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�() and ll �.p ll < 1. The Goldstine theorem II.A. 13 gives 9n + 1 E X* such that (1), (3) and (4) hold. Next we choose an + l E A which approximates F on 91 ! . . . , 9n + l so well that (3) gives w.
GelfandNaimark Theorem. Suppose that on a commutative, com plex Banach algebra X with unit we have an involution * such that (x + y ) * = x* + y * , ( x · y ) * = y * · x* , ( .Xx ) * = Xx * , x ** = x and llxx* II = l x l l2 • Then the map i is a linear multiplicative isometry from X onto C(M(X ) ) . This theorem implies in particular that Lao(n, J..L ) i s isometric t o a C(K) for some compact space K. Also any space C(K) as defined in 9 is isometric to a space C( K) for some compact space K. This can be found in any book on Banach algebras, e.g. Zelazko [1973] , and in many general texts on functional analysis, e.g. Rudin [1973] .
14
I.B. Examples of spaces and operators § 1 1 .
1 1 . Riesz Theorem. The dual of the space C(K), K compact, equals the space of all regular, Borel measures (scalar valued, but obviously not necessarily positive) on K. The duality is defined as J.L { f ) JK fdJ.L and the norm of the measure as a functional on C(K) equals its total variation. The space of all Borel, regular measures with finite total variation on a space K will be denoted M(K). So for compact K we have C(K)* = M(K). This is a classical theorem and can be found in most texts on measure theory and functional analysis, e.g. DunfordSchwartz [1958] or Edwards [1965] . =
Suppose that A c C(K) , K compact, is an algebra (i.e. sums, products and scalar multiples of functions from A are in A). Suppose also that if f E A then f E A (this condition is empty if we consider real scalars) and that 1 E A. If A separates points of K (i.e. for every k1 , k2 E K, k 1 f:. k2 there is an f E A such that f(kl ) f:. j(k2 )), then A is dense in C(K). If 1 f/. A but all the other assumptions are satisfied then A nfE , 1 (0) = {a} for some a E K and the closure of A equals {! E C(K): f(a) = 0}. If A does not separate points, then we can pass to the quotient space K/ ,...., where ,...., is the equivalence relation on K defined by k 1 ,...., k2 if and only if f(kl ) = j(k2 ) for all f E A. The space K/,...., is compact and the algebra in C(KJ "') corresponding to A separates points of K/ "' · This can be found in many texts on functional analysis, topology or approximation theory. A detailed discussion can be found in Edwards [1965] .
12. StoneWeierstrass Theorem.
A locally compact abelian group G is an abelian group which at the same time is a locally compact space and such that all group operations are jointly continuous. On each such group there exists a unique (up to multiplication by a scalar) invariant, Borel measure, i.e. a Borel measure m such that m(A) = m(A + g ) for every Borel set A and every g E G. This measure is called the Haar measure. Note that A+g = {h E G: h = a+g with a E A}. If J.L is any measure in M ( G) and f is a Borel function on G, then the convolution f* J.L is defined to be the function f * J.L (s) = J0 f(s  t)d�t(t) , provided that this integral exists. If f E Lp(G, m), 1 $; p $; oo then f * J.L E Lp(G, m) and ll f *J.LIIP $; 11�tllllfllp· Also if f E C(G) then f * J.L E C(G). We can define the convolution of two functions f and g by the formula f * g(s) = f0 f (s  t)g(t)dm(t) if this integral exists.
13.
l.B. Examples of spaces and operators § 14.
15
If g E £1 ( G, m ) this definition clearly coincides with the previous one after we identify 9 with the measure 9dm. If I E L1 ( G, m ) and 9 E L00 ( G) then I * g E C ( G) . We can also define the convolution of two measures J..L, v E M ( G) as the measure defined for A C G as f..L * v(A) = fa J..L(A t) dv(t) . One checks that (in all the above cases) the convolution is commutative and distributive, i.e. I* ( A1 91 + A292) = Ad * 91 + A2 I * 92 where A1, A2 are scalars. This means that every measure f..L defines an operator Tp.: Lp ( G) + Lp ( G) , 1 � p � oo and Tp.: C(G) + C(G) defined by Tp.( f ) = I * f..L · We have II Tp. ll � II J..L I I in all cases. When we consider Tp.: L1 ( G) + L1 ( G) or Tp.: C(G) + C(G) then we actually have II TP. II = II J..L I I · All this is basic harmonic analysis and can be found (in various degrees of generality) in any textbook on harmonic analysis, e.g. Rudin [1962a] . If G is a compact abelian group, then a character of G is a con tinuous group homeomorphism from G into '1'. The set of all characters of G is usually denoted by r. If we treat characters as complex valued functions on G then they form a complete orthonormal set in £ 2 ( G, m ) where m is normalized so that m ( G) = 1 (this is possible because the group is assumed to be compact) . The characters are also linearly dense in C ( G) . Given a measure f..L E M ( G) (or a function I E L1 ( G) ) we define its Fourier coefficients
14.
for The map ·: M ( G) + ioo(r) defined above is called the Fourier trans form. Since characters are linearly dense in C ( G) this map is onetoone, i.e. the Fourier coe fficients determine the measure. One easily checks that (J..L * v)" ( 'y) = [l,('y) v('y) . If f..L E M ( G) has Fourier coefficients [l,{'Y) then we will write f..L = L"Y E r [1,('Y) · 'Y and call this series the Fourier series of f..L even when we know nothing about the convergence. On the circle group 'I' the characters can be identified with the functions (ei nB) n E Z· We write [l,(n) for J"D' e i n8dJ..L (0) . Let us also recall here that a subset H C r is called a A(p) set, p > 1 , if there exists a constant K such that
for every sequence of scalars
(a"YhEH ·
I.B. Examples of spaces and operators § 1 5.
16
Let us also recall that a subset exists a constant K > 0 such that
H c r is called a Sidon set if there
for every sequence of scalars ( ay ) yEH· The details can be found e.g. in Rudin [1962a] .
15.
The Dirichlet kernel is defined as
n
'Dn(O) =
L
k=  n
e ik9 ,
n = 0, 1 , 2, . . .
Jt E M(11') and Jt = L::�: P,(k)eik!J then 'Dn * Jl = k L;":_nP,(k)ei !J. This means that the convolution with the Dirichlet ker
If we have
nel realizes the partial sum projection with respect to the Fourier series. We have r I'Dn(O) Idm(O) = 42 log(n + 1) + o( 1 ) .
I�r
1r
Note that this equals the norm of the operator f 1t f * 'Dn in the spaces £1 ( 11') and C(11') . This operator will sometimes be called the Dirichlet projection. For details see Zygmund [1968] , Katznelson [1968] p. 50, Kashin Saakian [1984] or any book on trigonometric series.
16.
The Fejer kernel is defined as
Fn(O)
1 =  n+ 1
) ik!J. t vk(O) = tn ( 1  J!1._ n+ 1 e
k=O
k=
One checks that Fn(O) ?: 0 for n = 0, 1 , 2, . . . and so J1r IFn(O) Idm(O) = JyFn(O)dm(O) = 1 . If f E L p( 11') , 1 � p < oo, or f E C(11') in the case p = oo, then II/ f * Fnllp + 0 as n + oo. The norm of the Fejer operator, f 1t f * Fn, is 1 in spaces Lp ( 11') , 1 � p � oo . The details can be found in Zygmund [1968] , Katznelson [1968] p. 1 2, KashinSaakian [1984] or in any book on trigonometric series.
17. The de la VallOOPoussin kernel is defined as Vn = 2F2nl Fnl· Clearly IIVnll1 � 3 and II/Vn* fliP+ 0 as n+ oo for f E L p( 11') , 1 � p < oo, or f E C(11'). This follows immediately from 16. The nice
17
I.B. Examples of spaces and operators § 1 8.
property of the de la VallOOPoussin kernel is that for f = L�n ajeii 9 we have Vn * f = f. All this follows easily from properties of the Fejer kernels. The details are in Zygmund [1968] and Katznelson [1968] .
18.
The Poisson kernel is defined as
+oo  oo
( 1  r2 ) Pr(8) = :�:::> l i l eijfJ = (1 2r  cos 8 + r2 )
for
0 < r < 1.
Clearly Pr (8) � 0 and J'lf Pr(8)dm(8) = 1 for 0 < r < 1. As for the Fejer kernel we have II/  f * Pr ll p  0 as r  1 for f E Lp(1l'), 1 � p < oo , or f E C(1l'). The important feature of the Poisson kernel is that for J.L E M(1l'), Pr*J.L can be treated as a function on D by h(rei 9 ) = Pr*J.L( 8) . Such a function is always harmonic i n D. Basic properties of the Poisson kernel can be found in Zygmund [1968] , Katznelson [1968] , Hoffmann [1962] , Duren [1970] , Koosis [1980] and in many other places.
