [Har]) to try this technique on divergent power series, i.e., formal power series. Let us consider a simple example, due to Euler. In Section 1.2 we have observed that the Euler differential equation x 2 y'
+y =x
admits a hyperfunction solution. On the other hand, if we try to find a formal power series solution we find
lex) = x L(-l)nn!x n, n~O
which converges only for x = O. Apply the Borel summation method of Lemma 5.1.1 to the formal series. We obtain I
F(z) = L(-l)nzn = - . 1+ z O n~
Applying formally the Borel formula of Lemma 7.1.1, we consider [(x)=x
100
e-tF(tx)dt=x
o
100 e-t--dt= I 100 e0
1 + tx
0
ds
S/ X _ _
I
+s
and we find an integral whose asymptotic development when x -4 0+ has already been considered by Euler and Poincare (see [01]). That is, if one develops the exponential term as a power series, one obtains [(x)
=
ro e-
Jo
s/ x
~ =x 1+ s
l! x 2
+ 2!
+ (_l)n-l(n -
x3
-
..•
I)! xn
+ Rn(x),
312
5. Summation Methods
:s
IRn(x)1
1
00
sne-s/x ds = n!xn+l.
That is, the Borel summation represents the hyperfunction solution f in an asymptotic sense. These elementary remarks show that, at least in this example, there is a relation between hyperfunctions, summability, and asymptotic developments. In fact, in recent years these ideas have been carried out to develop a very beautiful theory, which has already been used to solve several difficult problems. We refer the reader to [Ec] , [MR], [Mal], and [Si] for these new results. Our presentation owes much to [Lin]. EXERCISES
5.1.
1. Show that, a priori, it always makes sense to talk about the largest domain, starshaped with respect to the origin, into which a power series / admits an analytic continuation. 2. Verify that if fez) = Ln>oanz n is the Taylor series about z = 0 of the function Ll
=
n
{z: Re«z -
zdzd < OJ.
l:;;k~N
3. Show that if fez) = Ln>o anz n has radius of convergence R, 0 < R < entire function F(z) = Ln<:o(anln!)zn has exponential type exactly 11R.
00,
then the
4. Show that for a > 0, the set
U KZ) \{O}
U,o := (
zeB(,o.a)
is open, and that Uzo U 8(0, R) is star-shaped with respect to the origin (see proof of Proposition 5.1.4). 5. Justify the proof of Lemma 5.1.8, and prove Corollary 5.1.9.
6. Verify that Dp is a convex compact set, Mp(f) is open, as well as provide the proofs of statements Lemmas 5.1.12-5.1.14 and Propositions 5.1.15 and 5.1.16.
7. Let rp be a holomorphic function in the closed half-plane Re z ::: ct, a > O. Assume there is r, 0 :=: r < Jr, with the property that for every e > 0 there is a constant Ce > 0 such that Jr
IArgzl:=: (a) Let f3 :::
ct
2'
and m an integer so that m - 1 < f3 < m. Prove that for every integer
N:::m
~
n
L.Jrp(n)z =
n=m
1 rp(~)z~ e2rri~
_
1 d~,
YR
where YR is the contour described by the Figure 5.5. We denote in YR.
r R the arc of circle
5.1. Borel and Mittag-Leffler Summation Methods
313
R=N+1!2-~
m
m-l
Figure 5.5
(b) Let 1/10 E ]0, n/2[. Show that there are constants Ro for 0 < r < I, I; f3 + Reil/t, one has
~
=
-r)~ I < k P IqJ(l;)( e2r
-R(log(l/r)cos l/to-r-E)
0, kl > 0, k2 > 0, such that
if
11/II:so 1/10,
if
1/IO:SO 11/II:so
and qJ(I;)(-r)/; 1
(c) Choose
Let k
e21r1~ _
1
I
< k r Pe- R(1rsinl/to-r-E)
-
2
ro so that r < Jr sin 1/10 < Jr,
1/10,
0 < ro < 1, (-log ro) cos 1/10 < ro, I; ErR, one has
= max(k l , k 2 ) and show that for any 0 < r
i· = Jr sin 1/10.
-R(1r sin l/to-r-E)
.
Conclude that for 0 < r < 1
and that Lrp(n)(-r)"
=-
ni!m
=
(d) Let I; f3 + it, a > 0 very small, r Write (JI = (J - Jr and show that
I
rp(l;)z/; .
e27r1~ _
1-- r e .I fJ -0,1
1
1
1J+iOO
P-ioo
+a qJ(f3
(7)( r)1; qJ 2:il;
e
-=-
< (J < 2Jr - r - a, r > 0, and z
+ it).
el1rfJ "' _ e-J1r {J+trt
:so
rfJ exp { -It I
dl;. I
= re iO •
I
((71' - r + I~I (JI + e(t)) }
with e(t) --+ 0 as It I --+ 00. Given 0 < 11 < a, show there is A > 0 such that It I > A implies Jr - r + (t/ltIWI + e(t) > 11 and
I
1-
rp(l;)zl; P -~ltI e2r
314
5. Summation Methods
Conclude that the function rp(~ )z~
P+ ioo Z 1-+
1
is holomorphic in the angle (e) Let
t"
1
e 21ri \ -
P-ioo
< arg z < 2:rr -
d~
t"-
00
F(z) :=
L rp(n)z". "=0
Show that F is holomorphic for then F(z)
=
Izl
< e-'. Moreover, if m > 0 and m - 1 < f3 < m,
L rp(n)z" O~n~m-I
l
and for m < 0, -m - 1 < f3 < -m, F(z) = -
'"
~ l,on,o-m
rp( -n)
-- z"
rp(~ )Zl
P+ ioo
.~
2
fJ-ioo
l
e
7fI
rp(~)z~
P+ ioo
fJ-ioo
1
-
.
e21r1~
Conclude that F has an analytic continuation to the angle
t"
-
I
d{,
d~.
< arg Z < 2:rr -
t".
8. The following are essentially direct applications of the previous exercise: (a) Let 0 < eT < 2, 0 < f3 < I, and rp(z) := Z-I7'. Show that
L
n~O
zn nl7"
=1-
1fJ+ioo
fJ' -100
~-17~ z~
e21ri~ _
1 d{
in the angle :rreT /2 < arg z < 2:rr - :rreT /2. (b) Prove that for 0 < f3 < I 1
1 1+ioo Zl - = 1- 1 fJ-ioo 1- Z e 21ri \ -
I
d~
when z E C\[I, 00[. (c) Let K cc C\[I, oo[ and eTo > 0 be such that :rreTo :rreTO } K -C { -2- < argz < 2:rr - 2 .
Show that if 0 < f3 < 1 and 0
~
eT ~ eTo, then for z E K
z" 1 l L---= nl7" 1- z
fJ-ioo
";::0
Deduce that
fJ +ioo
(1 _ ~-17~)
e 21ril
-
1
z\d~.
zn I lim " 'n17" = 17-+0+~ 1-z n~O
uniformly on every compact subset of C\[I, 00[. (d) Let fez) Ln;::o anz" be a power series with finite nonzero radius of convergence. Show that the series
=
converges uniformly on every compact subset of D(f).
5.1. Borel and Mittag-Leffler Summation Methods
315
9. Let !p be a holomorphic function in 1C\(alo ... , aN j, aj E IC\Z, such that there is a value Ro > max/ laj I: I :::: j :::: N j with the property that for any e > 0 there is CE > 0 so that if Izl ::: Ro then l!p(z) I :::: CEeElzl. Use the results of this section and Exercise 5.1.8 to prove that: (a) The series fez) = Ln>orp(n)zn has an analytic continuation to IC\[I, 00[. (b) Let No E N be such that all aj satisfy Re(aj) > -No, and define . ~ ct>(z) := 2:n:1 ~ Res
(rp(~)z{
)
e27ri~ _ I' aJ
I~J~N
which depends on the choice of argz. Show that if z E 1C\[I, 00[,0 < argz < 2:n:, and -No - 1 < fJ < -No then the function f of part (a) satisfies fez)
()
~ !P -n _ ct>(z) _
=-
~
zn
l~j~N
(c) Let z E
I-
00, -1[,
1
1Hioo
= -(No + 1/2), and ~ = fJ+
fJ
(7)Z'
!P .'-'
, e 2m'-1 /3-100
d~.
it, then, for some e(f3) > 0,
-z){ I : : Izllle-7rltl+E(Il)JIl2+t2, l !p(~)( e 2m , - 1
with the additional property that if we let No -+ -00 then e(f3) -+ 0. Conclude that
11
1l+ ioo Il-ioo
!p(~)(-z){ d~1 e27rt{ -
1
::
21zlll {e E (lllJ211l1
Deduce that for Izl > 1 fez)
r
lill
10
e-7rt dt
rp(-n) = - 2: -zn
+
1
00
e- 7rt +E(lll../'ir
dt}.
IPI
ct>(z).
n~1
(d) Let n = /z E IC: Izl> I, z fj [I, choice of arg z in n, let fez) := -
ooD
rp(-n) 2: - zn
and 0 < argz < 2:n: for ZEn. With this
CI>(z)
(z En).
n:::l
Prove that for x > lone has F(x
+ iO) -
F(x - iO)
= -ct>(x) + ct>(e 27ri x) = 2:n:i 2: Res(!p(~)x~,aj). l,!!:j~N
Let FI (z) := F(z) + 2:n:i Ll~h,N Res(!p(~)z{, aj). Compare the boundary values FI (x ± iO) and F(x ± iO) for x > I. (e) Show that if rp is an entire function of exponential type zero, then FI F f, and this is a function holomorphic in IC\/lj, which vanishes at 00. (f) Let k E N*, rp(z) = Z-k. In this case,
= =
zn
fez)
. (1
2m Res
;;k
z~
e27ril; _
=:E kn ' n~J
0) _~
1'
-
2:n:i
(Izl > 1),
5. Summation Methods
316
where the
I{Ik
are defined by
~= eW
-
1
Then F(z) = (-1)
HI
!.. + '" r/ik( -I;') w'-
L
1 1 n k zn
= F(z) + 27ri
I.
zn
k,,1
-- -
k,,1
FI (z)
L..J
w
(27ri)k
(IOgZ)
k!
2m
--l{Ik
-.
,
(log Z)k-I (k _ I)! .
5.2. The LindelOf Indicator Function 5.2.1. Definitions. (1) The Lindelof indicator of order p > 0 for an entire function strictly less than p or of order p and finite type, is the function
=
hl,p(z)
f of order
log IfUZ)I) * ( limsup---tP
1-'>00
(2) The LindelOf indicator of order p > 0 of a subharmonic function u in C,
which is of order
:s p
and finite type, is the function hu,p(z)
=
. uCtZ»)* ( hmsup-'-+00 tP
5.2.2. Proposition. The indicator function h I.p of an entire function f ¢. 0 of order < p or of order p and finite type, is a homogeneous function of degree p, subharmonic and continuous. Proof. The proof is analogous to that of Propositions 1.1.15 and 1.1.16. The
only thing that requires modification is the proof of the continuity. Hence, we shall only give a detailed proof of the lemmas analogous to §§ 1.1.17 and 1.1.18.
5.2.3. Lemma. Let 0 1 < O2 , O2 h j ,p(eiiJ2 )
-
01 <
7r
I p.
Assume that hj,p(e iiJ ,)
:s h 2 • Let gCO) :=
Then,for every 0,0 1
:s hi
and
hi sinp(&2 - 0) + h 2 sinp(0 - ( 1) . sin P(02 - ( 1 )
:s 0 :s O2, hj.p(e iiJ )
:s gee).
Proof of Lemma 5.2.3. Let 8 > 0 and let g8 (0) = a p cos pO + b p sin pO be the trigonometric function that takes the values hj + 8 at 0 OJ. Define
=
/8(z) := !(z)e-i(a.+ib,)ZP
5.2. The Lindelof Indicator Function
317
The function 18 is of order p and finite type in the angle 01 ::: 0 ::: 02, and it converges to zero along the sides of this angle. By the Phragmen-Lindelof principle, I f.1 is uniformly bounded in the whole angle. It follows that ·8
81
h j,p(e' ) ::: g.(O),
:::
8 ::: 82 ,
o
and the conclusion of the lemma follows by letting 8 ---+ O.
Let us point out that if hj,p(e iIJ1 ) = hj,p(e i8,) = -00 and 82 - 0) < rr/p, then the proof of the lemma implies that hj.p(ei8 ) = -00 for 8 1 ::: 8 ::: 82 .
5.2.4. Lemma. Let 8 1 < 82 < 83,83 - 8 1 < rr / p. Let g(8) = a cos p8 be such that ht,p(e iIJ1 ) ~ g(8 1 ), ht.p(e ilh ) ~ g(82 ). Then
+ b sin p8
hj.p(ei83) ~ g(83). Proof of Lemma 5.2.4. We argue by contradiction. Let 8 > 0 be such that hj ,p(e i83 ) ~ g(83) - 8. Consider the auxiliary function
sinp(8 - 8)) . smp(83 - ( 1)
g.(8) := g(8) - 8 .
Then g.(8 1) = g(8), 8.(82) < g(82), and g.(83) = g(03) - 8. The previous lemma shows that
o
which is impossible.
The same proof shows that if h j .p (e i 8,) ~ g(02) and h j ,p(ei8J ) ::: g(83), then hj.p(eiIJI) ~ g(81).
The rest of the proof the continuity in Proposition 5.2.2 is analogous to that of Proposition 1.1.15. 0
5.2.5. Proposition. Under the same hypotheses as Proposition 5.2.2, for any e > 0 there is a constant C e > 0 such that,for all
If(Z)1 :::
C e exp(ht,p(z)
Z E
C,
+ elzI P ).
Proof. Let Izol = 1, and find 0 < a < 1 so that if z E B(zo, a), then we have hj,p(z) < ht,p(zo) + e/4. This is possible by Proposition 5.2.2. From the Hartogs lemma [BG, §4.4.40], for a given a, 0 < al < a, there is tl > 0 such that, for all t ~ t) and z E B(zo. al),
1
t P log If(tz)1 ~ hj.p(zo)
e
+ 2'
Clearly, we could choose a) sufficiently small so that for z have ht,p(zo) ~ hj,p(z) + e/2. Hence, I t P log If(tz)1 .::: hj,p(z)
+E
(for ( ~
(1, Z E
E
B(zo, al) we also
B(zo, al».
318
5. Summation Methods
It follows immediately that there is a constant C(E, zo) > 0 such that for every z E C* so that z/Izl E B(zo, ad one has
+ ElzI P ) + C(E, zo).
log If(z)1 :::: (ht,p(z)
0
It is easy to finish the proof using the compactness of the unit circle.
5.2.6. Proposition. Let 1/1 be a subharmonic function in C that is homogeneous of degree p, i.e., 1/I(tz) = t P 1/I(z) for all t > O. There is a constant A > 0 such that for every (), 0 :::: 8 :::: 2:rr, there is an entire function fo of order p and finite type such that '0
ht9,p(e' )
= 1/I(e''0 ),
Ifo(z)1 ::::
A(1
+ IzI 2 )eY,(z).
Proof. We can assume () = 0 and search for a function f = fo of the form fez) = g(z) - h(z)u(z), where g, u E Coo(C) and h(z) :=
II (1 _ Zj)'
j~l
2
For a given j ::: 1 and z E C such that ~ :::: Iz - 2j I :::: 4~
2j -< 2j
-
!4< -
4one has
Iz I -< 2j + !2 <- 2~ 2j •
Hence,
> -C .-
21+2
II
(3- 21.- k 4
j-3
1) > -C - 32
II 2
k
>
-
Cl
> 0
,
k=l
15k<j
where C:= TIk>I(1- 3/2k+ 1) > 0 and Cl > 0 is independent of j. Let cp E V(C), fJ :::: cp :::: 1, cp = 1 on Izl :::: cp = 0 for Izi ::: and define
4,
!,
g(z) :=
L cp(z -
2j ) exp(1/I(2 j
».
j~1
In fact, there is at most one nonzero term for each z, so g is a Coo function. Moreover, g is constant on every disk B(2 j , For the Coo function f = g - hu to be entire we must have
!).
o = af az
=
ag _ h au. az az
The function w(z)
=
_1_ { ~(Z)
ag (z) az
for z ¢
U B (2j, !) , j~1
for
Z E
U B (2j, !) ,
j~1
5.2. The Lindelof Indicator Function
319
is a Coo function. Furthermore, for some constant K > 0 we have Iw(z)1
:s K
Let p(z) := ( sup 1/I(z + 1~1~1/2
:! (Z)I·
1
~»)
+
+ 10g(1 + IzI2).
Then p is subhannonic in C and
:! r)
1
(lgl 2 + I
e- 2p dm
:s KI
< 00.
From Theorem 2.1.3 we deduce that there is a function v ov/oz = w, and
r
Iv(z)1 2
le (l + Iz12)2
e- 2p (z) dm < K2 <
E
Coo (C) satisfying
00.
-
It follows that the function 1 = g - hv is entire. To estimate it, we use that Izl 2j+I, then logO + x) x for x ::: 0, so that if 2 j
:s
log Ih(z)1
:s
:s
:s ~ log (I + ~n:s ~ k~1
log
(I + 2j+I-k)
l~k~j+1
+
~ ;k k~1
'~ " (h + 2 - ]) . = 1+ -2-(] log2. + I)(]. + I)
:s 1+ (log 2)
l~k~j+1
:s C2 (log2(l +
Izl)
+ I)
for some C2 > 0, and this estimation holds everywhere if C2 is sufficiently large. Let then
1
Iv(z)12 + Iz12)2
----'.,-'--:-::--::-e
e (l
It follows that if ql(z) := q(z)
-2q
+ log(l +
d
m < 00.
Iz1 2), then
11/12e-2q, dm <
00.
As we have done in Chapter 2, we have I/(z)1 =
I~ 7C
r
1
B(z,1)
1(~)dm(~)1 :s ~ 7C
r
1
B(z, I)
I/(nle-q'<~)eq,(~) dm(~)
5, Summation Methods
320
::::
Alexp(sup{ql(~):
Iz
-~I::::
:::: A2exp [log2(1 + Izl) + Therefore, using that
I))
(sup{1/I(z+~): I~I::::
1/1 is homogeneous of degree
p > 0, we have
limsuploglf(tz)1 :::: lim sup (sup {1/I (z+£): /--'>00
since
tP
t
/--'>00
nr],
I~I:::: ~})*:::: 1/I(z),
1/1 is upper semicontinuous. Hence, hi,p(z) :::: 1/I(z),
On the other hand, for z = I we can take t = 2 j , and we get f(2 j
= g(2 j
)
) -
v(2 j )h(2 j
)
=g(2 j ) = exp(1/I(2 j )). So, for those t log If(t)1 = 1/I(t)
= tP1/IO).
tP
It follows that hi,pO) 2: 1/1(1),
whence, hi,p(l) =
1/10). This concludes the proof of the proposition.
0
5.2.7. Lemma. Let u be a subharmonic function in C that is homogeneous of degree p > O. There is a decreasing sequence (U n )n2:1 of subharmonic functions, homogeneous of degree p, and Coo in C\{O}, such that Un converge toward u everywhere. Proof. We have shown in [BG, Chapter 4J that we can always approximate u by Coo subharmonic functions. The problem is to keep the homogeneity. This requires a slight modification of the usual construction. For that purpose, let q; be a standard radial function in C, i.e., q; E V(8(0, I», 0:::: q;:::: I, Icq;dm = I, q;e(z) = e- 2q;(Z/e), let
k k
ve(z):=
and define us(z):=
u(z - w)q;e(w)dm(w),
u(z -Izlw)q;,(w)dm(w).
It is easy to see that U e is Coo in C\ {OJ and it is homogeneous of degree p. We know [BG, Chapter 4J that the Ve are Coo, subharrnonic, and converge decreasingly to u as e -+ 0+. On the other hand, for Izl = 1, we have us(t) = vt(z),
so that ue(z) -+ u(z) decreasingly on Izl = 1 when e -+ 0+. The homogeneity now implies that U e -+ U decreasingly everywhere. (Note that for z = 0 we have u(O) = us(O) = 0.)
5.2. The Lindelof Indicator Function
321
The only thing to verify is that the for each r > 0 the function Ur(Z):=
121< u(z -Izlreiil)d(l
is a subharmonic function of z. If z Ur(z)
are subhannonic. Let us show first that
Ue
= re ia , we have
= 121< u(z =
1
2IC
= lim
e--+O
ze- ia re i9 )d(l
u(z - zre if3 )d{3
1
ve(z - zre if3 ) d{3.
2IC
0
Each of the last integrals is easily seen to be subhannonic by direct computatioll of the Laplacian. Since Ur is a decreasing limit, if follows that it is also q;e(r)r dr = 1, so that Us is an average subharmonic. Finally, q;e(r)r ~ 0 and of subhannonic functions. It foHows that U e is subhannonic. 0
f;
We are trying to prove that every subharmonic function that is homogeneous of degree p > 0 is the LindelOf indicator function h f.p of some entire function. Since we know that indicator functions are continuous, we need to show first that any such subharmonic function is also continuous. S.2.S. Lemma. Every subharmonic function u that is homogeneous of degree p > 0 is a continuous function in the whole complex plane. Proof. It is clear u is continuous at z
= O. Therefore,
its enough to prove that
in every angular sector of the form S9.a = {z E C*: I arg z -
(I
I<
ex <
~}
,
the function q;(I(z) = q;(z) := u(zlfp) is convex. The previous proposition allows us to assume first that u is Coo in C*. From the homogeneity of degree 1 of q; we conclude that oq; oq; x-+y-=q;,
ox
oy
a2 q;
a2 q;
ax 2 + Hence,
oy2
~
o.
5. Summation Methods
322
Therefore, for xy -:f; 0 we have a2qJ a2qJ (a 2qJ ) 2 ax2 ay2 ax8y
= O.
By continuity, this holds everywhere in the sector, so, (o2qJ/ox 2) (fPqJ/oy2) 2:: O. If we now consider qJ.(z):= qJ(z) + elzl2, its Hessian is now a strictly positive quadratic form, so that qJ. is strictly convex. Letting e ~ 0 we obtain that qJ is convex, when u is Coo in C·. In the general case, by Lemma 5.2.7 and what we have just proved, we have that qJ is a decreasing pointwise limit of convex functions in C·. Thus, qJ is also convex. This implies qJ is continuous, whence u is continuous. 0 For a given function qJ, subhannonic and homogeneous of degree p > 0, let us introduce the Banach space RI{J of all entire functions f in C such that
= 0(1)
If(z)le-I{J(Z)
with the norm IIflll{J = supzEIC If(z)e-l{J(z) I. Let qJn(z) := qJ(z) + IzlP In, then we also consider EI{J = nRl{Jn' n::;l
this space of holomorphic functions has countably many norms IIflll{Jn and it is a Frechet space with the topology defined by these norms. It follows from Proposition 5.2.5 that f E EI{J if and only if hf,p ~ qJ.
5.2.9. Theorem. Let 1/1 be a subharmonic function in C, homogeneous of degree p > O. There is an entire function f of order p and finite type such that h f,p =
1/1.
=
Proof. Consider a sequence (zn)n::;(, IZnl 1, which is dense in the unit circle and such that every point appears infinitely often in the sequence. Let Un be the following set of subhannonic functions, homogeneous of degree p: Un
= {u: u ~ 1/1 in C,
u(z)
~ 1/I(zn) - ~ in R (zn, ~) }.
Let us define a subhannonic function, homogeneous of degree p, by un(z) := (sup{u(z): u E Un})·.
We could appeal to a lemma of Choquet [Choq] to show that Un is the regularization of the supremum of a countable family of elements in Un, but in fact, it is easy to give a direct proof of the subhannonicity of Un. In fact, Un is upper semicontinuous, Un (z) ~ 1/1 (z) < 00, hence locally integrable and for any u E Un, U :5 Un. Therefore, for r > 0 fixed, u(Z) :5 A(u, z, r) :5 A(u n , z, r),
323
5.2. The Lindelof Indicator Function
and the last function is continuous as a function of z. Hence, sup u(z) ::: A(u n , z, r) UEUn
implies Un(z) ::: A(u n , z, r),
which is what we wanted to show. The fact that Un is homogeneous of degree p is clear, and it follows that Un E Un. So that Un = {U: U ::: Un}. To simplify the notation, let En = Eu n • We have the strict inclusion En ~ Ey"
since f E En implies that hU(z) ::: un(z) ::: 1/I(zn) - lin for z E B(zn, lin), and we know from Proposition 5.2.5 that there is a function In E Ey, such that hU(zn) = 1/I(zn)' Moreover, the inclusion map
is continuous and hence, by Proposition 1.4.11, En is a set of first category in Ey,. Thus,
I
we claim that hl,p = 1/1. If not, let I; be such that Choose e > 0 so that hl,p(I;) < 1/1(1;) - e, By continuity, there is a 8 > 0 so that hl,p(z) < 1/I(z) - el2 for all Z E B(I;, 8). There is a point zno of the sequence such that zno E B(I;, 8). Since the same point appears infinitely often, there will be an index n! ~ no such that Let
hl,p(I;) <
E Ey, \ Un>! En,
1/1(1;),
11;1= l.
and It follows that
EXERCISES
I
E
En,. This is a contradiction. The theorem is hence proved.
o
5.2.
1. Let rp be a subhannonic function, homogeneous of degree p > O. Let E", be the space defined in the text, Show it is a Frechet space. Prove that fEE", if and only if hl,p 5 rp. 2. Let f be a holomorphic function of order p and finite type in the angular region 91 5 argz 59,.,92 - 91 < 7rlp, and h = hl.p, the indicator function of f of order p. (a) Show that for 91 < 9 < 92 the following inequality is correct: h(9) - h(91 ) < h(92 )
sinp(9-91)
-
-
h(9 1)
sinp(92 -91 ) . (92-9) -9-1 ) seep +h(O,)smp - 2 - secp (9 -2 2
(0-9 -2-
1)
.
324
5. Summation Methods
(b) Let 92 - 91 ::: A < 7r/p and let k := max {lih(8)/(sinPA)21 : 81 ::: 9::: 92}. Show that for 8 1 < 8 < 92 h(8) - h(el ) h(8) - h(el) -'--"---'-'-'- < sinp(9-8d - sinp(92 -81)
Conclude that
8
t-+
h(9) - h(9d sinp(8 -8d
+ k (82 -
+ k(8 -
il) f7
•
el)
is an increasing function. Deduce the existence of the right and left derivatives and h~(ed. (c) Show that h~(91)::: h~(e}). (d) Show that if 90 is a local maximum of h and 18 - 80 1 ::: 7r / P then
h~(91)
h(8) ::: h(90 ) cos peo•
3. (a) Let
I
be an entire function of order 0 < p < 1 and finite type, so that I(z)
(ak
i= 0, c i= 0, m
k~1
::: 0). Show that if Co
colzl"'
(I -..:..)
= cz"' IT
ak
= Icl,
IT (1 - l:L) : : I/(z)1 ::: colZI"' ITk~1 (1 + -IIZI ) . k~1 lakl akl
(b) Let
II (z)
:= coz m
IT (1 __z_) , k~1
lakl
and let hI be the indicator function of order p of It- Show that hI (7r) (c) Let mer) = min{l/(z)l: Izl = r}, M(r) = maxll/(z)l: Izl = r}. Show that
= sup~ hI (0).
· logm(r) I· log III (r)1 I1m sup > 1m sup HOO logM(r) - ' .... 00 logl/l(-r)1 . log III (r)1 > I IIDSUP -
' .... 00
hI (ll')r P
hI (0) =- > COSlrp. h l (ll) -
(Use part (d) of the previous exercise.) (d) For E > 0 prove the existence of a strictly increasing sequence converging to 00 and such that
(rj)j~I' rj >
0,
(j ::: 1).
4. Let
I
be a holomorphic function of order p and finite type in the angular sector
a::: Argz ::: {J. Assume -7r < a < 0 < {J <
11.
(a) Show that for every y > 0 there is a sequence r. -+ 00 such that log I/(r.)1 ::: (hl(O) - y)r:. (b) Prove that if 0 <
~
«
1, then for 181 <
log I/(rej~)1 ::: (hl(e)
+
Arcsin(2e~)
y)r P :::
(hl(O)
and r :::
+
r(y,~)
2y)r P •
> 0 one has
5.2. The LindelOf Indicator Function (c) Let IPn(z) := I(rn
325
+ z)/I(rn ). Show that for Izl ::: 2e8r, log IC{ln(z)1 ::: 3y(rn + Izl)p(rn+1zi).
Using the Minimum Modulus Theorem ([BG], [Lev]), infer that, for 0 < w < I and H(t) := 2 + log(3e/2t), t > 0, the inequality log IIPn(z)1 2: -2yH (~) (rn
+ 2e8rn )p(r'+Mrnl
holds in Izl ::: 8rn , outside an exceptional set of finitely many disks, the sum of whose radii does not exceed w8rn • (d) Conclude that in the interval (1 - 8)rn ::: r ::: (l + 8)rn one has the inequality
(~)
log I/(r)1 2: {hj(O) - y - 3yH
(l
+ 2e8)2P }
r:,
with the possible exception of values of r lying in a finite collection of subintervals, the sum of their length is at most 2w8rn. (e) Since for rn ::: r ::: (1 + 8)rn, r: 2: (1 + 8)-P r P, and for 0 < y « 1, 0 < 8 « 1, hj(O) - y - 3yH
(~)
(1
+ 2e8)2p
2: (hj(O) - £)(1
+ 8)2 p ,
prove that log I/(r)1 2: (hj(O) - £)r P for Tn ::: r ::: (1
+ 8)rn' with the exception of a set of points r lying in a set of measure
::: 2wiJrn.
Prove the same result for any ray Arg z
= 8 E la, ,8[.
5. Let I be an entire function of order p and finite type, 1(0) :f. O. Let nCr) be the number of zeros of I in Izl < rand N(r) J~(n(t)/t)dt. (a) Show that
=
N(r) = _1_
rP
r
2r
log I/(rei6)1 d8 _ log 1/(0)1
211: TP Jo
rP
and conclude that N(r)
1
lim sup - - ::: -2 r--+oo rP 7r (b) Prove that lim sup -nCr) ::: -ep r~oo rP 211: *(c) For p > 0, let
ak
= 22k / p , k 2:
1 1
.
nCr)
h j (8)d8.
2 "
h j (8)d{}.
0
I, and
Show that I is an entire function of order p, h j hmsupr~oo rP
2 "
0
= 1 = -ep
211:
({})
= 1/ pe, and
1z"
h j (8)dB.
0
6. Let I be an entire function of order p, 0 < p < 1, 1(0) = 1. Assume the only zeros of I are simple, located at z = -rn, 0 < rJ < r2 < ... , and such that for some A > 0, net) - AlP as t ~ 00,
326
5. Summation Methods
(a) Show that for Log(f(z»
Z E
Ie\] -
00,
0[,
(1 + f) = 10f"\og (1 + ~) dn(t) = [n(t) Log (1 + :)]00 + z roo ~ dt = z roo ~ dt. t 0 10 t(t+z) 10 t(t+z) =L
n~1
Log
n
(b) Show that given e > 0 there is A > 0 such that t > A implies
1
z
1
00
o
n(t) - At P I dt < t(t+z) -
1A
Izl
n(t)
+ ().. + e)tP dt + elzl
1""
tlt+zl
0
0
tP - - dt. tlt+zl
(c) Prove that using the principal branch of zP
z
1'" o
tP ---dt=rrzPcosecrrp. l(t+Z)
(d) Conclude that for -rr < (} < rr one has Log(f(re i8 as r --+
» ~ eip rrA(cosec rrp)r 8
P
00.
5.3. The Fourier-Borel Transfonn of Order p of Analytic Functionals If T is an analytic functional, the entire function ~p(T) defined by (Z E C),
is called the Fourier-Borel (or Mittag-Leffler) transform of order p of T. One can see that 'l:'
s' ~n) z:= ~ r (1+n)
(T)( )
Up
""
n
(T
n~O
Z ,
p
and conclude without difficulty that ~p(T) is of order at most p and finite type. The case p = I coincides with the Fourier-Borel transform considered in Chapter I, hence we shall restrict ourselves to p > I in this section.
5.3.1. Definitions. (1) For a nonempty subset K of C let Hp.K(z) := sup Lp(~z) ~EK
be the p-support function of K. We say that K is p-convex if K = (~ E
(2) For'" # K
~
c:
Lp(~z) ::: Hp.dz)
for every
Z E
q.
C, its p-polar set is the set K*P :=
R
E
C:
Lp(~z)
< 1 for every Z E K}.
327
5.3. The Fourier-Borel Transform of Order p of Analytic Functionals
5.3.2. Proposition. Let A, B, Ai (i
E l) be nonempty subsets ofe.
(i) A S; B implies B*P S; A*p. (ii) A S; (A *p)*p and A *p «A *P)*P)*P. (iii) For a> 0, (aA)*P = (l/aP)A*P. (iv) If A is stable by multiplication by every a > 0, then
=
A*P
= (~ E c: Lp(~z) = 0,
Vz E A).
(v)
Proof Item (i) and the first part of (ii) are evident. From them it follows that «A*P)*P)*P S; A*P. Moreover, from the first part of (ii), A*P S; «A*P)*P). So (ii) holds. Part (iii) is a consequence of the homogeneity of Lp. It is clear that
R
E
C:
Lp(~z)
= 0, Vz
On the other hand, if ~
E
A) S; (~ E
c:
Lp(~z) <
Vz E
A)
A*P and aA S; A, then for a > 0 and
Z E
E
Lp(~az)
I,
= A*P. A we have
< 1,
so that
Since in part (iv), a > 0 is arbitrary, we obtain that Lp(~z) ::: 0 for every ~ E A*P, Z E A. As pointed out in Lemma 5.1.10, for p > 1, the function Lp ~ 0, hence we conclude that A*P S; R E C: Lp(~z) = 0, Vz E A}. From Ai E UjEI Aj and (i) we conclude that nEI A? ;2 (UiEI Ai )*P. Conversely, if ~ E nEI P and Z E UiEI Ai, then Z E Aj for some j E I and, hence, Lp(~z) < 1 since ~ E AjP. Therefore, ~ E (UiEI Ai)*p. 0
A7
Let us remark that the definition of p-convexity implies that every p-convex set contains the origin and that, for every K S; C one has
5.3.3. Proposition. (i) If K is p-convex, then K = (K*P)*P and, moreover, then K*P = {~ E C: Hp.d~) < 1).
if K
is also compact,
(ii) If K is compact and K = (K*P)*P, then K is p-convex. (iii) The intersection of an arbitrary number of p-convex sets if p-convex. (iv) If K =f:. (0 there is a smaller p-convex set containing K. It is called the
p-convex hull of K and denoted cVp(K). Moreover, cVp(K) =
R
E
C: Lp(zi;) ::: Hp.K(z),
Vz E
K}.
5. Summation Methods
328
(v) For any two closed p-convex sets KI and K 2, the inclusion KI 5; K2 is
equivalent to Hp,K ::::: H p,K2 ' (vi) For any nonempty bounded set K the function
Z r-+
Hp • K (z) is continuous.
Proof. (i) From the previous proposition we know that K 5; (K*P)*P. Assume that K is p-convex. If there was I; E (K*P)*P such that I; rf. K, then one could find Zo E C such that Lp(~zo) > Hp.K(zo), Hence Lp(l;zo) > 0 and
Or, in other words,
Therefore, E K*P
Zo
(Lp (I;zo» l/p
whence, as I;
E (K*P)*P,
,
we have Lp
(~ (Lp(;;O»I/P)
< I,
which leads to the contradiction
It foUows that K = (K*P)*P. Let K be p-convex and compact and let B := (I; E C: Hp,K(1;) < I}. If I; E K*P and Z E K, then Lp(zl;) < 1. By the continuity of Lp and the compacity of K, it foUows that sup{Lp(l;z): z E K} < 1, so that I; E B. Conversely, if I; E B, then for any Z E K, Lp(zt;) ::::: Hp.K(i;) < 1, hence ~ E K*P. (ii) Let K be a nonempty compact set such that K = (K*P)*P. To show that
K is p-convex it is enough to show that A := (I; E C: Lp(zt;) ::::: Hp,dz), Vz E K} ~ (K*P)*P.
Let t;o rf. (K*P)*P. There is then Zo E K*P such that Lp(zol;o) :::: 1. On the other hand, since Zo E K*P we have Lp(zot;) < 1 for aU i; E K. By the compacity of K it follows that H p.K (zo) < 1. Hence i;o rf. A. (iii) Let K = K;, aU K; being p-convex. Let A be the set defined in (ii). We need to show that A 5; K. Let i; E A, then for any z E C, i E /,
nEI
Lp(zi;) ::::: Hp.dz) ::::: Hp.K,(z).
Hence, i; E K; for all iE/, and so i; E K. (iv) The existence of a smallest p-convex set containing K follows from (iii). The set A defined earlier clearly contains K. Moreover, H p •A = H p •K , which will show that A is p-convex. In fact, for z E C, we have Hp.dz)::::: Hp.A(Z)
=
sup Lp(zi;) ::::: Hp.K(z), S"EA
5.3. The Fourier-Borel Transfonn of Order p of Analytic Functionals
329
by the very definition of A. Finally, if C is a p-convex set, K S; C, then for ~ E
A, Z E C, Lp(~z) ~
Hp,K(Z) ~ Hp,C
so that ~ E C, i.e" A S; C. This shows that A = cVp(K). (v) Assume Hp,K, S; H p,K2 and let ~ E K" then (Z E C),
which implies ~ E K 2 • (vi) We are going to prove something stronger, that Hp,K is locally a Lipschitz function. Let WI, W2 E C and define rp(t) := Lp(w, + t(W2 - WI»~, 0 ~ t ~ 1. If we assume that Lp(w,) = 0 and L p(W2) > 0, from the convexity of the region I Arg z I ~ rr /2p we conclude there is a unique to E [0, 1] such that rp(t) = {
0 Re{(wl
+ t(W2 -
WI»P)
for 0 < t < to, - for to < t ~ 1.