19.
The Hardy spaces Hp(D),
{
Hp(D) = f(z): f(z)
11/IIH, =
0 < p � oo , are defined as
is analytic in
D and
)
�
1 2� f(rei9 ) P d8 " 1 l ��� 271"
( 1
<
oo
}·
For 0 < p < 1 the Hardy space Hp( D) is a complete, linear metric space and for 1 � p � oo it is a Banach space. The limit limr1 f(rei 9 ) = f(ei 9 ), called the boundary value of f, exists for almost all 8 E 1l', and 11/II H, = 11/IIL , ('If) · This shows that we can identify Hp(D) with span{ e i nfJ} n?: O C Lp(1l'). This space of functions on 1l' will be denoted by Hp(1l'). If f E Hp(1l'),p � 1 then we can recover the original analytic function as f(re i 9 ) = f * Pr (8) where Pr denotes the Poisson kernel. This correspondence is also natural in terms of Fourier series. If f = E�= O anzn E Hp(D), p � 1 then the boundary value f E Lp(1l') has Fourier series E�= O anei nfJ. We will use the notation H�(D) or H�('l') for the subspace of Hp consisting of functions vanishing at 0 E D, or equivalently such that
Jy / = 0.
The details can be found in Hoffmann (1962] , Zygmund (1968], Katznel son (1968] , Duren (1970] or Koosis (1980] and in many other places.
18
I. B. Examples of spaces and operators §20.
20. There is an orthogonal projection from L2 ( T) onto H2 ( T) . In terms of the Fourier coefficients this projection can be written as 1?.(2:�: anein9 ) = L�= O anein 9 • This projection is called the Riesz pro jection. It is important to know how the Riesz projection acts on other Lp (1l') spaces. The basic fact is that 'R. is of weak type (11) (this is called Kolmogorov's theorem) , so it is a continuous projection from Lp ( T ) onto Hp (1l') for 1 < p < oo (see 7). We have II'R. II L.,L., � Cp2 (p  1)  1 . The Riesz projection 'R. is actually not continuous in Lt(T) and in L00(T) . This projection is clearly closely related to the Cauchy formula. If J.L E M (1l') then the Cauchy formula
2 _ 1 _ { .. dJ.L (eit ) f ( z ) = 211'i lo eit  z D. One easily checks that f (z) = 00 2: jL(n )z n Because 'R. is of weak type (11) we obtain that f (z) E n=O Hp (D) for p < 1. In a sense the Riesz projection is the Cauchy formula acting on T. As in 19 the details can be found in Hoffmann [1962] , Zygmund [1968] , Katznelson [1968] , Duren [1970] , Koosis [1980] and in many other places, e.g. a nice presentation of a proof that 'R. is bounded on Lp(1l') for 1 < p < oo can be found in LindenstraussTzafriri [1979] . defines an analytic function in .
21. Most of what has been said in 19 and 20 can be extended to polydiscs Dn and balls 1Bn in ern . For 0 < p � oo we define
{
Hp (Dn ) = f : f is analytic in D and II I IIH., = sup r< l
where
(
m is the probability Haar measure on yn,
{
)
Hp(1Bn ) = f : f is analytic in D and IIJIIH., = sup r< l
.!
{ IJ (rz) I P dm(z) " }yn
( Jsnr lf (rz) I Pdu(z) )
<
1
p
<
oo
oo
}
}
where u is the normalized rotation invariant measure on Sn. The boundary values limr 1 f(rz) = f(z) exist almost everywhere on yn or Sn , and we have IIIII H., = llfllp· Thus we can identity Hp(Dn ) +
19
l.B. Examples of spaces and operators §22.
with a subspace of Lp(P) denoted by Hp(lf'n) and we can identify Hp (ffin) with a subspace of Lp(§n) denoted by Hp(§n)· The natural orthogonal projection from L2 (P) onto H2 (P) extends to a bounded projection from Lp(P) onto Hp(P) for 1 < p < oo (apply the Riesz projection R in each variable separately) but is not of weak type (11). On the ball this natural orthogonal projection from L2 (Sn) onto H2 (§n), called the Cauchy projection and denoted by C , is of weak type (11), and so it is a continuous projection from Lp(§n) onto Hp(§n)· Using the Cauchy kernel for the ball mn we can express this projection as
C (f)( z ) = Jsr n (1 J(
for
and f E Lv (Sn)· This is simply the Cauchy formula for the ball ffin . The standard references for all this are Rudin [1969] and Rudin
[1980] .
L1 (1f') then we know (see
*
f(rei6 ) = ( ! Pr) (O) is harmonic in ID, so there exists a unique harmonic function j(rei 6 ) in ][) such that j(O) = 0 and f(rei 6 ) + if(rei6 ) is an analytic function. We know that f(rei 6 ) = 2:: � : r l n l ] (n)e i n(} so 22.
If f E
18) that
+oo
J(rei6 ) =  i L sgn n r l n l ] (n)ei n(} 00 
with the convention that sgn 0 = 0. One can show that limr+l j(rei 6 ) = f(ei6 ) exists almost everywhere on lf'. There is also a very useful integral representation of j(ei 6 ), namely
1 j(ei6 ) = P.V. 2 11'
1211" f(t 0
T
) cot T
2
T
d
.
The function j on 1f' is called the trigonometric (or harmonic) conjugate of f. There is a close connection between the trigonometric conjugation and the Riesz projection. Looking at Fourier coefficients we see that Rf = � J( 0) + ( ! + i j) . This shows that the map f �+ j is continuous on Lv (lf') for 1 < p < oo and of weak type (11). This is called Kolmogorov's theorem. This is a standard result in the theory of Fourier series. The details can be found in Zygmund [1968] , Katznelson [1968] , Koosis [1980] and many other books.
20
I.B. Examples of spaces and operators § 23.
23. One of the basic tools in the study of spaces Hp(DJ. 0 < p � oo is the socalled canonical factorization. For f E Hp(Dl let (zn )�= 1 denote the zeros of f each counted according to its multiplicity. The sequence (zn )�= 1 satisfies the Blaschke condition 2:: ::'= 1 (1  lzn l ) < oo. Conversely, if we have a sequence of points (with repetitions) (zn )�= 1 C such that 2:: ::'= 1 (1  l zn l ) < oo, then the following prcduct (called the Blaschke product)
D
B(z) =
00
IJ l zZnn l 1Zn ZZZn 
n= 1
converges almost uniformly in D and the function B(z) (also called the Blaschke product) has zeros exactly at the points (zn )�= J · Moreover From this we see that each II B IIoo = 1 and I B(ei 6 ) 1 = 1 a.e. on f E Hp(D) can be written as f = B · ft where ft is zerofree and 11/llv = ll !t llv and l f(ei 6 ) 1 = l ft (ei6 ) 1 a.e. on In order to factor ft we need to present the construction of an outer h � 0 be such that J'lf log h(t)dt > oo. function. Let h E L 1 It is known that for f E Hp ( ) , 0 < p � oo, 1!1 satisfies these assumptions. We take the harmonic conjugate of log h(t) and form the analytic function H(z) = exp(log h + ilog h) .
1I'.
(1I'),
1I'.
1I'
One checks that limr+1 H(rei 6 ) exists almost everywhere and has mod ulus h on Applying this to l f(ei 6 ) 1 , for f E Hp (ID) we get the outer function F(z). This function is in Hp (D) and en Y we have l f(ei 6 ) 1 = I F(ei6 ) 1 a.e. Thus we can write every f E Hv (D) aB f = B · F · l where B ( z) is the Blaschke product formed from the zeros of f (counting multiplicity) and F is an outer function with I F(ei 6 ) 1 = l f (e i6 ) 1 a.e. on The remaining factor I can be defined simply as f · (B ·F) 1 . It is an analytic function with ll l lloo = 1 and I I(ei 6 ) 1 = 1 a.e. on T. This factor is called a singular inner function (if it is not a constant) . In general an inner function is a function g in H00 ( ) such that l g (ei 6 i = 1 a.e. on Thus the Blaschke product is also an inner function, and a singular inner function is a zerofree inner function (nonconstant). The singular inner functions have the representation
1I'.
1I'.
1I'
1I'.
1 I(z) = exp  271'
!11' .e•i6 d + p, (O)  1r
z
e6z
where p, is a positive, singular measure on [  11' , 11' ] . The a)ove described factorization f = B · F · I is called the canonical factcrization. The
21
I. B. Examples of spaces and operators §24.
outer functions provide the tool to build analytic functions in Hp(D) with given modulus on 1l'. This will be used extensively in III.I. Detailed proofs of the canonical factorization can be found in Hoff mann [1962] , Duren [19 70] , Koosis [1980] and Garnett [1981] .