The function cp is hence continuous and differentiable except at to, where it has right and left derivatives. Clearly for 0 ~ t < to, for to < t ~ 1. It follows that Icp(1) - cp(O) I = IL p(W2) - Lp(w,)1 =
~ p(lwIi
=
11'
+ IW2i)P-'lw, -
cpl(t)dtl
w21·
=
If Lp(w,) L p(W2) 0, then cp(t) == 0, and if Lp(WI) > 0 and L p(W2) > 0, then cp is differentiable everywhere and the same inequality holds. Let now z E K, ~" ~2 E C, and set WI = z~" W2 = Z~2' We infer that
and so
hence with Cp(K) := pmax{lzIP:
Z E
K). It follows that:
IHp,d~l) - Hp,K(~2)1 ~ Cp(K)(Rll
+ 1~2I)P-ll~1 -
~21,
which shows that Hp,K is Lipschitz on any bounded subset of C. The continuity 0 is now obvious.
5. Summation Methods
330
5.3.4. Definition. Let E be a non empty subset of S2. The conjugate set defined by
E- := S2 \ where 1/00
{I;-:
Z
E
E
E is
},
= 0, 1/0 = 00. For simplicity 0 = S2, .52 = 0.
5.3.5. Proposition. For any E £ S2, subsets of S2, i E J, we have
E= E. Furthermore,for any family E; of
o
Proof. Left to the reader. Recall from Section 5.1 that for any
Kt 5.3.6. Lemma. For any
Z E
IC* we have
= ({ E C*: Lp(zg) ~ I} U {OJ. Z E
IC*,
if =
(W E IC: Lp(wz) < I}.
Proof. Let us define Az := (w E IC: Lp(wz) < 1). First, let us show that if £ At. If W ¢ Az, then Lp(wz) ~ l. hence wz #- O. Let 1" = I/(wz), ~ = n, then Lp(z/~) = Lp(I/1") ~ 1, which means ~ E Kf. Since w~ = 1"WZ = 1, we have w ¢ if. Conversely, to show that Az £ if, let w ¢ if. Then there is ~ E Kf such that ~w = 1. By definition, Lp(z/n ~ 1, so that Lp(wz) ~ 1, which means that W ¢ Az• 0 5.3.7. Lemma. Let 0
#- B £ C, then
U Kf = (B*P)-. ZEB
Proof. In fact, let G = UzeB Kf, then
{; =
(U Kf) -
=
ZEB
= (V E IC:
n if ZEB
Lp(~z)
< 1, Vz E B)
= B*P. Hence,
o
5.3. The Fourier-Borel Transform of Order p of Analytic Functionals
331
5.3.8. Lemma. If K is p-convex, then
K=
U
Kf·
zeK*P
Proof. From the previous lemma we have
U Kf = «K*P)*P)~, zEK;
while (K*P)*P
=
o
K by the p-convexity.
5.3.9. Lemma. Let 0 f=. K be a compact set in C, e > O. Then K,.p :=
{~ E IC: Lp(z~)
::: Hp.K(z)
+ elzl P ,
Vz
E
IC}
is a compact neighborhood of K.
o
Proof. Left to the reader.
5.3.10. Theorem. Let p > 1, K a compact nonempty set in C, and T E JIf'(K). Then, for every 8> 0 there is a constant C, > 0 such thatJor all ~ E C
Conversely, if f is an entire function for which there is a p-convex compact set K such that for every e > 0 (V~ E C)
for some C, > 0, then there is an analytic functional T carried by K so that 'Jp(T) = f· Moreover, if f(~) = 2:n>O an~n is the Taylor series of f, then the function \II(~) := 2:n>O i(l + n;p)ang n+ l , which is holomorphicfor I~ I sufficiently large. has an analytic continuation to the set K C , and T can be defined by (T,h) = _1_.j
2m
\II(~)h(~)d~,
hE JIf(K),
y
where y is any piecewise C l loop in K C , which has index I with respect to K and lies in the domain of holomorphy of h. Proof. Since T is carried by K, and, for there is a constant AE > 0 such that
8>
0, K E /2.p is a neighborhood of K.
l'Jp(T)(z) I ::: A, sup{lEp(z~)I: We also know that for every
8'
~ E
K E /2.p}.
> 0 there is BE' > 0 such that
and hence. sup IEp(z-nI::: BE' exp (Hp.X
+ ;lzlP + 8(8')lzIP)
.
5. Summation Methods
332
where 8(s') = S'SUP~EKEI2'P 1t;"IP. Choosing 8' > 0 so that 8(8') < 8/2 accomplishes the desired estimate. Conversely, consider the integral rpCt;) := p
1
00
e- t - P t p- 1f(tt;") dt.
It converges uniformly on any compact subset of K*P. In fact, if Q c c K*P, there is 0 :::: J.L < 1 and A > 0 such that Hp,KCt;") :::: J.L and It;" IP :::: A for any t;" E Q. Choose 8 > 0 so small that v := J.L + sA < 1, then for t ~ 0 and t;" E Q we have so that e-tP tP-1If(tt;")1 :::: Cetp-1e-(I-V)tP.
We conclude that K*P S; Bp(rp) = Mp(rp) (see Section 5.1). Therefore we have Kf S; Bp(rp) for any Z E K*P. It follows from Proposition 5.1.15 that rp has an analytic continuation to UZEK'P Kf = K. Now let y be a piecewise C 1 loop of index 1 with respect to K and contained in the domain of holomorphy of h E JIf(K). We know that rp admits the Taylor development
!.. L r
rp(t;") =
p
n",O
(1 + ~)
ant;"n
p
in a neighborhood of t;" = 0 (and hence 0 is an interior point of K). Therefore the function 1{fCt;) = (p/r;)cp(l/t;) is holomorphic in K, i.e., in KC, and its Laurent development at 00 is the one required above. This means that we can in fact define the action of an analytic functional T on h by (T, h) := _1_. 2m
ir1{f(t;")h(r;) dt;". y
This defines an analytic functional T carried by K and its Fourier-Borel transform of order p is given by Jp(T)(z)
=
_1_. 2T(l
Taking as y the circle 8B(0, R), R
J
(T)(z) =
I: n,m",O
p
»
(~1 27f1
= I:anz n
=
irEp (zt;)1{f (t;") dt;". y
1, we have r(l
y r(1
+ n/ p) ant;"n-m-l dl;) zn + m/ p)
fez),
n",O
as we wanted to prove.
o
5.3.11. Corollary. Let p ::: 1 and T an analytic functional in C, then there is a smallest p-convex carrier of T.
5.4. Analytic Functionals with Noncompact Carrier
333
Proof. For p = 1 it was done in Theorem 1.3.5. For p > 1, let K be the intersection of all p-convex carriers of T. If ~p(T) = Ln>oan~n, consider V(O = Ln>or(l +nlp)an/~n+l. Then V has an analytic co;tinuation to K C • Therefore, we can represent T by (T,h)
1 .1h(nV(~)d~, = -2 7rl
hE .ff(K),
y
where y is any piecewise C 1 loop in K C , of index 1 with respect to K, and can 0 be taken as close to aK as one wants. EXERCISE
5.3.
1. Use Definition 1.3.12 to show that for p = 1 and T BI (ZC(T)
G))
=
E
.ff(C),
~I(T).
Verify the analogous identity for p > 1.
5.4. Analytic Functionals with Noncompact Carrier In this section we indicate how some of the preceding results about analytic functionals (with compact carriers) can be extended to a more general setting. We follow the work of Morimoto [Mol], [M02]. Let L = [a, oo[ + i [k l • k z] be a closed half-strip of the complex plane, with a E JR, -00 < kl < k2 < 00. For e > 0, e' > 0, < k' < 00, we define the space
°
Q(L, e, k', £') := {f E .ff(ie)
n e(L£): sup If(z)e(k'+£')z I
where L£ = [a - e, oo[ + i[k l - e, k2 + e]. This space is a Banach space with the natural norm f f--+ sup If(z)e(k'+£')zl. If
°
<
£1
< e,
°
zELe
< £~ < e', the restriction map
Q(L, e, k', 8') -+ Q(L, e1, k', e~)
is a linear continuous injective map, which is also compact. We introduce the inductive limit [Hor]
-
Q(L, k'):= lim Q(L, e, k', e'). e,e'-)oO
One can show that it is equal (including topologies) to a countable inductive limit . Q L, -, k , - • Q(L, k' ):= hm ---+ n n n ..... oo
(1, 1)
5. Summation Methods
334
so that Q(L, k') is a DFS space. Denote Q'(L, k') the dual space of Q(L, k'). An element of this space is, by definition, an analytic functional carried by L and type ~ k'. Recall that a linear functional Son Q(L, k') is continuous if and only if, for every e, e' > 0 there is C = C(e, e') ~ 0 such that I(S, f)1 ~ C sup I/(z)e(k'H')zl ZELs
for every I E Q(L, e, k', ef). We shall frequently use the following version of the Phragmen-Lindelof principle: 5.4.1. Lemma. Let 0 < e < r, I alunction continuous on Lr and holomorphic such that there are n E N*, C > 0, with the property that in
L
sup I/(w)le- Ciwln <
00.
wELr
Then, the condition
sup I/(w)1
~ M
weL,\L,
implies
sup I/(w)1
~ M.
wELr
o
Proof. Left to the reader.
Let us introduce a new collection of spaces. For 0 < e < r, e' > 0, let us denote R(L, r, e, k', e') as the space of continuous functions F on Lr \Lt, which are holomorphic in (L r\Le)O and satisfy sup JF(w)e-(k'H')wl <
00,
weL,\L,
Considered with the natural norm it is a Banach space. Let R(L r , k', e') be the space of function F continuous on L" holomorphic in in and such that sup IF(w)e-(k'+e')WI <
00.
wELr
It is also a Banach space with the natural norm. Moreover, it can be considered as a subspace of R(L, r, e, k', e'). Lemma 5.4.1 allows us to conclude that it is a closed subspace. For 0 < el < e < r < rj, 0 < e; < e, we have the following commutative diagram of restrictions: R(L,rl, el, k', e') ---> R(L, r, e, k', e')
1
1
R(L r ], k', e')
------+-t
R(L r , k', e')
5.4. Analytic Functionals with Noncompact Carrier
335
The horizontal arrows are compact maps. We consider now the projective limits lim
R(C\L; k', e') :=
R(L, r, e, k', e')
(
= {F E.;tf(!C\L): if < e < r, then sup IF(w)e (k'+e')w 1< oo}. WEL, \L,
R(C; k', e'):= lim R(L" k', e') ~
'->00
= {F
E
.;tf(C): 'ir > 0 sup IF(w)e-(k'+e')wl < oo}. weLr
These two spaces are FS. Lemma 5.4.1 shows that the second one is a closed subspace of the first. If 0 < e~ < e', we have the commutative diagram of continuous linear injections R(C\L; k',
ell - -...,
R(C\L; k', e')
I
r
R(C; k', e;J - - - - . - . R(!C;k',e;J
This diagram induces a continuous linear mapping among the quotient spaces R(!C\L; k', e;J R(C\L; k', e') --+ ~~-----R(C; k', ell R(C; k', e')
~~~-~~
5.4.2. Lemma. The preceding map is injective. Proof. Let F E R(C\L; k',
ell n R(C; k', e'). For 0 <
sup
e < r, we have
IF(w)e-(k'+e;)WI < 00
WEL, \L,
as well as sup{!F(w)e-(k'+s')WI: WE L,} < 00. Consider the entire function I(w) := F(w)e-(k'+e;)w, e' - e~ > 0, which satisfies sup {e-(e'-e;)lwll/(w)!} <
00
weL,
because e-(e'-e;)lwll F(w)e-(k'+e;)w I ~ e-(e'-e;) RewIF(w)le-(k'+e;) Re w ~ IF(w)e-(k'+e')wl.
One can apply Lemma 5.4.1 to
I
to obtain
sup IF(w)e-(k'+S;)WI ~ weL r
sup
IF(w)le-(k'+s;)WI < 00,
weLr\Lc
which shows F E R(!C; k', ell and concludes the lemma.
o
5. Summation Methods
336
5.4.3. Definition. Let
H(C, k') = lim R(C\L; k', c') , +-- R(C k', c') E/~O
'
where the maps needed for the projective limit are the earlier maps (*). An element [F] of H (C, k') is determined by a family of functions (Fe' )e'~O, where Fe' E R(C\L; k', c') and if 0 < c; < c', then Fe'I - Fe' E R(C; k', c').
Let us assume f E Q(L, k') and [F] E H(C, k') are given. Choose co, c~ > 0 such that f E Q(L, co, k', c~), then for 0 < c < co, 0 < c' < eb, the integral
1
f(w)Fe,(w)dw,
Fe' E [F],
aL,
converges absolutely. Namely, there exists A 2: 0 such that If(w)1 :::: Ae-(k'+e~)Rew,
WE
Leo,
and, if we let B:=
sup
IF(w)e-(k'+e')WI <
00,
wEL:ze\L,
then IF(w)1 :::: Be(k'+e')Rew, If(w)F(w)1 :::: ABe-(e~-e)Rew,
WE
aLe,
WE
aLe.
This reasoning not only implies the absolute convergence of the integral, but it can be used to show that its value is independent of c and c'. In this way any [F] E H(C, k') defines a continuous linear functional on Q(L, k') denoted by Int([F)), (Int[F], j) :=
-1
f(w)Fe,(w) dw.
aL,
Furthermore, one can show that
H(C, k') [F)
~Q'(L, ~
k'),
Int([F]),
is a linear continuous map (see [Mol], [M02].) We are going to verify that the map Int is bijective by exhibiting the inverse map. 5.4.4. Lemma. For w
E
C\L, c' > 0 given, the function e(k'+e')(w-z) z~
is an element of Q(L, k').
w -z
5.4. Analytic Functionals with Noncompact Carrier
337
k -£ I
o
a-£
Figure 5.6
o
Proof. Left to the reader. 5.4.5. Proposition. Let f all Z E
t
E
Q(L,
80,
1
k',
8
0),0 <
8
<
0<
80,
8'
< 8~, then for
1 e(k'+s')(w-z) f(z) = - . few) dw. bn aL, w-z
Proof. One applies Cauchy's formula to the contour Yt suggested in the 0 Figure 5.6, and then let t ~ 00. 5.4.6. Definition. For S E Q'(L. k') its Cauchy transform is the family functions (58' )£'>0 defined by ,
1 \
S8'(W)
= --2 . lrt
5 of
e(k'+e')(w-z) )
Sz,
w-z
.
5.4.7. Proposition. Let 5 E Q'(L, k'). For any 8' > 0 the function 58' is holomorphic in C\L, and for every 8 > 0 it satisfies the inequality
sup
15.,(w)e-(k'+e')wl < 00.
w
Proof. The functional S can be represented by a convenient measure. Using Lemma 5.4.4 and the theorem of Morera it is easy to conclude that the function 5s' is holomorphic in C\L. Furthermore, for every 8 > 0, 8' > 0, there is a constant C = C(8, 8') ~ 0 such that for every f E Q(L, 8, k', 8') j(S, f)l ~ C sup If(z)e(k'+e')zl. zeL€
338
5. Summation Methods
For any w
E
L2e one clearly has sup zeL,
e-(k'+e')z 1
W -
e(k'+e')z
and thus, A
,
,
sup ISe,(w)e-(k+e)WI
= sup
W¢L2s
1
Z
W¢L2'
I
-2 rr
I(
I
~-, £
Sz,
e-(k'+e')z)1
z
w -
~ -C- , 2rr £
o
which proves the proposition.
5.4.8. Proposition. Let SEQ' (L. k'), 0 <
£"
<
£'.
The holomorphic function
WE
C\L,
admits an analytic continuation to an entire function satisfying the estimate sup IF(w)le-k'u-e'u+-e"u- <
00,
wee
where u = Re w, u+ = max(u, 0), u_ = min(u, 0). Proof. From Proposition 5.4.7 we conclude that for any
sup IF(w)le-k'u-e'u+-e"u- <
£
> 0
00.
w¢L,
Fix w
E
Le. Let
e(k'+e'g _ G(~) :=
then z
t-+
e(k'+e")~
~
,
G(w - z) is an entire function. Moreover, for any z E L2e we have IG(w - z)e(k'+s")zl ~ ~ (e(k'+S')u e (e"-e'Ha-2e)
In other words, z
t-+
+ e(k'+s")U) .
G(w - z) belongs to Q(L, 2£, k', £"). Thus, the function F(w)
1
= --2. (Sz, G(w 7f1
z»)
is defined for w E Le and, by the continuity of S, satisfies sup IF(w)1 ~ C\e
+ C2e(k'+e")u,
weLt
for some convenient constants C h C2. This concludes the proof.
o
We are now ready to give the following definition: 5.4.9. Definition. The Cauchy transform C(S) of S E Q'(L, k') is the element [S] E H (I(:, k') defined by the family S = (S.)e>O'
5.4. Analytic Functionals with Noncompact Carrier
5.4.10. Proposition. For S
339
Q'(L, k'), one has
E
S = Int([S]) = Int(C(S». In other words, Int oC = id on Q'(L, k').
Proof. Let f tation
E
Q(L,
80,
k', 8 0),0 <
I
f(z) = - .
1
8
<
<
8'
<
8
0, The integral represen-
eCk'+e')(w-z) f(w)
dw,
aLe
27r1
80,0
W -
Z
given by Proposition 5.4.5, actually converges in the topology of Q(L, k'). (Exercise for the reader, see [Mol), [Mo2).) One concludes that
r
(S, f) = -
f(W)Se,(w)dw,
JBLe
o
which is exactly the statement of the proposition.
5.4.11. Proposition. Co Int = idH(IC,k') .
Proof. Let [F) E H(C. k') defined by a family (Fe)po, and set S = Int([F]). From the definition of S we deduce that for every f E Q(L, 80, k', £0)' and every 0 < £ < £0, 0 < £' < 8 0,
r
(Se(w) - Fe(w»f(w)dw = O.
JBLe
wt-+ - - -
w-z
belongs to Q(L, £0, k', 80)' In fact, it is continuous in Leo' holomorphic in Leo (even holomorphic in L2eo ), and one can verify that _(W-Z)2
IWwELeo sup e
Hence, for all z (t)
E
I
e(k'+e')w < 00.
Z
L3eo \L2Eo
1
e-(w-z)2
A
dw = O.
(Se(w) - Fe,(w»
BL,o
W -
Z
Let us introduce an auxiliary function G, holomorphic in L4eo' by I G(z) := -2' 7r1
1
BL ..o
A
(Se'(w) - Fe,(w»
e-(w-z)2 W - Z
dw.
If z E L3eo \L 2to ' we can apply (t) and Cauchy's theorem to obtain G(z) = S6'(Z) - Ft,(z).
340
5. Summation Methods
Thus, the function Se' - Fe' admits an analytic continuation Ge' to L4Eo by G, hence G e' is an entire function. Moreover, one can prove without difficulty that le-(k'+<')zGe,(z)1 remains bounded in L3eo. It follows that G e, E R(C; k', s'), which finally proves that C(Int[F])
o
= [F].
We are now going to extend the Fourier-Borel transform to Q'(L, k').
5.4.12. Lemma. The function z HRe~ < -k'.
eZ~ belongs to Q(L, k')
if
and only
if
o
Proof. It is elementary and left to the reader.
5.4.13. Definition. The Fourier-Borel transform of S tion J'(S) defined in the half-plane Re ~ < -k' by
E
Q'(L, k') is the func-
J'(S)(~) = (Sz, eZ~).
5.4.14. Definition. The space Exp(] - 00, -k'[ + ilR) is the collection of all holomorphic functions F in the half-plane Re ~ < -k' such that IF(nle-Hdn-el~1 < 00
sup Re~:::-(k'+e')
for every s, s' > 0, where HL is the supporting function of the half-strip L. It is easy to see that letting ~ = ~ + i fJ
+oo
Hd~)
= supRe(~z) = { a~ - kIfJ a; - k2fJ ZED
if ~ > 0, if ~ .:::: 0, fJ ~ 0, if;':::: 0, fJ':::: O.
5.4.15. Proposition. The Fourier-Borel transform J' maps Q'(L, k') into Exp(] - 00, -kl[ + ilR). Proof. Let S E Q'(L, k'). hence for every E, s' > 0 there is a constant C C(E, s') ~ 0 such that for ~ =; + ifJ, ; .:::: -k' - s',
=
IJ'(S)(~)I = I(Sz, eZ~)1 .:::: C sup lez~e(k'+<')zl tELe
if fJ ~ 0, if fJ .:::: 0. This estimate is precisely of the form IJ'(S)(OI .::::
Therefore, J'(S)
E
ExpO -
c' eHL<~)+el~l, 00,
-k'[ + ilR).
Re ~ .:::: -(k' + E'). 0
We are now trying to prove that J' is a bijection between the two spaces considered above. Let F E Exp(] - 00, -k'[ + ilR) and fix two complex numbers ~o = ~o + ifJO,~' =;' + ifJ', such that ~o < -k', I~'I = 1,~' ::: O. The ray starting at
5.4. Analytic Functionals with Noncompact Carrier ~o
and direction
~'
is denoted
~o
+ ~'oo,
341
i.e.,
~o+~'oo = {So +t(: O:s t < oo}.
Let us introduce the function
F(w,
~o, ~') = _1_. 2m
{
F(r)e- WT dr
= ~ (Xi F(~o + t~)e-w(l;o+tl;') dt, 2m io
il;o+I;'oo
which is a sort of Laplace transform of F. The first thing to determine is for what values of w is the integral convergent. Due to the conditions on ~o and ~' there is 8' > 0, such that for all t :::: 0 /;0
so that for z
E ~o
+ tl;' :s /;0
+ ~'oo IF(z)1 :s
for some C
< -k' - £:',
= C(£:, 8') :::: O.
CeHL(z)+e1z l ,
It follows that if z
IF(z)e-ZWI
:s
= ~o + t~'
C'el((a-eW-(kl-e)~'-Re(I;'w)}.
Hence, if we consider the open half-plane
Ween := {w
E
C:
Re(~'w) > (a - 8)1;'
+ 811/1- k(l)'}
the integral converges absolutely and uniformly in We (~'), and thus the function F(w, ~o, will be holomorphic in this half-plane. It is also clear that if
n
Wen := {w E C: Re(~'w) > HL(~')}'
Figure 5.7
342
5. Summation Methods
then
W«() = U WEen· &>0
We summarize this fact in the following statement: 5.4.16. Lemma. For any ~O E C,~' E C, I~'I = 1, Re~' ::: 0, the function given by w ~ F(w, ~o, is holomorphic in the open half-plane
n
Wen.
Observe that we also have
U
Ween
= C\Le,
,,'1=1 Ref:::O
U Wen =C\L. WI=l Ref:::O
In order to gain information from these two identities we need to relate F(w, ~o, ~') for different values of ~'.
= I~"I = 1, Re~' ::: 0, Re~" ::: 0, then
5.4.17. Lemma. Let I~'I
Few, ~o, ~') for all w
E
=
Few, ~o, ~")
wen n W(~").
Proof. It is enough to show this identity for every w in We(~') n We(~"), for some s > 0. In the angular sector bounded by the two half-rays ~o + S' 00 (that is, (~o + ~'t : t E [0, oo]}) and ~o + ('00, the function r ~ F(r)e- Wr is of exponential type and decreases exponentially on the boundary. By the Phragmen-LindelOf theorem it also decreases exponentially in the sector. Thus, one can use Cauchy's theorem to obtain
r
F(-r)e- rW dr
=
il;o+I;'oo
r
F(r)e- rW dr
il;o+I;"oo
o
as we wanted to prove. 5.4.18. Lemma. Let Few, ~o) in C\L by
~o
= -(k' + s'). One can define a holomorphic function
Few, ~o)
=
F(w, ~o,;;')
for
wE
wen·
For every s > 0, this function satisfies
sup IFew, ~o)e-(k'+e')wl <
00.
w~L,
In particular, F(w. ~o) E R(C\L; k', s').
Proof. Left to the reader.
o
5.4. Analytic Functionals with Noncompact Carrier
343
5.4.19. Lemma. Let 0 < e" < e'. The function
F(w,
G(w) :=
-(k'
+ 8')) -
F(w, -(k'
+ e"»
has an analytic continuation to the whole plane. It satisfies the estimate IG(w)1 :::: Cek'u+e'u++e"u-
for some C 2: 0, where u ular,
= Re w,
u+
= max(u, 0),
u_
= min(u, 0).
In partic-
G E R(C; k', 8'). Proof. It is an immediate consequence of the fact that for C\L we have G(w)
=
I -.j 27r1
-(k'+e")
o
F(r:)e- rW dr:.
-(k'+e')
5.4.20. Definition. For F E Exp(] - 00, k'[ + ilR, L) we define for every 8' > 0 a function B(F)e' by B(F)e'(w)
= F(w, -(k' + e'»,
WE
C\L.
The family (B(F)e')e'>o detennines an element B(F) of H(C, k'), called Borel (or Laplace) transform of F. Denote B: Exp(] -
00,
-k'[
+ ilR, L)
~ H(C, k'),
F t-+ B(F).
5.4.21. Example. The function F(/;) =ezl; belongs to Exp(] - 00, -k'[ +ilR, L) if and only if Z E L. The Borel transfonn of F is the "Cauchy kernel" used in Definition 5.4.6, I e-(k'+e')(z-w) B(F)e'(w) = - . - - - 27r1
z-w
(w E C\L).
5.4.22. Proposition. The following identity holds:
= idExp(]_oo,-k'[+iIR,L)' 00, -k'[ + ilR, L), S = Int([B(F)]).
~o Int oB
Proof. Let F E Exp(] tion of Int we have, for Re /; < -k' - e',
~(S)(/;)
= -
r
JdL.
From the defini-
B(F)e,(w)e-WI; dw.
Decompose the path aLe into three parts following Figure 5.8. Denote 1;0 = -k' - e', and since in the portion I, we have 1m w k2, we can let Fe,(w, 1;0) = F(W,l;o, -i). Then
= k2 + e >
344
5. Summation Methods
W2.----+----~------------------
II
o WI = (a - e) wI ~__~____~~Ill________________
+ i(k l
-
e)
Figure 5.8
-1-.1 1
=
2ll"1
F(r)
(o-ioo
1 = --.
(le(~-r)w dW)
dr
I
e(~-r)Wl
F ( r ) - - dr. ~o-ioo t; - r
2ll" I
Similarly, if Imt; =1= 0 we have
re~w
In
F(w,t;o) dw
= re~w F(w,t;o, -1) dw In 1 =-
1
2ll"i
e«(-T)Wl -
e«(-T)W3
dr.
t; - r
(0- 1'00
Finally, we also have
r
Jm
e(W F(w,t;o)
dw =
r
Jm
e(W F(w,
= _1_. 2ll" 1
1
= -.
t;o. i) dw
r eW~ (1
JIll
1
2ll" 1
F(r)e- WT
e(,-T)Wl
F ( r ) - - dr. ,o+ioo t; - r
Thus, for Re l; < -k' - e', 1m l; =1= 0, we have ~(S)(t;)
1
= -. 2ll" I
1 c,
F(r)
e(~-r)Wl
l; -
dr) dw
(o+ioo
1
d"C "C
+ -2'
1
ll" I
where CJ, C2 are the paths suggested in Figure 5.9.
C2
e(~-")W3
F ( r ) - - dr. t; - r
5.4. Analytic Functionals with Noncompact Carrier
345
Figure 5.9
An application of Cauchy's theorem yields
1 11 1
e(~-r)WI {-F(~) F(r:)--dr:= ~ - r: 0
27ri
C1
27ri
e(~-r)W3 F ( r : ) - - dr: = c, ~ - r:
{O -F(~)
if if
Im~
if if
Im~ <
Im~
Im~
< 0, > 0,
0, > O.
Therefore, ~(S)(O
=
F(~)
whenever Re ~ < -k' - s'. This concludes the proof.
o
5.4.23. Proposition. The following diagram is commutative. Each arrow is a bijective linear map: Q'(L, k')
~ ,nt
C
~
•
ET:
-=.-k'[ +nR.L)
H(C,k').
(/njact, all the maps are topological isomorphisms [Mol] [Mo2].)
5. Summation Methods
346
Proof. We have already shown that C and Int are inverses of each other and J' oInt 0 B = id. It is enough to show B 0 J' = C to conclude the proof. Let J'(S)(I;), 1;0 = -(k' + e'). For WE Wen one has S E Q'(L, k'), F(I;)
=
F(w, 1;0) = -I . 2:;rl
1
~oH'oo
= _1_ / S 2:;ri \
F(r)e- Wr dr = -I. 2:;rl
f
1
(Sz, eZT)e- Wr dr
_oH'oo
er(z-w) dr) = __1_ / S
z, J~oH'OO
2:;ri \
z,
e-(k'+e')(Z-W»)
z-
W
0
= -Se'(W)'
We are now going to extend the results of Sections 4.1 and 4.2. Let 0:::: k' < 1, I; E e- L , then the function z ~ 1/(l -seZ) belongs to Q(L, k') (as the reader can easily verify). Thus, for any T E Q'(L, k') we can define a function G(T) by
_1_),
G(T)(I;) := / Tz , \ I -I;e z
which is hence defined for I; E Q (L) := C\r L and has the following properties:
5.4.24. Proposition. Let T E Q'(L, k'), 0 :::: kif < 1. Then: (i) Thefunction G(T) is holomorphic in Q(L). (ii) The Laurent development of G(T) at I; = 00 is given by
G(T)(l;) = -
L n:;:J
J'(T)(-n) ,
I;n
Proof. The first item is an immediate consequence of Morera's theorem. The second item follows from the development
which, for 11;1>
e-(a-£),
converges in Q(L. k').
o
If the half-strip L has width k2 - kJ < 2:;r, then Q (L) contains the angular sector A(-kJ, -k2 +:;r) := {I; E C*: - kJ < argl; < -k2 + 2:;r}. We are now going to describe the behavior of G(T) in this angular region.
5.4.25. Proposition. Let L = [a, oo[ + i[kJ, k 2] with k2 - kJ < 2:;r and let 0:::: k' < 1. If T E Q' (L, k'), then for every e such that 0 < 2e < 2:;r + kJ - k2 and every e' such that 0 < e f < 1 - k' there is a constant C ~ 0 with the property that C
IG(T)(~)I !S 1l;lk'+e'
5.4. Analytic Functionals with Noncompact Carrier
347
Figure 5.10
for every
Proof. From the continuity of T there is a constant C' e- L '/2. then
IG(T)(~)I =
1
I(Tz • -1--)1 ~ez
-
~
C'sup ZEL
I
~ c' sup _e_ _ ZEL'/2
e(k'+s'-l)z
I
e- Z -
(k' +s')z
~
0 such that if
I
1 - ~ez
I
~
~ C' (sup le- z _ ~ I-k'-S') zEL'/2
~
C' dist(~, e- L'/2)-k'-s' sup 11 _ zeLE/l
eZ~lk'+s'-I.
~
¢
348
5. Summation Methods
Since e- L '/2 S; A(-k2 - s12, -k\ + s12), for t; E Q(L f
)
we have
dist(t;. e- L '/2) ~ 1t;1 sin(sI2). On the other hand, for
Lf/2 and t; E A(-k\ + s, -k2 + 27r - s). we have
E
Z
_
s
larg(~e-)I ~
2(mod27r).
so that inf 11 - t;e=1
~
sin(sI2).
;eL E /2
Therefore. the inequality of the statement holds with
e = e'l sin(s 12).
0
Let us keep the hypotheses of the preceding proposition from now on. Denote 1io(Q(L). k') the space of all holomorphic functions q; in Q(L) such that: (i)
lim q;(t;) = 0; and
I~I-""
(ii) sup{lq;(t;g-k'+e'I: t; E A(-kl + s, -k2 + 27r - s)} < 00, whenever 0 < 2s < 27r + k. - k2, 0 < s' < 1 - k'. This space provided with the seminorms sup{Iq;(t;)I: 1t;1 > e-a+e} < sup{lq;(t;)t;k'+e' I: t;
E
A(-k\ + S, -k2 +
00,
27r - s)} <
00,
is a space of Frechet-Schwartz.
5.4.26. Proposition. The linear map G: Q'(L, k')
t-+
1io(Q(L). k')
is continuous.
Proof. It follows from the fact that the two sets {(l-t;e=)-I: 1t;1 ~ e- a+£},
{t;k'+e'O-t;ez)-I: t; E A(-k\ +s, -k2 +27r -s)}. 0
are bounded in Q(L, k').
5.4.27. Lemma. Let h E Q(L. k') and let s > 0, s' > 0, be sufficiently small so that 0 < 2s < 27r - k\ + k2' 0 < s' < 1 - k', and h E Q(L, s, k', s'). Then: (i) For p > 0 the function
h (z) = - 1 p 27ri
1
aL"
hew) dw 1 - e=-W
belonl:s to Q(L, 1) (and hence to Q(L, k'», with the contour aL e.p = aL f n {w: Re w ::: pl. (ii) For Z E I hew) h(z) = - . z-w dw. 2m iJL, I - e
If
1
5.4. Analytic Functionals with Noncompact Carrier
349
(iii) The limit lim hp = h
p-+oo
holds in the space Q(L, k').
Proof. (i) The fact that hp is holomorphic is clear. On the other hand, sup Ihp(z)eZI = sup ZeL,j2
ZeL,j2
1
~
iJL,
1
aL,
h(w)(e- Z - e-w)-I dwl p
Ih(w)1 dist(e- W , e L'/2)-1 dlwl <
00
p
because aLe,p is compact and the integrand is continuous, 0 (ii) Let Le(P) = Le n {w E
1
Denote by EF(P) z E ie
= Le n {w E
Re w 2:
.1
Since h
E
Q(L, k'
1
1
+ e')
pl. We only need to show that for
hew)
hm
iJE,(p)
p-+oo
1 - eZ-
dw
= O.
W
there is C 2: 0 such that
hew) I ---'---,"-- dw ~ C
iJE,(p) 1 - e'-
W
1
e-(~'+e')Rew R _
CiE,(p) 1 - e e .-p
dlwl,
and it is easy to see that the right-hand side converges to zero as P (iii) We need to show that if Ye.p = aLe \aLe,p, then
1 y,
For w
E Ye,p
hew)
1 _ eZ-W dw p->&p
~ 00.
. , 10
Q(L, k ).
p
we have e- w
E
A( -kl + e, -k2 + 2Jl'
inf le- z
- e-wl
- e), thus
2: e- Rew sin(s/2)
ZeL e j2
inf Il-ez - w l2: sin(s/2).
zeL ej 2
Hence, for some convenient constants C, C' 2: 0 sup le(k'+e'/2)Z {
lye,.
zeL'/2
= sup zeL'/2
11
Ye,.
p, we
1
~(~~w dwl e
h(w)(e- Z
-
e- w )-k'-e'/2(l - eZ -
W
)-I+k'+e'/2 dwl
350
5. Summation Methods
1 c'l
.:::: C
Ih(w)le(k'+e'/2)Rew dlwl
YE,p
-<
e-e'/2(Rew) dlwl p---,>&l
o.
D
Y~,P
5.4.28. Proposition. Let T E Q'(L, k'), h E Q(L, k'). Choose e, e' > 0 sufficiently small so that 0 < 2e < 21l' + kl - k2, 0 < e' < 1 - k', and h E Q(L, e, k', e'). Then
=~
(T, h)
21(1
r
laL,
G(T)(e-W)h(w)dw.
Proof. From part (i) of the last proposition we have _1. 21(1
r /T
In,
p \
z,
r
1
1 - eZ -
W
) h(w) dw = / Tz , _1. h(w) \ 21(1 laL, p 1 - e Z -
dW). W
As a consequence of part (iii) of the same proposition, the right-hand side converges to (T, h) when p ~ 00. It is also clear the left-hand side converges to
1. -2 1l'1
r
In,
G(T)(e-W)h(w)dw
by Proposition 5.4.25.
D
As a consequence of this proposition, we can recover T from G(T). We can apply this to obtain a generalization of the Carlson theorem (see Theorem 4.1.6). 5.4.29. Theorem. Let 0 .:::: k' < 1, L = [a, oo[ + i[k 1, kzl, k2 - kJ < 21(. If the function F E Exp(] - 00, -k'[ + ilR, L) is such that F( -n) = 0 for all n E N*, then F == O. Proof. There is T E Q'(L, k') such that ~(T) = F. The hypotheses and the D connectedness of Q(L) imply that G(T) == O. Thus, T = 0 and F == O. EXERCISES
1. For L
5.4.
= [a, co[ + j[-rr/2, rr/2], k' E JR, define a functional 1. (T, cp) := -2 rrl
r cp(~)
iaL,
T by
exp(e') d~
for E: > 0 sufficiently small and cP E Q(L, k'). (a) Show that T E Q'(L, k'). (b) Show that ~(t)(z) = -II r(1 - z) (use a Hankel integral for the function (c) Show that for any R E JR, M > 0, 8 > 0, there is c > 0 such that _1_ < cexp (-MRez l[,(z)1 -
for Rez ~ R (see [GrR], [01].) (d) Conclude that GT(W) = e l / w
-
1.
+ (~+8) IImzl) 2
n.
5.4. Analytic Funetionals with Noneompact Carrier
2. For L
351
= [a, oc[ + i[ -rr/2, rr/2], k' E JR, A > 0, let I (SA, cp) := -2'
rr I
(a) Show that SA
E
1
cp(~)
.
exp(A smh~) d~.
8L,
Q'(L, k').
(b) Show that ;j(S),)(z)
= -L,(A) (Jv
the Bessel function of order v).
(e) Show that for any R E JR, M > 0, e > 0, 8 > 0, there is c > 0 such that Ilz(A)1 for Re z
~
~
Cexp (-MRez
+
G· + e)
IImzl)
R (see [GrRJ, [01].)
(d) Show that
Gs.(w)
= Eln(A)W n = exp n~l
G(~-
w)) - ELm(A)w m. m~O
3. (a) Show that for cp E 'Ho(Q(L), k'), L = [a, oc[ + i[ -k, kJ (a E JR, k > 0), the integral M(cp)(z) := -1.
2rr I
makes sense and M(cp)
E
1
Exp(] - oc, -k']
(b) Assume further that 0
:s k
cp(w) ---;-:;:T dw
8 (e.p(-L,») W
+ iJR; L).
< rr, k' < 1.
c+ioo Imz
o
-5 -4 -3 -2-1
c- ;00
Figure 5.11
Rez
352
5. Summation Methods
For F E Exp(] - 00, -k']
+ iR; L) I
M-' (F)(w)
=
-~
{
we define
1,+ioo
F(z) _ -.-(-w)· dz
(k <
1
21 ,-ioo sm rrz 1
-~
21
I arg wi ::::: rr),
F(z) -.-(-w)- dz
r smrrz
where r is the path suggested by Figure 5.11 and -I < c < -k'. Show that for every t:: > 0, t::' > 0, there is a constant C > such that, with x and y Imz,
°
=
-w)' I exp«a
t::)x
+ (k + t::)lyl + x log Iwl + sinh2(rry» 1/2
(sin 2(rrx)
y arg(
-w»
= Re z
.
Conclude that the two integrals defining M -I (F)( w) are holomorphic functions in their respective regions. (Note, so far there is no claim that M- ' (F) is well defined.) (c) Prove that for every t:: > 0, t::' > 0, there is a constant D > 0 such that 1
1~
21
1'+ioo -.--(-w), F(z) dz I ::::: Dlwl-k'-E' ,-ioo slflrrz
if k
+ e ::::: I arg wi ::::: rr.
Conclude that the two integrals defining M- 1 (F)(w) coincide whenever k < I arg wi ::::: rr and Iwl > e- a • (d) Show that for Iwl > e- a M- ' (F)(w)
= - 2:::: F( -n)w" n~1
and deduce that M- 1(F) E .1fo(Q (L), k'). (The reader will find further developments in [Mer], [Mol], [Mo2], [MoYoI], [MoY02], [YoI]-[Y05].)
CHAPTER 6
Harmonic Analysis
The ongms of Hannonic Analysis lie in the work of Euler [Eu] and the Bemoullis who proposed to write periodic functions in terms of the exponentials einx , n E 2, in their study of the vibrating string. It is known that every COO-function which is 2Jr-periodic in the real line has an expansion of the form L::':oc aneinx (we remind the reader one can estimate these coefficients an very precisely, and that we do not need to restrict ourselves to COO-functions). It was the work of Fourier [Fo1 on heat conduction that showed, once and for all, the importance and the interest of such expansions, and since then they have been called Fourier expansions. It is clear that another way of saying that a function f is periodic with period T is to say that f satisfies the convolution equation (8, - 8)
*f
= O.
In this chapter we will consider the properties of solutions of convolution equations /J. * f = 0, where /J. is either a nonzero distribution of compact support in lR or a nonzero analytic functional in C. The solutions f of such an equation are called /J.-mean-periodic or simply mean-periodic if no confusion about the convolutor /J. can arise. In particular, we will discuss the existence of expansions for mean-periodic functions generalizing the Fourier series expansions of periodic functions. The name mean-periodic was introduced by Delsarte [Del] who observed that in Physics one usually only knows the average of a "quantity" f, i.e., a function f, and hence these quantities are only a determined modulo functions f with zero average over any interval of length T (T > 0 is fixed), i.e., /J.
I 1x+r/2
* f(x) = -
T
f(t)dt = 0,
x-r/2
where /J. = (l/T)X[-r/2.r/2j. This function f is periodic of period T (as it can be seen by differentiation of the above convolution equation) and has zero "mean" (average) over any interval of length T, whence the name mean-periodic.
353
6. Hannonic Analysis
354
6.1. Convolution Equations in R Recall that according to the Paley-Wiener theorem, F: £'(lR) ~ Ap(C) is an isomorphism where p(z) = I Imzl + Log(2 + Izl). As before, we denote by £(lR) the space of the COO-function on JR, with its usual Frechet topology. 6.1.1. Definition. Let 0 i= f.J- E £'(JR). We say that a function f E £(JR) is f.J-mean-periodic (or simply mean-periodic) if it satisfies the convolution equation
un
= {f E £(JR): f.J- * f 6.1.2. Proposition. The set of £(JR) which is invariant (under translations).
= O} is a closed subspace
Proof. We first recall that given h E JR one defines the translation operator Th by Th(f)(X) = f(x - h). To say that un is invariant under translations means that Th (un) ~ un for every h E lR. Since Th (f) 8h f, the fact that un is invariant is clear. The other property is also immediately apparent from the fact that the convolution operator f.J-*: £ (JR) ~ £ (IR) in continuous. 0
= *
6.1.3. Proposition. If f.J- i= 0 and f.J- i= c8a for any a E JR and c E C*, the space Ker f.J-* is a proper invariant subspace of £(IR).
un =
Proof. First we remark that un ~ Ker j1, where j1 is considered as a continuous linear functional on £(IR) (and not as a convolutor). This follows from the definition of convolution:
0= f.J-
* f(O)
= (j1, f).
Hence, if f.J- i= 0, then VJt i= £(IR). On the other hand, if f.J- is not of the form c8a , then the entire function Ff.J- has zeros [BG, §4.S.11J. Let ).. be such a zero and f(x) = exp(i)..x), then f.J-
Therefore
un i=
* f(x) = Ff.J-()..)f(x) = O. o
(OJ.
6.1.4. Remark. This proof shows that the exponential eih of frequency).. lies in VJt if and only if ).. is a zero of :Ff.J-. There are other simple functions in namely, the exponential monomials xke ih , when).. is a zero of FJ-t of multiplicity m > k. In fact, one has the identity
un,
k
(f.J-t, (x - t/eiA(X-t») = e iAX
L j=O
= eih
(~) x k- j (-i)j (fLt, tje- iAt ) J
t (~) j=O
J
x k-
j (-i)j (FJ-t)(j) ()..).
6.1. Convolution Equations in R
355
From this identity we see that if the multiplicity of the zero A of :Ff.,L is m, then xke iAX E 9J1 for 0 ~ k ~ m - 1. Conversely, if for some k, A we have xke iAx E 9J1, the polynomial on the right-hand side of this identity is identically
zero, hence (:Ff.,L)(j)()..) multiplicity m > k.
= 0 for j = 0, ... ,k. Therefore, ).. is a zero of :Ff.,L of
6.1.5. Proposition. Let 9J1 be a closed invariant subspace of £(l~), and let (!JJt).L = {f.,L E £'(l~): (/L, /) = 0, V f E 9J1}. Then (!)]l).L is a closed ideal of the convolution algebra £'(l~). Moreover, (9Jt).L = {f.,L E £'(l~): f.,L f = O,for all
*
f
9J1}. If'J is a closed ideal of £'(R), then (J).L = If E £(R): (it, f) = 0, Vf.,L E 'J} is a closed invariant subspace of £(R) which coincides with the subspace of £(R) given by If E £(R): f.,L * f = 0, Vf.,L E 'J}. Furthermore, under these conditions, «(m).Lf).L = 9J1 and (((J).Lf).L = 3. E
Proof. It is evident that (m).L is closed even if 9J1 were not closed. Once we show that (9Jt).L = {f.,L E £'(R): f.,L * f = 0, Vf E 9J1} is true, it becomes clear that (m).L is an ideal of £'(R). Let f.,L E (fUt).L, f E 9J1 then (f.,L
* J)(x) = (f.,Lt, f(x -
= (f.,Lt, (LAJ)f(t») = 0, ~ {f.,L E £'(l~): f.,L * f = 0, Vf E 9J1}. The other t»)
since LAJ) E 9J1. Hence (fUt).L inclusion is evident. We also have, for f.,L E J and f E (J).L, (f.,L,
rAfn = (f.,L * 'l"x(J)(0) = (f.,L * 8x * J)(O) = (f.,L * Ox, /) = 0,
*
since f.,L 8x E'J for every x E R Hence rAJ) E (J).L. The identification of (J).L is similar to the previous case. The final statement is a consequence of the Hahn-Banach theorem [Hor]. 0 Now let 0 i: f.,L E £'(R) and consider 9J1 = Ker f.,L*. We can identify (9Jt).L immediately as follows. Let I = :F«fUt).L). Then 1 is a closed ideal in the algebra Ap(C). From Remark 6.1.4 we know that if (A, m) E V(:Ff.,L), the multiplicity variety associated to :Ff.,L, then xke iAX E 9J1 for 0 ~ k ~ m - 1. Hence, if v E (9Jt).L we must have v * xkeO. x O. This implies that :Fv vanishes at the point A with multiplicity at least m. Therefore :Fv = ¢:F/L for some entire function ¢. In the terminology of Chapter 4, if we let J be the algebraic ideal in Ap(C) generated by :Ff.,L and let (:Ff.,L) be the algebraic ideal generated by :Ff.,L in Jt"'(e), we see that 1 = Ap(C) n (:Ff.,L). In other words, I = Jloc' If we define 9J10 as the invariant closed subspace generated by the exponentials monomials xke iAx , with 0 ~ k ~ m - I and (A,m) E V(:Ff.,L), then the preceding argument shows that 10 := :F(Cmo).L) coincides with Jloc' The reason is that to show that 1 = Jloc we only used the exponential monomials in 9J1. It follows, from Proposition 6.1.5 that 9J1o = 9J1. That is, the exponential polynomials solutions of f.,L * f = 0 are dense in the family of all solutions.
=
356
6. Harmonic Analysis
Lemma 6.1.8 and Theorem 6.1.9 extend this result to arbitrary closed invariant subspaces 9R. Before we proceed, let us point out that there is another way to obtain invariant subspaces. 6.1.6. Definition. Let f E £(JR). We denote by 'r(f) the closure of the space spanned by Tx (f), X E R This space is the smallest closed invariant subspace of £(JR) containing f. 6.1.7. Proposition. A function f E £' (JR) is fL-mean-periodic for at least one fL 1= 0, fL E £'(R), if and only lj'r(f) 1= £(R). Proof. It is clear from the previous Proposition 6.1.5 that 'r(f) = £(JR) if and only if the ideal ('I(f)').l is the zero ideal. If not, let 0 1= fL E ('r(fr ).l. Using 0 the same proposition we conclude that fL * f = O. We have just seen that the exponential polynomial solutions of a single convolution equation fL * f = 0 are dense in the family of all solutions of the same equation. Of course, we could not hope to obtain Fourier-type expansions of the solutions unless this density property were true. We will show in detail how to accomplish this expansion in Theorem 6.1.10 under some restrictions on fL. What is not immediately apparent is whether a given function could not have two different Fourier expansions. For instance, the reader could ask himself what happens with a function f which is both I-periodic and 2-periodic. How about I-periodic and J2-periodic? More generally, we could ask what happens with the solutions of systems of homogeneous convolution equations. Is it possible for such a system to have some nonzero solutions and, at the same time, have no exponential solutions? This problem is a particular case of the spectral analysis problem, which asks to decide whether given a nonzero closed invariant subspace 9R there is or there is not an exponential monomial xke iAx E 9R. Let us denote by 9Ro the closure of the span of the exponential monomials in 9R. The spectral analysis problem becomes: Does 9R 1= 0 imply 9Ro 1= O? From the previous Remark 6.1.4 and Proposition 6.1.3 we know that when 9R = Ker fL* then 9Ro 1= to} if and only if fL is not of the form coa , in which case 9R = 9Ro = to}. We also have seen for 9R = Ker fL* that 9R = 9Ro always hold. This question can be posed for arbitrary nonzero closed invariant subspaces 9R. The spectral synthesis problem for 9R is: Is
9Ro = 9)1?
If this holds for every closed invariant subspace 9)1, we say that the spectral synthesis property holds for £(JR). It is a famous result of L. Schwartz [Schwl] that the spectral synthesis property holds, in fact for £(JR). One can generalize these concepts to cORn) and find the surprising fact that the spectral synthesis property does not hold for cORn), n ~ 2 [Gul], [Gu2].
6.1. Convolution Equations in IR
357
We are sure the reader recognizes the relation between this question and the Wiener-Tauberian theorem for the algebra LI(JRn). We need a few more definitions before proceeding to prove Schwartz's theorem. Let 9J1 and 9J10 be as above, denote by I, 10 the closed ideals in Ap(C) given by F«mn.L) = I, 10 = F«Mo).L). Let V be the multiplicity variety associated to I, it is called the spectrum of 9J1. We recall also that Iloc is a closed ideal. For a function j, its spectrum is simply the spectrum of'I(f). 6.1.8. Lemma. 10 = I loc ' Proof. We remind the reader that Iloc = Ap(C) n I (V), where I (V) is the ideal in )f(C) of functions vanishing on V. First we want to show that Iloc s:; 10. Let xk e iAX E !mo. Then, by Remark 6.1.4, F p,(A) = 0 with multiplicity > k for every p, E (9Jt).L, since 9J10 S:;!m. It follows that (A, m) E V for some m > k. Hence, if F1) E Il oc , Fv will vanish at the point A with order at least m. The same remark shows that v * xke iAx = O. Therefore, v E (9Jto).L and Fv E 10. On the other hand, if Fv E 10 and (A, m) E V, we need to show that F1) vanishes at A with multiplicity at least m. If that were the case we could conclude that FlJ E I(V), hence Fv E I loc ' By the definition of V, for every p, E (M).L we have that Fp,(A) = 0 with order at least m, hence for 0 :s k :s m - I, p, xke iAX = O. Thus, xke iAx E «(9Jt).L)").L =!m. In other words, xke iAX E 9J10 for O:s k :s m - 1. This forces any Fv E 10 to vanish at Ato the order at least m. That is, 10 ~ I loc ' 0
*
From this lemma we can conclude that 9J1 admits spectral analysis if and only if Iloc "# Ap(C), which is equivalent to the statement that V oF 0. On the other hand, from the same lemma we can conclude that !m admits spectral synthesis if and only if I = I loc . Namely, I = I loc if and only if I = 10. Equivalently, if and only if, (9Jt).L = (9Jto).L. 6.1.9. Theorem. E(JR) has the spectral synthesis property. Proof. We follow the ideas of the proof given by Schwartz [Schwl]. Let !m be a closed invariant nonzero subspace of E(JR.) and let I be the corresponding ideal of Ap(C). In order to prove I = I loc we can apply Lemma 2.5.5 to reduce this problem to the case where the spectrum V of !m is empty. In this case, all we need to prove is that I = Ap(C). Let us first prove an auxiliary lemma.
6.1.10. Lemma. Let I be a closed ideal in Ap(C) which is invariant under differentiation. Then, either I = (O) or I = Ap(C). Proof of Lemma 6.1.10. Assume that I is a proper ideal of Ap(C) and let I be the proper ideal of E'(JR) such that F(I) = I. Then, from the properties of the Fourier transform, we conclude that I is invariant under multiplication
6. Hannonic Analysis
358
by the function x, hence, by multiplication by polynomials. Since I ¥- {O}, there is IL ¥- 0 E I. As we can find cp E V(lR) such that cp * IL ¥- 0, thus cp * IL E In V(R), we might as well assume that IL E V(R). Further, using a translation if necessary, we can assume IL(O) ¥- O. Let R > 0 be such that supp IL C [- R, R]. The Weierstrass approximation theorem allows us to find a sequence of polynomials Pn such that: ~
0 on [-R, R]; (ii) J~R Pn (x) dx = 1; and (iii) for any e > 0, Pn(x) -l> 0 uniformly in [-R, -e] U [e, R] as n (i) Pn
-l>
+00.
Let f E Il., thus T:yf E Il. for any y E R Since PnlL E I we have
I:
0= (PnJL, T:yf) =
Pn(X)IL(X)T:y (f)(x) dx
Therefore, 0 = -cyf(O) = f(-y) and so, f I = e'(lR), a contradiction.
-l>
IL (O)(T:y (f) (0».
== o. This proves that Il. = {O}
and 0
It is clear now that to conclude the proof of Theorem 6.1.9 we need to prove that if cP E I then its derivative cp' E I also. Recall from [BO, §4.6.l5] the Hadamard canonical product of cp is
cp(z) = zneaz+p
g[(
1 - :k) ezfz'r' ,
where n ~ 0 denotes the multiplicity of Z = 0 as a zero of cp and mk are the multiplicities of the distinct zeros Zk ¥- 0 of q;. As cp has order one, we have mk L -IZkl2 - <00 . k~l
Taking the logarithmic derivative of cp we see that cp'(Z) = cp(z)
~ +a + Emk k~l
Z
(_1_ + 2.), Z -
Zk
Zk
which converges locally uniformly in C\Z(cp). Hence (*)
ncp(z) cp I (z) = -
+ acp(z) + "'"' L.J mk
Z
k~l
(CP(Z) -Z - Zk
+ -CP(Z») Zk
and the series converges locally uniformly in Co Since V = ¢ we know that ¥- 0, cp(z)/z E I also. In fact, we recall from the proof of Lemma 2.5.5 the following identity. Let h E I be such that h(Zk) ¥- 0, then
cp(z)/(z - Zk) E I and, if n
cp(Z)
h(Zk) - h(z)
Z - Zk
h(Zk)(Z - Zk)
--=Cp(z)
cp(z) h(z)
+----, Z -
Zk h(Zk)
and the functions [h(Zk) - h(z)]/[h(zk)(z - Zk)], cp(z)/(z - Zk) E Ap(IC), while cp(z) and h(z)/ h(zk) E I. Thus cp(z)/(z - Zk) E I.
359
6.1. Convolution Equations in .IR.
Therefore, if we could prove that the series (*) converges in Ap (tC) it would follow that cp' E I. This convergence might not hold, so we use the trick of introducing an extra element of I in each term of the series. Namely, we rewrite (*) as I
ncp(z)
(**) rp (z) = - z
+ arp(z)
This is similar to the Mittag - Leffler procedure for the expansion of meromorphic functions. On the other hand,
mkCP(z)
lIZ) =mk--2" rp(z) (--+-+2' Z - Zk Zk zk Z - Zk zk z2
and there are constants A, B >
°
that
and
Therefore, when Iz - zkl
~
1,
m rp(z) z21 I k Z - Zk zf
and, for Iz - zkl
~
~
IzI 2Irp(z)1 Mk IZkl2
~
mk AeBp(z)
IZkl2
I, we have cp(z)
1(z - zd
1 ~ max{lrp/(s)I:
I~ -
zkl
~
I},
so that the inequality
also holds in the last case. Hence, the terms in the series (**) are bounded by mk/lzd AeBp(z), which implies the convergence of (**) in Ap(
= n-oo lim "'" L...-t
)"eA n
Pn.),,(x)eii..x,
360
6. Harmonic Analysis
where the limit is in the topology of £(lR). This is nothing other than the statement 9.no = 9.n. In fact, one can find an actual representation of f in the form of a series of exponential polynomials in 9.n o. These exponential polynomials turn out to be unique but the proof that this representation exists is rather delicate. It was originally given by Schwartz [Schwl], who used it to obtain the first proof of the Spectral Synthesis Theorem. We will restrict ourselves, for simplicity, to a special case of convolutors /L such that F /L is slowly decreasing and V is an interpolation variety. Our proof can be generalized to the case of arbitrary convolutors in £'(lR) and even further to the case of several variables, althought the spectral synthesis property is false for £(lRn) , n ~ 2. We refer the interested reader to [BT2], [BT3], [BYl], [Eh5], [Pa] , [BSt2].
6.1.11. Theorem. Let /L E £' OR) be such that F /L is slowly decreasing and let V = V (F/L) is an interpolation variety for Ap(C). Then, every COO-solution f of the convolution equation /L * f = 0 can be written in the form (A.m)EV O~j~m-l
(A,mJEV
The coefficients aA,j are uniquely determined. The last series is convergent in the topology of the space £(lR), i.e., absolutely and uniformly convergent on every bounded interval and the same is true for every derivative of this series, which then converges to the corresponding derivative of f. Moreover, the coefficients aA,j satisfy the estimates
2:
eBp(A) (
2:
j! laA.j
I)
<
+00
O~i~m-l
(A,m)EV
for any B > O. Conversely, any series of such exponential polynomial solutions, whose coefficients satisfy these estimates converges in £(~) to a function f solving the equation /L * f = O. Proof. We let I = I (V). This closed ideal coincides with the principal ideal generated by F/L in Ap(C) as F/L is slowly decreasing. Let p:Ap(C) -+ Ap(V) represent the restriction operator. Recall that if V = {(Zk. md: k ~ l} then, for ¢ E Ap(C),
With this notation Ap(V) is the space of sequences b = (b k ,/) such that, for some C > 0
IIblic = supe-CP(zd k",1
The condition that V is an interpolation variety means that p is surjective.
6.1. Convolution Equations in IR
361
The idea of the proof is the following. Consider the map ex = poF.
Using functional analysis we show that the transpose lex is a topological isomorphism between (Ap(V»' and its image, which turns out to be 9Jl = {f E E(IR): fL * f = 0). We also show that (Ap(V»' is a space of sequences a = (au). The duality bracket is (a, b) = Lk,/ak/bk/. Let us show that these statements are enough to obtain the Fourier representation of a function f solving fL * f = O. Namely, there would be a unique a = (ak,l) E (Ap(V»' such that f = lex (a). For x E IR we have
j)
Jex) = (ox,
= (ox, tex(a»)
= (ex(ox), a) =/ \
=
(-0
~
~
xl e- ixz ! I!
1
)
a
' ( k,l)
)
1
a k,/
k,i
= (p(e- ixz ), a)
Hence, f(x) =
(-0 . _ _ xle- 1xz ,. I' .
L Ak,lxleixzk k,l
where Au =ak,l(-i)lj/!. Let us proceed now to prove the above claims.
6.1.12. Lemma. (i) The space Ap(V) is an DFS space that can be represented as the inductive limit of the Banach spaces Ap,c(V) (c > 0), where
Ap«V) =
{b
= (bk,lh:::]
:
IIbli c =
O:::i::om,-l
supe-CP(zd ( k
L
O::i~m,-l
Ibk,ll) <
+oo}.
(ii) The dual space (Ap(V»' can be identified to the space of the sequences a = (ak./h:::] such that,for every B > 0, O:::i:::m,-l
{;eBP(z<>
C::OI~-] lak,ll) <
00.
Proof. The space Ap(V) was described, as a set, as the union of the sequences of spaces Ap,c(V) in Chapter 2 (§2.7.1). We have only to show that the quotient map _ A (C) p:
T
--+ Ap(V)
becomes a topological isomorphism when Ap(V) is described as the inductive limit of Ap,c(V), It is clear that the spaces Ap,c(V) are Banach spaces and that,
6. Hannonic Analysis
362
if CI > C2, the canonical injection Ap,c, (V) ~ Ap,Cl (V) is a compact map. It is also evident that if we restrict the value of C to a strictly increasing sequence converging to +00 the corresponding .c~-topology is stronger that the topology of the coordinatewise convergence for Ap(V), Then this topology is the same as the one obtained using all values of c. It follows that this topology is DFS. Since the map p is continuous and surjective from Ap(C) to Ap(V), and Ap(C) is also a DFS space, it follows from the Open Mapping Theorem that p is a topological isomorphism. To prove the second part, let us show that any sequence a = (ak,l) satisfying the estimates from (ii) defines a continuous linear functional on Ap(V) by b
~ (a, b) = L
L
aklbkl .
k!,:IO:;:/:;:mk- 1
In fact, we have I(a, b)1
:s L
L
laklbktl
k!': 1 0:;:1 :;:m,-I
:s IIbll e LeCP(Z;) k!': I
L
lak.li:S const.llbll e .
0:;:/ :;:m,-I
This shows the continuity on each space Ap,c(V), Conversely, given an element v E (Ap(V))', we define akJ = (v, E k,/),
where is the Let B fJk,1 = mk
:s
E k,/ is the sequence in Ap(V) whose only entry different from zero one whose index is (k, I), and the value for this entry is exactly one. > 0 be fixed and define bkl = eBp(Zk l (hi, where fJkl = Qkil ak/ if akl =1= 0, 0, otherwise. Since there are positive constants A 1 and A2 such that AI p(Zk) + A2 for every k (see Exercise 2.2.9) then Ibkl! = eBp(Zklmk
L
:s
(A1P(Zk)
+ A 2)e Bp (Zkl,
O:;:/:;:mk-I
Choose any c > B, then it follows that b = (bk,/) E Ap,c(V). (It is this reasoning that allows maxI lakJ I and L:/ lak,11 be interchangeable in the definition of (Ap(V»'.) Therefore, all the finite sequences
satisfy Since v is continuous on Ap(V) we have I(v, ben»)! :::: Mcllb(nlil c :::: Mcllbllc <
+00.
6.1. Convolution Equations in lR.
363
On the other hand,
=
L
L
eBp(:d
1::;I::;n
lak.d
O::;I::;ml-1
by definition of f3kl. Thus
L
L
eBp(:d
k~1
lak,d < +00.
O~/::;ml-1
This shows that the sequence a = (ak.!) satisfies the desired estimates. It also follows that, for any h E Ap(V), we have (t!, b) =
L L k:::1
aklbkl = (a, b),
O::;I::;ml-1
o
since the last series is absolutely convergent.
6.1.13. Lemma. (i) la: (Ap(V»)' ~ [OR) is a topological isomorphism onto its image. (ii) 1m la = (Kera)1- = (f E [(JR.): /-l j = OJ.
*
Proof. We know that a: ['(JR.) ~ Ap(V) is a continuous surjective map between two DFS spaces. It follows from Section 1.4 that 1m la is closed and la is a topological isomorphism from (Ap(V»' onto 1m la. The equality 1m la = (Kera)1- is a result from functional analysis (see [Hor]). On the other hand, F(Kera) = I, hence Kera = /-l * ['(JR.). From Proposition 6.1.5 we get o (Kera)1- = (f E [(JR.): /-l * j = OJ. As we observed at the beginning of the proof, when /-l satisfies the hypotheses of the theorem Lemmas 6.1.12 and 6.1.13 imply that for any /-l-mean-periodic function f there is a unique sequence a E (Ap(V»' such that f(x) =
L L k:::1
Aklx1e ix :1 ,
O::;/~mk-1
where Au = ak./( -0' / I!. To see that the series converges in [(JR.) one notes that (:x)m(xleiXZk)
=( L
(7)1(l-I) ... (I-J+I)XI-J(iZk)m-j)eixzk.
0::;) ~lDf(l,m)
For Ixl :::: R and IZkl ities:
~
1, whenever k
~
ko we obtain the following inequal-
6. Harmonic Analysis
364
:s m! eR Le(R+m)p(z') k~
(
lak./l)
L
<
+00
Os/sm,-I
1
Since
(m J. )
::::om!,
RI-j
~
___ <e R
L
(1- j)! -
and
This shows that the differentiated series converges absolutely and uniformly over all compact subsets of lR. This concludes the proof of Theorem 6.1.11. 0 Given a JL-mean-periodic function f it would be interesting to have a simple formula computing the coefficients Akl of the above development of f (at least when JL satisfies the hypothesis of Theorem 6.1.10). For that purpose we need to construct distributions JLk,/ such that if (r, s) -::j:. (k, I), if (r, s) = (k, I). where V = V(FJL) = {(Zk. mk), k
~
I}, O:S 1::::0 mk - 1, and 0::::0 s ::::0 m, - 1.
6.1.14. Lemma. Given an arbitrary 0
-::j:. JL E £' (IR) one can find explicit distributions JLk.l satisfying the above properties. Moreover, if [a, III is the smallest closed interval containing the support of JL. then supp(JLk./) <; [cr, Ill.
Proof. From the formula (v, xqe- iAx ) = (-i)q(Fv)(q)()..),
which holds for every v E £'(IR), we conclude that FJLk,/ must vanish to the order m, at z, for every r -::j:. k and the Taylor expansion at Zk is given by il
FJLk,/(Z) = T!(z - Zk)
We note that for any polynomial Pk.1 (z) than mb the function
1
+ O«z
= L j aj (z
m
- Zk) '). - zd j of degree strictly less
Pk,/(Z)FJL(z) (z - Zk)m,
is entire, belongs to Ap(C), and has the required order of vanishing at z, for Pk,l we obtain
r -::j:. k. We claim that for a convenient choice of the polynomial
6.1. Convolution Equations in R
365
the Fourier transform of the distributions J.Lk,1 we are looking for, To verify the claim it is enough to observe that the Taylor development of F J.L about Zk is FJ.L(z)
=
(z - zd mk
(FJ.L)(mkl(zd
mkl
+ O((z -
zk)mk+l).
It is immediate from this expression that there is only one polynomial Pk,l that satisfies all conditions. In fact, it coincides with Pk,/(Z)
= i l (z
- zkl mk
{PrinCiPal part of ((Z
~lzd /
at
FJ.L(Z»)
Z
= Zk} .
By the Paley-Wiener theorem we conclude that the support of J.Lk.l is also contained in [a,l'n These distributions are of the form Pk,/(f
:x)
Tb
where FTk(Z) = FJ.L(z)j(z - Zk)m k • This last distribution is easy to compute. We leave as an exercise to the reader to show that (Tk,1/I)
=
'm,
(m: _ 1)! (/L x , e- iZkX
100 x
(s -
x)mk-IeizkS1/f(.~)ds).
The distribution Tk thus defined has support in [a, for 0 ::: j ::: mk - 1. 6.1.15. Corollary. Let I E
f:l] because
FJ.L(j)(Zk) =
0 0
g(~) and let J.L E E'(~), Assume that
I(x) =
L L
Ok/X l / zkx ,
k:"IO:"I:"mk-1 where V(FJ.L) = {(Zbmk)} and the series converges in the topology oIE(~). Then the coefficients akl can be computed by the lormula ak/ = (J.Lk,/,
f).
In particular, if F J.L is slowly decreasing and V (FJ.L) is an interpolation variety then b = (l! akl) E (Ap(V»'.
Proof. Due to the convergence of the series, the function I is J.L-mean-periodic because that is true for each term. The formula for the coefficients is an immediate consequence of Lemma 6.1.14. Finally, if J.L satisfies the hypotheses of Theorem 6.1.11, then we know that the coefficients of I must satisfy the prop0 erty (I! akl) E (Ap(V»'. 6.1.16. Remark. It is clear that this corollary holds even if the series converges in E(~) only after some summation procedure has been applied. In fact, the theorem of Schwartz [Schwl], [BTl], [BT2] shows that every J.L-mean-periodic function has such a series representation which is convergent after grouping terms and Abelian summation. It is also clear that if J.L is a distribution of order N, one needs the convergence to be in the topology of eN (~). This is the case since all the distributions J.Lk.1
366
6. Hannonic Analysis
have order at most N. Moreover, one can prove that the expansion theorems hold in V'(IR), i.e., if /L * f = 0, f E V'(IR), then f has a Fourier expansion of the same kind as before which converges in V'(IR).
6.1.17. Corollary. Let /L E £'(IR) satisfy the hypotheses of Theorem 6.1.11, sUPP(/L) S; [a, In a < {J. Let f E £(IR) be /L-mean-periodic which vanishes identically on [- fJ, -a l. Then f is identically zero in IR. Proof. It is a known fact of the theory of distributions ([Schw2, Theorem XXVIII]) that if v E £'(IR) has supp(v) 5; [a, fJl, then (v, j) = O. From the previous corollary, the coefficients ak,! = (/Lk.!, j) = O. Thus f = o. 0
6.1.18. Remark. It is also clear from the proof and Remark 6.1.16 that the restrictions on /L of F /L being slowly decreasing and V an interpolation variety are not necessary. It is also clear that f is detennined by its values on any interval of length L = fJ - a (if L > 0). When L = 0, /L is an ordinary differential operator with constant coefficients (up to a translation). In this case, one needs the values of f and a certain number of derivatives at one point to vanish. We will assume that L > 0 unless stated otherwise. 6.1.19. Proposition. Let f E £(IR) be the solution of two convolution equations /L * f = v * f = 0, and assume that f has two representations f(x) = (>..,m)eV(FJL)
(a,n)eV(Fv)
which converge in £(IR). Then each a in the second series for which Qa ¢. 0 is a zero of F/L and moreover Qa = Pal and a similar statement holds for ).. in the first series.
Proof. If Q,,(x) contains the nonzero tenn aa,IX'e iax let Va = Va,o be the distribution defined in Lemma 6.1.14. It is easy to see that Va
* f(x) =
Va
*(
L
Qp(x)e iPX ) = Qa(x)e iax .
(,B,n)eV(Fv)
Hence /L * (Qa (x)e iax )
= (/L * Va * f)(x) = 0,
which implies that F/L(a) = O. In the same way the zeros).. of F/L which appear in the first representation must be zeros of Fv. The fact that Per = Qa is due to the uniqueness of the coefficients in Corollary 6.1.15. 0
6.1.20. Corollary. If f has a convergent representation f(x) =
L
P>..(x)e iAX
A
in terms of v-mean-periodic exponential polynomials, then the only nonzero coefficients are those P>..(x)e iAX in 'r(f). In other words, P}" is different from zero if
6.1. Convolution Equations in lR.
367
and only if A. belongs to the spectrum of f with multiplicity bigger than or equal to (degree PA) + 1. Proof. It is enough to observe that in the previous proposition one does not need to know that f has a representation with respect to J1, to conclude that F J1, vanishes at A to the order (deg PA) + 1 when J1, * f = O. D
6.1.21. Examples. (l) If a function f E t' (lR) is periodic of period. > 0, then of the convolution equation J1,
* f(x) = «8, -
80)
* f)(x) = f(x -.) -
f(x)
f
is the solution
= O.
The variety V of F J1,(z) = e- in - 1 is exactly the collection of points 2krr I., k E Z, every point with multiplicity one. The function F J1, is slowly decreasing since it is an exponential polynomial and V is an interpolation variety. The distributions J1,k J1,k,O are given by
= (J1,b t) = i Pk,O (8, -8 0 , e- i(2rrk/,)x = -iPk,o
l'
1
00
t(s)e i(2rrk/')SdS)
t(s)e i(2rrk/,)sds.
Here, Pk,O = 1/(FJ1,)'(zk) = 1/(-iT) and the coefficient ak of ak
=
(J1,b /) =
~
r
• io
11'
=_ •
fC-s)e i(2rrk/,)sds =
~
•
1°
f is given by
fCs)e-2rrikS/Tds
-T
f(s)e-2rriks/rds.
0
This is the classical Fourier coefficient of f(x)
=
f, f is represented by the series
00
L
ake2rriks/T,
k=-oo
and the estimates from Theorem 6.1.11 are 00
L
eBp(Zk)lakl <
00
k=-oo
for every B > O. Here P(Zk) = log(2 + 1(2rri/.)kI 2), hence we have ktoo
(2 + 12;i
k1
2 )
B lakl <
which is equivalent to lakl = OClkl- N ), as Ikl (2) Let J1, = 2:f=o CkOak) so that J1, * f = of the ordinary differential equation with the equation CN f(N)(X) + ... + co/(x) = O.
00,
~ 00, for every N >
o.
0 means that f is a solution constant coefficients, namely, The characteristic equation is
6. Hannonic Analysis
368
Ft-t(')..) = 2:~=o Ck(i')..)k = 0 and the exponential monomial solutions are x j eiAX if ').. is a root of F t-t of multiplicity bigger than j. This is the classical result of
Euler. (3) As a final example let us consider the difference-differential equation f'ex) = f(x
+ 1).
This equation corresponds to convolution with t-t = 80- 8_ 1 , its Fourier transform is Ft-t(z) = iz - eiz and (Ft-t)'(z) = i(t - eiz ). The zeros of Ft-t are all simple as we can thus see easily. Let V = V(Ft-t) = {ztJ, since (Ft-t)'(Zk) = i + Zb then V is an interpolation variety by Proposition 2.7.8. In fact we can describe reasonably well the roots Zk as follows. Let i z = pe i8 , then the equation defining V becomes {
-n < () k E Z.
p cos () = log p, p sin () = () + 2kn,
:s n,
The first equation implies that cos e > 0 for p > 1, this restricts () to the interval -n /2 < e < n /2. Furthermore, we can see that p2 _ (log pf {
tan()
= «() + 2krr)2, = e + 2kn. logp
From the first equation Pk ::::::: 2lkln. Hence, the second equation becomes tane :::::::
2krr , log21kln
which shows that for k ~ +00 one has a root ()k ~ n /2 and for k ~ -00 one has a root ()k ~ -n /2. Therefore, one can improve the estimate of Pk to Pk ::::::: 21kln
.
n
+ (slgnk)2"'
The corresponding roots Zk are such that
and arg Zk ::::::: Arctan (
2krr ) - ~. log21kln 2
It follows that the Zk lie in the lower half-plane. For k ~ +00, arg Zk ~ 0 and, for k ~ -00, argzk ~ -n. Furthermore, one can see IImzkl = Loglzkl hence p(zt> ::::::: Log IZk I : : : : log 2lkln. From these considerations we obtain that
any COO-solution of this difference-differential equation has the form 00
f(x)
=L
akeiz,x
k=-oo
with lakl = O(lkl- N ) for all N > O. We can also give an explicit formula to compute the coefficients ak' We have, after a simple computation, and setting
369
6.1. Convolution Equations in lR J-Lk
= J-Lk.O, 'Z
I +Zk
=. (I
{1°e "I(-s)ds -
Zk·
v
a/.. = (J-L/.., f) = -.-Zk,
[
-I
t
+ Zk)(IZk)N io
'}
I
- . 1(0) IZk
e-iz,s I(N)(s)ds _ !(N)
(0)] .
IZk
The last line is obtained using the equation and integration by parts. It shows the coefficients have the correct rate of decay. We have seen some properties of Fourier development and uniqueness for solutions of a homogeneous convolution equation and one could ask when is the inhomogeneous equation J-L * 1 = g solvable.
6.1.22. Proposition. Let 0 i= J-L is surjective
if and only if FJ-L
E £'OR). Then the operator J-L*: £(lR) -+ £(lR) is invertible.
Prooj. The surjectivity of J-L* is equivalent to the injectivity and closed range of the transpose operator IH: £'(lR) -+ £'(lR). By Fourier transformation the
last statement is equivalent to the fact that the multiplication operator by Fji in Ap(C) -+ Ap(C) must be injective and have closed range. The injectivity is obvious since J-L i= and the range is the principal ideal j generated by F ji in Ap(C). Since FJ-L is invertible then 1 = (FJ-L)Ap(C) is closed. Since 1 is closed if and only if j is closed we have proved the surjectivity. Conversely, if J-L* is surjective, then j is closed, and by the spectral synthesis property 1 = Iloe, which is precisely the definition of the invertibility for F J-L. 0
°
Before continuing the discussion of inhomogeneous convolution equations, let us consider the relation between the two concepts, slowly decreasing and invertibility for J-L E £' (lR). Let us recall that in Definition 2.2.13 we have defined 1 = F J-L to be slowly decreasing if and only if there exist constants e > 0, c > 0, and A I > 0, such that the connected components Oa of the set
SClII, e, c)
= {z E C:
are relatively compact and, for every z,
11(z)1
W E
< ee- cP (:)}
00' we have
p(Z) :::: AIP(w).