24. The following inequality due to R.E.A.C. Paley will be very useful. Suppose that f = E:=o an z n E H1 ( D) . Then
Since this inequality will be used later to prove Grothendieck's The orem III.F.7. which is the basis for many results presented in chapters III.F I, I have decided to present sketches of the proofs here.
Proof A. (Paley [1933] )
It follows from 23 that one can write
1 1 f(z) = (E:=o bn zn ) (E:=o en zn ) where fE:=o lbn l 2 ) 1 2 (E:=o l en l 2 ) 1 2 = ll f ll 1 · From this we infer that a2 k = L s ,r: s+ r= 2 k c8 br, so
00
00
� 4 L lcs l 2 L lbs l 2 = 4 11 ! 11 � s=O s=O Proof B. (Smith [1983] ) define inductively
Let
o:k
=
10 1 a2 k (E� 1 l a 2• l 2 )  1 / 2
91 = o: 1e2i9 k k 9k = 21 o:k e2 i9 + (1  lo:k l 2 )9k  1  21 o:k e  2 i9 9k2  1 · 
and
22

I. B. Examples of spaces and operators §25.
D we have I� + (1  I W)a1
I
a;b � 1 we see that IIYk I � 1. One checks that for l � k we have Yk (2 ) = (1  la1 l 2 )(1  l a2 l 2 ) · · · (1 ln1  1 l 2 )a 1 and Yk (s) = 0 for s � 0 and s =F 21 . Since L� 1 1ak l 2 � 100 1 this product is bounded away from 0, so 2 .. ll f ll 1 � 2 Yk f k 2 = L (1  l a1 1 ) · · · (1  l a 1 1 l 2 ) a1 a 2 t 1=1
Since for a, b
E
� �1 l
�c Since this holds for all
(t, I a,. 1 ')"'
k we have the claim.
We will also use the following inequality due to Hardy. If f = then I: :'= o � 1r ll f l h · The proof can be found in Duren [1970) p . 48, Hoffmann [1962) p. 70, Katznelson [1968] p. 91. The inequalities of Paley and Hardy fail for £ 1 ('1'). This is one possible way to show that the Riesz projection is not continuous on
25.
l: :'=o anzn E
H1(D)
(�a+!)
£ 1 ('1').
A(D)
D
The disc algebra is the space of all functions analytic in which admit a continuous extension to D. We can identify with the subspace span{ e i n9 }n�o C C('l'). This is the space of all functions f E C( ) such that j(n) = 0 for n =  1, 2, . . . . It will be denoted by The symbols Ao(D) and Ao('l') will denote the subspace of A(D) consisting of functions vanishing at 0 E D. The annihilator of C C(T) is described as follows.
26.
T A(T). A('I')
F.M. Riesz Theorem. If J.L E M('l') and JT fdJ..L = 0 for all f E A('l') then J..L is absolutely continuous with respect to the Lebesgue measure. Consequently df..L = hdm with h E
Hf('I').
The proof can be found in Hoffmann and Katznelson [1968) p. 89.
[1980]
D, Hp (D) Hp (D),
[1962] ,
Duren
[1970) ,
Koosis
and z E then the map f �+ f(z) defines a If f E linear functional on 0 < p � oo. The norm of this functional is estimated in the following simple inequality
27.
I.B. Examples of spaces and operators §28.
The proof is in Duren
[1970]
23
p. 36.
V E ([n is an open set, then we can define the Bergman space Bv (V), 0 < p as the space of all holomorphic functions in Lp (V ) , where on V we have 2ndimensional Lebesgue measure. We will be mostly interested in the case V = D or sometimes V = IBn and V = Dn . In all those cases Bv (V) is complete. Also the point evaluations are continuous and we have inequalities of the form: for every compact set K V there exists a constant C = CK such that l f (z) l C l f l v for z E K. We will also discuss the ball algebra A ( IBd ) which is the space If
28.
:::;
oo
:S:
C
of all functions analytic in IBd C ([d which have a continuous extension to IBd. The polydisc algebra A(Dd ) is the space of functions analytic in Dd which admit a continuous extension to Dd . The space A(Dd ) can d) . be naturally identified with a subspace of A good reference is Rudin [1980] where the case IBn (so in particular D) is discussed in detail. Also Axler [1988] discusses in detail. For the polydisc the standard reference is Rudin [1969] .
C(1I'
f
Bp (D) (K
f
, p) , we say 29. If is a scalar valued function on a metric space that satisfies the Holder condition of order a, 0 < a :::; 1 (for a = 1 the term Lipschitz condition is also used) if
k K
l f (k) l 'Po:U) Lipo:(K,
Clearly for every E the quantity is a norm and + the corresponding Banach space of all functions satisfying the Holder condition of order a is denoted by p) . We will consider only being or [0,1] with the natural metric. One can also replace the function in the definition by a more general function.
1I',R. o: t
K
Now we want to discuss the Sobolev norms which measure the smoothness of functions of several variables. The theory is usually done in a much more general setting, but we restrict our attention to functions on s = 1, 2, . . . . Let = 0, 1, 2, . . . be the set of all multiindices A= such that the j ' s are nonnegative integers and p A) = z::; = l :S: Each A E defines a partial derivative denoted OA . We define the Sobolev space for k � 0, s � 1, to be the space of all functions on such that oA is continuous for all A E The norm which makes into a Banach space can be defined as
30.
1I'8 , (i1. i2 , . . . , is ) ij k.
D(k), k i D(k) k 8 C (1I' ) ys k 8 f C (1I' )
11 ! 11 � = sup { i i &Aflloo : A E D(k) } .
(
D(k).
24
I.B. Examples of spaces and operators §31 .
If we want to extend this definition for p < oo we encounter some techni cal problems, so to avoid them, we first define the norm on functions as
11 / 1 1 ;
= ( AED(k) L I 8A f l � )
c=
1
p.
By the Sobolev space w; (1fB ) we understand the completion of the functions under this norm. Quite often we will treat as a subspace of where V is the disjoint union of copies of11' 8 • (Here the modulus signs represent cardinality.) The natural isometric injection is defined by the rule that on the torus 1rs which is a subset of V with the index E is simply We can also treat as a subspace of E) , the space of continuous, Evalued functions on 11'8 where E is the Banach space The natural injection maps � The same operations work also for w; ( 11'8 ) and we can treat w; (11'8 ) as a subspace of Lv (V) (V the same as above) or a subspace of Lp ( 11's ; The standard reference for the theory of Sobolev spaces is Adams [1975] . It is also presented in Stein [1970] and SteinWeiss [1971] .
c=
Ck (1fB)
C(V) I D (k) l j: Ck (11'8 ) + C(V) j(f) A D(k) 8A f· kC (1fB) 8 C(11' ; t'�(k) l . f (8A f(t)) AED(k)· f1D(k) l ) . c=
31. We have defined w; ( 11'8 ) as the completion of the functions under a certain norm. Nevertheless it is interesting to know up to what degree the elements of w; ( 11'8 ) are really functions. This is answered by the socalled embedding theorems. We will use the simplest one.
c=
functions ex The identity on tends to a continuous, linear operator from w; ( 11'8 ) into w;  m ( 11'8 ) where 0 :=:; m :=:; and mp < s and p :=:; q :=:; (s �:T.v ) . In particular w; ( 11'8 ) embeds into Lq ( 11'8 ) if p :S q :S s �lv .
Sobolev Embedding Theorem.
k
( )
The proof can be found in Adams [1975] or SteinWeiss [1971] .
ein1 91 ein2 92 eins 9s
32. We will also use the general multiplier theorem on 11'8 • Let us recall that the characters of the group 11'8 are when ( n t , . . . , n 8 ) E 7L8 • A bounded function m(�) , for � E 7L8 , defines an operator on £2 ( 11'8 ) , called a multiplier and defined by •
.
• .
The following theorem describes sufficient conditions for a multiplier to act on Lp (1I'8 ) , 1 < p < oo and to be of weak type (11) .
25
I. B. Examples of spaces and operators §32.
Multiplier Theorem. (1) r.p is bounded, (2) for every multiindex with
k
>
� we have
Suppose that r.p is a function on
R8 such that
A E D(k) (the description of D(k) is in 30)
nr sup R2 A 1 8Am(x) l dx < oo O
for
zs
)
00 .
c
1 < p < oo and of weak type (11).
R8
This theorem is a well known HormanderMihlin type multiplier theorem. It is usually formulated on and in even greater general ity, so I was unable to locate the exact reference. It follows e.g. from Theorem 4.4 of Chapter XIV of Torchinsky [1986] together with an easy observation that a map bounded on some < 1 and some q > 1 is of weak type (11) (see FollandStein [1982] Th. 3.37). Also the proof in GarciaCuervaRubio de Francia [1985] pp. 21014 gives it.