We have also proved in Proposition 2.2.14 that if 1 is slowly decreasing then the ideal 1 = 1 . Ap(
6.1.23. Theorem. Let 0 i= J-L E £'(lR), 1 = FJ-L, 1 = 1 Ap(C). Then the following properties are equivalent: (1) f is slowly decreasing.
370
6. Hannonic Analysis
(2) There exists a constant a 2: 1 such that for each x that: (i) Ix - x'i ::: a log(2 + Ix I); (ii)
If(x')1 2:
(a
E IR
there is x'
E IR
so
+ Ix'l)-a.
(3) There exists a constant C > 0 such that for every
Z E
C there is z'
E
C so
that: (i) (4) (5)
(6) (7) (8)
Iz - z'l :::
Cp(z);
(ii) If (z') I 2: e-Cp(z'). If v E £'(IR), Fv/f E )f(C), then Fv/f E Ap(C). I = i toc (i.e., f is invertible). I is closed. 11-*: £(IR) ~ £(IR) is surjective. For every subset B <; £'(IR), if 11- * B is hounded in £' (IR), then B is bounded in £'(IR).
Proof. We will first show that conditions (1), (2), and (3) are equivalent. Then the proof of Proposition 6.1.2 shows that (6) implies (7) and (7) implies that ih: £' (IR) ~ £' (lR) is an isomorphism onto its image. The definition of bounded sets in a topological vector space shows that this last condition implies that whenever B S; £' (IR) verifies that jJ, * B is bounded in £' (IR), then B must also be bounded in £' (IR). Since v j--')- is an isomorphism of £' (IR) onto itself then condition (7) implies (8). Finally, to conclude the proof we will show that (8) implies (2).
v
6.1.24. Lemma. Conditions (1), (2), and (3) are equivalent. Proof of Lemma 6.1.24. Since f = FIJ- is slowly decreasing there are constants e > 0, c > 0, Al > 0, A2 > 0, with the properties imposed earlier. Let a > 0 be such that '>Ir 2: O.
Assume x E IR and Ifex)1 < (a + Ixl)-I, then x E S(lfl, e, c). Let 0 be the connected component of S (I f I, e, c) containing the point x and let x + i Y be a point in 80 n (x + iIR). Then
Iyl :::
p(x
+ iy)
::: Atp(x) + A 2 ,
that is, and
If(x + iy)1 =
e-cp(x+iy)
2: 8(2 + Ixl)-Y
for convenient 8 > 0, y > O. From the Minimum Modulus Theorem [BG, §4.S.14J or Lemma 2.2.11 above, we have that for any R > 0 there is a value r, R/4 ::: r :5 R/2, such that mip.
1~-(x+ly)l=r
log If(nl 2: 9y log(2 + Ix\)
+ 9log8 -
Slog M,
6.1. Convolution Equations in
~
371
where M =
max
1~-(x+iY)I::::2eR
I/(~)I.
Let us choose R = 51YI, then for any ~ in B(x + iy, 2eR) we have that ::s 311yl, IRe~l::s Ixl + 301yl. Thus, for any such~,
Ilm~1
1/(s)1 ::s
::s (2 + Ixl)BI
C1eCzp(n
for a convenient constant Bl > 0, since p(~)
+ log(2 + 611yl + Ixl) ::s 31Allog(2 + Ixl) + 31A 2 + log(2 + 61A 1 log(2 + Ixl) + 61A2) ::s B o log(2 + Ixi). ::s 311yl
Therefore there exists
KO
> 0 such that
min
l~-(x+iy)l=r
log 1/(s)1
~ KO
log(2 + Ixl)
and since r ~ ~IYI there is x' E JR so that lx' - (x that lx' - xl ::s r ::s ~IYI. Thus Ix' - xl ::s ~(Al log(2
+ Ixl + A 2)
+ iy)1 =
r, then it follows
::s a log(2 + Ixl)
and I/(x')1 ~ (2
+ Ixi)-Ko
:::: (a
+ Ix'I)-a
for a > 0 conveniently chosen. Hence, we have just proved that (1) implies (2). To show (2) implies (3) given a point z = x + iy, using (2) choose x' E JR, lx' - xl ::s a log(2 + lxI), and I/(x') ~ (a + Ix'l)-a. We can assume that a:::: 1, thus Iz - x'i ::s a log (2 + Ixl) + Iyl ::s ap(z). Therefore z' = x' satisfies (3). We are going to prove that (3) implies (1). Given a point z we know there exists a point z' such that Iz - z'l ::s Cp(z) and I/(z')1 ~ e-Cp(z') for some C ~ 1. Let R = 51z - z'l, then for I~ - z'l = 2eR we have I Im~1 ::s I Imz'l
+ I~ -
+ 281z :5 11m zl + 291z -
z'l ::s I Imz'l
z'l z'l :5 30Cp(z).
Similarly, we have I~I ::s Iz'l
+ I~ -
z'l ::s Iz'l
+ 2eR ::s ::s
+ 281z - z'l Izl + 291z - z'l ::s Iz'l
Izl
+ 29Cp(z).
Hence,
+ log(2 + I~I) ::s 30Cp(z) + log(2 + Izl + 29Cp(z»
peS) = I Im~1
::s A1P(z)
for some Al > O. Applying the Minimum Modulus Theorem as done earlier we obtain a circle r of center z' and radius r ~ Rj4 so that Z E Int(r) and on r we have an inequality of the form
372
6. Harmonic Analysis
for a convenient choice of e > 0, c > O. Moreover, the same computation shows there is A2 > 0 such that for any two points w, Wi of Int(r) one has pew) s A2P(W'). 0 6.1.25. Lemma. If / is not slowly decreasing there is an unbounded/amily (gj)j in Ap(C) such that the/amity (jgj)j is bounded.
Proof of Lemma 6.1.25. Since / is not slowly decreasing then condition (2) cannot be satisfied for any choice of a ::: 1. Hence, for any j E N* there is Xj E JR. so that: (i) IXj I ::: e3j ; and (ii) for any X ElR, Ix-xjl sjloglxjl implies
For each j let k the function
1/(x)l:::: Ixl-J.
= [j log Ix) 11. Recall that sinc z = sin z/ z and let us define
hJ(z) = (sinc(rrz/j»j
=.r ((irrX[-j/rr,)/rr1)*j)
(z),
where the power represents repeated convolution. The family that will work is given by First, we observe that all the functions h j are of exponential type rr. Hence, the same is true for the gj. On the real axis, we have the following properties: (a) hj(O) = 1; (b) Ihj(x)1 :::: 1 (x E JR.); and (c) Ihj(x)1 :::: (j/rrlxl)j for Ixi
'# O.
We claim that the sequence (gj) is not bounded in A p (C). For that, it is enough to show that there is no value n > 0 such that Igj (Xj) I :::: IXj In for every j ::: 1. In fact, Igj(Xj)1 = eklhj(O)1
= ek ::: ~Ixjlj. e
We want to show now that the sequence /gj is bounded in Ap(C). All these functions are of exponential type:::: rr + (type of f). From the PhragmenLindelOf theorem we conclude that it is enough to prove that / gj are uniformly bounded on JR. by a function of the form (2 + Ixl)N for some N. For Ix - Xj I :::: j log IXj I we have from (b) that Igj(x)1 = eklhj(x - x})1 :::: ek ::::
On the other hand, for Ix -
x}
I :::: j log IXj I we have
IXj I
:::: Ix I + j log Ix} I
Ix)I}.
6.1. Convolution Equations in R
373
while
so that and Therefore, for the same range of x, Ix - x] I :::: j log IXj I, one has that I/(x)g] (x)1
For the complementary range,
::::
Ix -
IXjlJlxl- j :::: 2 j ::::
IxI-
Xj I ::: k. using (c) one has
I/(x)gj(x)1 :::: I/(x)l/n i ::::
I/(x)1 ::::
(2
+ Ixl)N
for some N. This concludes the proof that the family (f gj )j is bounded in Ap(C) while (gj)j is not. 0 The last lemma shows that condition (8) implies concludes the proof of Theorem 6.1.23.
I is slowly decreasing. This 0
6.1.26. Remark. Using the same type of argument the reader can find in [Eh3] a few other important equivalences of the slowly decreasing condition. For instance, I is slowly decreasing if and only if: (9) {t
* V' (~) = V' (~);
(IO) {t*: V(~) -+ V(~) is an isomorphism onto a closed subspace of (11) for any v E £' (~), if {t * v E V(~), then v E V(~). These equivalences are also valid in [H02] for details.
~n.
V(~);
The reader should consult [Eh3] or
For {t slowly decreasing such that cv(supp {t) = [a,,8] let us consider the Cauchy Problem {
{t*1
=g,
II[a, ,8] = h,
where g E £(~), h E £([a, ,8]). From Corollary 6.1.15 we know that if V(F{t) is an interpolation variety for Ap(C) then (*) has at most one solution I E £(~). (Remark 6.1.16 implies that the uniqueness is true for any {t f. 0.) The obstruction to find a solution is only to be able to solve the simpler problem { {t
* 10 = 0,
101[a,,8]
= ho,
for ho E £([a, ,8]). Namely, to solve the Cauchy problem (*) we choose an arbitrary solution II of {t * 11 = g, which exists by Proposition 6.1.21. If f is
6. Hannonic Analysis
374
a solution of the Cauchy problem (*), then 10 = I - It is a solution of (**) with ho = h - (fll[a, .8]). Conversely, if (**) is solvable we take I = 10 + II to solve (*). It is not true that (*) and (**) are always solvable when /.t is slowly decreasing. The first difficulty is that (**) requires obvious compatibility conditions for ho and its derivatives: for any integer n ::: 0,
0= (/.t * lrin»(O) = (/.t, (fo)(n») = (/.t, h~n»). The corresponding compatibility conditions for (*) are g(n)(o) = (/.t
* I(n»)(o) =
(/.t, (h(n») = (/.t, hen»).
The above procedure of reducing (*) to (**) when F/.t is slowly decreasing preserves the compatibility conditions. In any case, it is not generally true that (**) can be solved. If every h satisfying 0 (/.t, hbn») can be extended to a solution of (**) one says that /.t is hyperbolic. The necessary and sufficient condition for hyperbolicity is the following:
=
6.1.27. Theorem. A slowly decreasing distribution /.t such that V is an interpolation variety is hyperbolic if and only if there exists a constant C > 0 such that:
lor every zero
Zk
01 F /.t.
6.1.28. Remark. The theorem is valid without the assumption that V(F/.t) is an interpolation variety [Eh3]. The proof is similar to the one given below except for the grouping of terms. Proof. Let 001 = Ker /.t* and E = {4> E [([a, .8]): (/.t, 4>(n») = 0, Vn EN}. Since the map I ~ fl[a,.8] is an injective continuous map from 001 into E, it is surjective if and only if it is a topological isomorphism. Hence, if /.t is hyperbolic and e > 0 is fixed, there are C I > 0 and N EN such that for any f E 001 I/(x)1 ~ C I
sup xE(a-e,/l+eJ
Let
Zk
be a zero of Fp, and I(x) sup a-e:",x:",/l+e
sup 1/(j)(y)l. yE(a.IlJ O:",j:",N
= e- iXZk . We obtain immediately
I/(X)I=eXlmZk~CI(2+lzki)Nexp(
sup YlmZ k). yE[a,IlJ
From this inequality one can easily conclude that eel Imzkl ~
C 1(2 + IZk I)N ,
which is the condition we wanted to obtain. Note that we only need to use the extension to an open interval of length> L fJ - a. We continue the proof with the following lemma:
=
6.1. Convolution Equations in R
375
6.1.29. Lemma. Let be 0"# IL E £'(lR), cv(supp IL) = [a, ,8], a < ,8,:FIL slowly decreasing, V (:FIL) = V = {(Zk, md} an interpolation variety satisfying 11m Zk I ::: C log(2 + IZk I)
(k:::: l).
If .rIL(O) = 0 we denote Zo = 0, if not the index k runs only through k :::: 1. Then,for every S E [' OR), there exist two distributions So and SI, such that SI is of the form Lk>O S I.k, the series is convergent in [' (JR). Each S I,k is a linear combination of th; derivatives ILk:] for a convenient q E N (q = 0 for k = 0), so that S = IL
* So + SI
cv(supp SI) ~ [a,,8] and the ILk,1 are the distributions defined in Lemma 6.1.14. Furthermore, for any B > 0, q can be chosen to depend only on B so that the map S ~ SI is linear and continuous from F-I(Ap,B(C» into £(",lll(JR) (the distributions with support in [a, ,8].) Finally,for R > 0 and any x E JR, Ixl ::: R, the coefficients of ILk,/ for the distributions S/,k which correspond to S = Ox are linear combinations of x j e- iXZI , 0 ::: j ::: mk - 1.
Proof of Lemma 6.1.29. From the proof of Theorem 2.2.10 we obtain disjoint disks B(Zko rk), 0 < rk ::: such that on aB(zk, rd we have
!,
IFIL(~)I
The condition on the zeros
Zk
:::: £1 exp(-C1P(Zk». implies that this inequality can be replaced by
IFIL(~)I ::::
£
(1
+ IZkl)N
for some £ > 0 and N E N. Let q be a nonnegative integer to be chosen later. For those o ¢ B(zko rk) we define a function
Zk
such that
This function is a rational function, linear combination of the functions (z - zd- I , I :s I ::: mko hence the function
1/Ik(Z) = zq:F IL(Z)
376 F f..L(0)
6. Harmonic Analysis
= 0 and we can then reduce ourselves to have at most one exceptional = 0.) We claim that the series
value, Zo
L 1/Ik(Z)
g(z) = -
k~O
converges in Ap(C) and defines a distribution with support in [a, In In fact, for each B > 0, q can be chosen so that the map FS t-+ g is a linear continuous map from Ap.B(C) into F(e[".tl](~»' Namely, for Iz - zki :::: 2rb if k is not an exceptional value, then
l¢k(Z)1 ~
AeBp(Zk)
+ IZkl)N
(1
elzkl q
where IFS(z)1 ~ AeBp(z), M = B(e
~ AI(2 + IZkl)M-q,
+ 2) + N. Choosing q = M + 2 we obtain
11/Idz) I ~ A I (2 + IZkl)-2Izl q IFf..L(z)l. 11/Ik(Z)1
~ A2
(l
+ Izl)q+rexp(H(Imz» (2 + IZkl)2
•
where H is the supporting function of [a. ,8] and
+ Izl)' exp(H(Imz».
IFf..L(z) I ~ const.(1
For the exceptional value of k = O. we obtain the same estimate for 1/Iko with q =0. We need to estimate 1/Ik inside the disk B(Zko 2rk). There is a k > 0 such that the weight function w(z) = H(Imz) + (q + r) log(2 + Izl) satisfies Iw(z) - w(w)1 ~ K
if Iz - wi ~ I. Therefore, by the maximum principle, the above estimate for 1/11. holds inside the disk B(Zko 2rk) up to a small modification. Since the series
f;
1
(2
+ IZk 1)2
< +00.
it follows that: which proves the claim. We will show now that g interpolates the values of FS on the variety V. If we consider a zero Zk of F f..L, then all the functions 1/Ij. j =I k. are multiples of Ffl. in a neighborhood of Zk. Therefore. it is enough to show that Ffl. + 1/11. vanishes at Zk with multiplicity at least mk. Consider a point w E B(Zko rk)\{zd. Let 0 < e < Iw - zkl, then letting y = a(B(Zko rk)\B(zk. e» we have FS(w) wqFfl.(w)
=
-1-1 2:rri
FS(~)
y ~q.Ffl.(~)(~ - w)
d~.
6.1. Convolution Equations in IR
377
hence
For Iz - zkl > rk one has
therefore, for those z, l/1k(Z)
r
= zqFIL(z) 2rci
1a8(Zk.e)
FS(~) d~. ~qFIL(~)(~ - z)
Since the two sides in this expression are holomorphic in C\B(Zb £) they coincide at the point w. Hence, the above expression (t) for FS(w) can be rewritten as with hk(w)
=
wq 2rci
r 188(2ko")
FS(~)
~qFIL(~)(~ - w)
d~ ,
which is holomorphic in B(zk, rk). This proves that FS - g is a multiple of FIL in Jt'(C). Since FIL is slowly decreasing we conclude there is a function h E Ap(C) such that FS = g + FIL' h. So
Let SI E E'(lR) with cV(SUPPSl) 5; [a,,8) be such that FS 1 = g and define E £'(lR) by FSo = h, then we have S=IL*SO+SI.
To conclude the proof we first observe that, from the expression of 4Jk as a linear combination of 1/[(z - zd j ], 1 ~ j ~ rnk, it follows that each l/1k is a linear combination of F«d/dx)q ILk.!), 0 ~ I ~ rnk - 1, for k ::: 1, and similarly, of F(ILo.,), 0 ~ I ~ rna - 1, if Zo = 0 is a zero of FIL. Here, the ILk,/ are the distributions introduced in Lemma 6.1.14. Furthermore, if S = 8x for some x E JR., the computation of the residues that define 4Jk shows that the coefficients of l/[(z - zk)i] are linear combinations of x'e- iXZk with 0 ~ I ~ rnk - 1. We note that the coefficients of these linear combinations depend on q, which can be taken to be the same for all x, Ixl ~ R. This concludes the proof of Lemma 6.1.29. 0 Now let us return to the proof of Theorem 6.1.27. We need to show that the inequalities on the zeros of F IL imply the hyperbolicity. To see how to proceed,
6. Hannonic Analysis
378
= 11- * So + Sl. then f(x) = (ox * nCO) = (f * 11- * So)(O) + (Sl * ncO) = (Sl * ncO) = (SI. f)· Therefore. if we start with h E E. we define an extension f of h to the whole suppose
f
E
!m and let 8x
real line by f(x) = (SI. h).
* So + SI. First. we need to show that this definition of f Ox = 11-
is independent of the choice
of decomposition of Ox. In fact. let Ox = 11-
* So + SI = 11- * To + TI
with cv(supp Sl) !:; [cr. p] and cv(supp TI ) 5; [cr. ,B]. Then TI - SI = 11-
* (So -
To)
and, from the support theorem. [cr. p]
+ cv(supp(So -
To»
= cv(supp(TI -
SI» 5; [cr.
Therefore. cv(supp(So - To» = {O}, which indicates that So - To for some polynomial P. Hence (TI-SI.h) = \11-*P
Pl.
= P(d/dx)(8o)
(~) (oo),h)
= \11-. P ( -
:x) (h»)
= 0
by the compatibility conditions defining E. Note that. in particular, if x E [cr.,B] it follows that f(x) = hex)
since we can take So = 0 and SI = Ox in this case. Second. we are going to show that the function f is of class Coo. If we fix R > 0, then the decomposition of Ox can be made so that R: S ~ SI given by Lemma 6.1.29 is linear and continuous. It is a general fact from functional analysis that f is Coo since the map x ~ Ox is Coo for Ixl < R. For the sake of completeness let us show the existence of the first derivative. For x fixed and o "" It I small we have 1 v v /1 ) t(f(x + t) - f(x» = \ t[R(ox+t) - R(ox)]. h
The limit
. OX+I - Ox 11m
1-+00
t
= .'
Ox
6.1. Convolution Equations in R
379
is valid in the sense of distributions, hence
1 v lim -(f(x t
1..... 00
+ t) -
v
I
f(x» = ("R.(ox)' h).
Repeating this procedure we see that j
= ("R.(o~n», h).
To conclude the proof of this theorem we need to verify that the function f is a solution of /1- * f = O. This follows from the last part of Lemma 6.1.29 which tells us that (x) is given by an infinite series of the form
I
L L k
Ck/xle- iXZk
O:::;/:::;mk-1
with coefficients Ckl independent of x as long as Ixl < R, and the series converges in £ (] - R, R [). This representation shows that (/1- * f) (x) = 0 if Ixl < R - Max(lal, 1.81). Since R is arbitrary, f E rot. 0 6.1.30. Example. We will give a general method to construct slowly decreasing functions ¢ in Ap(1C) with V(¢) an interpolating variety and ¢ not hyperbolic. The method consists of moving some of the zeros of the function sin 1r Z away = (logn)2. We choose a subsequence A = (nk) of from the real axis. Let integers::: 3 and a sequence (On)neA of positive numbers such that:
en
(1) n - On ::: e~ for n E A; 00
(2)
e
L...!!!..
< +00;
k=IOnk (3) nk + onk < nk+l - Onk+'·
For instance, we can take On = .;n and nk = k6 for k ::: ko where ko is a convenient integer. (The construction holds for other sequences Note that en::: 1 and Lk>l(enJnk)2 < 00.) We define -
en.
f(z) =
[!! (1- :~)llg (1- C:ienY)] . n~O
ie
This function has zeros at the integers n E Z*\A and at the points ±(n + n ) for n EA. Therefore, it is a function of exponential type by Lindelof's theorem (see [Lev, Chap 1, Theorem 15]). To show that f E Ap(1C) it is enough to show that it grows at most like a power of x for x E lR.. This fact and the fact that f is slowly decreasing will be a consequence of the following lemma: 6.1.31. Lemma. Let n (1)
E
A and On, en ::: 1 as above. For x
E
R we have:
6. Harmonic Analysis
380
(2)
I: : : On we have
When Ix - n
x 21 < 11 - (-n +-i Sn ) (3) For
Ix - nl :::: On
we
n 2 ---x 1 -
2
1 (-n-.)21 .
n2 + S;
X+l
have
X- )21 < n2- ( 1+2sn)21 1- (x+i)21 -. 11- ( 2 n + ien - n + e; On n Proof. (1) We have
I1 -
x I - In-x+isnl > In-xl > n n+ien - In+isnl - In+isnl-In+ienl
I1 -xIn
the same inequality holds when we replace x by -x. Multiplying the two inequalities we obtain (1). (2) Remark that because en :::: 1,
II - n +xiSn! = In-x+iBnl<snln-x-il In+isnl In+isnl = ,n:isn(nll-X;il· Hence, replacing x by -x and multiplying the inequalities
2 11 - ( -n +-XiSn- )21 -< -n2-n+-S~sn21 1 -
(x+i)21 -n
.
From the condition (1) we obtain that x :::: Bn, so that
(3) We have 11
(_x_)! n+ien
= In -x + iBnl = In -x -
i'll +
::::: In : iS nl!l- (X_;_i) I
(1 + -I~-~-:I)
In+isnl
::::: In:i sn I 11 -
In+iBnl
C;
i)
111 + 2 ::
_i(_sn_+_l_) ! n-x-i
I·
In the last line we have used that Ix - n I :::: On and en :::: 1. Applying now the same argument as in the previous parts we obtain
X.- )21 ~ -2--2 n2 ( 1 + 2Bn)21 1 11 - ( n + IBn n + Bn On
(X+i)21 -n
.
o
6.1. Convolution Equations in lR
6.1.32. Corollary. For x
381
JR we have
E
If(x)1 ~ C(lxl
+ 1)1 sin;rr(x + i)l·
Thus, f is of at most linear growth on the real axis. Proof. It is easy to show that for x
E
JR,
The function f being even we can limit ourselves to x ~ O. Consider first the case where, for some n E A, we have Ix - nl ~ 8n • By property (3) of the sequence A we have Ix - kl > 8k for any k E 1\ \{n}. Therefore, using the definition of f and the inequalities (2) and (3) of the previous lemma we have If(x)1 =
n 11- (~rl' n 1
m¢A
m
kEA\{n)
1 - (_x. k + 18k
)21.1 1 _ (~)21 n+ len
m#O
<
n 11 - (x+i)21' m
m¢A
IT kEA \{n)
(--/-z) (1+ 28k )21 1 _ (xti)21 + k
8k
8k
m#O
n 2 x - -2I x l 11n 2 + 8~
X +1
(
- -.
n
)21 .
We have all the terms of the infinite product expansion of Isin ;rr(x + i) I/(;rr Ix + i J) and the numerical constants form a convergent infinite product. Therefore If(x)1 ~ Cixll sin;rr(x
+ 01
~
C.lxl
+ C2.
In the case where 0 ~ x f/ UnEA[n - 8n , n + 8n l, the above proof works expect that there is no need to consider one separate term. In fact, for these x we have If(x)1
~ C' I Si~;~ ~ i)l.
Altogether, we obtain the required inequality
o
If(x)1 ~ CI sin;rr(x + i)I(lxl + I) ~ C'(lxl + 1).
6.1.33. Corollary. There is a constant A > 0 such that for every x If(x)1
~A
1
Sinx;rrx
E
JR we have
I·
Hence, f is slowly decreasing in Ap(C). Proof. From the definition of obtain If(x)1 ~ rr nEA
f and the first inequality in Lemma 6.1.31, we
- - 2 )rrl 1 - (X)21 = Al ISin;rrxl -( -n2n2 + 8 n#O n ;rrx n
6. Harmonic Analysis
382
with A J > 0, since the infinite product is convergent. Given any point x E lR there is y E lR such that Ix - Y I ::: 1 and I sin 7T Y I = 1. This is clearly sufficient to show that f is slowly decreasing. 0 It is clear from Corollaries 6.1.32 and 6.1.33 that f is the Fourier transform of a distribution J-L which satisfies cv (supp J-L) 5; [-7T, 7T]. A simple application of the Squires theorem (Remark 2.6.24) shows that V (f) is an interpolation variety. (We give a direct proof below.) Clearly the zeros of f of the form = n + iSn, with n E fl., do not satisfy
I Im(n + isn)1
::: C log(2 + In
+ iSn I)
since this would imply log2 n ::: CJlog(2 + n) for n E fl., n -+ 00. Therefore J-L is not a hyperbolic distribution. We proceed now to estimate If'(zk)1 from below for any zero Zk of f. It is clear that all zeros of f are simple. It is easy to see that
f'(zd =
_3... II Zk I#
(1-
(Zk)2) . zJ
Consider first the case where Zk is a real zero of f, by the first part of Lemma 6.1.26 we see that if we replace every zero n + iSn, n E fl., by n we obtain
since, up to the constant C2 = I1neA [n 2 / (n 2 + s~)], we have evaluated the derivative of sin7Tz/7Tz at the integer Zk. Let us now consider a zero of the form n + ien. The only terms that have not been considered in Lemma 6.1.31 are of the form 11
For k
E
(nk ++ ~8n)21. 18k
fI. \{n} we compare 11 - [(n k+iek
k
k+is k
i
k-n
1 I. I nil k l+i(edk)
> 1- -
-
Similarly,
n/ kl. We have
I 11 _ ~ II-kIII + (6k- 6 I
1 - n + i en =
1
+ ien)/(k + i8k)]1 with 11 -
n + i6 I I1+nil 1 I I1+--k+i8k k l+i(ek/k) n
>
n)
6.1. Convolution Equations in IR
383 LkEA (sd k)2
On the other hand the convergence of the series is a constant A such that
implies there
I~ II 11+(~fl~A
independently of n. Therefore
II
kEA\[n)
For k
E
r
II - (n ++ I>.!. k
i Sn
iEk
- A
II
kEA\{n)
II - (~f I· k
N*\A we have, by the same method,
It follows that
If'(n
+ iBn)1
~ Aln! iEnl gil - Gfl k~n
C,
>---
In + iBnl with C, > O. Therefore, for every zero Zk of f we have 1f'(Zk)1 ~ D/lzkl for a convenient choice of D > O. This proves that V(j) is an interpolating variety.
o 6.1.34. Remark. In the paper [Eh3, p. 333] one finds an example of /-L E £' (~) such that all zeros of F /-L are real but /-L fails to be hyperbolic because F /-L is not slowly decreasing. In this example F/-L(z) =
II (cos ; )d
n
n:::::l
n
for the choice dn = n! log2 n. 6.1.35. Example. Any difference-differential operation is a hyperbolic operator. That is, we consider the finite sum (/-L
* f)(x) =
L.ak,d(j)(x j,k
Tk)'
In this case F/-L(z) = Lak,jijzje-irkz. j,k
The zeros Zk of F /-L satisfy the inequality C log(2 + IZk!) by the results of P6lya described in Chapter 3. We also know from Chapter 3 that F/-L is slowly decreasing. On the other hand, the variety V(F/-L) is not
11m zkl
~
6. Hannonic Analysis
384
necessarily an interpolation variety. This is the meaning of Remark 6.1.28. For instance, if Ff.,L(z) = cosz· COSAZ, then V(Ff.,L) is an interpolation variety if and only if A is not a Liouville number. This F f.,L corresponds to the difference operator (f.,L
* f)(x)
=
i
[/(x
+ 1 + A) + I(x + 1 -
A)
+ I(x - 1 + A) + I(x -
1 - A)] .
The previous example and the conjecture of Ehrenpreis mentioned in Chapter 3 are also related to the concept of an almost periodic function. Let us recall [Kat] that a function I continuous on IR is called almost-periodic if for every E > 0 there is a number L = L(E, f) > 0 such that for every interval] ~ IR of length L there is a E ] with the property that III - 'l'aflloo :s E. L is called an almostperiod. Note that I is automatically bounded. One can prove that there is a collection A ~ IR of frequencies and coefficients (cAhEA such that for x E IR I(x)
= L cAe iAx AEA
and
L IcAI <
00.
AEA
Conversely, any such series of exponentials is almost periodic. The set A of frequencies could be dense in lR, but it is clearly countable. If I were also mean-periodic, then A would have to be a discrete set, it would coincide with the spectrum of I and every point of the spectrum will be simple. Kahane [Kahl] has proved that every bounded mean-periodic function is necessarily almost periodic. The conjecture of Ehrenpreis can be interpreted as follows: let f.,L be a difference-differential operator with algebraic coefficients and involving only translations with algebraic steps. Assume further that each zero ak of F f.,L is real and simple, then every f.,L-mean-periodic function must be almost periodic. (Clearly, this extends to the case where f.,L is slowly decreasing, V(Ff.,L) is an interpolation variety, all terms are simple and real.) The theory of this section can be easily extended to the Beurling algebras Ew(lR) [Bj] and similar algebras of Coo functions where one imposes restrictions on the growth of the derivatives ([Grud2], [Ni2] , [B5t2]). EXERCISES
6.1.
Unless explicitly mentioned every function belongs to £(lR), convergence is understood in the sense of £(lR), and every distribution belongs to £'(lR). 1. Show that if P is a polynomial, P(x)e ih is JL-mean-periodic if and only if xkei)'x is JL-mean-periodic for 0 ::: k ::: deg P. 2. Show that the spectrum '!(f) of a function f is either empty, discrete, or the whole complex: plane. 3. Let M be (closed) invariant subspace of £(lR). Show that (d/dx)(M) ~ M, JL * M ~ M. If P is polynomial and P(x)ei).x EM, conclude that xkei)'x EM for 0::: k ::: deg P. 4. Find all finite dimensional invariant subspaces of £(lR). 5. An invariant subspace M is minimal if M =1= to} and there is no proper invariant subspace N ~ M. Find all the minimal invariant subspaces of £(R). Show every invariant subspace contains a minimal subspace.
6.2. Convolution Equations in C
385
6. Show that if a function ["!- 0 then there is an exponential e iA' which can be obtained as a limit of finite sums of the form L aj [(x - Xj). 7. Let [be a I-periodic function, [(x) = L':"'""ane21rinx. What can be said about the coefficients an if I is also periodic of period 1/2? 8. Let
I
be I-periodic and also r-periodic, r
It Q. Find [.
9. (a) Let I(x) = ex2 . Show that for any polynomial P, P(x)e X2 E'I(f). (b) Show that [ is not mean-periodic.
#- I ELI (l~) n £(l~), show [ cannot be mean-periodic. Is it possible when n £(lR.), for some p, I < p :5 2? 11. Show that if 0 #- I E V(lR.). then every continuous function in 1R. can be locally
10. Let 0
I
E U(lR.)
uniformly approximated by linear combinations of translates of I. Generalize to the case [
E C~(lR.).
12. Let I be JL-mean-periodic, I (x) = Ln.k an,kxk eiAnx. Show that for any polynomial P, P (x) I (x) is also mean-periodic, Find its Fourier expansion. What about the function g = (XJL) * I? Show that any derivative and any primitive of I is mean-periodic. 13. Is I(x)
= eX
mean-periodic?
14. Let 0 #- I E £(1R.) be such that II(x)1 :5 A(l further that IE LP(lR.) for some pEW. Show Consider supp(:F(fP».)
+ Ixl)N
for some A, N :::: O. Assume
I cannot be mean-periodic.
(Hint:
15. Let JL E £'(1R.) be slowly decreasing, the zeros of :FJL all real and simple, V(:FJL) an interpolation variety. Using Theorem 6.1.11 show that every IE £(1R.) which is JL-mean-periodic can be written as I (x) = Lk Ckeia,x, Lk ICk I < 00. (In fact, the estimations on Ck are even better than this one.) In particular, I is almost periodic. 16. Show that if
I
E
C(lIt) is almost periodic then tEL 00 (R).
'17. Using Theorem 6.1.11 show that if JL = Ol+). + 01_). + 0),-1 + 0-1-1. and A is a Liouville number, then there exist I E £(1R.) which are JL-mean-periodic but not almost periodic. 18. Let I E £(1R.) (f E C(1R.) is enough) be mean-periodic with respect to a measure JL of support in la, b]. Define 1+ = Ix[o,oo[, 1- = 1- 1+· Show that if g := 1+ * JL then: (a) g = - 1- * JL and suppg ~ la. b]; (b) the functions .cb(g) = G, .cb(JL) = M (.c b is the bilateral Laplace transform) are entire functions. Show that if v is another measure of compact support such that v * I = 0, h := 1+ * v, H = .cb(h), N = .cb(v), then the meromorphic functions G / M and H / N coincide. (This meromorphic function is called the Fourier-Carleman transform of t.)
6.2. Convolution Equations in C We recall that the convolution of an analytic functional T in C and an entire function f is given by T
* I(z) =
(T~, I(l;
+ z»).
6. Harmonic Analysis
386
We note that this definition is slightly different from the one used in the previous section for distributions. Regretfully, both terminologies are standard. We will therefore adhere to the definition (*) in this section. We also recall from Chapter 2 that we can convolve analytic functionals T, S as follows: (T
* S, f)
= (S, T
* f)
(f E JIf(C))
and that the Fourier-Borel transformation ~(T)(z) = (T~, e~Z)
(z E C)
is a topological isomorphism between the algebra JIf' (C) of analytic functionals and the algebra Exp(C). The considerations from the previous sections about the relations between translations, invariants, (closed) subspaces of JIf(C), and closed ideals of Exp(C) hold verbatim. We summarize them in the following statements: A subspace !.JJ1 of the space JIf(C) is said to be invariant if!.JJ1 is closed and "l"h(!.JJ1) ~!.JJ1 for all hE C, where "l"h(f)(Z) = fez + h). A subspace !.JJ1 is invariant under differentiation if !.JJ1 is closed and f' E !.JJ1 for every f E !.JJ1. It is clear that if !.JJ1 is an invariant subspace of JIf(C), then it is invariant under differentiation but the converse is also true because f(n)(z)h n f(z+h) = I
L
n.
n:::O
is a series convergent in the topology of JIf(C). We leave it to the reader to verify that this equivalence does not hold in the space £(lR). 1fT E JIf'(C) then KerT* is an invariant subspace. Moreover, if~(T)(a) = 0 then eaz E Ker T*, hence Ker T* "" 0 in and only if T "" Coa(c "" 0, a E C) and Ker T* "" JIf(C) if and only if T "" 0. The following is an easy result: 6.2.1. Proposition. !.JJ1 is an invariant subspace of JIf(C) if and only if the subspace!.JJ1-L := (T E JIf'(C): (T, f) = 0, "If E ~Jl} is a closed ideal of JIf'(C). Equivalently, let I = ~(!.JJ1-L) then !.JJ1 is invariant if and only if I is a closed ideal of Exp(C). If J is a closed ideal of JIf'(C) we observe that J-L = (h E JIf(C): (T, h) = 0, E J} coincides with the invariant subspace !.JJ1 given by
"IT
!.JJ1 =
n
Ker T*.
TEl
We also note that zkeaz
E
!.JJ1 if and only if there exists mEN, m > k such that
(a, m) E V(l),
It follows that
zj eaz E
!.JJ1 for all 0
I = ~(!.JJ1-L).
:s j :s k.
6.2.2. Definition. An invariant subspace !.JJ1 "" {OJ admits spectral analysis if there exists a E C such that eaz E !.JJ1. !.JJ1 admits spectral synthesis if and only if [ = [loco
6.2. Convolution Equations in C
387
6.2.3. Theorem. Every nonzero invariant subspace 9J1 of Jr(C) admits spectral analysis and spectral synthesis. Proof. This follows from Chapter 2 where we have showed that every nontrivial 0 closed ideal of Exp(C) is localizable. 6.2.4. Definition. A function f E Jr(C) is mean-periodic if there exists T #- 0, T E Jr' (C), such that T * f = O. We also say that f is T -mean-periodic in this case. 6.2.5. Lemma. A function f is mean-periodic if and only if the invariant subspace 'ref) generated by all 'fh (f), h E C, is different from Jr(C). Every T -mean-periodic function admits a unique expansion in terms of the exponential-monomials zke,tZ, k :::: n, with (a, m) E V(J(T)). We shall now proceed to give a proof of this statement, which though similar to the proof of Theorem 6.1.11, requires in the general case to introduce the concept of series convergent after grouping of tenns. 6.2.6. Theorem. Let V = V(J(T)) = {(ak, mkk:d, lad:::: la21 :::: . ". Either this sequence is finite or there exists a sequenceofindicesk 1 = I < k2 < ... such that any T -mean-periodic function f in Jr(C) admits the following expansion convergent in )fCC) (that is the series in n converges absolutely and uniformly on every compact of C):
where Pk is a polynomial of degree < mk. (If the sequences (ak) is finite then (**) is a finite sum.) Moreover, the function f is identically zero if and only if all the polynomials Pk are zero.