Hp ,P
Lq ,
Part II Basic concepts of Banach space Theory II.A. Weak Topologies In this chapter we discuss two topologies which are weaker than the norm topology. They play an important role, both in the general theory and in applications. These topologies allow easy use of compactness arguments, and so they are helpful in existence proofs. This is particularly true of the w* topology which can be defined on any dual space. We also introduce and study the class of reflexive spaces.
(
1. On each Banach space X there exists a weak topology {or a X, X * ) topology or wtopology) . For each point xo E X its basis of neighbour hoods is defined as
U(xo; c, xi , . . . , x�) = { x e X: l xj (x)  xj (xo) l < c
for
j = 1,
. . . ,n
}
where xi , . . . , x� is an arbitrary finite set in X* and c is an arbitrary positive number. Obviously this defines a locally convex topology on X. A sequence (xn )�=l C X such that for some x E X, Xn + x in the a X, X* )topology is said to be weakly convergent. We write this convergence as Xn � X. This is clearly equivalent to the condition that for every x* E X* the sequence x* (xn) + x* (x) as n + oo. A sequence (xn )�=l C X such that for every x* E X* the scalar sequence x* (xn) is convergent is said to be weakly Cauchy.
(
Xn = {1 , . . . , 1, 0, 0, . . . ) . Then Xn ..___..., n is weakly Cauchy because for every x* = (�j )� 1 E c0 = £ 1 we have x* (xn) = E;=l �j + E�1 �j as n + oo . On the other hand {xn )�=l is not weakly convergent, because {1, 1, 1, . . . ) fj CQ . 2 Examples.
{a)
Let X
= eo
and
times
Usually it is quite difficult to describe the weak topology o n the whole space. Quite often it is easier to describe this topology when restricted to the unit ball of X.
28
II.A. Weak Topologies §3.
B
(b) On the unit ball of eo the a(co , t'l )topology coincides with the topology of pointwise (i.e. coordinatewise) convergence (i.e. the Ty chonoff topology on n:=l [1, 1] restricted to the unit ball of eo ) . Clearly every open set in the topology of pointwise convergence is open in the a(eo , t'1 )topology. Conversely, given a neighbourhood n U(xo; c, xi , . . . , x�) with ll xo ll � 1 we fix N such that L �N l xi (k) i < 1e0 for j = 1, . . . , n and put M = m i = l , ... , n ll xj ll . One checks that {x E eo : ll x ll � 1 and l x(k)  xo(k) l < 1 C:M for k = 1, . . . , N } C n U ( x o ; c, xi , . . . , x �). On the other hand the a( eo , t'1 )topology and the topology of pointwise convergence do not coincide on the whole space Co · To see this let us consider Xn = 2 n en for n = 1, 2, . . . . Clearly E t'1 we have Xn converges pointwise to zero, but for x* = n 2
B
a.x
B
( ) �= l
X n ¢ U(O; 1, x*).
For more examples see Exercises
1,2,3.
3 Lemma. A weakly Cauchy sequence is normbounded. Proof: For a given weakly Cauchy sequence (x n )�=l C X we define an operator T: X* + c by T(x* ) = (x* ( xn ))�=l · By the closed graph theorem I.A.6, the operator T is continuous so IIT II = supn llxn ll < oo.a 4 Theorem. (Mazur)
If A is a convex set in X then the norm closure r
A of A equals its a(X, X*)
closure
.
Clearly A c r . On the other hand if X E r\A then by the HahnBanach theorem I.A.lO there exists x* E X such that x* (x) > sup{x* (a): a E A } = sup{x* (a): a E A } . This implies that • x ¢ r.
Proof:
Suppose Xn � x. Then, by Theorem 4 for each j we have conv{xn } �=j · Thus we have the following useful
5.
xE
Corollary. If Xn� x then there exists a sequence of convex combinations
Yi = E��] AkXk such
we
that II Yi
 xll
. 0.
a
6. On a dual space X* can introduce the w* topology or a(X* , X) topology. The basis of neighbourhoods of a point x 0 E X* is given by
U(xo; c, X b
, , .
.
j = 1, . . . , n} an arbitrary
{x* E X* : I x * (xi )  x(;(xi ) l < c for Xn is an arbitrary finite subset of X and c is
•
, Xn) =
where x 1 positive number. Clearly this defines a locally convex topology on .
•
•
X* .
29
II. A. Weak Topologies § 7.
7 Examples. (a) On £00 = li the w*topology restricted to the unit ball coincides with the topology of pointwise convergence. The argument is the same as in Example 2b) .
(b) On lp and Lp [O, 1] for 1 < p < topology coincide.
oo
the wtopology and the w*
8 Remark. Let T: X + Y be a continuous linear operator. Then T is continuous from (X, a( X, X* ) ) into (Y, a(Y, Y* ) ) and T* is continuous II from (Y* , a(Y* , Y)) into (X* , a(X* , X)). 9. The great usefulness of the w* topology stems from the fact that it allows us to use compactness. We have Theorem. (AJaoglu)
The closed unit ball Bx · of X* is a(X * , X)
compact.
Proof: Let Kt for t � 0 denote the set of scalars of absolute value � t. We define a map cp: Bx · + P = H�: e x Kll x ll by cp(x* ) = {x* (x) }x eX · By the Tychonoff theorem P is a compact space when equipped with the product topology. The very definitions of w* topology and product topology give that cp is a homeomorphic embedding. The proof is finished once we show that cp(Bx · ) is closed in P. If Po = Po (x) E P\cp(Bx · ) then either
(a) for some Xt , X 2 such that Xt
=
>.x 2 we have Po (xt ) =f. APo (x 2 )
or (b) for some X1 , X 2 , x3 such that Xt + x2 Po (x3) ·
=
X3 we have po (xt ) +Po (X 2 ) =f.
In case (a) we define U = {p E P: Jp(x t )  Po (xt ) l < 8 and Jp(x 2 )  Po (x 2 ) 1 < 8} where 8 = ! I Po (Xt )  >.po (x 2 ) J . Case (b) is treated analogously. In both cases we get an open set U C P with Po E U and U n cp(Bx · ) = 0. Thus cp(Bx · ) is closed. II Let X be a Banach space. There exists a canonical embedding i: X + X** given by i(x) (x* ) = x* (x) . When X = eo one easily sees that i: co + £00 is the identity map. For X = f.p, 1 < p < oo we have X** = X and i is the identity. Thus the following is not very surprising. 10.
Proposition.
The map
i: X + X**
is a linear isometry.
II.A. Weak Topologies § 1 1 .
30
Proof:
We have i(ax1 + ,8x 2 ) (x * )
x * ( ax1 + .8x 2 ) = ax * (xi ) + ,8x * (x 2 ) = [ai(x1 ) + ,8i(x 2 )] (x* ) . =
Since this holds for arbitrary scalars a, ,8 and arbitrary x1 , x 2 E X and x* E X* we see that i is a linear map. From the HahnBanach theorem I.A.9 we get, for x E X, , ll i(x) ll
=
l xsup• l $ 1 l i(x) (x * ) l l xsup• l $ 1 l x* (x) l =
thus i is an isometry.
=
ll x ll ; •
Quite often we identify X with i(X) and treat X simply as a sub space of X** . Note that this identification identifies the a(X, X*) topology with the a( X** , X* )topology restricted to i(X). In other words i is a homeomorphism of (X, I!( X, X*)) onto i(X) with the a( X** , X* )topology. 11. One can view the a( X, X*)topology on X as the weakest topology such that all normcontinuous linear functionals are still continuous. Analogously the a( X* , X)topology is the weakest topology which makes all functionals in i(X) C X** continuous. It is a useful fact that these are all the functionals which are continuous in those topologies. We have the following
(a) The space of all continuous linear Eunctionals on (X, a( X, X*)) equals X* .
Proposition.
(b) The space of all continuous linear Eunctionals on (X* , a( X* , X)) equals X.
Proof: Part (a) is easy since the a(X, X* )topology is weaker than the norm topology. The proof of (b) is as follows: Let r.p be any linear funct ional on X* continuous in a(X* , X) . Then {x* E X* : l r.p(x* ) l < 1 } :::> {x* E X * : l xj (x* ) l < c j = 1 , 2, . . . , n} for some c > 0 and some x1 , . . . , Xj E X*. The application of the following algebraic lemma com pletes the proof. ,
12 Lemma.
•
•
•
II.A. Weak Topologies § 1 3.
(b) ker <po
:::>
31
n
ker <{Jj . =l jn
Proof: (a)=9 (b) is obvious. To prove (b)=9(a) let K denote the scalar field and consider a map 1r: X  Kn defined by 1r(x) = (r.pj (x)) j=l · Since ker 1r = nj=1 ker r.pi the condition (b) implies that r.po induces a linear form A on Kn , i.e. r.po (x) = A1r(x) . Thus there exists a sequence a of scalars (ai ) J=l such that r.po (x) = Ej=1 ai r.pi (x) . 13. We know that eo is pointwise dense in c0* = £00 • This suggests the following Theorem. (Goldstine) The dosed unit ball of X is a(X** , X* )dense in the closed unit ball of X** .
Note that we identify X with its canonical embedding into X** ( cf. 10) . Proof: Let V denote the a(X** , X* )closure of Bx and assume that there exists x** E X** with ll x** ll � 1 and x** f/. V. From Theo rem 9 we infer that V is a( X** , X* )compact. Thus (use the Hahn Banach theorem I.A. 10) there exists a continuous linear functional r.p on (X** , a(X** , X* ) ) such that r.p(x**) > sup{r.p(v): v E V}. Proposition 1 1 shows that r.p(x**) = x** (x0) for some x0 E X* . We have
ll x0 11 = sup { l x0 (x) l : x E X, ll x ll � 1 } � sup{ l v(x0 ) 1 : v E V} = sup{ l r.p(v) l : v E V} < r.p(x ** ) = x ** (x 0 ) � ll x ** ll · ll xo ll � ll xo ll · This contradiction completes the proof.
a
14. We now turn to the study of a class of spaces for which the duality theory is particularly simple. The space X is said to be reflexive if i(X) = X** where i is the canonical embedding. The spaces lp and Lp, 1 < p < oo, are reflexive while £1 . £1 . co, C(K) , L00 are not, unless they are finite dimensional. Theorem. The following conditions on the Banach space X are equiv alent
(a) X is reflexive;
II.A. Weak Topologies § 1 5.
32 ( b ) X* is reflexive;
( c ) Bx is a ( X, X* ) compact;
( d ) every subspace of X is reflexive;
( e ) every quotient space of X is reflexive.
( a) => ( c ) . Bx with the a ( X, X* ) topology is homeomorphic to Bx·· with the a ( X** , X* ) topology which is compact by Theorem 9. ( c ) => ( a) . We get that i ( Bx ) is a compact subset of X** with the a ( X** , X* ) topology. Thus by the above Theorem 13 we have i ( Bx ) = Bx•• , so also i ( X ) = X** . ( d ) => ( a) . Obvious. ( a) => ( d ) . If Y is a normclosed subspace of X then the Hahn Banach theorem yields that Y is a ( X, X* ) closed. Thus By is a ( X, X* ) compact. But on Y the a ( X, X* ) topology coincides with a ( Y, Y* ) so the implication (( c ) => ( a)) gives that Y is reflexive. ( a) => ( b ) . If X is reflexive then the a ( X*, X ) and a ( X* , X** ) topologies coincide on X* , so Theorem 9 gives that Bx· is a ( X* , X** ) compact. Thus ( c ) => ( a) gives that X* is reflexive. ( b ) => ( a) . If X* is reflexive then using the implication ( a) => ( b ) we get that X** is reflexive so by ( a) => ( d ) X is reflexive. ( e ) {:} ( a) . We have to note that the dual of a quotient space of X is a a subspace of X* and use a previous equivalences. Proof:
We would like to conclude this chapter with the following useful observation. 15.
Proposition. able.
( a)
If X is separable then (Bx· , a (X* , X)) is metriz
If X* is separable then (Bx , a (X, X* )) is metrizable. Proof: ( a) Let (xn)�=l be a dense set in Bx . We define a metric on Bx· by p(xi , x;) = nL=l T n l xi (xn)  x;(xn)i. Clearly p is a a ( X*, X ) continuous metric. Let U = {x* E Bx· : i x *(x) < c} for a fixed x E X and x0 E X*. Changing c we can assume x0(x)i x E Bx . If we take such that Xn is close enough to x then U ::J {x* E Bx· : ix* (xn)  x0 (xn) i < �} ::J {x* E Bx· : p(x0 x* ) < � } . ( b)
00
n
,
33
II.A. Weak Topologies §Notes.
Taking finite intersections we infer that every a(X * , X)open set in Bx· contains a ball in metric p, so p defines the a( X * , X)topology on Bx· . (b) is an immediate consequence of (a) and the remarks made after Proposition 10. a Notes and
remarks.
Weak convergence of sequences in some special instances was used at the beginning of the century by D. Hilbert and M. Riesz. The general theory is presented in Banach [1932] . Banach, however, did not use the notion of weak topology, he relied exclusively on sequential arguments. This resulted in unnecessary separability assumptions in many theorems. The weak topology defined by weak neighbourhoods appears (for algebras of operators on Hilbert spaces) already in von Neumann [1930] . In the thirties it became clear that general topological notions are needed in order to establish nonseparable theorems. Theorem 9 for separable X is in Banach [1932] and the general form was announced by several authors with the proof first published by Alaoglu [1940] . Theorem 4 and Corollary 5 are from Mazur [1933] and Theorem 13 was first proved in Goldstine [1938] . The notion of reflexive space is implicit in Banach [1932] . Reflexive spaces, under the name 'regular spaces', were an object of intensive study in the late thirties and early forties. Theorem 1 4 summarizes some of this work. The implication ( c) =? ( a) for separable X is in Banach [1932] chapter XI Th. 13. It is interesting to note that he did not realize the converse, despite having a sequential version of Theorem 9. The equivalence ( a ) <=? (b) was known to Plessner before 1936 (see Lusternik [1936] ) . The equivalence ( a ) <=? ( c) was proved in Smulian[1939] , Bourbaki [1938] and Kakutani [1939] . As is clear from the above the content of this section is classical. The reader will find much more in DunfordSchwartz [1958] . This material is also covered in most textbooks on functional analysis.
Exercises 1. 2. 3.
Show that, if X is an infinite dimensional Banach space then the a( X, X* )topology on X is not metrizable.
Let r be an uncountable set. Show that the weak topology on the unit ball of is not metrizable.
£2 (f) Let fn E C(K), ll fn ll ::5 1 for
n = 1 , 2, . . . . Show that fn �O if and only if fn (k) > 0 for every k E
K.
34 4. 5.
6.
7.
8.
II.A. Weak Topologies §Exercises
(x�)�=l
Find a Banach space X and a sequence c X* , such that tends w* to 0 but every convex 1 for n 1 , 2, . . . and combination of 's has norm 1 .
l ! x� l =
= x�
x�
Show that on the unit ball o f H00 (D) the topology of uniform conver gence on compact subsets of D coincides with the C!(H00 , (Ld H1 )) topology.
=
Let T: H00 (D) + A(D) be defined by Tl ( z) 1( � ) . Show that T maps w* convergent sequences onto normconvergent sequences but is not continuous from the C!(H00 , (Ll / H1 ) )topology into the norm topology.
=
l an l (cpn)�=l
(an)�= l (qn)�=l qn+ l 3qn · = + an qnt + �Pn))
(Riesz products) Let be a sequence of real numbers such that ::; 1 for n 1 , 2, . . . . Let be a sequence of nat ural numbers such that Show that for every sequence � the products n : 1 (1 cos( converge in the C!(M(Y) , C(Y))topology as N + oo to a positive measure J.L. Show also that
fi,(qn) = ei'l'n an. Suppose that ( an ) �� oo is a sequence of numbers such that lim infN+oo II L �N aneinOI I < Show for some p, 1 p < P that if 1 < p < then there exists a function I E Lp(Y) such that }(n) = an for n = 0, ±1, ±2, . . . , and if p = 1 then there exists a measure E M(Y) such that fl(n) = an for n = 0, ±1, ±2, . . . . Let G be a compact abelian group with dual group r. For g E G let 19 ( / ) ( h ) = l( h g ) . Suppose that T: L1 (G) + L 1 (G) is a linear operator such that I9T = T 9 for all g E G. Show that there exists a measure on G such that TI = I * 
::;
oo
oo,
oo .
J.L
9.

J.L
/
J.L
10. Show that every positive, harmonic function in D is a Poisson inte gral of a positive measure on Y. This means (show this) that every analytic function in D with values in a half plane is in Hp(D) for all p < l.
II.B. Isomorphisms , Bases , P rojections
In this chapter we discuss a diverse set of topics all centred on the concept of isomorphism, the basic equivalence relation between Banach spaces. We discuss projections and closely related direct sum decompositions of a Banach space. We present the 'decomposition method' which allows easy proofs of the existence of isomorphisms between Banach spaces. The concept of Schauder basis is introduced and some fundamental results about blockbasic sequences are proved. 1. We will say that two Banach spaces X and Y are isomorphic (in symbols X Y) if there exists an invertible operator I (called an isomorphism) from X onto Y. An operator X . Y such that � for some > 0 and all E X is called an isomorphic embedding (or simply embedding) . Clearly then T(X) is a closed subspace of Y and X is isomorphic to T(X). An isometry is an operator I: X Y such that for all E X. Two Banach spaces X and Y are = isometric (in symbols X £':! Y) if there exists an isometry from X onto Y.