Proof. Let us denote J(T) by and recall that there is a sequence 0 < Rl < R2 < .. " with Rn + I :::: Rn+l :::: 2Rn, n :::: I, such that
for some e > 0 and A > 0 (see proof of Proposition 2.7.14). It follows that the principal ideal (
6. Hannonic Analysis
388
Let us denote by Vn = {(ak' mk): kn ::: k < kn+d the family of (ab md such that ak belongs to the annulus Cn, Cn := {~
E
C: Rn- I <
I~I <
Rn}
(n
~
E
V
2),
while C 1 is the disk C 1 = {~ E C: I~I < Rd. In this case we have a canonical interpolation formula. Namely, if g is a holomorphic function on Cn and continuous up to the boundary, we consider
In(g)(~) = _1_. 2m
r
g(w)
Jac.
for ~ E Cn. It is easy to see that In(g) is also holomorphic on Cn. It is clear, using the residue formula, that In (g) depends only on the restriction Pn (g) to the multiplicity variety Vn • We have the estimate:
2;e IIgll"" ee-
Aoe BORn AR• = CeBRnllgll"",
IIIn(g)lIoo:::
where
en is the length of 8C
n,
and
B
and A are given by (t) and
I
It is clear that if a = Pn (g), then lIali n ::: IIln(g)lI"" ::: CeBRnliali n •
In fact, if h is anotherfunction with Pn (h) = a then In (g) = In (h), this implies the second inequality. In future, we shall write In(a) for In(g). Remark that if Vn = 0, then A(Vn) is the trivial vector space {OJ. We define the space A.(V) of sequences with grouping in the following way: given a E A(V) we can write it in a unique way as a sequence (an)n:::1 with an E A(Vn), then A.(V) := {a = (an)n:::1 E A(V): 3D > 0: lIaliD := sup lIan line-DR. < +oo}.
It is clear that if g is an entire function such that Ig(~)1 S KeLI~I,
then peg)
E A.(V)
with: IIp(g)IID S KC,
D
= L+B,
so that, the map p of Exp(C)/(
6.2. Convolution Equations in C
389
Let a = (ank:), lIaliD < +00, for some D > O. Choose a sequence (hn)n?! with hn E .Jf(Cn ) satisfying
= an,
Pn(hn)
IIhnll oo S 211a1lDe DRn .
en
Since the components of the set S(I4>I, e, A) are contained in Un» we can apply the Semi-Local Interpolation Theorem 2.6.4 and obtain h E Exp(C) such that p(h)
= a,
Ih(nl S Mlla"De(D+N)I~I,
where M > 0, N > 0 depend only on 4>. In order to obtain the Fourier expansion of the T -mean-periodic function f we need to identify the strong dual A~(V) of the space A*(V). It is easy to see that any element b E A~(V) can be written as a sequence b = (bn)n?) , bn E A'(Vn ), with the dual norm IIbnll~ =max{(an,bn): IIanli n S I},
where the duality bracket is (an' bn ) =
L
XjYj,
!~j~dn
when we identify A(Vn ) with Cd. by enumeration of the values g(l)(ak)/ I!, an = Pn (g). Hence, for every D > 0 we have
"bn"~ :=
L
IIbn"~eDRn < +00
n?!
and
(a, b)
= L(an , bn }. n?;l
From these considerations it follows that to every f E Jf"(C) such that = 0 corresponds to a unique element bU) E A~(V) in such a way that for S E .Jf1(C) T
*f
(~)
(S, f) =
(p(~(S»,
b).
(The verification of this last identity is done the same way as in Theorem 6.1.11.) Moreover, the continuity of f as a linear functional on .Jfl(e) tells us that the convergence of the series that are implicit on the right-hand side of (~) are uniform for any bounded family of analytic functionals S. In particular, we can take S = 8z , z belonging to a compact subset of C. Then zj eakZ ) p(~(8z» = ( k?~
-y-
O~J:5mk
6. Hannonic Analysis
390
and
where
o It is clear that the choice of grouping is not unique, though one can construct examples showing that they are necessary for the convergence of the series [Leo]. One natural way to group terms is to use the connected components of S(I<1>I, s, A), this observation and basically the same proof as that of Theorem 6.2.6 can be used to prove the expansion of Coo JL-mean-periodic functions in R when JL E £' (R) is just slowly decreasing. In order to find an explicit formula for the coefficients bk,t//! we need to construct analytic functionals Tk •j such that: p(:J(Tk,l»j.j =
(~(Tk,I»(j)
.I
1
(aj) = -/I81,j . 8k,i.
J. . Let fJ be a holomorphic function near w = 0 which vanishes at w = 0 with order exactly m E N*. Then 1 A-m+I A-I - =A-m -+- + ... +-+ .... O(w) wm wm- I W
Denote the principal part of wi /fJ(w) at w = 0 by
[(J7~)L
'
then, for 0 ::::: 1 < m, Wi] [--
(J(w)
0
_ I -w
=
I W
(A_ A-+ --+ ... +-w
I- I )
m
wl
m
(1 {A_ (J(w) -
I
~
I
A_I }) + A-I+l Wl-I + ... + --;:;;- + ...
WI
= O(w)
+ HI (w),
where HI is holomorphic near w fJ(w)
= O. Hence,
[~] = (f(w) 0
wi
+ fJ(w)HI(w),
6.2. Convolution Equations in
rc
391
for w near 0, and the order of vanishing of the term () HI at the origin is bigger or equal to m. Therefore, when O::s j < m,
We define Tk,l by the formula
~(Tk,l)(~) := (~) [(~!;~kt] a, ' where ['la, represents the principal part, as done above. Clearly the right-hand side of this expression is an entire function of exponential type, namely it coincides with times a rational function. Besides, it is obvious that is vanishes at every a;, for i "# k, with multiplicity at least mi. Therefore this analytic functional Tk,l has the properties we were looking for. As we did in the previous section we can write more explicitly the functional Tk,l. First we define a functional Tk by the formula
(Tko f) = ( T"
ea,z (mk - I)!
1 z
f(w)(z - w)m,-le-a,w dw ) .
Zo
One can see that this formula is independent of the choice of base point zoo Changing Zo amounts to adding (Tz, ea,z P(z)} where P is a polynomial of
degree less than or equal to mk - 1, but this quantity is zero because ak is a zero of multiplicity mk of <1>. Using the same observation one obtains by integration by parts ~(Tk)(O = (r., - ak )m k
It is also easy to see that if K is the minimal convex compact carrier of T then K is a convex compact carrier of Tko for instance, by taking Zo E K on the explicit integral formula. Now one can consider the polynomial Pk,l of degree 1 P (r).- (r -a )mk k.l.,
.-.,
k
[(tI!(O -ad] a,'
then
Moreover,
6.2.7. Proposition. Any mean-periodic function f Fourier representation of the form
"# 0 in
Je(C) has a unique
392
6. Hannonic Analysis
where the series (in n) is uniformly and absolutely convergent on every compact subset of C, the CXk are among the zeros of some function
and Tk,o
* zl ef!z
= (:f3) 1 (Tk,o
=
C)
L
* ef!Z) = (~) I (J(Tk,o)(fJ)ef!Z)
(J(Tk,o»IS) (f3)zl-s ef!z.
O~s~1
If f3 = CXj i- CXb since I < mj, we have (J(Tk,o»(s)(CXj) = 0, and if f3 = CXk then (J(Tk,O»(S) (CXk) = 0 for 0 < s :::: I :::: mk. Hence, if
j
i- k,
Tk,o* Pk(z)ea,Z = Pk(z)e"'z.
Therefore, if
f is also S-mean-periodic we have S * ho
* f = S * Pk(z)e"'Z = Tk•O* S * f
= O.
This implies that Pk(z)e"'Z is S-mean-periodic and therefore ~(S) must vanish at CXk with multiplicity strictly bigger than dk • For the same reason if the expansion of f with respect to S is of the form
f('J
~ ~ Cf;,., Q,(M") .
with QI i- 0, J(S)(f3I) = 0 with multiplicity VI strictly bigger than deg QI = lh, then J(T) must also vanish at f31 with multiplicity strictly bigger than deg QI. Assume now that there is a frequency CXk in (ll) which does not appear in (llll). Consider Sk.O defined by J(Sk,O)(~) = J(S)(~)/(~ - CXk)v" then
o
~ S.., * f ~ S,.o' (~y Pj(,Je"")
6.2. Convolution Equations in C
393
In other words,
o=
L L Sk,O * (P (z)eajz ) j
n
j
= Sk,O
* Pk(z)e ak ; =
PkCz)ea,Z,
which is impossible. Therefore the frequencies that appear in both expansions are exactly the same, On the other hand, if ctk is a frequency that appears in (~) then Qk(z)eO:kZ = Sk,O
*f
= Pk(z)eO:,z
this concludes the proof of the proposition.
D
Now we would like to make more explicit the estimates of the coefficients bk,i/ l! of a mean-periodic entire function. For that purpose one needs to find
norms equivalent to the norms II . lin used in the proof of Theorem 6.2.6. This depends on the fact that given PnCg) there is another natural interpolation formula, Newton's interpolation formula. We proceed to recall some properties of divided differences and of Newton's formula. (See [Os].) 6.2.8. Definition. Let f E Jf"(Q) and Zo, ZI, ... , Zm be distinct points in Q. The mth divided difference of the function f at the points Zo, ... ,Zm is
~mf(zo"",zm):=
L O:::;j:::;m
f(Zj)II(Zj-zk)-1 ki)
and the Newton polynomial of f of degree m is PmCz):=
L
~j fCzo,···, Zj)(z - zo)··· (z - Zj_I).
O:::;j:::;m
One can also define
~m
f by recurrence as follows: Llof(z):= fez),
A il
If(
) ._ f(zl) - f(zo)
Zo, ZI .-
ZI - Zo
Consider now the function g (z) = ~ 1f (zo, z) for z ~2 f
(zo,
Zl,
.
¥- Zo,
then
Z2) := ~ 1g(zlo Z2),
and so on. We can verify that this inductive definition coincides with the previously given formula. One advantage of the formula for ~ m f given in Definition 6.2.9 is that one can see immediately that divided differences are symmetric on the arguments Zo, Zl, •.. , Zn of the Newton polynomial. One can show that Pm interpolates the values of f, namely Pm(Zk)
In fact, for arbitrary z
E
= fCzd,
Q\ {zo,
Zl, .•. ,
0
::s k ::s m.
zm} one has
fez) - Pm(Z) = ~m+1 t(zo, ... , Zm, z)(z - Zo) ... (z - Zm)
which is an exact expansion of the remainder.
6. Harmonic Analysis
394
The divided differences t"m f can be still defined by continuity when some of the points coincide. For instance, for Z\, ... , Zk such that Zj of. Zj for i of. j, each repeated mj + 1 times, m = mj + ... + mk + k - 1 t:;.m f(ZI,"" ZI, Z2, ... , Z2, ... , Zk, ... , zd 1 am'+"'+ml = ml m2 mk~k-lf(Zl, ml!m2!···mk! aZ I aZ2 ··.azk
•.• ,Zk)'
The interpolation formula has to be interpreted to mean that Pm interpolates the values of f and its derivatives, and the remainder formula remains true. The reader will find other expressions for divided differences and applications to interpolation in [Os], [MiT]. For our purposes we need to relate the size of the function f, the Newton polynomial Pm, and the corresponding divided differences, that is, the coefficients of the Newton polynomial. 6.2.9. Lemma. Let f be a bounded holomorphic function in Zo •.... Zm in Q o = {z E Q: d(z, QC) > 8}. Then It:;.m f(zo • ... , zm)1
:s
2m 8m
Q,
8 > 0, and
IIflloo.
Proof. It is clear that it is enough to prove the statement when the points are distinct. We proceed by induction on m. For m = 0 it is evident. For m :::: 1 let us define g(z) = fez) - f(zo) , z of. zoo { Z - Zo g(zo) = f'(zo),
in other words, g(z) = t:;.l /(zo. z). Clearly g is holomorphic in Q and if we have Iz - zol :::: 8 then Ig(z)1
:s 211~lIoo.
By the maximum principle, this is true everywhere in m-I
It:;.
g(z\' ...• zm)l:s
Q.
By induction
2m- I llglloo 8m- 1 •
Since t"m /(zo, ZI .... , zm) = t:;.m-lg(ZI, ... , zm) the induction step is correct.
o
the number of zeros counted with multiplicity of 4> = Denote by z8, .... Z~._I these points, each of them repeated according to their multiplicity. It is clear that the divided differences t:;. j (g) for Pn (g) = an E A (Vn ) are independent of the choice of g. For this reason, we denote t:;.j (an) := t:;.j (g)(zo' ... , zj), Let us denote by
Vn
J(T) in the annulus
en.
395
6.2. Convolution Equations in IC
o~ j
~ Vn - 1. The diameter 8n of C n is 2R n , using this notation, we introduce a new norm in A(Vn )
Let Pn be the Newton polynomial of degree
L
Pn(z) =
Vn -
1 given by
L~)(an)(z - zo)'" (z -
Z}_I)'
O::sj:Sv,,-l
then Pn(Pn) = an and, for z
E
Cn, we have
L
IPn(z)1 ~ Illanlll n
o,;-jlz - zol" 'Iz - z}_11
O::sj::Svn - l
~ 1110.111. (,,~_, 1) ~ v.llla.III., Since
Vn ~
KoRn for a constant KO > 0, which depends only on <1>, we have lIanll n ~ IIPnil oo ~ KoRnlllanllln.
To obtain an inequality in the opposite direction, given an E A (Vn) we consider the sequence a = (a m)mo::l E A.(V) with am = 0 if m i= n. Then, for D > 0 fixed Iiallo = lIanline-ORn. From the proof of Theorem 6.2.7 we know there is h p(h) =
Ih(nl We apply Lemma 6.2.10 with
~
n=
E
Exp(
a, Mlialloe(D+N)lsl. B(O, 5Rn), 8 = 20n, then
IL~)(an)1 = I~jh(zo' ... ,z})1 ~ ~ max (28 n )1 ISI::S5 R
Ih(nl
n
~
M
M
8n
8n
) Iia II Oe 5 (O+N)R n ~ } II an IIn e (40+N)R n.
Choose D = ~ to obtain
Illanlll n ~ Me(N+l)R'lIanli n. We conclude that we could have defined the topology of A.(V) using the norms Illalllo = sup Illanlllne- DR, and the dual space with the corresponding dual norms Illbnlll~
= max{(an, bn ):
Illanlll n ~ I}.
To make completely precise the estimates of the coefficients bk,/ / I! of the Fourier-expansion of a mean-periodic function we still need to determine the norm Illbnlll~.
396
6. Hannonic Analysis
Let us assume first that the points zS •...• z~n -I are distinct and consider the map
J: an = (ao •...• avn-I)
t--i>
t. = (t.(O)(a) • ...• t.(v.-I)(a».
Its inverse can be computed very simply, we just observe that
L
ak = Pn(zZ) =
t.j(a)(z~ - z3)'" (z~ -
z1-1)
O:U:svn-1
L
=
t. j
(a)(z~
- z3)'"
(z~
- z1-1)'
O<:j<:k
In other words, the matrix representing J- I is
o
If we let
Cn
(z1 - zo)
o o
(z2' - zS)
(z2' - zS)(z2' - z7)
o ...
o ...
0)
~
~
:::
0
.
= (co •...• Cn-I) = t(J-I)(bn ), we can conclude that Illbnlll~ =
L
ICjI8;i.
O::::j~vn-t
In fact, if V denotes the diagonal matrix of entries 8~, we have Illanlll n = IVJanl oo .
For U E
= max{[ujl : 0 S
j S Vn - I} and lull
= Lj:,,~1 IUil.
Illbnlll~ = max{{an• bn): IVJani oo S I}
= max{{rIV-Iu. bn): Iuloo S I} = max{(u, t(,D-I) t(J-I)bn ): lul oo S I} = max{(u, V-lc n ): Iuloo S I} = IV-IcniJ,
which is precisely the above identity. In case there are multiple zeros the principle is the same but the matrix J- 1 is more complicated.
6.2.10. Remark. The most important point is the observation that except for the factor 8;i the computation of the norm IlIbnlll~ is entirely computed from the knowledge of the zeros zJ and the values bu. We also note that, in the definition of the norm III . III Dwe could replace the weight exp( - D Rn) by exp( - D IzJ [) for an arbitrary j and. up to a change of D, the norms are equivalent. This depends on the fact that the weight p(z) = Izi does not change too much in the region en. As an application of the previous theory we will study the entire solutions of a difference-differential equation with constant coefficients.
397
6.2. Convolution Equations in C
Let T =
L
(-l)lak.l8~),
l~k~p l~l~ql
where
ak.l,
ok are complex numbers, OJ :f:. Ok if j :f:. k. Then
(T
* /)(z) =
L
+ Ok)
akt/(I)(z
l~k~p l~l~ql
and
with
QkU;) =
L
ak,le,
1~I~ql
As we pointed out after the proof of Theorem 6.2.6, the groupings in the Fourier-expansion of solutions of the equation T * / = 0 are really determined by the connected components Qn, n ::: 1, of the sets S(II, e, A). In Proposition 3.1.31 we have shown that one can choose e > 0 and A > 0 such that the diameter of every component is bounded by 1. It follows that the number of zeros of in any such component is bounded by some constant v > 0, independent of the component. It is then easy to see that the norms III . Illn can be defined without introducing the diameter 8n in the definition. The topology of the corresponding spaces are absolutely equivalent. In other words, the function / has the expansion
where In is the set of indices of distinct zeros
ak
of that are in Q n and
ilz,
"~ bkl
Pk(Z) =
I
O~I~ml
with
Vn
= I:kE/n mk :::: v. The estimate of the coefficients becomes
L IIlbnlll~eDltJnl < +00 n~l
for every D > 0, where f3n is one of ak in Q n and
IIlbnlll~
=
L
ICjl,
O~j~vn-I
where Cn = I(J-l)(b n ). Furthermore, we recall that in this case we have a large amount of information about the asymptotic distribution of the frequencies ak (see Chapter 3).
6. Harmonic Analysis
398
We would like to study in more detail the case where
= (ea~ - 1)(eb~ - 1).
In this case the zeros of
27ri fh+-IZ.. r In fact, there are always double zeros when r E Q, for instance, if a = b all zeros are double. In any case, if f is an entire function satisfying the equation C:j:)
fez
+ a + b) -
fez
then fez)
+ a) -
=L
fez
+ b) + fez)
= 0,
Pn(z)ect,Z ,
n::>:1
where
L
IbnleDlct.1 <
+00
n::>:1
is obviously equivalent to the condition that for every D > 0 IbnleDI",1 = 0(1).
(If the degree of Pn is 1, Pn (z) = bn.o + bn, IZ just replace Ibn I by Ibn.ol + Ibn. I I·) If alb fJ. Q then all the zeros are simple, with the possible exception of ~ = 0, and they lie in one of two sequences ~
27rik
= --
a
or
~
27ril
= -
b
(k E IZ., I E IZ.).
Since lea~ -
11 ::: CCd (~, (27r i I a )IZ.)),
where d(~, (27rila)lZ.) = inf(1,~, (27rila)lZ.) (see Theorem 3.1.22) and the corresponding inequality for eb~ - 1 also holds, it follows that in the case there are constants e > 0 and A > 0 such that (:j::j:)
27rik 27rill --- >ee -A(lkl+ll1l Ia b-
(for all k, IE IZ.),
then, we can find el > 0 and Al > 0 such that the disks B(a, ele- Ailctl ), a E V, are disjoint and 1
6.2. Convolution Equations in IC
399
fez) =
Coz
+ CIZ + L
bneCL,z
no:: I
with
lim log Ibn I = -00 (0 < lall S la21 S ... ). Ian I For instance, the condition (:j::j:) is satisfied if alb is not real. Among the real numbers alb that satisfy (:j::j:) we have all the irrational algebraic numbers. Since, in this case, C8 n ..... oo
ml ;:: niH lba- -;;
for every 8 > 0 and some C8 > O. On the other hand, if for every A > 0 there are infinitely many rationals min such that
I~ - ~ I s e-A(lnl+1m/),
(§)
then (:j::j:) is not satisfied and we represent any solution f(z)
f of (:j:) in the form
= COZ + CI + LbkeClkZ + L(b,.oeCL/OZ + bU eCLItZ ), ko::1
'0::1
where the two infinite series are convergent in ;fCC), the first one corresponds to the frequencies ak that can be isolated and the second one to the pair al.O and a',1 that have to be grouped together because (:j::j:) fails. The coefficients bk satisfy the condition lim log Ibkl = k ..... oo
-00,
lakl
and the coefficient pairs (b"o, bl,!) satisfy the two conditions I . log Ib"o + bl,!l 1m
hoo
la',ol
=-00
and lim log la"o - a"I1lb" Ii = / ..... 00
-00.
la',ol
Note that the failure of (:j::j:) is exactly the condition . log la"o - a',11 11m 1..... 00
la"ol
= -00.
Moreover, the conditions on b',1 and on b"o are actually symmetric (just use that Ib',ol s I - b/,ll + Ib"o + bl,!D· It is very easy to see that the groupings are necessary in this case where (§) holds. The two conditions on the coefficients b"o and hI,! are satisfied by taking b"o = 1 and hi, I = -1. If the series was convergent without grouping, the general term of this series should be bounded at every point. Let us suppose for simplicity that a > 0, b > 0, and set z = -i, then Ib/,oe-Cl/,oi I = e2nm // a ,
m,
which is unbounded for a certain sequence -+ +00. This example which shows that groupings may be necessary is due to Leont'ev.
400
6. Hannonic Analysis
6.2.11. Remark. It follows from the proof of Theorem 6.2.6 that a sufficient condition on the analytic functional T for the convergence of the Fourier series representations f (z) = L:k> I Pk (z )eUkZ without groupings is that for some c > 0, A > 0, each component-of S(l4>I, c, A) contains only one zero of 4>. This happens precisely when V is an interpolation variety so that also the series
'""' ck,).zj eak ' ~ k;o>:l
O:Sj<mk
is absolutely and uniformly convergent over compact sets. In the previous formulation we are really grouping all the terms (L:o:sj<mk Ck,jzj)eUk = Pk(z)e UkZ together. We know from Section 2.4thatifT E Jf"'(C)\{O}, then T * f = g,g E Jf"(C), is always solvable in Jf"(C). It is clear how to solve explicitly an equation of the form T * f = zme uz but it is not so clear how to find an explicit solution for an arbitrary entire function g. Nevertheless, there is a method, due to Gelfond and Leont'ev, which we will explain in detail in the following section for the very simple equation fez + 1) - fez) = g(z). It consists of the following: Given g(z) = L:n;o>:o(an/n!)zn, there is an increasing sequence Rn > such that the function
°
fez) = P(z)
+ LanBn(z, Rn) n;o>:O
is an entire function solving the equation T * f = g, where P is a polynomial of degree strictly less than m, m is the order of vanishing of 4> = J(T) at z = 0, and Bn(z, Rn) = - 1
2lri
1
lul=Rn
e zu - du -u n 4>(u)
(see [Ge], Chapter 5, Section 7, p. 358). Observe that the map g ~ f we have just defined is not in general a linear operator since the choice of Rn depends on g. Another application of Theorem 6.2.7 is the analysis of overdetermined systems of convolution equations. For instance, let us consider the system (*)
{S
*f = g T*f =h
where g, h E Jf"(C) are given and we are looking for an entire solution is clearly a compatibility conditions, namely,
f. There
T *g = S *h. Is this condition sufficient for solvability of the system (*)? To simplify we consider only the case where 'J(T) and 'J(S) have no common zeros. We leave the general case to the exercises.
6.2. Convolution Equations in C
401
6.2.12. Lemma. Assume that ~(T) and ~(S) have no common zeros. The necessary and sufficient condition for the previous system to have a solution for every pair satisfying the compatibility condition is the existence of S I. TI E J't' (C) such that SI * S + TI * T = 8. Proof. Consider R := {(g. h) E (J't(C»2: S * h - T closed subspace of (-*'(C)) 2 . Let
*g =
O}. Then R is a
a: J't(C)""""* R. aU) := (S
* J. T * f).
This map is injective as a consequence of the hypothesis that J(T) and ~(S) have no common zeros (see Proposition 6.2.7). Therefore a is surjective if and only if a is an isomorphism. and this is equivalent to the fact that la is also an isomorphism. By the Hahn-Banach theorem every element of the dual space R' can be represented as a pair (SI. T I ). SI. TI E J't'(C). Therefore there exist SI. TI such that On the other hand. (la(S], Td. f)
= (SI. Tl), a(f)} = (SI. TI). (S * f. T * f)} = (SI. S * f) + (Tl • T * f) = (SI * S + TI * T. f).
i.e .• la(SI. TI) = SI
* S + TI * T = 8.
Conversely, if this convolution equation has a solution then we can choose
f = SI * g + TI * h. which one can verify is a solution of the system because g and h satisfy the compatibility condition. 0
*
The equation S1 * S + TI T = 8 is sometimes called the Dezout equation. It is equivalent to finding functions 1. 1/11 E Exp(C) such that 1 ~(S)
+ 1/11 ~(T)
= 1.
We have already considered this last problem in Lemma 2.6.15 and we can state the following corollary:
6.2.13. Corollary. Under the assumptions of Lemma 6.2.12, the necessary and sufficient condition for the existence of a solution f of the previous overdetermined system for every pair satisfying the compatibility condition is the existence of constants e > 0, C > 0, such that IJ(S)(~)I
+ 13'(T)(~)1
(strongly coprime condition)
::: ee-Cl~1
(~ E IC)
402
6. Hannonic Analysis
We would like to consider next the case in which J(T) and J(S) do not have common zeros but they do not satisfy the strongly coprime condition. The question is: What extra conditions should g and h satisfy in order that the system is solvable? To simplify the analysis let us assume all the zeros of J(S) and J(T) are simple. Since the convolution operator S* is surjective, we can find a function G E Jf'(C) such that S * G = g. Consider F =
f -
G, then letting H := h - T * G we have
* F = 0, T * F = H. S
The compatibility condition becomes S * H = S * h - S * T * G = S * h - T * g = O. Hence F and H are S-mean-periodic. Let V J(S), then F(z)
= {ad
be the variety of zeros of
= Lake,,"kz, k;::1
H(z) = Lbke,,"Az, k;::1
where these series may require groupings to be convergent. Therefore, H(z)
= T * F(z) = LakJ(T)(ak)e,,"kz k;::!
is the Fourier-expansion of the S-mean-periodic function H = T uniqueness of the coefficients in the expansion we conclude that
* F.
By the
bk
ak = J(T)(ak).
(0)
The extra necessary conditions on the functions g and h appear now as a condition on the sequence (bd so that the sequence of the coefficients (ak) defined by (0) must satisfy the growth conditions discussed before Remark 6.2.11. For instance, if V is an interpolation variety for Exp(C), then the condition for solvability is lim log lak I = lakl
-00.
k-H'O
That is, lim
log Ibkl -log IJ(T)(ak)1
k~oo
= -00.
lakl This condition is definitely stronger than the minimal condition lim log lakl = lakl
k~oo
-00
6.2. Convolution Equations in C
because we have assumed that particular we have
403 ~(T)
and
~(S)
are not strongly coprime, and in
. -log 1~(T)(adl hmsup k-+oo
lakl
= +00.
If V is not an interpolation variety and we need groupings we can make a similar analysis using the norms III . 11111' EXERCISES
6.2.
I. Verify the details of the example of Leont'ev given in the text about groupings for any solution I E Jt'(C) of the equation J1. * f = 0, J1. = 8o+ b - 8a - 8b + 80 • Consider also the case alb is not real.
2. Study the system S * I = g, T * 1= h, as given in the text, in the cases (a) there are no common zeros but some zeros of Ij'(S), Ij'(T) are multiple; (b) there are common zeros to Ij'(S) and 1j'(T); and (c) there are no common zeros but the variety of zeros of Ij'(S) is not an interpolation variety. 3. This exercise is a self-contained treatment of results from the recent manuscript [BezGr]. (a) Show that Jt'(C) is a reflexive Frechet space and that the transposed map ir: (Exp(C»' ~ Jt'(C) is a topological isomorphism such that irlJt"(C) Ij'. Deduce that one can introduce a convolution product in (Exp(C)' by means of the formulas:
=
* /)(z) := (R~, I(z + n) (Exp(C)' and I E Exp(C), which satisfies R * IE Exp(C), and (RI * R2, f) := (RI, R2 * f) (R
for R
E
for R I , R2
E
(Exp(C)', f
E
Exp(C). Show that Ij'I(RI
* R2 ) = \j'(R I)\j'(R2 ).
=
=
Moreover, if g ir(R) and c E C, show that ir(eC~ R)(z) g(z + c). (b) Show that I E X(C) is T -mean periodic (T E X'(C) if and only if Ij'(T)R = 0 in the sense of (Exp(C)', where R E (Exp(C)' is such that Ij't(R) I. Show also that for R E (Exp(C»" g E Exp(C), one has R * g = 0 in Exp(C) if and only if the product pT = 0 in X'(C), where p = ir(R) and Ij'(T) = g. For pEN and R E (Exp(C)', define the derivative R(p) by (R(p), g) := (-l)P(R, g(p») for any g E Exp(C). Show that R(p) = 8&P) * R, where 80 is the Dirac delta at the origin. (c) Show that for any am E C, a E C, R E (Exp(C»', 1= ir(R) we have
=
Show also that if Qrn(z) := 1l'kC,):=
:L05
m
am.qz q E C[z], and
2: 2: O:5m::sM lc:5q::sNm
C-l)q
(k)am.q(ma)q-kema~.
404
6. Hannonic Analysis
then
=
(d) Let a, fJ E C* be linearly independent over JR, let P(z) = E:=o amz n and Q(z) I:~=o bnz n, be two complex polynomials such that Go -f 0, b o ::/= O. Show that there are y > O. R > O. with the property that IP(e"')1
+ IQ(eP:)1
2: y
Izl 2: R.
whenever ~ E
Deduce the existence of a nonzero polynomial P(e":)
C[z) such that the two functions
Q(e P:) h(z):= - - .
I,(z):= - - . ~(z)
~(z)
are entire functions of exponential type without common zeros and show there are C > O. satisfying
f.
> 0,
=E IC.
for all
Prove that under these conditions any entire function rp satisfying the system of equations M
L
N
amrp(z
+ rna)
= L
bnrp(z
+ nfJ} == 0
m=O
must be an exponential polynomial. (e) Let a, fJ E iC be JR.-linearly independent, ao • .... am E C, ao I: 0, Qo • ...• QN be polynomials in C[z), QN ::/= O. The purpose of the following question is to show that if I E Jt" (iC) satisfies the system of equations M
N
Laml(z
L
+ rna) == 0,
m=O
Qn(z)/(z
+ nfJ) == O.
n=O
then I must be an exponential polynomial. Denote P(z) = E:=oamz n • F(l;} = P(e"'). R E (Exp(IC)', such that ~(R) = I. and let Jrk(l;) be the sums of exponentials introduced in part (c). (ed Show that R verifies the equations (in (Exp(lC)') N
(e2) Let L
E
FR = LJrkR(k) = o. k=O N* be arbitrary. p the column vector (R(N+L-I). R(N+L-2) •... , R'. R)
with entries in (Exp(IC)'. By differentiating (N - I) times the functional F Rand (L - 1) times E~=OJrkR(k) find a (N + L) x (N + L) upper triangular matrix A with entries in Exp(1C) of the fonn L
A=
o F
o such that Ap = O.
N
o
F
6.3. The Equation I(z (e~)
+ 1) - I
405
Deduce from (e2) that for all 0 ::: j ::: N 7T~ FN RCj)
+L
- 1
= O.
Conclude that (7T£ FN)(j) R = 0 for all 0::: j ::: N + L - I. (e4) Use (e) to prove by induction an kEN that (FN)(I)7T~(I+1l/2 R
= O.
(Note that TTN appears raised to a power, not a derivative, in this equation.) (es) Show that there is an integer k such that F N , (FN)', ... , (FN)(I) have no common zeros, and there are polynomials Uo, ... , U. such that
L• Uj(e<X~)(FN)Cj)(l;) == 1. Conclude that 7T~(k+1)/2 R = 0, and use the equation F R exponential polynomial.
6.3. The Equation fez
+ 1) -
=0
to deduce that
I
is an
fez) = g(z)
In the previous section we have studied convolution equations for entire functions in the complex plane. In particular, the simplest equation, that of periodicity: 1 (z + I) - 1 (z) = O. We have also mentioned the existence of "explicit" solutions to the inhomogeneous equations /-L * 1 = g, without going into the details. In this section we consider the very simple equation l(z + 1) - l(z) = g(z) for meromorphic functions in the plane, though the methods are elementary the results are interesting and not so well known. One can think that this equation gives another illustration of one of the main themes of this book, describe the functions that are close to being periodic. The theory developed in the last section, as well as the classical theory of Fourier series, show that if 1 is an entire function of period 1, that is 1 (z + 1) - 1 (z) = 0 for all Z E C, then 1 (z) = LnEf: ane2rrinz, and the series converges absolutely and uniformly on every horizontal strip 11m zl :::: A < 00. We now proceed to give a different proof of this fact, which will also allow us to consider also periodic meromorphic functions. The key step is the following simple definition and lemma: 6.3.1. Definition. A merom orphic function 1 in the complex plane is said to be I-periodic if for any z E cC such that neither z or z + I is a pole of 1, 1 satisfies the equation l(z + 1) - l(z) = O. In this section we shall just say that 1 is periodic. It is clear that if Zo is a zero of a periodic function, then any Zt E Zo + IE is also a zero and of the same mUltiplicity. Clearly, the same statements hold for the poles of f. Note that we could also consider periodic functions with essential singularities, we will not do that here.
406
6. Harmonic Analysis
Let us recall from [BG] that we denote by C* = C\{O}, and by Log~, for i; E C\] - 00, 0], the principal branch of the logarithm, i.e., the principal branch of the argument of i; is taken to be -rr < Arg i; < rr. By arg i;, log i; , we denote arbitrary branches of the argument and logarithm of i; E C*. Let exp: C ~ C* be the map z 1-* e 27riz • this map is the universal covering map of C*. We can relate periodic merom orphic functions in C to meromorphic functions in C* by means of the following lemma:
6.3.2. Lemma. Let f,' C ~ S2 be a p~riodic meromorphic function, then there is a unique meromorphic function F,' C* ~ S2 such that f = F expo Moreover, if Pc/), P(F) denote their respective polar sets, then P(F) = exp(P(f» and the order of the corresponding poles is the same. 0
Proof. Since the group of automorphisms of the covering map exp is generated by the transformation z 1-* z + 1, and f being periodic is exactly the same as saying that f is invariant under this group, then F is uniquely defined by passage to the quotient. As exp is a local biholomorphic map, the other assertions follow.
o Note that F is explicitly given by
F(i;)
(_1_. 10gi;) 2m
=f
for any i; ¢. P(F), the choice of branch of the logarithm is irrelevant.
6.3.3. Corollary. Any meromorphic periodic function f can be represented as f = g/ h, with g, h entire periodic functions without any common zeros, h ¢ O. Proof. Let F(i;) = f«I/2rri) log i;) as in the previous lemma. F is meromorphic in C*, hence by the Weierstrass factorization theorem F = G / H, with G, H holomorphic functions in Co, without any common zeros, H ¢ 0 (see, e.g., [BG, Corollary 3.3.2]). Clearly g(z) = G(e 27riz ), h(z) = H(e 27riz ), satisfy 0 the statement of the corollary. As an immediate consequence we obtain the following result:
6.3.4. Proposition. Any entire periodic function f admits a Fourier series expansian fez) =
L
cne27rinz,
nEZ
which is uniformly and absolutely convergent on every horizontal strip of finite width. Moreover, the coefficients are determined by Cn
=
1
Z0
ZO
+1
f(z)e-27rinZ dz
6.3. The Equation I(z
+ 1) -
I(z) = g(z)
407
for any Zo E C, n E Z. Any meromorphic periodic function g admits a representation as a quotient of two Fourier series
each of the series with the same kind of convergence properties as above.
Proof. It is clear it is enough to prove the assertions for periodic entire functions. Let FU;) = f(0/2lri) log~), then F is holomorphic in e. It admits a Laurent expansion about ~ = 0,
which converges absolutely and uniformly on every annulus 0 < r :::: 00. Its coefficients can be computed by Cauchy's formula
c n -
-1-1 2 . lrl
F(~) d rn+l~'
0< p <
I~
I ::::
R <
00.
1;I=p"
Letting p = leZJTiZO I, ~ = e 2JTiz , the chain rule implies that Cn
=
1
f(z)e-2JTinZ dz,
[zo.=o+l]
where [zo, Zo + 1] denotes the horizontal segment from Zo to Zo + 1. Since f is entire, the integral has the same value if we take any path starting at Zo and ending at Zo + 1. The properties of the exponential map imply that the Fourier series fez) = F(e 2JTiz ) = cne2JTiz
L
nEZ
is absolutely and uniformly convergent on any horizontal strip of finite width IImzl::::A
+ 1) -
fez)
= g(z),
where g is an entire function. It follows from the general theory of the last section that there are always entire solutions to this equation. Clearly any two solutions differ by an entire periodic function and we have just studied them. We are going to give here an explicit construction of a solution to (*). The original idea of this construction is due to Guichard, Appell, and Hurwitz a century ago,
6. Harmonic Analysis
408
we refer the reader to [Hul], [Ap] and the beautiful monographs of Whittaker [Whi] and Gelfond [Ge] for other related results. We start by solving (*) in a few particular cases. We denote by 6g, a sum of g, any particular solution of the equation (*). Then g is called the difference of f. It is similar to the concept of primitive and derivative. For instance, it is easy to verify that one can take
61 = z,
z(z - 1)
6z =
2
.
We leave other examples to the exercises. The one example important to us is the following, let a > 1 be a fixed number and consider the entire function a Z (fixing a determination of log a). Then it is easy to see that
aZ 6a z = - - . a-I
Clearly the same result holds for any a E IC\ to, I}. In particular, if we let a = e t , t ¢ 27riZ, 6e lz = e 'z j(et - 1), where the sum is with respect to the variable z. Consider the auxiliary function
te tz
B(z, t) = - - = t6e
Iz
e' - 1
tz
= 6(te ),
which satisfies B(z
+ 1, t) -
B(z, t)
= te tz
and it is holomorphic, for every z fixed, for It I < 27r. (This was the reason of the multiplication by t.) If we expand B(z, t) in a Taylor series in t about t = 0, we have
It I < 27r. It is very easy to convince ourselves that Bn (z) is a polynomial of degree exactly n (usually called the Bernoulli polynomial), but more interesting, by comparison of the two sides of (**) we see that Bn(z
+ 1) -
Bn(z)
= nzn-I,
so that 6z n
°
=
Bn+1 (z)
n+l
for n ~ 0. We also note that if < p < 27r, then Cauchy's formula for the Taylor series coefficients gives the representation n! /, e tz dt --t 27ri 111=p e - 1 t n
B(z)=n
(0 < p < 27r).