T:
,....,
cJ J x l
c
x
I I(x)J I l x l
co . c c0 c.
I J Tx l .
x
+
c
2 Examples. (a) The identity embeds eo into and into l00 • The operator I: defined by I((�i )� 1 ) = (6 �i+ 1 )� 1 is an isomor phism between and Obviously is not isomorphic to l00 since eo is separable while l00 is not. (b) If (S1, p,) is a measure space which contains an infinite se quence of disjoint sets (Aj )� 1 with 0 < p,(Aj ) < oo, then lp em beds into Lp(n, p,), 1 :5 p :5 oo. The embedding is given by T(�i ) =
co
Lj �j P, (Aj )  ; XA;
(c) The space Lp (1R) is isometric to Lp [O, 1] for every p, 1 :5 p :5 oo. The isometry I: Lp [O, 1] . Lp (1R) can be defined as
I(f)(x) =
1
( tan 1r (x  21 )) (1r ( 1 + tan2 (1r (x  21 ))))
.!
P .
More generally if (S1, p,) is a afinite measure space then Lp(S1, p,), 0
:5
p :5 oo, is isometric to Lp on a probability measure space. In order to see this let us take a sequence of disjoint subsets (Ai)� 1 of S1 with 0 <
36
II.B. Isomorphisms, Bases, Projections §3.
J.L (Aj ) < oo for j = 1, 2, . . . and 0 = U�1 Aj . Let us define a measure J.L l on 0 by J.Ll (A) = L:�1 2 j J.L (Aj )  1 J.L (A n Aj ) · One checks that J.L1 is a probability measure on 0 and that the map I: Lp (O, J.L ) + Lp (O, J.L l ) defined by 00
I( f) =
Ll.
j=l
XA; .
2j fp J.L (Aj ) t
is an isometry onto.
Let X1 and X be two closed subspaces of a Banach X both of codimension 1 . 2Then X1 X2 . Proof: If X1 = X2 there is nothing to prove. Otherwise, the space Xo = xl n x2 has codimension 2 in X so there are X l E xl and X2 E x2 such that l x 1 l = l x2 l = 1 and X1 = Xo + ..\x 1 and X2 = Xo + ..\x 2 . We define T: xl x2 by the following rule: T I Xo = id and T(x l) = X 2 . a
3 Proposition.
space
,....,
+
Now we will give examples of rather nonobvious isomorphic (even isometric) embeddings. 4 Theorem. (BanachMazur)
Every separable Banach space isometric to a subspace of C ( Ll.) , where Ll. is the Cantor set.
X is
Proof: Let B* be the closed unit ball of X* with w* topology. By II.A.9 and II.A. 15 it is a compact metric space so by the well known AlexandroffHausdorff theorem (see Kuratowski [1968] 4§41.VI or Lacey [1974] §6.3) there exists a continuous map cp: Ll. � B*. We define C(Ll.) by
T: X +
T (x ) ( ) = t5
cp
x
( t5 ) ( ) for t5 E Ll.. a
It is easy to check that it is an isometric embedding.
We can replace C(Ll.) by C[O, 1] in Theorem 4. To see this, it is enough to find an isometric embedding of C(Ll.) into C[O, 1] . Such an embedding can be realized as an extension operator; to each I E C(Ll.) we assign F E C[O, 1] such that F I Ll. = I and F is linear on each interval of [O , 1] \ Ll..
(xn)�=l
x X,
X
of a Banach space is called 5. The sequence of elements a Schauder basis (or simply a basis) if, for every E there exists
37
Il.B. Isomorphisms, Bases, Projections §6.
)�
E:
a unique sequence of scalars ( an = l such that x = = l anXn· Let us emphasize that the sign '=' means here that the series = l anXn converges to x in the norm of X. Sometimes we will consider bases indexed by countable sets other than the natural numbers, but then the order of summation has to be specified. Let us recall (I.A.20) that an idempotent operator P: X + X, i.e. an operator such that P2 = P, is called a projection. Obviously P acts as the identity on P(X), so P(X) is a closed subspace of X. A subspace Y C X such that there exists a projection P: X + X with P(X) = Y is called complemented. Bases and projections are connected as follows:
)�
E:
6 Proposition. Let ( xn = l be a basis in X. For each N we define a partial sum projection PN : X + X by PN = l anXn = = l anXn . Then sup N II PN II < oo.
( E:
) E�
)�
The number sup N I l PN II is called the basis constant of the basis
( xn = l and is denoted by bc( x n) ·
The linearity of PN and the identity P'J. = PN are obvious. The real difficulty is that we do not know that the PN 's are continuous. We define a new norm on X by l l lxl l l = supN II PNxll · We want to show that (X, 1 1 1 · 1 1 1 ) is complete. Once this is established the closed graph theorem I.A.6 will give a C such that l l lxl l l :::; C llxll for x E X, i.e. sup N II PN II :::; C. Let us take (y k ) C X which is Cauchy in I l l · I l l · This implies that there are ZN E X such that II PN ( Y k )  ZN II + 0 as k + oo, uniformly in N. The sequence ( zN = l is Cauchy in II · II because, given c > 0, if k is fixed so that II PN (Y k )  ZN II < � for all N we have Proof:
)�
II ZN  zM II :::; 3c
2
+ II PN (Yk )  PM (Yk ) II < c
for N and M big enough. Let z E X be such that li z  ZN II + 0 as N + oo. Since every linear operator on a finite dimensional space is continuous and dim PM (X) = M we have PN ( ZM ) = PN (lim PM (y k )) = lim PN PM ( Y k ) k
k
= lipt Pmin(N,M) (Y k ) = Zmin(N,M) ·
38
II. B. Isomorphisms, Bases, Projections § 7.
ZN r::= 1 anXn,
(an)�= 1 such that PN (z) ZN so
z r::= 1 anXn ·
This shows that there exists a sequence of scalars = so that = This gives
=
a
If (xn)�= 1 is a basis in the space X and for x E:= 1 anXn we define x�(x)· an,2 then (x�)�= 1 are continuous lin ear functionals and l xn l l x� l ::::; bc(xn) · The functionals x�, called biorthogonal or coefficient functionals, uniquely determined by the conditions x�(xm) 6n, m · Thus if (xn)�= 1 is a basis in X , then for each x E X we have x E:= 1 x�(x)xn, where (x�)�= 1 are biorthogonal functionals and the series converges in norm. The partial sum projection PN admits the representation PN (x) r::= 1 x�(x)xn. 7 Corollary.
=
=
are
=
a
=
=
The following routine but useful proposition gives an equivalence between bases and some systems of projections. 8.
If(xn)�= 1 is a basis in X then its partial sum projections for all x E X we have PN (x) + x as N + dim PN ( X ) N,
PropoSition. satisfy
(a) (b)
oo ,
=
)�= 1 2 , (xn)�= 1
Conversely, if we are given a sequence of projections (PN satisfying (a) , (b) , (c) above then any sequence of nonzero vectors such that E (X) and for n = 3, . . . is a basis E n ker a in X .
X1 p1
Xn Pn(X)
Pn 1
9. Any complete orthogonal system in a separable Hilbert space is a basis. Also the standard unit vectors in the spaces lp, 1 ::::; p < oo and in eo form a basis. Now we will discuss more interesting examples of bases. We start with the Haar system, which is one of the most important orthonormal systems. Each number n = 1, 2, 3, . . . we represent as n =
39
Il.B. Isomorphisms, Bases, Projections § 1 0.
2i + k with j = 0, 1, 2, . . . and k = 0, 1, . . . , 2i  1. We define the Haar functions on [0, 1] , {hn)�=O as follows: ho ( t ) =
{;
1;
for k2 i :::; t < { 2k + 1)2  i  1 ,  2� hn (t) = 2 for { 2k + 1)2 i  1 t < (k + 1)2 i ,
:::;
otherwise.
It is imperative that the reader draws the picture and understands how the supports of consecutive Haar functions are located. 10 Proposition.
(a) The Haar system
is
an orthonormal system in
£2 [0, 1] . { b) The Haar system is a basis in Lp [O , 1] , 1 :::; p < oo .
(a) This is obvious once one realizes that J; hn (t) dt = 0 for m > n , hn (t) · hm(t) equals either zero or a constant times hm . { b) Let FN , N = 2i + k, j = 0, 1, 2, . . . , k = 0, 1, . . . , 2i  1, be the space of functions which are constant on intervals of the family Proof.
n
= 1, 2, . . and that for .
SN = {(s2  i  I , {s + 1)2  i  1 ): s = 0, 1 . . . , 2k + 1 , and (s2 i , (s + 1)2  i ) s = k + 1, . . . , 2i  1}. Since h n E FN for n = 0, 1, . . . , N and dim FN = N + 1 we infer that span {hn ) �=O = FN . This description implies that span {hn ) �=O is dense in Lp [O , 1] , 1 :::; p < oo. One also easily sees that the projection PN ( f)
=
L I I I  1 I f(t) dtxl
IESN
I
is a partial sum projection. Applying the Holder inequality we have
p 1
I PN !I I � = L i I f (t) dt i I XI :::; L I I I 1 P I J I P/q I I J I P = L I I J I P = 1 ! 1 � So I PN I = 1 for N = 0, 1, 2, . . . and Proposition 8 completes the proof. P IESN
IESN
a
I I I p
0
I
I
IESN I
40
II.B. Isomorphisms, Bases, Projections § 1 1 .
Let us consider the trigonometric system (e i n9 )�� oo · We know (see I.B.20) that the Riesz projection 'R( f) = E n > o j(n)ei nB is bounded in Lp (T), 1 < p < oo. Since the operator of �ultiplication by e i nB , In (f)(O) = einB f(O), is an isometry from Lp (T) into Lp ('I') we infer that for oo < N < M < oo, the map SN,M = IM (I  'R)IN  M'RL N is a bounded operator on Lp (T), 1 < p < oo. One checks that SN ,M(f) = 11.
E nM= N f(n)em9 . A
•
Trigonometric polynomials are dense in Lp (T) so Proposition 8 gives that the characters (ei n B )�� oo ordered 0, 1,  1, 2,  2, . . . form a basis in Lp (T), 1 < p < oo . The FaberSchauder basis in C[O, 1] . Actually we will describe a whole class of bases in terms of partial sum projections. Let (tj )� 1 be a sequence of distinct points of [0,1] with t 1 = 0, t 2 = 1 and such that {tj }� 1 = [0, 1] . We define projections in C[O, 1] as 12.
Pt (f)(t) =
/ (0),
PN (f) is a piecewise linear function with nodes
at tj , j = 1, 2, . . . , N and PN (f)(ti ) = f(tj), j = 1, 2, . . . , N. Clearly the projections PN , N = 1, 2, . . . satisfy conditions (a) , (b) , (c) of Proposition 8. The classical FaberSchauder system is defined as r.p 1 ( t) = 1, r.pn (t) = ll hn  2 11 1 1 J; hn  2 (s)ds for n = 2, 3, . . . , where (hj )�0 are the Haar functions. The FaberSchauder system corresponds to the sequence (tj)� 1 being naturally ordered dyadic points, i.e. 0 , 1, 1/2, 1/4, 3/4,
1/8, 3/8, . . . .
13. A system (x n )�=l C X is called a basic sequence if it is a basis in its closed linear span. Obviously every subsequence of a basis is a basic sequence. Also if (x n ) is a basic sequence in X and i: X + Y is an isomorphic embedding then (i( xn )) is a basic sequence in Y. We can also note that the Haar system is a basic sequence in £00 [0, 1] . Two basic sequences (x n )�=l in X and ( Yn )�=l in Y are called equivalent if the series E:= l an Xn converges if and only if E:=l an Yn converges. This is equivalent to saying that there exists an isomorphism be tween span( xn)�=l and span(yn)�= l which sends onto Yn for n = 1, 2, 3, . . . .
Xn
Now we will present two useful ways to find basic sequences. 14. One is a smallperturbation argument and the other involves the so called blockbasic sequences. They are of fundamental importance in
41
II.B. Isomorphisms, Ba.ses, Projections § 1 5.
Banach space theory and in many applications. Let us start with the perturbation arguments. They are ba.sed on the following well known
U: X + X and l idx  U l < 1. Then U is invertible. The series L::'=0 (idx U) n is absolutely convergent and defines
Lemma. Let Proof:
u 1 •
a The following proposition sums up the most useful perturbation arguments.
(xn)�= 1
(x�)�= 1
x�(xm) 8n,m · (Yn )�= 1
15 Proposition. Let be a ba.sic sequence in X and let be its biorthogonal functionals, i.e. = (a) C X is a sequence of vectors such that is a ba.sic sequence equiv = < 1 , then alent to is complemented by a projection with (b) If, moreover, span < is complemented in X. then span y
If (Yn)�=1 2::::'= 1 l xn  Ynl l l x� l 8 (xn)�= 1 . (xn)�= 1 1 ( n)�= 1 I PI 8
T:
(xn)�= 1
P
+
T(x)
T(xn) Yn· 2::::'= 1 x�(x) Yn· (1) l x  T(x) l :S nL= 1 l x�(x)l l Yn  Xn l :S 8 l x l we infer that (1  8) l x l ::; I T (x) l ::; (1 + 8) l x l so T is an isomorphic embedding. This shows that ( Yn)�= 1 is a ba.sic sequence equivalent to (xn)�= 1 · (b) Let us define A: X + X by A idx  P + TP where T is defined above. By (1) I I  A l = l i P  TP I ::; I P I 8 < 1, so Lemma 14 yields that A is invertible. One ea.sily checks that APA  1 is a projection Proof:
(a) Let us define Obviously
=
span Since
X by
00
=
onto span(yn)�= 1 ·
(xn)�= 1
a
Un (an)
anXn (nk ) (uk )/:'= 1
16. If is a ba.sic sequence in X then a sequence of non zero vectors of the form = I:�:�\ for a strictly increasing a sequence of scalars is called a blockba.sic sequence of integers and is a ba.sic sequence and sequence (block ba.sis for short) . Clearly
be(uk ) :S bc(xn) ·
17 Proposition. Let biorthogonal functionals
(xn)�= 1 be a ba.sis in X with l xn l 1 and (x�)�= 1 . Let (zk )� 1 be a sequence in X such =
42
Il.B. Isomorphisms, Bases, Projections § 1 9.
that
(2) l zkl l :2: 8 > 0 for k = 1, 2 ,3, . . . (3) k+limoo x�(zk ) = 0 for n = 1, 2 , 3, . . . Then there exists a subsequence (zn . ) which is equivalent to a block basic sequence of (xn)· Proof: Let us put K = bc(xn) and let Pn denote the partial sum projections of the basis (xn) · Fix < 3 /4 8(K + 1)  1 . Inductively we choose two increasing sequences of integers n8 and N8 in such a way that (4) I PN.  1 (zn.) l < 'f/4 s , 8 (5) l zn.  PN.(Zn.) l < 'f/4 • We start with n1 = 1 and N1 chosen to satisfy (5). This is possible since Pn(x)  x as n for x E X. Having n and N8 for = 1, 2 , 3, . . . we pick nr+l so big that (3) ensures that8 (4) holds. Next we find Nr+l so big that (5) holds for Zn r+ 1 • The sequence U 8 = PN.Zn. PN _ zn. is a blockbasic sequence. We can find biorthogonal functionals u: .such1 that l 2u: l  :5s 2K I us l  1 :5 2K(8  'f//4)  1 (Corollary 7). Since l zn.  us I :5 .,.,4 we infer from Proposition 15 (a) that (zn. ) is the "'
oo
, r
s
a
desired subsequence.
Corollary. Every weakly null sequence has a basic subsequence.
(xn)�= 1 in X with l xn l = 1
(xn) 4
1]
Proof: Since span is separable we can assume that X is separable as well. Using Theorem we can treat X as a subspace of C[O, which has a basis ( 12) . Now the corollary follows from Proposition 17. a
1, . . . Xn
Ej= 1 x :L:j= 1 1, . . . , Xn (xj)j= 1
Xn
be subspaces of a Banach space X. We say that X 19. Let X is a direct sum of Xj 's and write X = Xi or X = X1 E9 · · · E9 if every E X has a unique representation = Xj with Xj E Xj . If we are given Banach spaces X we can form their direct sum X= Xi as follows: X is the set of ntuples of elements with Xj E Xi , with coordinatewise algebraic operations and product topology. We can introduce a norm in X in many equivalent ways. The most = ( i Y. common are lpsums, 1 :5 p :5 oo , where I
x :L:j= 1
Ej=1 I x l i P 1 The direct sum with such a norm will be denoted ( Ej= 1 Xj ) . P (xi ) l i P
43
II.B. Isomorphisms, Bases, Projections §20.
20. The following proposition states the easy but fundamental prop erties of these constructions:
(a) The space X is a direct sum of its subspaces X1 and X2 if and only if there exists a projection P: X + X such that X1 = P(X) and X2 = ker P. {b) IfT; : X; + }j for j = 1 , . . . , n, then we can define an operator
Proposition.
by the formula T((x; ) j= 1 ) = (T; (x; )) j=l · (c) If in (b) every T; is an isomorphism then T is.