Let us assume g(z) = L:n2:0 gnzn, then, at least formally,
6.3. The Equation fez
+ 1) -
fez) = g(z)
409
If the series were to converge unifonnly over compact sets, then it would represent a solution of (*). Whether it does or not, depends on the behavior of the coefficients gn of g. If we only make the assumption that g is entire, all we have is the Hadamard estimate lim
n.... OO
\Ijg,;T =
O.
On the other hand, we could modity each Bn by adding a periodic function, and, as in the Mittag-Leffler expansions of meromorphic functions, force the series to converge to a sum of g. One way to do this is to observe what happens to the integral defining Bn if we make p larger. For instance, if 2Jr < p < 4Jr, then we have to add the residues at t = ±2Jr i, so that n! 2Jri
1,
e tz dt ( ---=Bn(z)+n!
Itl=p e t - 1 t n
e- 27riz (-2Jri)n
2Kiz
e +(2Jri)n --) .
In other words, we obtain Bn(z) plus a periodic trigonometric polynomial. In general, let
1,
n! etz- dt Bn, k(Z)-- 2 ' t Jrl Itl=p e - 1 tn'
if 2Jr(k - 1) < p < 21fk,
k E N*. It is clear that B
n,k+l
(z) = B
n,k
(z)
+ (2Jrik)n n! (e 2Kikz + (_I)n e -2Kkz)
and Bn •l = Bn. Thus, we also have 6z n = Bn+l,k(Z)
and we could choose any sequence kn 6g(z)
E
N* and, at least fonnally,
= L n ~ 1 Bn+I,kn+1 (z). n~O
6.3.5. Proposition. For any entire function g(z) = Ln~o gnzn. the series f(z) = ""' ~ n gn + 1 Bn+1,n+I(Z) n~O
converges locally uniformly and it defines an entire junction solving the difference equation f(z + I) - f(z) = g(z).
Proof. We need to find estimates for Bn+l,n+l so that the fonnal series of f converges. Let us choose p = (2n + l)Jr in the integral representation of Bn+l.n+l' Note that there is a constant c > 0 independent of n such that for It I = (2n + I)Jr let - 11 ~ c,
as one can reduce this inequality to the Lojasiewicz inequality for sin z. Hence, using Stirling's fonnula, we can prove the existence of A > 0 such that for
410
6. Harmonic Analysis
all z, n IB
(z)1 < n+1.n+l
-
(n
+ I)!
c«2n + 1)Jl')n+l
< Ae(2n+l)rr(lzl+1).
e(2n+l)rrlzl
-
This bound, together with the Hadamard estimate for gn, shows that the series defining I converges absolutely and uniformly on any disk of C. 0 6.3.6. Remarks. (1) If g is a function of infraexponential type, i.e.,
lim 10gM(g, r) = 0, r
r~OO
where M(g, r) = max{lg(z)l: Izl :s r}, then one can prove that the series
is locally uniformly convergent (see Chapter 4). (2) In general, one can choose kn in the last proof so that for any given s > 0, e > 1 there are constants c > 0, ro > 0, 8 = (1 + s) flog e, so that the function I defined in Proposition 6.3.5 satisfies I/(z)l:s CM(g,er)~,
Izl :s r.
In particular, I is of the same order as g, and of finite (resp., minimal) type if g is of finite (resp., minimal) type. (See [Ge, Chapter 5].) (3) In case g is a merom orphic function, the case we shall consider next, one can also find a merom orphic sum I with good estimates on the Nevalinna characteristic T(r, f). (See [Whi, Theorem 4).) We want \!O find now the sum of a meromorphic function. We observe that the difference equation (*) can be used to find two formal solutions as follows: I(x) = -g(x)
+ I(x + 1) =
-g(x) - g(x
+ 1) + I(x + 2)
= .. "
so that one is lead to consider the "right formal sum" Ir(x) = -
L g(x + n). n~O
If this series converges in a certain region, it is a solution of (*) in that region.
Similarly, we have I(x) = g(x)
+ I(x
- 1) = g(x)
+ g(x -
which leads to the "left formal sum" hex) =
L g(x n~O
as a candidate for a solution.
n)
1)
+ I(x + 2)
= "',
6.3. The Equation f(z
+ 1) -
f (z)
= g(z)
411
Returning to the equation (*) with g meromorphic, let us assume that there is real number p so that all the poles of g lie in the half-plane Re z < p. The idea is to try to find polynomials Yo, YI, ... so that the series !PI(Z):= (Yo(z) - g(z)}
+ (YI(Z) -
g(z
+ I)} + ...
converges to a meromorphic function in the whole plane. (Note that this function is obtained by applying the Mittag-Leffler procedure to the right formal sum of g.) In order to choose the Clj, let kEN be the smallest value such that p - k < O. Choose Yo = ... = Yk-I = O. The poles of g(z + n) lie in the halfplane Re z < p - n, hence for any n ::: k we have that g(z + n) is holomorphic in a neighborhood of the disk Izl : : : n - p and n - p > O. Choose Yn as the partial sum of the Taylor series of g(z + n) about z = 0 such that Ig(z
+ n) -
for
Yn(z)1 :::::: Z-n
Izl:::::: n -
p.
For any R > 0 we have that the series L
(Yn(z) - g(z
+ n)}
converges absolutely and uniformly on Izl : : : R, hence it defines a holomorphic function in Izl < R. It follows that the series !PI converges to a meromorphic function in C. Moreover, outside a discrete set, !PI (z
+ 1) -
fIJI (z) = g(z) - Yo(z)
+ L{Yn(Z + 1) -
Yn+1 (z)},
n;:,O
and the series converges locally uniformly. Thus, we have
where 1/11 is an entire function. From Proposition 6.3.5 we know we can choose an entire function fh = 61/11, then
is a merom orphic sum of g. If g had all its poles to the right of a vertical line we could have modified in the same way the left formal sum of g to obtain a meromorphic sum of g. In general, choose any p E lR. such that the line Re z = p contains no poles of g. The Mittag-Leffler theorem (see [BG]) allows us to find a meromorphic function g 1, whose poles lie to the right of that vertical line and whose principal parts at their poles coincide with those of g. Then g = gl + g2, where the poles of g2 lie to the left of the same line. Thus, from the previous construction we have two meromorphic functions II, h such that 11 = 6g l , h = 6g2 • Clearly, we can take 6g = I := 11 + fz. Let us remark that this construction of a sum of the meromorphic function g has a special feature, if all the poles of g are simple, then the same is true for this particular sum I. Moreover, if the residues of g at those poles are integral,
6. Hannonic Analysis
412
the same is true for f. The importance of this feature lies in the following application. Recall that for a meromorphic rp, its logarithmic derivative rp'lrp has only simple poles with integral residues. Conversely, for any merom orphic function cf> with this property we can find rp merom orphic such that cf> = rp' / rp. These two remarks together tell us that given a meromorphic function 1/1 there is another merom orphic rp such that rp'lrp = 6(1/1'/1/1).
That is, rp'(z + 1) rp'(z) 1/I'(z) -'------'-- =rp(z
+ 1)
rp(z)
1/I(z)
By integration we obtain that rp(z + 1) = 1/I(z). rp(z)
Retracing the steps we have shown that given a merom orphic function 1/1 one can find merom orphic function rp such that (**) is satisfied. Clearly we can multiply rp by an arbitrary periodic meromorphic function and obtain a new solution of (**). All solutions are obtained this way. Being meromorphic, rp is of the form e f PI I P2 , where f is entire and PI, P2 are Weierstrass canonical products, but following the technique of the right and left formal sums used to solve (**) one can anticipate the form of PI and P2 • We shall illustrate this below in the case of the r -function. We can now state the main theorem of this section. 6.3.7. Theorem. Let ao, al be a pair of meromorphic functions in the plane such that neither is identically zero. For any meromorphic function g in C there is a meromorphic solution f of the linear difference equation aICz)f(z
+ 1) + ao(z)f(z)
= g(z)
(z E C).
Proof. From the previous remarks we know there exists a nontrivial meromorphic function fl such that
/I(z
+ 1)
h (z)
=
al (z) ao(z)
The equation (t) now becomes h(z
+ l)f(z + 1) -
fl (z) fl(z)f(z) = ---g(z). ao(z)
We have also shown there is a meromorphic
h
such that
/I (z) h(z + 1) - h(z) = ---gCz). aoCz)
Letting fCz)
= h(z)//ICz)
we obtain a solution to Ct).
o
6.3. The Equation f(z
+ 1) -
413
I(z) = g(z)
As a title of example let us consider the equation f(z
(!)
+ 1) =
zf(z).
If we assume f(1) = I, then f(n) = (n - 1)! for n EN so that f interpolates the factorials. One of the solutions to this equation is Euler's [,-function. Following the previous method we let q;(z) := l'(z)lf(z) so that: q;(z
1
+ I) -
q;(z) = -.
z
The right formal sum of this equation is 1 L:n~O + n
q;r(z) = -
Z
which has its poles at the points z = 0, -I, -2, .... From here we conclude that a solution of the equation (!) is given by et(z)
I(z)=
00
z
IT
n=l
z
(I + -) en
' z/ n
where 1jr is entire and the denominator is the Weierstrass canonical product with zeros at z 0, -1, -2, .... Letting
=
fm (z)
m! exp (1jr(Z)
et(z)
= Z
IT
=
Z
m
+ _) e-z/n
(I
+ n~ zln)
---:--~---:-:':"""":,---:""':"""
z(z
+ 1) ... (z + m)
n
n=l
we have f(z) = limm ... oo Im(z) exists, locally uniformly in C\{O, -1, -2, ... }, and zf(z) = lim zfm(z) f(z + 1) m ... OO fm(z + I)
oo + m + I) exp [1jr(Z) -1jr(z + I) -
= lim (z m...
=
(I
Ji.moo + m:
n=l
n
1)exp [1jr(Z) -1jr(z + +~og(m + ~ ~
The limit y = lim
t !.]
m"""""""oo
1)
[~!. ~n
-
log(m
+
1)-
) ].
1)] ~ 0.557
n=l
is the Euler-Mascheroni constant. Inserting this value and using equation (!) we obtain zf(z) 1 = f(z + 1) = exp[1/t(z) - y,(z
+ 1) -
y).
6. Hannonic Analysis
414
In other words,
1/1 satisfies a difference equation of the fonn
+ 1) -
1/I(z
for some k we have
E
= -y + 2rrik,
1/I(z)
Z. The simplest solution is 1/I(z)
= -yz. With this choice of 1/1
Hence, f(l)
= =
lim fm(l)
m-->oo
lim exp
m ..... oo
[1/1(1) + ~ .!. -log(m + 1)] ~ n n=l
= m-->oo lim exp [-y +
t .!. n=l
n
-log(m
+ 1)] = 1.
This f satisfies the extra condition fO) = 1, so that it is the correct choice. This solution is exactly the r -function. r(z) =
g(1
[eYZ z
+ ~) e- z / n
]-1
We conclude this section studying a sort of unexpected relation between difference equations and analytic continuation, which is hinted at in the proof of Lemma 6.3.1. Consider a function F which is holomorphic in the simply connected region C\] - 00, 0] and assume it admits an analytic continuation along any path in C*. Thus, we can talk about F(~) for any ~ E C* as a multivalued holomorphic function, or, what is the same, a holomorphic function on the Riemann surface of log ~. Let us denote for the rest of this section .Ij the vector space fonned by these multivalued functions F. It follows that for; < 0 the values F(; ± iO) = lim~ ..... o+ F(; ± I rJ) are well-defined but the jump is not necessarily equal to zero. Let ~(n
:= F(;
+ iO) -
F(; - iO)
(; =1= 0).
This function is identically zero for; > 0, continuous for; < 0, and it represents a distribution on lR if and only if there is k ::: 0 such that ITJlk F(; + iTJ) remains bounded in a punctured neighborhood of ~ = 0 (see [BO, Proposition 3.6.12]). In general ~ is a hyperfunction, as explained in Chapter 1. Introducing a change of variables ~ = e 21riz (or unifonnizing parameter z), let us define f by F(O = fez), in other words,
f
(~IOg~) = F(n 2m
Thus, f is an entire function of z which satisfies, for x = Re z the difference equation fez + 1) - fez) = ~(e21fiz) = ~(_e-2"'Y).
= - t, y = 1m z,
6.3. The Equation f(z
+ 1) -
f(z)
= K(Z)
415
Note that with this notation it is natural to rewrite the original equation as F(e2rri~) - F(n = rp(~)
(~ < 0).
(Equations of the fonn F(q~) - F(~) = rp(~), 0 < q, often arise in Number Theory and are related to the theta functions mentioned at the beginning of this section.) Let us consider a special example, rp(~) = Log I~I for ~ < O. Then we have rp(_e- 2rry ) = -21fY for Y E JR.. By an easy inspection we see that f(z) = 1fiz 2 satisfies f(z + 1) - f(z) = -2rry (Rez =
-!).
Hence,
F(~) =
-4 1 . Log2~, 1f1
in C\] - 00,0]. Since rp E V'(JR.), this answer could also have obtained with the help of the Cauchy transfonn, as suggested by the general theory in Chapter 1 and in [BG, §3.6]. After these preliminaries, let us consider a type of question raised by Hurwitz [Hu2, Vol. 2, p. 752]. The homotopy group 1f, (C*) of C* is isomorphic to Z and it is generated by the unit circle IS I = 1, oriented counterclockwise. This group acts on Sj by sending a function F to the function obtained by analytic continuation along the unit circle. This is precisely what we have denoted by F(e2rri~). In this case F(e2rrin~), n EN, means the new function in Sj obtained from F by analytic continuation along the unit circle traversed n times counterclockwise, and for n < 0, traversed Inl times clockwise. Since the space Sj is very large it is natural to consider subspaces invariant under the action of rr,(C*). Typically they are defined by functional equations. For instance, the equation F(e2rri~) = F(~)
describes all the functions in Sj that are single valued, that is, the collection of all Laurent series
convergent in C*. Observe that if F E Sj, then its derivative F' also belongs to Sj. So that in Sj we have three types of operators acting: D: F(~) ~ F'(~); A: F(~) ~ A(~)F(~), multiplication by A E Jf'(C*); and r: F(~) ~ F(e2rri~), "monodromy" operator. It is not hard to see that r commutes with D and with A. For this reason Hurwitz
considered the following simple-looking equation in 1918: DF = r F, that is, (H) F'(~) = F(e 2rri The solutions of such an equation, if there are any, also fonn an invariant subspace for the action of 1f( (C*). It is clear that if we assume F is single
n.
416
6. Harmonic Analysis
valued, then there is essentially only one solution for (H), namely F(l;) = eel;, for any e E C*. Hurwitz seemed to believe this was the only solution. Note that when using the uniformizing parameter l; = e 21riz this equation becomes essentially more complicated than those in Theorem 6.3.7. Namely, d 2 . - fez) = 271:ie 1rlZ fez
(H*)
+ 1),
dz where fez) = F(e 21riz ) as earlier. In 1975, Naftalevich [Na] used the method of the formal sums to obtain many nontrivial solutions of (H*), i.e., different from e exp(e Z1riZ ). Earlier Hans Lewy had found the explicit solution
t)
(_1_.
Fo(O = ('" e-I;t exp Log2 dt, Jo 47rl which is well defined and holomorphic for Re l; > O. To show that this function belongs to the class n, let us consider for 0 E IR the auxiliary functions Fo(l;):=
l
1
ooe ;9
o
exp(-l;t
+ -.logZt)dt. 471:1
In the integral defining Fo we have t = re iO , 0 < r < 00. If l; = peiO/, p > 0, then the integrand is absolutely convergent when cos(O + a) > 0, in particular, if -71:/2 - 0 < a < 71:/2 - O. It follows that Fe is holomorphic in the half-plane -71: /2 - 0 < arg l; < 71: /2 - (). Note that if 0 < ()2 - 0, < 71:, then the functions Fe] and F~ have a common domain of definition, and applying Cauchy's theorem they can be seen to coincide there. It follows that the collection (Fe)eEIR defines a single function FEn. Moreover, for Re l; > 0 we have F(e Z1ri l;) = F21r (l;). This last integral can be computed from the observation that for r > 0 _1_. log2(reio) =
471:1 For
e=
~(Logr + iO)2 47rl
=
~ LogZ 471:1
() 02 r+ - L o g r - - . 271: 471:i
271:, we get
so that F21r(l;) = -
roo re-I;r exp (_1_. Log2 r)
Jo
dr.
471:1
On the other hand, F6(i;) can be computed differentiating under the integral sign F6(l;)
=
roo ~(e-I;r) exp (_1_. Log 2 r) dr, dl; 471:1
Jo
re-I;r exp (~ Log2 r) dr, Jo 471:1 which shows that Hans Lewy's function Fo is indeed a solution to the Hurwitz equation (H). = _
roo
6.3. The Equation I(z
+ I) -
I(z) = g(z)
417
It is not clear that this solution is obtainable by Naftalevich's method. It has been shown recently [BSe] that Fo generates all the solutions to (H). Namely, any such solution F E 5) can be written in one and only one way in the form
F(~) = ce{
+ LCnF(e2Jrin~), nEZ
where the series converges locally uniformly in the Riemann surface of log ~ . This example indicates that it is natural to consider a wider class of equations for multivalued holomorphic equations. Their natural name should be monodromic differential equations, e.g., an equation of the form
L
An.k(~)Dkrn F(~) = D
n.k
for F E 5), where the (finite) sum has coefficients A n.k E .)t"(IC*). More generally, we could consider infinite order differential operators or consider other domains like IC\{D, l},IC\{al, ... , ad, IC\Z, etc. We remind the reader that the case IC\ {D, I} is intimately related to the hypergeometric functions. Here the group Jrl (IC\ {D, I}) has two generators, and Riemann showed that if a multivalued function F has the property that at any point ~ =1= D, I, the number of linearly independent branches of F is exactly 2, then F satisfies a hypergeometric differential equation (see [BG, Exercise 5.15.2], [Tr] [Pool]). The domain IC\Z appears in the closely related theory of resurgence [Ec] , [Ram], [Mal]. EXERCISES
6.3.
I. Prove the converse of Lemma 6.3.2. Show also that if I is periodic merom orphic and F(?;) = 1((1/2rri) log ?;), the order of the corresponding zeros for I and F coincide. When is I entire? When is the function F entire?
2. Construct a periodic function I holomorphic in C except for a discrete set of points, where it has essential singularities. 3. Recall that a trigonometric polynomial of degree m is a function of the form E:-m cne Z1ri n= with either C- m or em is :j:- O. Show that it has the form I(z) = ao/2+ E:=I(akcosmkz +bk sin2rrkz). Show that an entire periodic function I is a trigonometric polynomial if and only if I is a function of exponential type (Le., lim sup, ~oo (log M (r, f) / r) < 00). Could it be of infraexponential type if I '" constant? Could it be of type < 2rr if I '" constant? I(z) =
4. (a) Let I be a periodic meromorphic function such that limx~oo I(x + iy) = a for every y E R. Show that I(z) == a. (b) Let I be a periodic meromorphic function such that the limits limy->±oo I(x + iy) a± (possibly infinite) exist and are independent of x E [0, 1], then I is a rational function of e 2rriz •
=
5. Verify that the following sums are correct: (a)
(5
) ( nZ) = (nZ +l '
where
(Z)=Z(Z-I)"'(Z-n+l) n
-
n!
•
n
E
No
418
6. Hannonic Analysis
(b) 6e 2rr ;n,
= ze2rr;n,.
n E N. cos(z - 1/2) (c)6sinz=- 2sin(1/2) . (d) u(z)(6v)(z
+ I) + v(z)(6u)(z) = v(z)(6u)(z + I) + u(z)(6v)(z).
6. Compute the first five Bernoulli polynomials B.(z). 7. Use right formal sums to find the following sums: 1
(b)6 z (z+I); What happens with the left formal solutions in these cases? 8. Does Theorem 6.3.7 hold when either
ao
or al are identically zero (but not both)?
*9. Let g be an entire function of infraexponential type, g(z) = ~n>O gnzn its Taylor series about z = O. Show that the series I(z) = ~n>O[gn/(n + 1»)Bn+;(z) converges to a sum of g and it is also of infraexponential type. 10. Study the solvability of the equation I(z done in the proof of Proposition 6.3.5.
+ 1) -
al(z)
= g(z), a E iC\(O, Ij. as
11. (a) Show that a solution of (z - 1)/(z + 1) = 2z(z - 2)/(z) is the function I(z) = 2'[r(z)/(z - 2»). Find all the solutions of this equation. (b) Let r(z) be a rational function. Find all the meromorphic solutions of the equation I(z + 1) = r(z)/(z) with the help of the r -function. 12. Let 1/I(z) = r'(z)j rez). Show that 1_ = (_l)m 1/I<m)(z), 6_ zm+1 m!
mEN.
Determine 6(z - a)-m. Verify that 610g z = Log r (z) (at least for z E )0,00[.) 13. Show that any function F E 5) admits the expansion ~.>o an logn S', absolutely and locally uniformly convergent in the Riemann surface of log f 14. For a
E
C solve the equation F(e 2rr ;S')
= aF(S),
FE 5).
15. For IE .1t'(IC), D = djdz, a E C, one defines eOv I(z) := ~n>o(an /n!)Dn I(z). (al Show that eOv I(z) = I(z + a). (b) Fix P E .1t'(1C) and define operators Land K in the space .1t'(iC) by the formulas LI := e- v 1- e P DI. KI := e-v(e v D/). For an entire function ((J define U((J to be the formal series U((J = ((J + K((J + K2rp + .... Show that formally
LUrp(z)
= ((J(z + I).
16. (a) Let I(z) = ~n>O ane 2rr ;nz be such that an 2: 0 for all n E N. Show that if I is not a trigonometric polynomial, then I is a function of infinite order. (b) Let I be an entire function, WE IC* is called an asymptotic period if the order of growth of the function z ~ I(z + W) - I(z) is strictly less than that of I. For I of infinite order, this difference is assumed to be of finite order ([WhiJ). Let I(z) = ~n=1 [e 2rr ;.!, j(n!)!]. Show that I is of infinite order but that for every W E Q*, f (z + w) - f (z) is a function of order 1. (That is, every nonzero rational is an asymptotic period.)
6.4. Differential Operators of Infinite Order
419
(c) Use Exercises 9 and 3 of this section to show that if f is a function of infraexponential type it cannot have any asymptotic period.
6.4. Differential Operators of Infinite Order Let T
E )If' (C)
be an analytic functional, if
~(T)({)
=L
an {",
n~O
then (T
* f)(z) = L
anf(n)(z).
n~O
In other words, one could write T = L(-l)n ano (n). n~O
We are interested in a particular class of operators of this kind, those for which ~(T) is an entire function of exponential type zero, that is, T is carried by {O}. As we shall see, these operators have a number of interesting properties, not shared by general convolution operators. 6.4.1. Definition. An operator of the fonn (T, f) = Lanf(n)(O), n~O
where the coefficients an satisfy lim sup n yTa;;T = 0 n-+OO
is called a differential operator of infinite order. The function
(!!...) (f)(z) = L anf(n)(z). dz n~O
Recall that we denote Exp({O}) the space of entire functions of exponential type zero. 6.4.2. Proposition. If n is an open set of C and ct>(djdz) an infinite order differential operator, then ct>
(:z ):~(n) ~ ~(n)
420
6. Harmonic Analysis
is a continuous operator. The same is true for this operator acting on the space Oa = Je({a)), a E C. Proof. The proposition is clearly correct once we know it holds for disks. On the other hand. Propositions 2.4.2 and 2.4.3 have already shown this for any 0 open convex set Q. If L is a linear continuous operator in 0 0 • it is automatically a continuous operator from Je(C) into Je(B(O. r» for some r > O. by the definition of the topology of 00. Therefore. it makes sense to say L commutes with translations 1:a • at least for small lal. In fact. Je(C) is dense in 0 0 so that L is completely detennined by its action on Je(C). then. if lal < rand f E Je(IC). the function 1:a (L(f)(z) = L(f)(z - a) is holomorphic for Izl < r - lal, that L commutes with 1:a means that L(1:a (f»(z)
= 1:a (L(f)(z).
Izl
< r
-101.
Let us consider
L: 0 0 --+ 00 linear continuous and such that L commutes with translations. We can define it also as an operator from Oa --+ Oa by
Now, f ~ L(f)(O) defines an analytic functional T. We want to show that F(T) = CI> is an entire function of exponential type zero and L(f) = T * f. The last identity is an immediate consequence of the definitions. On the other hand. the continuity of L: 0 0 --+ 0 0 implies that for every e > 0 there exists a constant Ce > 0 such that I(T,
f}1
= IL(f)(O)1 :::: Ce sup If(z)l. 1,I:::e
In particular, ICI>(OI
= I(T., eZ{}1 :::: C e exp(sup Iz~1) = Ceeel~l. 1.I:::e
This proves the following proposition:
6.4.3. Proposition. Any linear continuous operator L: 00 --+ 0 0 which commutes with translations is an infinite order differential operator. We now proceed to give a detailed proof of a result already mentioned in Chapter 2, §4.
6.4.4. Theorem. Let Q be a convex open set in C (resp. K convex compact) and CI>(dJdz) =I- 0 an infinite order differential operator. Then: (i) CI>
(:z)
(AP(O» = AP(O).
6.4. Differential Operators of Infinite Order
(ii) 4>
(:z)
421
(JIf(K» = JIf(K).
(iii) If M = Ker{(d/dz): JIf(S1) ~ JIf(S1)} and Mo = span{zkell!Z EM}, then Mo = M. (iv) Similar to (iii) for JIf(K). Proof. (i) The transpose of the map h E JIf(Q) ~ (d/dz)(h) E JIf(S1) is given by S 1-+ T * S from JIf'(S1) into JIf'(S1), where :F(T) = <1>. To prove the surjectivity of (d/dz) it is necessary and sufficient to show that its transpose is injective and has closed image. Since :F(T S) = :F(S), the injectivity is obvious and we only have to prove that the ideal Exp(S1) is
*
closed in Exp(S1). Due to the properties of the space Exp(S1), it is enough to prove that, for every f E Exp(Q), which is of the form f
= lim
n-->oo
fn E Exp(S1) (the limit in the topology of Exp(Q», then f =
=
From [BG, §4.5.5) (or the following lemmas), g is an entire function of exponential type. We need to show that the indicator function of g is of the form He, the supporting function of C, for some compact convex subset C of S1. This depends on the following lemmas: 6.4.5. Lemma. Let C be a convex compact subset of C, let S E JIf' (C) be an analytic functional carried by C, f = :F(S). If is an entire function of exponential type zero and g = fI is an entire function, then there is an analytic functional R carried by C such that g = :F(R). We shall prove first a lemma due to Martineau, which is a variation on a result of V. Avanissian (see [BG, §4.5.5.).) Let us recall the notation for the area average A(, z, r) = (l/Jrr2) i8(z.r)
~
(
1 +1..)2 -1..log 14>(0)1
+
(1 - C~ r) A
(s(1
+ A)lzl + DE)·
Proof of Lemma 6.4.6. Let R = (1 + A)lzl. Since is of exponential type zero, given s > 0, there exist DE E lR such that log 1<J>(u)1 :s slul + DE' Hence, the
422
6. Hannonic Analysis
subhannonic function U t-+
log leIl(u)1 - sR - De
is negative in lui::: R. On the other hand, B(z, >"Izl) ~ B(O, R);
therefore, R2 A (log leIll - 8R - De, 0, R) ::: (>"lzI)2 A (log leIll- sR - De, z, >"Izl).
By subhannonicity, log leIl(O)1 - sR - D. ::: A (log leIll- sR - De, 0, R). Thus.
In other words. >..2 A (log leIll. z, >"Izl) ~ (1 + >..)2 log Iell (0) I + (>..2 - (1 + >..)2)(sR + De),
o
which implies the stated inequality.
Proof of Lemma 6.4.5. Let 8 > 0 be given. We are going to show that Ig(z)1 ::: A. exp(HcCz)
+ 81zl)
for some A. > O. Up to translation we can assume that eIl(O) =I- O. The hypothesis implies that for every Sl > 0 there exists Ee , > D such that log If(w)1 ::: HcCw) + sllwl + E.,. Let M := max{HcCu): lui::: I}, z =I- 0, and>.. > 0, then max
WEB(z,Alzl)
HcCw)::: HcCz)
+ >..Mlzi.
Since log Ig I is subhannonic we have log Ig(z)1 ::: A (log Igl, z, >"Izl) = A (log If I. z. >"Izl) - A (log leIll. z. >"Izl). For 81 > D and S2 > 0, to be fixed later, there exist constants E., and Dez given by Lemma 6.4.6 such that: log Ig(z)1 ::: HcCz) + >..Mlzi
+ 81(1 + >")Izl + E e,
1+>..)2 log Iell (D) I + ((1+>..)2) - ( ->..->..- 1 (82(1 + >")Izl + D.z )· Let us choose A > 0, 81 > 0,
S2
> 0, such that
>"M < s/3, 81(l +),,) < 8/3.
6.4. Differential Operators of Infinite Order
423
(C: rA
1) (1 + A)82 < 8/3.
Let 1+A)2 log 1ct>(0)1 + ((1+A)2) Ne = - ( -A-A- 1
DB2
+ Eel'
then log Ig(z)1
~
Hc(z) + 81z1 + N e ·
In other words, there is R an analytic functional carried by C such that the function g = F(R). 0 From Lemma 6.4.5 we can now conclude that part (i) of the theorem is correct. (ii) Since (Jf(K»' is the space of analytic functionals carried by K the proof of part (i) works verbatim. (iii) Let JJ. E Jf'(Q) be orthogonal to Mo, this implies that J(JJ.) is divisible by ct>, J(JJ.) = ct> . h, h is an entire function. By Lemma 6.4.5 h E Exp(Q). Hence there is R E Jf'(Q) with J(R) = h, thus JJ. = T *R.
If
f
EM, then (JJ., f) = (T
* R, f) =
(R, T
therefore Mo = M. (iv) The same proof as (iii).
* f) =
0,
o
We shall now prove that any solution f E Jf(Q) of a homogeneous differential equation of infinite order has a Fourier-expansion convergent in Jf(Q) of the same kind as that found in the previous sections. It is clear that once we have proved this result for an arbitrary open convex set Q it also holds in the space Va. This Fourier representation goes back to the work of Valiron [Val], Gelfond [Ge], Dickson [Di2]. On the other hand, the novelty of our proof is that it obtains estimates for the coefficients of the expansion and it is essentially the same as Theorem 6.2.7 up to technical details. 6.4.7. Lemma. Let ct> i: 0 be an entire function of exponential type zero and let 0< 8 ~ ~. There exists R = Re > 0 such that for any z with Izl ~ R there are constants 0 < a, 8/8 < A < 8/4 so thatfor any t; such that Aizi - a ~ It; - zl ~ Aizi one has while exp(-28Izl)
~ a ~
1.
Proof. Let us recall a property of entire functions of exponential type zero (see [Bo, Corollary 3.7.3, p. 52]), given l3 > 0 and TJ > 0 there exists ro > 0 and a measurable set Ea ~ [0, +oo[ such that if r ~ ro m(Ea
n [0, rD
~ TJr
424
6. Harmonic Analysis
and if r ::: ro, r
~
£8, then
log 1<1>(01 ::: -8r,
I~I
= r.
We shall choose convenient 8 > 0, 1'} > 0, later. The first condition on 1'} is that 41'}(1 +.<:) <.<:, under this condition, if r > 21'0. then m(£8
n [r, I' + '<:1']) :::::
Therefore, there is a value s such that .<: r + -r < s < r 8 - -
e
gr.
.<:
+ -r 4
and 10gl
-8s
for I~I = s. For z with Izi = r. let Z' = (s/r)z. We are going to apply the Minimum Modulus Theorem ([BG. §4.S.14] and the previous Lemma 2.2.11) to the function
<1>(0 1/1 (0 = (Zl)
in the disk B
=
B(zl. (.<:/2)r). For this purpose we have to estimate
M =
max
11;-:'19"
11/I(~)1.
Since is of exponential type zero there is a constant D8 > log 1<1>(01 :::::81~I+D8
(~ E
°such that
C),
we conclude that 10gM ::::: 48r
+ D 8•
Let H = 1/4s. then the Minimum Modulus Theorem says that. outside a finite collection of disks Cj in B such that the sum of their diameters is less than or equal to 4H er the function 1/1 satisfies log 11/1(01) -
(2 +
log
;~ )
log M.
Let us remark that the disk B(z, (.<:/4)r) ~ B. Since the sum of the diameters of the Cj is 4Her < (.<:/8)r. there exist A, £/8 < A < .<:/4. such that aB(z, Alzl)
n Ci = VI
for all j. On this circle we have log I({)I ::: log 1<1> (zl)1 - (2 + log(60»)(48r
+ D 8)
::: -8r(13+410g60) - (3Iog60)D8 ::: -308r - 8Da.
425
6.4. Differential Operators of Infinite Order
Let 8 = e/60 and r, >
°
such that
e
"2rl = 8D& + log 2. Hence if r ?: max(r" 2ro) we have 1(nl ?: 2e- er whenever Iz - ~I = Ar. In order to obtain an estimate in the annulus, let us point out that the derivative <1>' is also of exponential type zero, thus there is a value D; such that
I'(~)I s
exp GRI
+ D~)
(~ E
C).
(I
therefore I'(~)I :::: exp((l + A)(e/4)r + (D; + 1» for U E B(O, + ),,)r Let w = z + te i9 with Ar - u S tSAr + u, then, for ~ = z + Arei9 1(w) - (~)I :::: u sup{I'(z
S u exp
+ re ill ) I:
).r - u S
t'
S ).r
+ 1).
+ u}
((I + A)~r + (D~ + 1)) S u exp (ir + D~ + 1) .
We want
u exp (ir
+ D~ + I)
S e-er.
If this condition is satisfied we have 1(w)1 ?: I(~)I- I(~) - (w)1 ?: e-er. On the other hand, for r ?: r2
= 2(D; + 1)/e >
0, we have that the choice
u .'- exp (-3er 2 - D'e -
1)
will satisfy u ?: exp( - 2r). Thus to conclude the proof it is enough to choose R ?: max(rl, 2ro, r2). 0
=
In order to define the space A * (V) for V V (<1», we first note that the number of zeros n (r) of <1>, counted with multiplicity, in the disk iJ (0, r) satisfies lim nCr) = 0. r
r-+oo
If we denote the zeros repeated according to multiplicity by Iztl ~ IZ21 :::: .. " either they form a finite sequence or
.
IZkl
k->oo
k
lim -
(Zn)n~I'
with
=+00.
Moreover, taking the constant Re from Lemma 6.4.7 bigger if necessary we can assume nCr) S er Assume has infinitely many zeros, otherwise is a polynomial.
6. Harmonic Analysis
426
Let us denote V = {Cab mk), k::: I}, latl ::: la21 ::: "', with distinct ak. Let en = 2- n and Rn = max(Re., 2R n_ l ) as given by Lemma 6.4.7. We denote by kl.l the first index k such that lakl ::: R I . Let Au > 0 associated by Lemma 6.4.7 to au = akl,1 and e = el. Denote Bl,I:= B(al,l,Al,Ilal.tI).
We let hi be the collection of indices such that ak E Bu. If there is any k > k l • 1 so that ak is in the annulus RI ::: Izi < R2 and it does not belong to Bl,I, we let k1,2 be the first such index and consider the corresponding disk with Ak'2 chosen with respect al.2 := akl.2 and e = el. We let J1.2 be the collection of indices of the ak E B1,2 \ B 1,1. We continue in this fashion until we exhaust all the zeros of <1> in this annulus. We obtain in this way disks BI,}. 1 ::: j ::: N], and corresponding disjoint index sets II,}, 1 ::: j ::: NI, with II.} the collection of indices of the ak belonging to the set BI,} \ (
U Bl,l). I::'OI::'O}-I
Observe that if ak does not belong to U} BI,} and R3 > lakl ::: R2, then the index k > kl,N 1 ' We let k2,1 the first such index (if there is any such zero). Denote with A2,1 chosen with respect to a2,1 := ak2,1 and e2 according to Lemma 6.4.7. One can construct in this way a double indexed sequence of zeros (al,}), 1 ::: j ::: NI, disjoint index sets II,} and closed disks B I ,}. Let 10 denote the collection of indices such that lakl < RI and also ak ¢ UI::'O}::'ON 1 B I ,}.
This construction exhausts all the zeros of <1> and defines the groupings that we need in the definition of III . III. Let us also point out that BI,} never intersects Iz I > RI +2. Let us now denote by VI,} the set of zeros aj of <1> counted according to their multiplicity when i E .ft.}, and let VI,} be the total number of points in Vt,}. Remark that we have VI.} :::
n(la/.}I(l + AI,}»
~ 2sdal,jl.
Hoping that there is no confusion with the previous notation in Section 6.3, we denote a E A (V) as
We recall that
6.4. Differential Operators of Infinite Order
427
where Ol,} is the diameter of BI,} Ol,j
= 2AI,jlal,jl.
Let (Kp)p':.l be an exhaustion of ()) by convex compact sets, and let Hp = H Kp ' then we denote
With the last definition in mind we define A*(V) := {a E A(V): 3p ~ 1: Illalllp < +oo}
considered with the natural inductive limit topology. For further reference, we let Mp = sUPlul::"l Hp(u) and let f/p > 0 be such that (z E
q,
that is, Kp+l contains the f/p-neighborhood of Kp.
6.4.8. Theorem. Let <1> be a nonzero entire function of exponential type zero, then Exp(Q) p <1> Exp(Q) ~A*(V) is an isomorphism. Moreover, every f a Fourier expansion fez)
=
E .K(Q)
L(L I,j
aEl,
such that <1>(d/dz)(f) = 0 has
Pa(Z)e(1Z) , J
where Pa is a polynomial of degree < ma = multiplicity of a as a zero of <1>. The series on the indices (I, J) converges absolutely and uniformly on compact subsets of Q. The frequencies for which Pa '" 0 and the corresponding coefficients of the polynomial Pa depend only on the function f, and the coefficients satisfy estimates of the same kind as in Theorem 6.2.7. Proof. Given g E Exp(Q) there exists p is a constant Ce ~ 0 such that
~
1 so that for every 0 < s < 1 there (z
Consider a then
= peg) = (al,}) l~i(al,})I8f,j .::::
Eq.
and apply Lemma 6.2.10 to the set B(ak"j' Ol,}),
i
max{lg(z)l: Iz - al,jl
:s Ot,j}'
For Iz - al,} I .:::: Ol,j we have Hp(z)
+ slzl
+ sla/,JI + (Mp + s)o/,J .:::: Hp(a/,J) + la/,J I(e + 2A./,J (Mp + 1». .:::: Hp(al,j)
428
6. Harmonic Analysis
On the other hand, we can find l(p) 2: 1 such that for I 2: l(p), we have 2A l ,j(Mp
Choose
S
+ 1) ::::
Sl 2(Mp
Yip
+ 1) :::: 2'
= Yip /2 then
IIlal.jllll,j = max , I~i (al,j)8i) :::: C, exp(Hp(al.j)
+ Yiplal,ji)
:::: C, exp(Hp+l(al,j». Since there are only finitely many indices l, j with I < I (p) there is a constant Bp > 0 such that
with
This implies the continuity of the restriction map p. To prove the surjectivity and the openness of p, we observe first that given a E A*(V) such that Illalllp < +00, if Pl,j denotes the Newton interpolation polynomial such that Pl,j(Pl,j) = al,j. then IP/.j(z)1 :::: vl,jeHp(U'J)lllalllp
for
Z E Bl,j'
We recall that Vl,j :::: 2SJ\al,jl,
It is also clear that for z E Bl,j' one has lal,jl ::::
Izi + Al,jlal,jl
::::
Izi + klal.jl,
Hence For the same reason, Therefore, for z in Bl,j. IPI,j(z)1 :::: IIlalllpeHp(z)+,/(3+Mp/2)lzl.
We need to construct disjoint open sets dinate to them, so that the Coo function
Ul,j
and cut-off functions
()I,j
subor-
o = L Ol,j PI,j I,j
coincides with PI,j on a neighborhood of any a E VI,j and there are good estimates for iJ() jiJz. to be able to apply the idea of the semilocal interpolation Theorem 2.6.4. The difficulty here is the need to estimate, very precisely, the
429
6.4. Differential Operators of Infinite Order
constants so that at the end of the process we obtain a function of Exp(Q). We introduce the disjoint open sets
BI.1' U1.2 := B1,2\BI.1' UI.I :=
and an open set Uo which is the union of small disjoint disks surrounding each with k E Jo and which is also disjoint from the U/. j . We need to estimate the size of a region inside the aU/. j where we have good lower estimates for
ak
exp(-2eliaul) :::: al,1 and 1ct>(~)1
::::
e-Etlaul,
whenever Aulaul - au :::: I~ - al.Ii :::: Au/aul· For U2 , I we have the subregion union of the two open subsets: (i) AI.liau/ < /~ - al.Ii < Aulal,) 1+ al,)
(ii) Al.2ial,2/ - au <
/~
-
ad
(~ E U2.)). (~ E
U2,)).
The construction of these regions for the general region U/,j is similar. We shall call these regions collars. We need to determine what estimates we have on these collars and their size. For this purpose, we recall that if two disks B/,j and Bk,i, with I :::: k, intersect, the indices k and I differ at most by 1. Therefore, at a point, and we have /a/,j / :::: 21~ I + a/,j < 4Iak,; 1+ 1
and Hence, for such a point Ict>(~)/
2:
e-E/1a/,jl
2:
e- E /(4 Ia k ,M\)Ict>(.;)/
2: e-SB.lak,de-I/2.
6. Hannonic Analysis
430
For the same reason the width of that portion of the collar is the inequality
al.j
which satisfies
Therefore we can find functions B/,j E V(VI.j) which are identically 1 except on the collar, 0 ~ fll.j ~ 1, and 1
a:~j (Z)I ~ CeI6etla',JI
for some constant C independent of I and j. Furthermore, the Coo function of compact support (l/ef>)(afll,j/(Jz) satisfies 1_ ael,j ef>(z) az
_ 1
(Z)I -< eI/2Ce24etlal,jl <- el/2Ce48etlzl.
Since fI = ~/.j flU PI,j we are searching for a function \II such that g =
e + \lief>
belongs to Exp(Q). For this reason, we choose \II as a solution of the equation 1 ae
(J\II az
(*)
= -q;- az = -
" 1 ael,j L..J I• j ' PI,jq;- az .
In VI,j, one has the estimate 1
~Z) :~ (z) 1 ~ e I / 2er IlalllpeHp(z)+e,(51+Mp/2)lzl.
For a convenient choice for
I:::: i(p).
Thus, up to a multiplicative constant Dp , which depends only on p, we have everywhere 1
ef>~Z) :~ (Z)I ~ DplllalllpeHp(z)+'/p/2IZI.
From Theorem 2.6.3, with the subharrnonic weight qJ(z) = 2Hp(z)
+ I1plzl + 2 log (1 + IzI 2 ),
we use the fact that
[I ~Z) :~
2
(Z)1 e-q;(z)dm
~ D~lllalll; [
(1
~ D~lllalll~ to choose \II so that it solves (*) and
J
11/I(z)12 e-q;(z)dm < lD' Illall1 2 . + Iz12)2 - 2 P P
(1
C
:~j2)2
6.4. Differential Operators of Infinite Order
431
Therefore, the holomorphic entire function g defined above, satisfies peg) = a and Ig(z)1 2 ::: 2(161(z)1 2 + It(z)1 2 Iq,(z)1 2 ) ::: 2(Dp II lall l;e2Hp(z)+~plzl
+ It (z) 12C;eelzl).
Now, we introduce CPo(z) = cp(z)
and we obtain
+ 210g(l + Iz12) + slzl
J
Ig(z)1 2e-'Po(z)dm ::: 2(D;
+ C;D~)IIlalll;.
IC
Let epl(Z)
= max lepo(u + z)1 ::: epo(z) + E p , luI:::: I
by the mean value property, we obtain Ig(z)1 2
:::
Bp.ee'Po(Z)IIlalll;
for some convenient constant Bp.e which depends on p and
E.
Hence,
Ig(z)1 ::: B~:;IIlalilpeHp(z)+[(e+~p)/2Jlzl+2Iog(l+IZI2). We can choose
E
= '1p/2 and find a new constant Bp > 0 such that Ig(z)1 ::: BpillalilpeHp(z)+~plzl ::: BpillalilpeHp+,(z).
This concludes the proof of the isomorphism between Exp(O)/q, Exp(O) and A*(V).
The rest of the proof of Theorem 6.2.7 and subsequent considerations, including estimates of the coefficients of the Fourier expansion hold verbatim. The only thing that requires a minor remark is the fact that if! has two Fourier-expansions converging in $(0) the coefficients are the same because the operators Tk used in the proof of Proposition 6.2.8 are also infinite order differential operators in this case, since F(Tk)(Z) = q,(z)/[(z - ad m,] are entire functions of exponential type zero. 0 As in the case of mean-periodic functions, if the variety V of 41 is an interpolation variety for the space Exp({O}) of entire functions of exponential type zero, then the Fourier representation of an arbitrary solution of equation q,(dldz)! = 0 converges in )f(0) without groupings. One can prove that the condition for V = (ab mk) being an interpolation variety in the space Exp({O}) is that for every s > 0 there is a constant Ae > 0 such that (k
~
1).
6. Harmonic Analysis
432
(See [BLVJ.) The following theorem ofP6lya-Levinson [Levs, Theorem XXXI, p. 92] furnishes a very simple geometrical condition on the ak. when mk = 1 for all k, for V = (akk~1 to be an interpolation variety in Exp({O}).
6.4.9. Theorem. Let that:
(ak)'\~l
be a sequence of nonzero complex numbers such
lim ~ = o. k--+oo ak (ii) There is a constant y > 0 such that for any k, j E N* (i)
lak - ajl ?: ylk - jl. (iii) There exists e E [0, 21l' [ such that limk--+oo adlak I = e iO . Then, the function
II (1 _z:) . k~1
ak
is of exponential type zero and satisfies, for every s > 0,
(a) (b)
\
\_1_\
O(e s1zl ),
whenever
= O(eSlakl)
'
Iz - akl
?:~,
Vk?: 1;
k ?: 1.
We remark that condition (iii) is missing in the statement of this theorem in [Levs], [Bo, p. 146], but it is used in the proofs given there. Letting a2k+1 = -a2k. one can easily see that some condition on arg ak seems necessary to ensure (b) for the function
lakl = O(ee1aki )
(k?: 1),
then there is a way to choose integers nk ?: 0 so that
g(z) :=
L
~
k?: 1,
g(-ak) = O. Instead of the proof of Theorem 6.4.9 we shall give an important generalization of this theorem, due to Vidras [ViI], [Vi2]. Namely, that the condition (iii) is not necessary for the conclusion that the sequence {ad is an interpolation variety.
433
6.4. Differential Operators of Infinite Order
=
6.4.10. Theorem. Let Z {ak} k': \ be a sequence of complex numbers satisfying the following two properties: (i) adk -+ 00 as k -+ 00. (ii) There is c > 0 so that for any n, k E N*: Ian - akl =.:: cln - kl. Then, there exists an entire function F of exponential type zero, with simple zeros {Zm };;'=l which include all the ab and such that "Is > 0 one has
(a) l/W(reioz)1
=
O(eer) whenever Ire iO
-
zml =.:: 1/8c for all mEN as
r -+ 00.
(b) l/IF'(zm)1 = O(eBlzml) as m -+ 00.
In particular, not only the sequence {Zk}bl is interpolating, but this property also holds for the larger sequence {zm};;'=l of zeros of F.
The proof of Theorem 6.4.10 consists of a sequence of lemmas, some of which already appeared in [Levs]. The main idea to get around the problem that we have no control on the arguments arg ak is the following grouping critierion:
6.4.11. Definition. Let Z = {an}~l be a sequence of complex numbers satisfying the properties (i) and (ii) of the statement in Theorem 6.4.10. Then we say that am is paired to an, n #- m, if and only if (iii)
We say that am has the property P if there exists n so that {am, an} fonns a pair (i.e., it satisfies condition (iii». An immediate consequence of the definition is that any am is paired to at most one element of the sequence Z. Indeed, assume that am is paired to ak and a p . The property (ii) implies that
c
~
clk - pi
~
lak - apl
~ lak +aml
< c /2
+ lam +apl
+ c /2 =
c,
which is a contradiction. Thus we are able to separate the tenns of the sequence Z into two disjoint groups Z\ = {an: an not satisfying the property P}, Z2 = {am: am has the property Pl. The following lemma is a simple generalization of Levinson's Theorem 6.4.9 given in [LevsJ. The main differences are that we do not assume that ReAn > 0 or that 1m An/An -+ 00 as n -+ 00.
6.4.12. Lemma. Let A \ = {An }~I be a sequence of nonzero complex numbers satisfying the properties (i) and (ii) of Theorem 6.4.9. Assume further that no point
434
6. Hannonic Analysis
of A I has the property 'P and that there is a constant () satisfying 0 ::::: () < rr 12, so that for every n E N either
I arg An I ::::: ()
or
I arg An - rr I ::::: ().
Then there exists an even entire function Fl of exponential type zero, vanishing ±An and satisfying "Ie > 0: only at the points z
=
(a) I/IFI(reia)1 = O(e") whenever Ire ia (b) I/IF{(An)1 = O(eEIAnl) as n ~ 00.
± A.nl
::: c/8 as r ~ 00.
Proof Let us define FI (z) = 0':'1 [1 - (z2/A;)]. The hypothesis (i) is equivalent to the fact that the counting function nA, (r) = o(r), thus FI E Exp({O)) [Lev]. Since no point of A I has the property 'P, every zero of FI is simple. In order to show that FI has the other desired properties, we first note that since FI is an even function it is enough to prove (a) and (b) for Re z ::: 0 and, given o < e < only for r = Izl sufficiently large. Following Levinson, assume that z ¢ A I, and divide the terms of the sequence into three disjoint sets as follows:
!,
A = {An E AI: IAnl ::::: (1 - e)lzl),
B = {An E AI: (1 - o5')lzl < IAnl < (1 - o5')lzlL C
= {An E AI:
(I + o5')lzl ::::: IAnl}.
Let us denote FA(Z) := OAffEA[l - (z2/A~)J and define similarly FB, Fe. Then FI = FAFBFe ,
and we can divide the proof in several steps. Step 1. Assume An E B, then we have (1 - e)lzl < IAnl < (l + e)lzl, and therefore Izl (l+o5')lzl 1+05' < < - - <3. IAn I IAn I - I - 05' We are trying to estimate IFB(z)1 from below, and since every An appears squared, we may assume that Re An > O. Hence Iz + An I ::: Re(z + An) ::: Re An = IAn I cos () > O. So that ----~~=
11 -
z2/A~1
IA~I (1 +o5')lzl < ~------~ Iz - Anllz + Ani - cos(}lz - Ani
Let N be an index that minimizes Iz - An I. Then, as
we have
2 IAN - Ani
---<--The term AN could be in any of the three sets A, B, or C; to simplify the notation we will assume that AN E B, which is the worst case, but the reasoning
6.4. Differential Operators of Infinite Order
435
that follows is not affected by this assumption. Denote by L the number of the tenns in 8 distinct from AN. Let 8(z) be the finite product B~)=
2
n (1 -A~z) )...EB.nf'N
.
Thus, 18(z)l:::::
:::::
(1 + 8)lzl ::::: ((1 + 8)l z l) II 1 II )...EB,ni-N cos () Iz - An I cos () >".EB,mIN Iz - An I L
Z ( (l+8)I I)L L cos() 2
>..nJLN IAN -
Ani
On the other hand, since IAN - Ani::: clN - nl, we have ----<--IAN - Ani - clN - nl
Therefore,
II
IB(Z)I«(1+8)lzl)L(~)L -
cos()
C
1
)...EB,ni-NIN-nl
Observe that for every n there are at most two tenns equal to IN - nl. Hence, the last product has at least L] distinct tenns, so that
[!
II
IN - nl ::: ([L/2]!)2.
)...EB,ni-N
And the estimate of 8 is reduced to 18(z)l:::::
(0
+ 8)IZI)L (2/c)L cos ()
([~]!) -2 ::::: 2
(_4_)L IzlLe3Le-LlogL c cos ()
On the other hand, for Izllarge enough, L ::::: nA «(1 + 8)lz/) ::::: 281zl, thus, using the inequality x 10g(1/x) < valid for 0 < x < I, one sees that j
XI/2
L Izl -log - < Izl L so that
(L)I/2 < .../2 Izl
-
v
Iee
,
436
6. Harmonic Analysis
for a convenient constant J.l > O. The last point of this step of the proof is to bound the term involving AN. From (*) we obtain > 1( 1-~)1 A1 -
cos8lz-ANI 2~1
'
so that, altogether,
for a convenient constant v > O. Step 2. This time we are going to estimate !FA (z)1 from below. Similarly to the previous step, let a be the value of the index n that minimizes IAn - zl among An EA. We see, as earlier, that when An E A we have
1
----.,,.........~
11-z2/A~1
IAnf (1 + 8)lzl = IZ-Anllz+Anl-lz-Anlcos8 < ----2(1 - 8)lzl 21z1 < ---'--'--- IAa - Ani cose - cia - nl cos 8 <
Let us denote by A(z), as in the previous step, the product of the terms 1/[1 - (z2/A~)] for An EA. n =1= a, and by M denote the number of terms of this product. It is easy to see that we have M ::: f:lzl and, as earlier,
IA(z)1 <
2(1- 8)lzl < (_2_)M IzlM 1 - AnII;
2 ::: ( - cos8
)M IzlM
1
([M/2]!)
2:::-r Me 3M exp[Mlnlzl-MlnM],
for some constant -r > O. Again we conclude that for a convenient constant (1 > 0 and all Izi » 0 we have
On the other hand, we have that IAal ::: (1- .c:)lzl and, hence, we also have Iz - Aal ::: Izl -IAal. Thus, 1 II - z2/A~1
IAal2 IAal 1 IAal < < - - --'--'-Aallz + Aal - cos81z - Aal - cose Izl-IAal 1 (l - .c:)lzl 1 (1 - .c:)lzl 1 (1 - e) <-=------- cose Izl- (1 - .c:)lzl cose Bizi cose.c:
= Iz -
Combining this inequality and the bound for A, we obtain for some constant, TJ > 0, the lower estimate
valid for Izi
» o.
437
6.4. Differential Operators of Infinite Order
Step 3. Finally we have to consider the function Fe. When An E C one has < I/O + s) < 1, so that
Izl/IAnl
/1-£t/=I-£t. IAnl2 IAnl 2 Thus, if r = Izl, rn = IAnl, we have
IIe (1- rnr~) :S II /1- Anz~ / = IFC(z)l. C
It is immediate that '" 1 L 2" :S e rn
1
00
(l+e)r
dnA, (u) 2:S
1
00
(I+c)r
U
2n/'q (u) 3 du, U
where the last inequality was obtained after integration by parts. Note that
r
21
00
(1+e)r
2nA, (u) nA, (u) du
Since nA, (u)/u ~ 0, we obtain that r2 2:e(1/r;) < er for r large enough. In order to profit from the last observation we need a small side remark. Let f3 = fJ(s) > 1 be defined by the equation (1 - s)fJ + logs = O. Note that sfJ(e) ~ 0 when s ~ O. Then, for 0 :S x :s 1 - s, the inequality 1/(1 - x) :s sf3 x holds. Therefore, since (r/rn)2 :S r/rn :S 1 - s, we have
I]
(1 -
~2/r;) :S exp
(fJ r2
(~ r~) )
This inequality implies that IFc(z)1 :::: e-2/lclzl.
This concludes the third step of the proof. It is clear that, after convenient renormalization of the constants, we have shown that given s > 0 there are constants rc > and y > 0 such that for Izl :::: re we have ylz - ANle- elzl :S !FI(z)1 :S eelzl.
°
Since F{(AN) = limz->AN F1(z)/(z - AN), the two properties (a) and (b) follow.
o
This concludes the proof of Lemma 6.4.12.
6.4.13. Lemma. Let A2 = {An }~I be a sequence of complex numbers satisfying the properties (i) and (ii) of Theorem 6.4.9. If every term An has the property P then limr->oo 2: IAn I9(l/An ) exists. Proof. By the Cauchy criterion we have to show that for 0 <
rl
< r2
438
6. Harmonic Analysis
Let us now observe that, because the counting function nA,(r) > 0 if r > 0 is sufficiently large then
= oCr),
given
8
~
~ r::oIA n l9+cl2
~ < nA,(r) < A r -
8.
n
This implies that adding the other element of the pair, if it is not already included in the sum (**), does not change the nature of the problem. On the other hand, if Am, An is a pair we have 1
1 1I Am +An
=
lAm + An I IAmAnl
c
1
~2IAmAnl'
Moreover, IAmAnl ~ IAmI2. Hence, there is a constant C > 0 such that
Since the series L:I(l/IAn I2 ) is convergent, the lemma is completely proved.
o
The main point of the next lemma is that we do not make any assumption on the arguments of the sequence.
=
11.2 {An l:'1 be a sequence of nonzero complex numbers satisfying properties (i) and (ii) of Theorem 6.4.9 and also assume that every element has the property P. Then there exists an entire function F2 E Exp({O}) vanishing exactly at the points of 11.2 and so that for every 8 > 0 the following estimates hold:
6.4.14. Lemma. Let
(a)
(b)
1.
iF2(re,B) I
= O(eer),
1 ----,.__ = iF2(A n ) I
whenever
O(esIAnl)
as n -+
Ire iB
-
An I ~
!c
as r -+
00.
00.
Proof. If the genus of the sequence were zero, we could use a simpler construction of F2 , but we prefer to give a unified formula, irrespective of the genus. Using Lemma 6.4.13 define w:= limr--> 00 LIAn I9(l/An ). Let us consider the function F2(Z) =
e- wz
IT (1 - ~) n=1
An
II
e z/An := lim e- wz r-->oo IAn l::or
(1 - ~) e An
zlAn ,
where the infinite product is defined by the limit of the finite products. The reason is that the IAn I are not necessarily increasing. Note that in the genus zero case, this product is not the Weierstrass canonical product [Lev], while this is true when the genus is 1. Nevertheless, the usual estimates show that F2 E Exp({O}) (see [Lev, Theorem 15, p. 26].) Thus, we need to concentrate in
439
6.4. Differential Operators of Infinite Order
proving lower bounds for Fz, in a way altogether similar to Lemma 6.4.12, so that we shall be able to skip the details when they are the same. Given 0 < s < ~, we fix z ¢ Az, with Izl» 1. Divide the terms of the sequence into three disjoint sets as before A = {An E A2:
B
IAnl :5
= {An E A2: (1 -
(1 -
s)lzl),
s)lzl < IAnl < (1 + s)lzil,
C = {An E Az: (1 +s)lzl :5I AnlJ.
Step 1. This is practically identical to Step 1 in the proof of Lemma 6.4.12. We denote by N the index that minimizes Iz - An I, and assume AN E B without loss of generality. We let L be the number of terms in B district from AN. We obtain
II AnEB.n#N
1
< (1
11-z/AIn
II
+ 8)LlzIL2L
AnEB.n#N
As was done in Lemma 6.4.12, we can now find a constant Izl » 1 we have
Iz -
1
IA N -AI' n K
2: 1 such that for
II 11 - : I < eelz l .
ANle-Kv'Elzl <
n
An EB
Step 2. It is clear that the same proof used in Step 2 of Lemma 6.4.12 yields the following estimate:
II 11 - : An EA
for some convenient value
K'
12: e-K'v'Elzl, n
> O.
Step 3. Let us now assume that Izi is sufficiently large so that 81z1 > e/2. It follows that if an element of C is such that the other member of the pair is not in C, then it must lie in B. Let us consider here the subset D of elements of C such that the other member of the pair is not in C. We shall denote by E the remaining portion of C. It is clear thaHn E D implies that IAn I :5 (l + 8) Iz I + e /2, hence, D is finite, and if we denote by P the number of its elements, then
for Izl » 1. Let As denote an element of D that minimizes Iz - An I among An (1 + 8)lzl :5 IAsl :5 (1 + 2B)lzl, so that I
IAsl IAsl < lAs - zi - IA.I-izi -
---- = --- <
II - z/Asl
(1
E
D. Then
(l+2Blzl 1+28 :5 - - . + 8)lzl-lzl s
On the other hand, TIAnED.n#s 11 - z/Anl can be estimated from below as in Step 2 of Lemma 6.4.12. As a consequence, we obtain that there is p > 0 such
440
6. Harmonic Analysis
that for large Izl
Step 4. In this part, we are going to estimate the product of all exponential factors with frequencies Am ED. As a consequence of the estimate of P in the previous step, we have
P
1 2:--< Am ED
lAm I - (1
+ 8)lzl
<8
-
,
and thus,
IAmIIED e
Z/Am
I :::: e-elzl.
Step 5. In this part we take care of the exponentials of the terms which are "near" the origin, that is, with frequencies An E A U B. From the definition w we have that for Iz I » 1
I L : -WI::::8. A,EAUB
n
Therefore,
le
II
wz
e- ZIA , I : : e-elzl.
A,EAUB
Step 6. Finally, we are left with the task of estimating the infinite product over the set E. In this case,
Izl 1 -<--<1. IAnl - 1 + 8 Hence, w.e can apply [Lev, Lemma 3, p. 11] to obtain log
III (1 -~) e-ZIA,I :::: -12elz1 roo n£;t) dt. 1 An JIZ I 2
£
Since, for 1:::: Izl
»
8Izl-2 /2. Thus,
1, we have n£(t) :::: 81. we can estimate the integral by
II} (1- :J
eZIA'I:::: e-6eelzl.
The collection of estimates we have obtained concludes the proof of Lemma 6.4.14. 0
6.4.15. Proposition. Let A = {An }~l be a sequence of nonzero complex numbers with the following properties: (1) There is a constant 0::::
1argAn
-
rrl :::: e.
e<
rr/2 such thatJor every n EN, I argAnl ::::
e or
6.4. Differential Operators of Infinite Order
441
(2) Anln ~ 00 as n ~ 00. (3) There is a constant c > 0 so that for any n, k E N*, IAn - Akl ::: cln - kl. Then, there is an entire function F E Exp( {O}) vanishing at all the An, whose zero set is denoted by {~m} ~= I' and such that Ve > 0 (i) I/IF(z)1 = O(e*l) whenever Iz - ~nl (ii) I/IF/(~m)1 O(e&l~rnl) as m ~ 00.
=
::::
~c, as
Izl
~
00.
Proof. We introduce the pairing of Definition 6.4.11, then the terms of the sequence
{An}~1
are divided into two sets: A I = {An: An does not have the property P},
A2 = {An: A" has the property P}. We apply now Lemmas 6.4.12 and 6.4.14 to obtain functions FI and F2, corresponding to the two sequences. The first one vanishes also at all the points of -AI {-An IAn E Ad. Let us remark that -AI is disjoint from A, so that the function F := FI F2 has simple zeros at the points of the sequence A* := A U (-AI)' Moreover, it is easy to verify that, with the possible exceptions of points ~m E A * with I~m I < c14, the disks centered at the ~m of radius c/8 are disjoint. This is enough to conclude that F satisfies the required properties.
=
o We are now ready to prove Vidras' theorem.
Proof of Theorem 6.4.10. Let us assume first that no
ak
is zero. Divide the
sequence Z into two subsequences as follows:
ZI
:= {an: I arg a,,1
Z2 := {an:
~
<
::: ~ } U {an: I arg an I ::: rr
I arg an I <
rr -
~}
U {an:
-
~}
~ + rr
<
:
I arg an I <
2rr -
~}
.
Consider now two new subsequences of Z, Al consists of all points of ZI together with those points of Z2 that can be paired to a point in Z I. A2 contains the remaining points of A. We keep the indices of the points in these subsequences the same as in the original sequence in order to preserve property (ii). Up to a rotation, Proposition 6.4.15 can be applied to the new sequences and we obtain thus two functions FI and F2 whose product F = FI F2 has the required properties. If z = 0 is a point of Z we just multiply the previous function F ~z.
0
Further generalizations of the theorem can be found in [BLY] (see also [OR] for related results). For the applications of infinite order differential operators to the question of overconvergence of Dirichlet series we need first a generalization of Theorem 6.4.8 to convolution operators.
442
6. Hannonic Analysis
Let T be an analytic functional whose minimal compact convex carrier is K, then, for any open convex set Q we have that
is a continuous map. This map is surjective for arbitrary Q if and only if the entire function defined by ~(T) = is of completely regular growth. We refer the reader to Section 2.4 for the details.
6.4.16. Definition. Let Z
= Z(
multiplicities) of and d(z) decreasing if:
(i) There exists an integer m ::: 1 such that for every e > 0 there is a constant C e > 0 for which 1(z)1 ::: Ce(d(z»m exp(Hdz) - elzl). (ii) For every £ > 0 there exist At, Be > 0 such that the diameters connected components Se.n of the set
St:= (z
E
on.t
of the
C: d(z) < Atexp(-elzl)}
satisfy
Clearly any exponential-polynomial is regularly decreasing. It is easy to see that the condition (i) implies that
is surjective for every Q and therefore, any regularly decreasing function has completely regular growth. We leave it as an exercise to the reader that in fact (i) implies that if h E Exp(Q + K) and h/ = g is entire, then g E Exp(Q).
6.4.17. Proposition. Let T be an analytic functional with minimal compact convex carrier K, = ~(T) regularly decreasing, and Q a convex open set. For any / E Jf(Q + K) such that T * / = 0 there is a Fourier representation o/the/orm (z E Q
+ K),
where V(
443
6.4. Differential Operators of Infinite Order
following: we choose 8 n = 2- n and let 0 < An We add to the definition of Rn the inequalities Rn ~
8;1
=
A E, ::::: 1 and Bn = BEn> O.
max(Bn - log An).
For ak such that Rn < lak I < Rn+ 1 we considerthe component Su, that contains and this way we can construct sets Un.} and corresponding an.) as in the proof of Theorem 6.4.8. Since it is hard to obtain an estimate of the number v n .) of zeros of cI> in Un.) better than const. Ian.) I, we are compelled to modify slightly the norms 111·lIln.):
ak
Illan,jlll".) = max 16i(an.j)(40n.dl, I
where on,) is the diameter of Un,j, which can be estimated in terms of the quantities Bn, and i runs over the sequence i = 0, ...• Vn,j - 1. Let (Kp)p?1 be an exhaustion by compact convex sets of Q and define lila Illp
= sup{liial.) 1IIl,j exp( -
Using the fact that the convex sets Kp HKp+K
we can prove that
+K
H Kp+K (aU))}'
exhaust Q
+K
and
= HKp + Hko
Exp(Q + K) p cI> Exp(Q) ----+A.(V)
is a topological isomorphism. The rest of the proof is the same as in Theorem 6.4.8. 0 Henceforth, we will assume that Q and K are such that
r = r9 =
{z: I Argzl <
(J
<
~}
C Q
+K
C {z: Rez >
OJ.
6.4.18. Lemma. Assume further that for some 0 < 0 < B there are at most finitely many zeros of cI> in the set
r; := {z E C*: B + ~ -
I) :::::
Argz : : : 3; - + o}. (J
Then every solution f in ;f(Q + K) of the equation T function.
*f
= 0 is an entire
Proof. We can use the same groupings and the same definition of the norms III· 1111.j to define a norm in A(V) by Nda)
= sup{lllal,jllll,j exp(-Hdal,j»},
where L is an arbitrary compact convex subset of C. The space A •• (V)
= (a E A(V): 3L.
Nda) < +oo}
is isomorphic to Exp(C)/cI> Exp(C) by Theorem 6.2.7. Therefore, to see that any f E ;f(Q + K) for which T * f = 0, is entire, it is enough to see that the two spaces A.(V) (given by the proof of Theorem 6.4.11)
444
6. Hannonic Analysis
Figure 6.1
and A**(V) are isomorphic. In fact, this means the dual spaces coincide, and so the coefficients of the Fourier expansion of f in Q + K are also the coefficients of the Fourier expansion of some entire function f, T * f = O. It is clear, that f is the analytic continuation of f to the whole plane. In order to prove this isomorphism all we need to show is that, for every compact convex set L in C, we can find L I, a compact convex subset of Q + K such that for all a E Z(<1» n (r;t. The reader can convince himself that it is enough to take a convenient closed 0 vertical segment in r e (sufficiently far to the right). (See Figure 6.1.) If we do not have the extra assumption of Lemma 6.4.18 we can still conclude the following:
ro
= {z E C*: rr/2+e < Argz < 3rr/2-e}. Assume 6.4.19. Lemma. Let Z(<1» ~ ro u (r;)C for some 0 < 8 < e. Let f E Jf"(Q + K), T * f = 0, be such that for some 8 > 0 the polynomial coefficients POI of f in the Fourier expansion vanish for all a in f o. Then f is an entire function. Proof. This time we only have that every entire solution of the convolution equation is a solution in Q + K but the converse does not necessarily hold. For the f of the statement what we have to show is that the sequence b E A. (V)' that corresponds to f belongs to A •• (V). For al.j E we have bl,j = O. For
fo'
445
6.4. Differential Operators of Infinite Order
the other indices (l, j) the construction of the compact L 1 in the previous lemma still works, that is, Hdal,j):S HL,(al,j)
r;.
if al,j ¢ Remarking that the groupings can be made in such a way that no points in ro and (r;y are in the same group, the last inequality shows that b is a linear continuous linear functional on A •• (V). 0
6.4.20. Lemma. Let
P(z) = II (1 _z:)ml k:::l
ak
be afunction of exponential type zero such that the variety V (P) is an interpolation variety for the class Exp({O)). Assume further that there is an angle rp such that lim ~ =e irp , k.... oo lak I
Let S = {z E C': a < arg z < t3}, t3 - a < 7r, and e > O. Let f be a holomorphic function in So = S U B(O, e) satisfying the equation P
(:z) f =
O.
Then f is holomorphic in the region S obtained by translating So in the direction perpendicular to e- irp Srp = U(So + i~e-irp). /ielR
In particular, if neither ±ie-irp E S, then f is holomorphic in a half-plane containing So.
SUB(O,E)
Figure 6.2
S
446
6. Hannonic Analysis
Proof. Since P(d/dz)(f) = 0 and V(P) is an interpolation variety we have that
+ Lqk(z)e-a,Z,
f(z) = Lqk(z)ea,Z
where both series are convergent in S. Let 1/I(d/dz) be another infinite order differential operator then
is also holomorphic in S,. Now, for any cx L:O:oJ:od Cj zj we have
'" U,) while, if r
E
(q (, ),"')
~
(,E,
c]
E
C and any given polynomial q(z) =
J.;] (0 '" ,»
(a ),]-,) c"',
C, then
q(H r leo',',)
~ (,E:] ,~] (0 r' ,]-''"') ,"'.
Therefore, if
(O:::s k we have q(z
Fix
~ E
+ r)ea(z+r)
(q(z)e az ).
lR and define the sequence ak.' =
(ie-irpn' I!
.
exp(jcxk~e-lrp)
Since cxdlcxkl -+ ±eirp when k find D6 such that
so that
(:z)
= 1/1
:::sd)
-+
L
±oo, respectively, then, for any 8> 0 we can
ICXk.ll:::s
el~le6Ia'I+D•.
O:ol<m, As V (P) is an interpolation variety for Exp({O}), there is an entire function
1/1 of exponential type zero such that
1/F (I) (CXk) I!
= ak./·
447
6.4. Differential Operators of Infinite Order
Let U~ =
S:
{Z E
Z
+ i~e-itp
which is a nonempty open set. Then, for Z E
U~
E
S},
we have
hence fez + i~e-itp) has an analytic continuation from U~ to the whole set S8' Letting ~ vary in lR we obtain the lemma. 0 6.4.21. Theorem. Let T satisfy the hypotheses of Proposition 6.4.17 and Lemma 6.4.19, ro c Q + K C {z E C: Re z > OJ. Assume further that the zeros of
ro
k-+oo
ro
laj,kl
and
. laj,kl I1m -- = k
00.
k--'>oo
Moreover, assume that the functions
are entire functions of exponential type zero whose varieties V (Pj ) are interpolation varieties for the space Exp({O}). Under these conditions, any function f E .}f'(Q + K) that satisfies the equation T * f = 0 and has an analytic continuation to a disk B(O, e) for some e > 0, has also an analytic continuation to the open set
where S = ro. In particular, f has an analytic continuation to the cone ro - ie. Proof. We can write f = f+
+ f-, f+ =
L
qa(z)e aZ ,
ae(r;)C
f-
= L qa(z)e'l/Z, aero
where both series are convergent in Q + K. From Lemma 6.4.19 we know that f + is an entire function and therefore, the function f _ has an analytic continuation to B(O, e).
448
6. Hannonic Analysis
The function
f- satisfies the infinite order differential equation PI
(:z) (:z) ... P2
Pr
(:J
f - = O.
Let
which is holomorphic in s,=reUB(O,B)
and satisfies the equation PI
(:z)
gl
=0.
Therefore, by Lemma 6.4.20, gl is holomorphic at least in the half-plane S"'I' Due to the surjectivity of an infinite order differential operator, there is a function hIE .1!f(S"'I) such that
Then
P2 The function
f- -
(:z) ... (:z) Pr
(f- - hI) =
o.
hI is still holomorphic in So. Let now
Thus,
so there is h2 holomorphic in
S'P2
satisfying
In other words,
Continuing in this fashion we get functions (h j) I sj 9-1 holomorphic in that
S"" such
6.4. Ditferential Operators of Infinite Order
449
Applying Lemma 6.4.20 once more we obtain that
hi := f- - hi - h2 - ... - h r is holomorphic in
S"".
Hence,
f-
= hi + ... + hr
I
is holomorphic in
o Theorem 6.4.15 appears for the first time in [BSt 1]. The classical Fabry Gap Theorem (see [Levs], [Die)) is a particular case of Theorem 6.4.15. It states that if 0 < AI < A2 < ... is a sequence which satisfies the conditions: (i) limn_oo(An/n) = 00. (ii) For some constant C > 0 one has An+1 - An ~ C, then, every Dirichlet series Anz , f(z)
= Lanen~l
which has abscissa of convergence point on the line Rez = {Ya.
{Ya,
-00
<
{Ya
In fact, one recognizes that f is a solution in Re z > differential equation P(d/dz)f = 0 with
< +00, has no regular
(Ya
of the infinite order
By Theorem 6.4.9, V(P) is an interpolation variety, hence if f has a regular point zo, Rezo = {Ya, f will be holomorphic in Rez > {Ya - B for some 8 > O. It will satisfy the same infinite order differential equation in this bigger domain. By the representation theorem and uniqueness of the Fourier representation, the Dirichlet series will converge absolutely in Re z > {Ya - B, contradicting the definition of {Ya' Lemma 6.4.19 is the crucial ingredient in the proof of the Fabry Gap Theorem. One can find some particular cases of this lemma in the work of P6lya, Bernstein, and Levinson [Levs], [Berns], [Die], and more recently, using microlocal analysis in [Kawl], [Kaw2]. If the set S of the lemma is an arbitrary open convex set our proof provides some sort of enlargement to a set St. A very precise description of this set and a proof of this lemma without any conditions on the infinite order differential operator appears in the work of [Se], [Ao]. We show now that the ideas exposed in this section can also be used to provide an elementary proof of the general version of Lemma 6.4.19, namely, Theorem 6.4.24. 6.4.22. Lemma. Let be a nonzero entire function of exponential type zero and K, L two convex compact sets such that K £ Land (~ E
Z()).
450
6. Hannonic Analysis
Then,for every entire function f E Exp(L) and every 6> 0 one can find entire functions g and h, such that g E Exp(K,), h E Exp(L,), and
f
= cJ>h
+ g,
where K., L. are the open 6-neighhorhoods of K and L, respectively. Proof. One can certainly assume that K #- L and that cJ> is not a polynomial. (In the latter case it is enough to take g a polynomial that interpolates the values of f on V(cJ».) In the general case, when Z (cJ» is infinite, we use the construction of the sets U,.j from the proof of Theorem 6.4.8 and corresponding standard functions e,.j. Introduce now the function
e:= L(O"d). ',j
For
~ E
U"j we have that
IHd~) - HdCX',j) I :::: M L81,j :::: MLT1Icxl,jl,
where ML = max{Hdu): lui:::: l}, With a similar definition for MK we also have IHK(~) - Hdcxl.j) I :::: M K81,j :::: MKT1Icxl,jl. On the other hand, given Y/ > 0, for I :::: lo(Y/)
»
1, we have
Hdcxl.j):::: Hdcxl.j)+Y/lcxl,jl by hypothesis.
We also have that for Y/ > 0 there is a constant log If(~)1 :::: Hd~) Hence, for I :::: lo(Y/) and
~ E
log If(~)1 :::: HK(cx,.j) ::::
+ Y/I~I + F~
F~
:::: 0 so that (~ E
IC).
Ul,j we have
+ Icx/.jl {2Y/ + ~(ML + MK + Y/)} + F~
HK(~) + RI {4Y/ + ~(2ML + 4MK + 2Y/)} + F~.
Moreover, we have that for some constant Co :::: 0
I ael j I 48 log I- - - - . : (n < - I I~I cJ>(n a~ - 2
+ Co
(~ E
IC).
Given 6 > Owecanchoose Y/ = 6/16 and 1\(6) :::: 10 (6/16) so that if 1 :::: 1\(6), then 1 6 21 (2ML + 4ML + 2y/ + 48) :::: 4' Hence, there is a constant C \ :::: 0 for which 10gIB(~)1 :::: HK(~)
E
+ 21~1 + C\
6.4. Differential Operators of Infinite Order
and log
451
1<1>:~) :; (~)I ~ HdO + ~I~I + C
1•
We can conclude that there is a Coo function 1/1 such that the function
= e + 1/1<1>
g
is an entire function satisfying the inequality log Ig(OI ~ HdO
e
+ 21~1 + 4 log (I + R12) + C2
for some C2 ::=: O. Namely, if p(z) := 2HK (z)
+ elzl + 210g(l + Iz12)
there is a solution 1/1 of the equation
oe
01/1
I
oz
<1> OZ
-=---
satisfying
r
r 1_1_ (Z)1
11/I(z)1 2 e-Pdm < 1 ae 2e-Pdm < +00 + Iz12)2 - 2 Jc <1>(z) az . We obtain the desired inequality for log Igl in the same way as it was done Jc (I
in the proof of Theorem 6.4.6. Define the function h by
f
h =
and apply Lemma 6.4.5 to show that h
-g II> E
Exp(Ls).
o
We shaH now consider some constructions of [Ki] that will help us to describe the largest domain Q into which all solutions in n of an infinite order differential operator will have an analytic continuation. Let A be a nonempty set of complex numbers and let U be a convex open set in Co For every convex compact K in U we set YA(K):=
nlz
E
C: Re(z~) ~ HK(~)}
(EA
and KccU
where the union takes place over all compact convex subsets of U. Let us note that r A (U) is a convex open set. In fact, each YA (K) is clearly a closed convex set, and by taking an exhaustion (Kn)n~l of U by compact convex sets we see that r A (U) is an increasing union of convex sets and hence convex itself. Moreover, if e > 0 is sufficiently small so that Ke = K + B(O, e) ~ U, then YACKo) ;2 YA(K)
which is a neighborhood of YA(K).
+ B(O, e) =
(YA(K»e,
452
6. Hannonic Analysis
Let us assume now that A is an unbounded set, and let, for r 8
A •r
n
(K) =
{z E C: Re(zs) :::;: Hds)
~
0,
+ r},
,eA 8
A (K)
= U8
A ,r(K),
fi':O
8 A (U)
U
=
8
A (K).
KccU
These three sets are convex and 8 A (U) is open. In fact, for K C C U let such that Ke CC U. Then, for every r ~ 0,
e = eK >
°
8 A ,r(Ke) ;2 8 A ,r(K)
+ B(O, e),
Hence,
therefore, o
-"-
(where X denotes the interior of X, while X is the interior of the closure) and KcCU
KCCU
Note that we have shown that
where K runs over the family of convex compact subsets of U. Given a sequence of complex numbers A = (Sj)ji':l such that lim ISj I = +00, )-+00 we denote by .B(A) the subset of the unit circle 'IT' of points S for which there is a subsequence (Sj,hi':l with
S = lim
k-+oo
lL, lSi, I
We denote by a(A) the closed cone with vertex at the origin generated by the directions in .B(A). 6.4.23. Lemma. Let A = (Sj)ji':l, limj-+oo ISjl = +00, and let U be an open convex set in C. Then Proof. It is easy to see that
r A(U) ~ E>A(U) ~ ra(A)(U),
6.4. Differential Operators of Infinite Order
453
If we translate the set U keeping the set of directions peA) fixed, then 8 A (U) and r alA) (U) are translated by the same amount. Therefore, we can assume that o E U. Let K be a compact convex subset of U with 0 E k, then there is a constant )...k such that for I < )... < )...K, the set )"'K is a neighborhood of K still contained in U. We fix such a ).... Observe that there is an £ > 0 such that the assumptions (~E
q
because 0 E k. From this inequality, it is easy to see that there exists /3 > 0 such that the assumptions Z
E
K,
I;
and
E a(A),
imply that Hence, if ~ E
a(A),
then,
Denote
-
a(A, 8) := {I) E C: 3~ E a(A) with I~
1)1 :::: 811)1}.
The inequality (*) implies that Ya(A)(K) ~ Ya(A.8)(K).
It is also clear from the definition of peA) that there exists r > 0 such that Ar := A
n (B(O, r)t
~ a(A, 8).
It is then immediate that
Hence, Ya(A)(K) ~
n
)"'8 A(K) = 8
A
(K)
A>l
and
o
We finally have here the theorem about analytic continuation of solutions of a differential operator of infinite order due to Sebbar and Aoki (see also [BGV]).
6.4.24. Theorem. Let be a transcendental entire function of exponential type zero and let r2 be an open convex set in Co Every solution f E Jot'(r2) of the equation
454
6. Hannonic Analysis
has an analytic continuation F to the open convex set
Q=
ra(Z(
which is still a solution in Q of the same equation. Proof Let us denote by M(Q)(M(Q), resp.), the kernel of
is an injective continuous map between two FS spaces. We want to prove it is surjective. The spectral synthesis property tells us that the image of R is dense in M(Q). (In fact, we know the exponential polynomials in M(Q) are dense in M(Q).) It follows that 'R: M'(Q) -+ M'(Q) is injective, thus we only need to prove Im(' R) is closed, and under all the above conditions, it is exactly the same as proving that IRis surjective. Let T E M'(Q). By the Hahn-Banach theorem we can assume T E Jf"(Q), and therefore it has a minimal convex compact carrier L cc Q. From the definition of Q and Lemma 6.4.23 we have
Lee
0z(
U
=
0 Z (
KccU
by the identity (t) preceding Lemma 6.4.23. Taking an increasing exhaustion of Q, (Kj)j~l' by convex compact sets, we will have L contained in an increasing union of open sets, so there is K convex compact subset of Q such that E>z(
= U 8z(4))(Kn ). neJli
We claim that there is no E N such that (**).
Namely, the sets An
:= 0 Z (4))(K n )
form an increasing sequence of convex sets so that
We leave the elementary proof of this last assertion to the reader. Let us translate what does the inclusion (**) mean. By definition, Hd~) ::: HK(~)
for every
~ E
+ no
Z(
~
L, if not we replace L by
cv(K U L), which is a still compact convex subset of Q and (~ E
C);
6.4. Differential Operators of Infinite Order
455
hence. (~ E
Z(
We are thus in the situation of Lemma 6.4.22 so that for a convenient e > 0 we have lY(T) =
cc n.
and We want to show that tR(TdM(O» = /.t := TIM(n). From the above decomposition of 1Y(T) we have T S* T2 + T1•
=
=
=
where S L::n::;:o(-l)nano(n) if (~) L::n::::oan~n. Accordingly. for F E Jf'(n). we have (T. F)
Denote by v we have
E
=
(h
(:z ) F)
+ (T1• (FIQ»)·
M'(Q) the restriction of T, to M(O). For every F E M(n)
(tR(v). F)
= (v. R(F») = (T,. (FlO») = (T. F) = (/.t. F).
Thus. tR(v) = /.t and tR is surjective. As we said before. this proves that every function in M(O) extends to n. In fact. it provides a sort of formula to compute F(zo). for Zo E from FIQ. We let T = ozo. then F(zo) (ozo. F) (T,. FIQ). always assuming F E M(n).
=
6.4.25. Corollary. Let Q. functions h E Jf'(n) and g
n.
=
n. and be as in Theorem 6.4.24. Assume that the E
Jf'(Q) are such that
(:z)
g = hlQ·
Then g has an analytic continuation G to
n satisfying the equation
Proof. The operator (djdz) is surjective in Jf'(Q). hence there is Go such that
so that
(~) (g -
(GoIQ»
=0
in
Q.
E
Je(n)
6. Hannonic Analysis
456
n,
By the previous theorem, g - (GoIQ) has an analytic continuation Gl to then we let G = G 1 + Go. 0 The set
n admits slightly different characterization [Ki], namely n = ra(A)(Q) = (
n
(z E iC: Re(z() :5
HQ«()}) .
sEfl(A)
In comparison with the Fabry Gap Theorem 6.4.15, one can see that if the sequence Ao = (akh~t. where the ak are nonzero complex numbers such that
. lakl I1m k-->oo
k
=00
(but not necessarily distinct), and
. rr 3rr} .B(Ao) S;; { e''P: B + "2 :5 rp :5 2" - B for some 0 < B < rr /2, then any function
which satisfies the equation
with
has an analytic continuation to the cone r e - s. In fact, .B(Z(
rr rr rr 3rr} . < - - B or B + - < rp < - - (J • - { e''P: - -2 + B < - rp -2 2- - 2
C
Observe that the set -.B(Ao) does not really intervene in the construction of the domain nO,t since
r -.8(Ao) (re) = C. The domain nO,e, where all the solutions
f extend, can be found as follows:
ne,t = Ura(Z(t/l»(Q) = Urp(Ao)(Q), Q
where the family of
Q
Q
runs over ail open convex sets Q £
re U 8(0, s).
Q
such that
6.4. Differential Operators of Infinite Order
457
I I
I
I
Figure 6.3
For instance if,
re n aB(O, 8) is the arc e'} e, and we take Q = r/l' {sei
where 0 <
e' = e'(e, 8)
<
e'
<
I{J
<
8
then
Q9,. ;2 rp(Ao) ;2 r9 - 8. EXERCISES
6.4.
"I. Use the Minimum Modulus Theorem to prove the density result used in the proof of Lemma 6.4.7. Namely, let ct> "# 0 be an entire function of exponential type zero and denote mer) := min{log 1ct>(~)I: I~I = r}. Then -mer)
= o(r),
except on a set of linear density zero. That is, given (i > 0, if E8 := (r ::: 0: mer) < -(ir}, for any '1 > 0, there is ro = ro(8, Tj) such that m)(E8 n [0, TIT for any r ::: o. (Here m) denotes the Lebesgue measure on JR.)
rn :::
"2. From the construction of the components v,.j in the text preceding Theorem 6.4.8, conclude the following result holds for any entire function ct> "# 0 of exponential type zero: Let Z = (z: ct>(z) OJ, then for any e > 0 there is a value R. > 0 such that
=
dist(z, Z) ::: elzl
implies
1ct>(z)l::: e-· 1zl •
(Hint: For R; » 1 any connected component of the set S(ct>, e) = (z E IC: 1(z)1 < exp( -e Iz /)} that contains any point with Iz I ::: must have diameter not exceeding
R;
458
6. Hannonic Analysis
Elzl/2. Assume dist(z, Z) ~ eiz I, Izl ~ RE , z E S(<1>, E), to conclude that log 1<1>(~)1 is harmonic in the component that contains z. Use the minimum principle for harmonic functions to conclude the proof.) Note that this result can be used to give a slightly simpler proof of Lemma 6.4.16. 3. Let <1>(z) = cos.fi, Z = (z E CC: <1>(z) from Theorem 6.4.24, in the case
= OJ. Find the asymptotic cone a(z) and f2 (0
~
() < 2JT)
and
n = 8(0. 1).
6.5. Deconvolution The problem of solving an equation of the form
Jl*[=g, where Jl is a measure with compact support in R appears often in engineering and other applications of mathematics. As we know. this equation does not have a unique solution in the spaces & OR), V' (lR), not even if we assume [ E L 00 OR). For instance, the Fourier transform :F1-t could have a real zero a and then 1-t * eiar = O. For that reason, it is customary to restrict the domain of the [ considered, to a space like L 2 (JR.) or L 1 (JR.), so that there is at least uniqueness of the solution. In fact, if [ ELI (JR.) then the Fourier transform :F[ is a continuous function which tends to zero at infinity, thus Jl * [ = 0 implies (~ E
JR.).
Since F1-t = 0 is a discrete set, it follows that :F[(0 is identically zero. Therefore, [ = 0 a.e. Nevertheless, the operator ELI (JR.) ~ Jl
[
*[
ELI (JR.)
fails to have a continuous inverse in general. This is essentially due to the presence of zeros of FJl. Recall that FJl has no zeros only if 1-t = c8a , for some c '# 0, a E R The same kind of reasoning applies to other function spaces like L 2 (JR.). Given that this problem is very important, there is a large literature about how to find approximate inverses. They are usually called regularization methods [Mor]. The interpretation of the equation 1-t * [ = g is that [ represents an unknown signal, Jl a measuring device, and g the output signal, i.e., the data. In some situations, it is possible to multiplex the signal, in other words, to use several measuring devices JlI,"" Jl". We have therefore a system of convolution equations. fJ.1
* [ = glt···, fJ." * [ = g".
If we could find measures or even distributions of compact support such that VI
* JlI
+ ... + V" * Jl" = 8.
VI, .•. , V"
6.5. Deconvolution
459
Then we can solve the system (*) very simply,
f
=
V(
* gl + ... + Vn * gn'
This inverse formula will be continuous in £ (R) and also in some Sobolev spaces [Ad]. The problem, of course, is to find out whether (**) has a solution and, moreover, to find the deconvolvers Vl, ••. , lin explicitly. Let us consider only the case n = 2 for simplicity. From now on, we use indistinctly il or F IL to denote the Fourier transform of a distribution or measure. The equation (**) is equivalent, via the Fourier transform, to the Bezout equation
and we are looking for a solution pair (ill, il2) E F(£'(R» = Ap(C) for the weight p(z) = I Imzl + log(2 + Izl). From Hormander's Theorem 2.8.15 (see also Theorem 2.8.16) we know that the necessary and sufficient condition or the solvability of (* * *) is that there are constants 8 > 0, C > 0, such that: (~ E
C).
In the engineering literature this condition is sometimes called "strong coprimeness." Generally speaking, this is simply a condition of how separated are the zeros of ill from those of il2' We would now like to construct explicitly deconvolvers VI, V2 for distributions IL I, 1i2, which satisfy a number of conditions that are often found in concrete examples. We point out that most of the limitations that we are going to impose in Proposition 6.5.1 can be eliminated by paying the penalty of more complicated formulas for the deconvolvers. 6.5.1. Proposition. Let ILl> IL2 (i)
E
£'(R) be such that
ill (0), il2(0)
=1= 0 and there exists a constant T > 0 so that all zeros of ild12 lie in the logarithmic strip
I Im~1 :5 T log(2 +
I~I)·
(ii) All the zeros of ilj are simple and they are constants
that
ilj(n
=0
implies
lilj(nl:::
(1
+81~I)L
8
> 0, L > 0, such
(j ::; 1,2).
(iii) There are constants 8 > 0, M > 0, such that A
ILI(~) = 0
and
A
implies
IIL2(~)1 ~ (1
8
+ I~I)M
460
6. Hannonic Analysis
(iv) There is an increasing sequence of piecewise C l Jordan curves (Yn)n;::l (i.e., Int(Yn) S; Int(Yn+I)), and a sequence (rn)n;::1 o/positive real numbers, converging to infinity, such that (a) l(Yn) :;; length of Yn = O(rn ). (b) There exist two strictly positive constants CI > C2, so that
(c) There exist two positive constants)... and N, so that (~ E
Yn, n
~
1).
belong to F(£'(lR» and ~q ILl (ngl (~)
+ ~q IL2(ng2(~) + ILl (nIL2(~)p(~) =
1,
where P is a polynomial of degree at most q - 1, P(~)
= Res
+ ... + u ( ~q-l + u~q-2 ~.
q- l
u q ILl (u)/Jdu)
)
,u =0 .
Proof. Let For
Z
E
1=
Int(Yn) we can rewrite the Cauchy formula as follows:
_1_1_1_ = -1-1 2ni
Yn ~ -
z
2ni
0(0 -O(z) d~ - z)
Yn O(n(~
+ O(z) 2ni
1
d~
Yn e(~)(~
. - z)
We observe that for any z fixed, this formula is valid as long as n is sufficiently large (i.e., n ~ n z ). Let us now compute by the residue theorem the first integral:
-1-1 2ni
Yn
e(~) e(~)(~
e(z) d - z)
~
- ~ (z)' (z)P(z) - /1-1 /1-2 q
~
()
+z /1-1 z
'"
~
IL2(Z)
Rq' (R)~I(R)Z- R P-2(fJ)=O I' ILl I' /1-2 I' I' tlElnl(Yn)
6.5. Deconvolution
461
Let A > 0 be such that Il1j(z)1
s
A(l
+ Izl)AeAI'mzl
(j = 1, 2, z E iC).
Then, if Iz - .81 ::: 1 we have
I:~z~ I s
1112(z)1
s
A(1
+ Izl)AeAI'mzl
and, if Iz - .81 < 1, then
I s Iwl=l max 1112(.8 + w)1 s Ae I112(z) z -.8 S A 2 e A (3
+ Izl)AeAI'mzl
A (2
+ 1.8I)AeAI,mPI (z
E
IC).
Therefore, there is a constant B > 0 such that (z E iC)
and (Z E
iC).
On the other hand,
and, similarly,
for a convenient positive constant C. As q was chosen so that q ::: L + M + 2, the series defining gj converge uniformly and absolutely in the whole plane, since
1
.L
(1
+ lal)2
.L
(1
+ 1.81)2
<
+00,
<
+00,
III (a)=O
and 1
1l2(P)=O
because the functions I1j are entire functions of order 1. Therefore the functions gj are entire and they satisfy the inequality (z E C).
462
6. Hannonic Analysis
To finish the proof we have to show the other integral goes to zero as the integer n -+ +00. For n sufficiently large so that clrn > Izl, we can use hypothesis (iv) to obtain e(z)
r
I
d~
I 2Jri Jrn e(~)(~ -
1 Izl) (Clrn)q-N
le(z)ll(Yn)
z) ::: A2Jr(clrn
-
+ 1.
which tends to zero thanks to the choice q ::: N
o
Condition (i) is the condition that both /11 and /12 are hyperbolic distributions. In this case the inequalities (ii) are exactly the condition that their zero sets are interpolation varieties for F(e' (1R.». Moreover, condition (iii) is exactly the requirement that strong coprimeness holds on the set Z([11[12). In particular, the proposition says that if the distributions are hyperbolic, the zeros of [11 and [12 are simple, and the varieties V ([1d and V ([12) are interpolation varieties, then the strong coprimeness on the zero set Z([11[12) implies strong coprimeness everywhere. We also note that if we define distributions VI. V2 by
+ !P(~)[12(~)' = ~qg2(n + !p(n[1I(~)'
VI(~) = ~qgl(n V2(n
then VI
* /11 + V2 * /12 = 8.
Moreover, due to the explicit formulas from Lemma 6.1.14 the distributions VI and V2 are very explicit. It is enough to recall that
\ J- I
(:~~~) , 1/r )
= i \ (/1I)x, e- iax
1
00
e ias 1/r(s)
dS)
and that multiplication by ~ corresponds to differentiation. Observe also that the proposition shows that when supp(/1j) S; [- A, A] then supp(Vj) S; [-A, A) for the Vj just described. Clearly this condition is not true for arbitrary deconvolvers since for an arbitrary distribution a, VI
+ a * /12
and
a
V2 -
* /11
are also deconvolvers. Assuming hyperbolicity, the zeros are simple, and assuming that V([1d and V ([12) are interpolation varieties, then there is a simple condition which implies (iv). Namely, assume that [11 and [12 satisfy a Lojasiewicz-type inequality
{
+ I~I)-N" (l + 1~I)-N2,
1[11(~)1 ::: CIJ(~, Z([1I»m'(l 1[12(~)1 ::: C2J(~, Z([12»m
2
for some positive constants C lo C2, ml, m2, N I , N2. In this case, for every outside a convenient logarithmic strip, one has ~
)
~
1/11(~ . /12(~)1 ::: (l
C
+ 1~I)N,+N2
~
6.5. Deconvolution
463
and the Jordan curves Yn can be constructed as follows. The distribution f-LI * f-L2 is invertible, since each of them is surjective in £(lR), therefore, the function illil2 is slowly decreasing. For r E lR fixed, Ir!large, we apply the Minimum Modulus Theorem to construct an arc of a circle centered at r of radius between 2T1og(2 + Irl) and 4T1og(2 + Irl) on which the function !1/ill(~)il2(nl is at most CI!rIQ for some a > 0. The intersection of this circle with the logarithmic strip of hypothesis (i) will have two connected components, we choose the outermost. We do the same for the point -r, and we complete a Jordan curve by adding two semicircles lying outside the logarithmic strip. Let us now see that under the assumption (.co), the crucial condition (iii) is just a separation condition between the sets Z (ill) and Z (il2)' It is the condition they are well separated (see Section 3.1), that is, there are constants s > 0, N > 0, such that for every a E Z (ill) and fJ E Z (il2) I
la - fJ I ~ s -(1-+-la-!+-!fJ-I)"'7:"N . In fact, if ill (a) = 0, then
d(a, Z(il2»
s ~ (2 + 21al)N
and, by (.co), A
1f-L2(a)! ~ (1
C3
+ l(1)N+N2
for a convenient constant C3 > 0. Conversely, if the zero sets are not well separated there must be sequences an -+ 00, fJn -+ 00, of zeros of ill, il2, respectively, such that I
Ian - fJn I ~ (I
+ Ian! + !fJn I)n .
Since il; is also in F(£'(lR», it follows that A
B(2
+ Ian !)A
1f-L2 (an) ! ~ (l + lan!)n
for some A > 0, B > 0, fixed. Hence condition (iii) cannot be satisfied. In fact, in general, the condition (.co) cannot be expected to be true for il, even for exponential polynomials, unless the origin belongs to the interior of cv(supp f-L). For instance, for f-L = 8) - 83 we have il(t)
= e-il; -
e- 3i l;
=
e- 2i l; (eii; _ e-il;)
= 2ie- 2i i; sint· We have
(t
E
while !e- 2i l; I = e2lm l;, so that (.co) cannot hold for 1m { < O. Multiplying fl by exp(2i{) solves the problem.
6. Hannonic Analysis
464
The Lojasiewicz condition that one could expect to be true would be
If (.cd holds, then the function (sin A~)jL(n will satisfy conditions (.co). In this case, we could apply the proposition to the products (sin A~) . ill (~). Let us now consider some typical examples for which Proposition 6.5.1 applies. 6.5.2. Examples. (1) Let I-Lj = XI-aj,a,] (aj > 0, j = 1,2), then sinaj~
A
I-Lj(~) = -~A
Z(I-Lj)
=
(~ E C),
7t * -Z .
aj
These functions satisfy (.co) and the only condition we need to verify is whether the zeros are well separated. That means there must exist constants E > 0 and N > 0 such that
I:~ - :: I ~ for all p, q
E
(1
+
IP7t/al~+
Iq7t/a2I)N
Z*. Equivalently,
I:~ -~I ~ (Ipl :lql)N' This is exactly the arithmetical condition that the quotient al/a2 is badly approximated by rational numbers. In particular al a2
r/Q.
We remind the reader that the theorem of Roth [BaJ implies that if al/a2 E Q\Q (that is, an irrational algebraic number) then, for every 1/ > 0, there is 8 > 0 such that
I:~ -~I ~ (IPI+~ql?+~'
For a quadratic irrational we can take 1/ = O. In any case, we can take N = 3 in the earlier inequality whenever ada2 E Q\Q. (2) Let us we consider a distribution I-L of the form I-L =
XI-a,a]
+ a,
where a is a C 2 function in IR with supp(a) ~ [-a, aJ. One can see that satisfies the Lojasiewicz condition (.co) because 18(01 < -
C
(1+1~1)2
eallm~l.
il
6.5. Deconvolution In fact, for some
465 K
> 0,
1t1(~)1 ~ d(~, Z(I1»)(1
+ I~I)-I
for I~ I sufficiently large, since zeros of 11 are asymptotic to (rr /a)Z' and they are simple for I~ I » 1. The multiple zeros will only complicate the formulas but their existence does not affect Proposition 6.5.1. When we have two distributions of this kind we only need to worry about their zero sets being disjoint and the separation condition holds outside some big disk (and therefore everywhere). Let
fJ-2 =
X[-a,.a,)
+ U2,
where al > 0, a2 > 0, and there exist constants 8 > 0, N > integers) such that a, pi 8
la2 - q
°(not necessarily
~ (Ipl + Iql)N
for every p, q E Z', and assume that eQjllml;'1
laj(nl ~
K (1
+ 1~I)Mj'
j = 1,2.
If we impose the differentiability condition
Mj > N
+1
then, it follows that the zeros of 111 and 112 are well separated outside some big disk. It is hard to give an analytic condition that implies the remaining zeros are distinct. In practice, one has to explicitly verify this numerically. (3) The last simple example occurs when fJ-I and fJ-2 are difference-differential operators. In this case, all the conditions of the proposition reduce to verify that the zeros of the exponential polynomials 111 and 112 are well separated. This problem is related to the Ehrenpreis conjecture, which was already discussed in Section 3.1. There is another convolution problem of some importance, the situation in which the function 1 in the system of convolution equations (*) is defined in a fixed finite interval. In this case, even the uniqueness of the solution 1 may fail for strongly coprime convolvers fJ-I, fJ-2. To be more specific let us assume that we consider the system fJ-I
*1
=gl,
* 1 = g2, R, RD, R > 0, cv (supp (/Lj » = [-aj, aj], fJ-2
1 E £' (] -
°
< aj < R. The functions R - aj[, and not defined anywhere else. We shall now see that the main condition for the uniqueness of a possible solution 1 is the very simple requirement
gj are COO in] - R
+ aj,
al
+a2 < R.
466
6. Hannonic Analysis
This is just the one-dimensional version of a theorem in [BOl]. We impose a number of simplifying hypotheses to make the proof simple. 6.5.3. Proposition. Let J-LI be a hyperbolic distribution, assume that cV(SUPP(J-LI)) = [-ai, ad, J-L2 is a distribution with cV(SUPP(J-L2» = [-a2, a2], o < ai, 0 < a2, ill and il2 without any common zeros, and
Let f E t' (] - R, R [) satisfy the pair of equations J-Lj then f
= 0 in ] -
*f
= 0
R, R[.
We remark that when R < 00, which we can assume, this proposition is not a consequence of Proposition 6.5.1 (even assuming the extra conditions required there). The reason is that the deconvolvers VJ, V2, in general, will not have support at a point, hence one cannot obtain information about f in the whole interval] - R, R[.
Proof of Proposition 6.5.3. Let V (ill) = {(Ilk, md, k we conclude that f(x) = Pk(x)e iakX
~
1); from Lemma 6.1.24
L
k:::1
in t'(] - R, RD. Let us denote by t'k the distributions given by (J-Llh,O of Lemma 6.1.14, then cV(SUPP(t'k)) = [-ai, ad and
(t'k
* f)(x) =
Pk(x)e iakX
On the other hand, since il2(ak) #; 0,
J-L2
* (Pk(x)e
iakX )
= Qk(x)eic>v,
where Qk is a polynomial with deg(Qk) Ixl < R - al - a2 we have
Qk(x)e iakX = (J-L2 therefore, Qk ] - R, R[.
= 0,
= deg(Pk).
* Tk * f)(x) =
and so Pk
= O.
Tk
Since al
+ a2
< R, for
* (J-L2 * f)(x) = 0;
This shows that
f
is identically zero in 0
It is clear that the proof we have just given cannot work if al + a2 > R, the reason being that the convolution J-L2 * t'k has too large a support. In fact, the result could be false as the following example shows: Let
467
6.5. Deconvolution
and let R be such that
../2n
< R <
n(1 +../2).
Let b = R - ../in < n and let rp =1= 0 be a Coo function with compact support in the open interval lb, n[. We can extend it to R as a periodic function of period 2n. Therefore we can expand in a Fourier series (x) = LCneinx, neZ
<1>1] - n, n[ = rp.
From Example 6.5.2 we know that 1 IJL2(n) I
(n E Z).
:::: C(l + n 2 )
Accordingly, the sequence
has the decrease for every kEN and the function I(x) = L
bneinx
nEZ
is now a nonzero Coo periodic function of period 2n. We consider 1 restricted to the interval] - R, R[. It satisfies f-tl (f-t2
*1 =0,
* f)(x) =
L
bnJL2(n)e inx
nEZ
But R - ../in
=b f-t2
(Ixl = (x) = rp(x) and rp == 0 in the interval 1-
* 1 (x) = 0
for
< R-
v'2n).
b, b[. In other words,
Ix I < R - v'2n.
This shows Proposition 6.5.3 cannot be true, in general, if al + a2 > R. When al + a2 < R one can also reconstruct explicitly the function 1 in ] - R, R[ from the data f-tl * f in ] - R +aj, R - aj[' j = 1,2. The main elements of such a reconstruction procedure have in fact been presented in this chapter and we refer the interested reader to [BGY], where this is done in the case of n variables. We would like to conclude this section with the statement of some theorems about holomorphic and harmonic functions in the plane whose proofs are analog to the theorems about Fourier representation of mean-periodic functions, Theorem 6.1.11 and Proposition 6.5.1. The extra ingredients are the elements of
6. Hannonic Analysis
468
hannonic analysis of functions of several variables and of holomorphic functions of several variables. For the proof, we refer to [BST], [BZ], [BOl], [BG2], and we also recommend the very lively reports on this subject by Zalcman [Za2] and [Za3]. 6.5.4. Examples. (I) Generalization of Morera's Theorem. Let f be a continuous function in the disk B(O, R) and let To be a closed triangle (or a square, or a simple polygon, etc.) such that
To S; B(O, R/2). Assume that
r
Jar
f(z)dz
=0
for every T S; B(O, R) which is congruent to To (i.e., T is obtained from To by a transfonnation of the fonn z 1-* ei 9 z + a). The conclusion is that f is holomorphic in B(O, R). (2) Generalization of the Mean-Value Characterization of Harmonic Functions. Let us recall that if f is a continuous function in B(O, R), then f is hannonic in B(O, R) if and only if for every B(z, r) S; B(O, R)
f(z)=)...(f,z,r)=_l 21t
r
21t
Jo
f(z+reili)dfJ.
For ro > 0 fixed, it is possible to find a continuous function f(z) = A(f, z, ro)
and rl
f in C such that
(z E C)
f is not hannonic anywhere. Surprisingly, it is possible to choose two radii
> 0 and r2 > 0 so that the mean value property with respect to them implies
hannonicity. More precisely, consider the set of positive numbers E given by E =
{~1: .jiiJo(~j) ~
1 = 0 (j = 1,2) and
!!. > o}. ~
(Here Jo is the Bessel function.) We observe that 1 E E, just take ~l = ~2; in general, the numbers ~j are complex and we only consider those whose quotients are real and positive. Delsarte has shown that this set is finite [De2]. Now let rl > 0, r2 > 0, be such that
0.. r2
and
¢ E
469
6.5. Deconvolution
Under these two conditions, if / E C(B(O, R)) satisfies the mean value properties /(z) = A(f, z, rl) for Izl
= A(f, z, rz)
for
Izl <
R - rz,
we can conclude that / is harmonic in B(O, R). These two examples were included to give the reader an inkling of an area of current research in integral geometry in which the reader will find a profitable use of the subjects covered in this volume and the preceding one. EXERCISES
6.5.
1. Let a > 0 and consider the lattice Za generated by a. Show that Za is dense in JR if and only if a is irrational. 2. Use the previous exercise to give a direct proof that if a" a2 > 0, al/a2 ¢ Q, then the only continuous function in JR, such that J~~j f(x + t) dt = 0 for all x E JR, j = 1,2, is the zero function. How does this relate to the deconvolution problem?
=
=
3. Let J.l-I X[-a.a], a > 0, J.l-2(X) eiOx X[-a,a] (x), 0 < 0 < 7r. Are there solutions of the Bezout equation J.l-I * VI + J.l-2 * V2 = 8, with V" V2 E E'(JR)? 4. Let al/a2 be badly approximated by rationals, J.l-j = XI-aj,aj] * X[-aj,aj]' j = 1,2. Find explicitly VI, V2 E E' (JR) solving the Bezout equation. (Hint: Although the zeros of ilj are multiple, the proof of Proposition 6.5.1 can be applied anyway.)
References
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Notation
We follow the standard notation from [BO]. Here the reader will find the extra symbols introduced and the first page where they appear. A(Q) A(V), Ap(V) Ap(Q), Ap,B(Q)
39 119
A2(C)
74 120 151 170 178 219
Ap,oo(V) Ap(Q) Ap(C) Ap,o(C)
AEP B(Q)
b(!) B B(f) = B\(f) Bp(f) Bn
E Exp({O})
330 419
109
36 36 63 300 310 408
Ir
2
f*
13
:FI ~p(T)
51 58 326
G(T)
354
~(T)
~p,
HP
Jf"o(QC)
2 58 64
HK hf, hi h/,p Hp,K H(C, k')
65 316 326 336
hoc(f\,,··,lm)
122
K(T) K(T)
D(f) Dp 81R
102 203 299 309 237
Kz KPz K*P z
54 58 299 308 326
E, Eo, E*, Eo Ep
199 303
c'(f) Lp
60 306
C(T) cn(T) cv p(K) C(S)
2i dO
58 99 327 338
479
480 M(/1)
v:n
M(f), M1(f) Mp(f)
Illanllln
In Q(K)
P(/1) p(z,r)
P(f)
Pr R(L, r, e, k', 6'), R(L r , k', 6') R(C\L; k', e'), R(C; k', 6')
Notation
5 266 299 310 396 37 354 3 120 219 119
S(f, e, C)
6g
115 408
<J:(f)
5 54 356
U(T)
54
T(f)
t
V(f) V(cp, /1, H) Wp(Q)
119 217 3
334 335
Z(f) Z(f)
263 278
Index
Analytic functional 51 carried by L of type ~k' 334 analytic Radon measure 18 asymptotically exponential polynomial, AEP 219 Bernoulli polynomials, Bn 408 Bezout equation 459 Blaschke condition, Blaschke product 14 Borel polygon, B(f), B1 (f) 301 Borel summation method 303 Borel transform, 13(f), 13p (f) 63, 300 boundary values of a hyperfunction, b(j) 46 C -transform, C (T) 63 carrier 52 Cartwright's theorem 274 Cauchy problem for convolution equations 373 Cauchy transform, 55 completely regular growth 88 conductor ideal 166 conjugate set, E 330 convex carrier, K(T) 58 corona 26 Corona Theorem 26
t
Deconvolution 458 degree of exponential polynomial, do 208 difference of a merom orphic function 408 differential operator of infinite order 419 discrete Laplace transform 263 divided differences 393 doubling condition (for a weight) 129
Exhaustive ideal 160 exponential polynomial 198 coefficients 198 degree, do 208 frequencies 198 irreducible 201 normalized 201 P6lya polygon, P(f) 219 (real) diameter, OR 237 ring of exponential polynomial, E 199 simple 201 exponential sums 199 ring of exponential sums, Eo 199 exponential type zero, Exp({Oj) 419 Fantappie decomposition 101 transform 96, 99 Fock space A2(C) 74 Fourier coefficients, en (T) 99 Fourier transform, ~ 82 Fourier-Borel transform, ~ 58 of order p, ~p 326 G-transform, G(T) 260 Gelfand transform 23 Hankel determinants 281 Hardy spaces, ~ P, H P 3 holomorphic hyperfunctions, A(Q) 38 holomorphically convex carrier, K(T) 54 hull, KQ 53 Hormander algebras, Ap(Q) 109 hyperbolic distribution 374 hyperbolic variety 374 hyperfunctions, 13(Q) 36 481
482
Index
Indicator function, h / 65 of order p, h/. p 316 regularized, hj 65 infraexponentia1 type 71 inner function 19 integral valued entire functions 278 interpolation variety 120 invariant subspace 354 invertible functions in Ap(n) 122 irreducible functions in the ring Eo 201 jointly invertible functions in Ap(n)
122
Laplace transfonn, C(f) 60 Laplace-Beltrami operator, t. 102 LindelOf indicator function, h/ 65 Lindelof theorem 17 local ideal hJC 122 localizable ideal 167 logarithmic half strip, V(rp, 11-, H) 217 Lojasiewicz inequality 210 Maximal function, M (f) 5 mean periodic function 354 Mellin transfonn, M 260 Minimum Modulus Theorem 125 Mittag-Leffler function of index p, Ep 303 star, Mp 310 transfonn of order p, Bp 310 monodromic differential equations 417 multiplicity variety V, V (f) 119 restriction map, p 119 space of holomorphic functions in V, A(V), Ap(V)
119
Newton polynomial 393 series 271 Nullstellensatz 153
Poisson transfonn, P (11-) 3 P61ya polygon, P(f) 219 P6lya representation 63
Real diameter of an exponential polynomial,,s1R 237 regularly decreasing function of exponential type 442 relative zero-measure set 87 F. and M. Riesz Theorem 18 p-convex 326 p-convex hull, cv p(K) 327 p-polar set, K*P 326 p-support function of K, H p • K 326
Semilocal Interpolation Theorem 115 slow (tempered) growth 102 slowly decreasing function 129 spectral analysis and synthesis problems 356 spectrum of an invariant subspace 357 star of holomorphy, D(f) 299 Stolz region 8 strongly coprime 401 sum of a function (5 g 408
Variety
119
Weak interpolation variety 120 weight 110 weighted spaces, Ap(n), Wp(n) 110 Wigert - Leau - P6lya -Carlson Theorem 263
Z transfonn 263 Z transfonn 278