X2
{b) and (c) are obvious. For (a) note that for x = x1 + x2 with X1 E X1 and E X2 we define P(x) = X 1 . P is continuous by the closed graph theorem. Conversely given a projection P we write x = Px + (x  Px), and Px E P{X) and (x  Px) E ker P. The a decomposition is unique.
Proof: Parts
Quite often we will need infinite direct sums. Since a basis provides an infinite direct sum decomposition of a space into onedimensional subspaces, we see that, unlike the finite case, the form of the norm on the whole sum is crucial. In practice we will consider only lpsums. They are formed as follows: Given a sequence of Banach spaces {X; )� 1 we put for 1 � p < oo 21.
00
00
l.
( J�=l xi) p = { (x;)� l : x; E X; and ll (x;) II P = ( J�=l l x; II P) " < oo } . We also define
( jtx i) =l
and
00
=
{{x; )� 1 : x; E X; and
l (x; ) l oo = s�J p l x;l l < oo}
= { (x;)� l E ( fx;) ( Jfx;) l x;l l  o } . =l J =l 0
00
:
In all the above sums the algebraic operations are performed coordi natewise. It is routine to check that all the above sums are Banach spaces.
44
Il.B. Isomorphisms, Bases, Projections §22.
Very similarly to the case of lp spaces one checks that
( L:xi 00 ) ( I:00 x; ) , ( jL:xi 00=1 ) 0 ( jI:00=1 x; ) 1 if *
*
p
j= 1
=
=
1 � p < oo
q
j= 1
1
1
and  +  = 1 . p
q
In particular if the X; 's are all reflexive and 1 < p < oo then E�1 X is reflexive.
i)
(
P
p�
(
(
(a) Lp [O, 1] is isometric to E:= 1 Lp [O, 1]
22 Examples. 00
(
(b) lp is isometric to E:= 1 lp
)
)
P
for 1 � p �
oo
)
P
for 1 �
and co is isometric
to E:= 1 co . 0 (c) C(�) is isomorphic to E:= 1 C(�) where � is the Cantor 0 set. To see (a) is relatively easy. We put An = [(n + 1)  1 , n  1 ] and identifying f E Lp [O, 1] with the sequence ( !IAn)�= 1 we see Lp [O, 1] = E:= 1 Lp(An )) p. Since Lp(An ) � Lp [O, 1] we have the claim. (b) is analogous. In order to prove (c) we start with the same idea. Let us fix an increasing sequence tn /' 1 such that t1 < 0, t 2 > 0 and � n (tn , tn+ l ) = � n is homeomorphic to �. Identifying the nth summand of E:= 1 C(�) with C( �n ) we can identify E:= 1 C(�) with 0 0 the subspace Co (�) = { ! E C(�) : / (1) = 0}. So we have to prove that Co (�) ....., C(�). This follows from the following
(
)
(
)
(
23 Lemma. morphic to
Co .
(
)
The space C(�) contains a complemented subspace iso
With �n as above we put en = sup �n· The map P f = E:= 1 [/ (en)  / (1)]x.6., is a projection from C(�) onto a subspace of
Proof:
Co (�) of functions constant on each �n· One easily checks that this a subspace is isometric to CO · Now we can write C(�) ....., Z $ co . Since C0 (�) has codimension 1 in C(�) Proposition 3 gives that Co (�) is isomorphic to Z $ V where
45
Il.B. Isomorpliisms, Bases, Projections §24.
V C eo , V we have
=
{(ai )� 1 E eo: a1 Co (a)
=
0}. Since V is clearly isometric to
"' z a1 "' z a1 "' v
eo
C(a).
Co
a
24. Arguments like that used in 22 . ( c ) are used quite often in proving that two Banach spaces are isomorphic. In the above case it is possible to write an explicit formula for the isomorphism, but in more complicated cases it is practically impossible. The following theorem is the most common variant of the 'decomposition method' which is the elaboration of such arguments.
X Y
X Y Y X. X X) P X "' Y. Let us write X "' Z $ Y and Y "' V $ X. Then
Theorem. Let and be two Banach spaces such that is iso morphic to a complemented subspace of and is isomorphic to a complemented subspace of Assume moreover that is isomorphic to E:'= 1 for some p with 1 � p � oo or 0. Then
(
Proof:
and also
X "' ( � x)P "' ( � za1Y) P "' ( � z) P $ ( � y)P "' ( � z)P $ ( � y) P $Y "'X$Y. So X "' Y. One easily checks that each step is justisfied.
Notes
and Remarks.
a
Much of the material in this chapter is folklore of Banach space theory. The notion of isomorphic embedding appeared in the early 30's, and var ious cases of embedding and impossibility of embedding were studied by Banach and his collaborators. Those results are summarized in Banach [1932] where in particular Theorem 4 is proved and in BanachMazur [1933] . The comparison between Banach [1931] and Banach [1932] shows the crystallization of the notion. The general notion of a Schauder basis was introduced in Schauder [1927] . By now the subject has grown enormously. An encyclopaedic
46
ll.B. Isomorphisms, Bases, Projections §Exercises
treatment is in Singer [1970] and [1981] and a very good, concise pre sentation in LindenstraussTzafriri [1977] . The Haar system, introduced in Haar [1910] , is one of the most important orthonormal systems (see KashinSaakian [1984] ). It is also a prototype of the general notion of martingale difference sequence. Haar studied his system in order to show that the expansion with respect to this system of every f E C[O, 1] is uniformly convergent {Exercise 7{a) ) . Proposition l O. (b) was proved by Schauder [1928] . The classical FaberSchauder system was introduced and shown to be a basis in C[O, 1] in Faber [1910] . The general case was independently treated in Schauder [1927] . The notion of a basic sequence and a block basis emerged in the late 50's in the work of Pelczyiiski and Bessaga. The main ideas of Proposition 1 5 are in KreinMilmanRutman [1940] . They were rediscovered and applied in BessagaPelczyD.ski [1958] . This paper contains also Proposition 1 7 and Corollary 1 8. Direct sums appear already in Banach [1932] . Theorem 24 is taken from PelczyD.ski [1960] but germs of such ideas are due to Borsuk (see Borsuk [1933] and Banach [1932] chap XI. §7) .
Exercises 1.
Show that, if X1 and X2 are closed subspaces of a Banach space X , both of codimension n , then xl x2 . A norm 1 1 1 · 1 1 1 on a Banach space {X, 11 · 11) is called an equivalent norm if for some C > c > 0 we have cllxll ::;; l l l x l l l ::;; C ll x ll for all .....,
2.
x E X.
3.
(a) Suppose Y c X and d ( Y, Z) < c {here d ( Y, Z) is the Banach Mazur distance defined in II.E.6) for some Banach space Z. Construct an equivalent norm 1 1 1 · 1 1 1 on X such that c 1 ll x ll ::;; l l l x l l l ::;; ll x ll for all x E X and {Y, 1 1 1 · 1 1 1) is isometric to Z. (b) For f E C[O, 1] put I I Ifi l l = ll f ll oo + ll f ll2 · Show that I l l · I l l is an equivalent norm on C[O, 1] and £ 1 is not isometric to any subspace of (C[O, 1] , 1 1 1 · 1 1 1 ) . (c) A norm is strictly convex if for all x, y, with x f. y and ll x ll = II Y II = 1 we have ll x + Yll < 2. Show that every separable Banach space has an equivalent strictly convex norm. {d) If r is a set of continuum cardinality then ioo (r) does not have any equivalent strictly convex norm. Suppose that (xn)�= l is a basis in X. Show that the biorthogonal functionals (x;'.)�= l form a basic sequence in X* . This need not be a basis (even if X* is separable) .
Il.B. Isomorphisms, Bases, Projections §Exercises
47
C X be a bounded sequence which is not norm rela tively compact. Show that there are sequences nk and mk such that (xn,.  Xm ,. ) � 1 is a basic sequence.
4. Let (xn)�= l 5.
Suppose X and Y are subspaces of a Banach space Z, such that X n Y = {0}. Show that, if the algebraic sum X + Y is not closed in Z, then there exist a basic sequence (xn)�= l C X and a basic sequence ( Yn )�= l C Y such that ll xn ll = llYn I = 1 for n = 1, 2, . . . and ll xn  Yn ll < 4  n for n = 1 , 2, . . . . In particular X and Y have isomorphic subspaces.
6.
Prove the following. (a) The space C[O, 1] has a basis { !n )�= l consisting of polynomials. {b) There exists a sequence of polynomials which is a basis in all Lv [O, 1] , 1 � p < oo.
7.
(a) Show that for I E C[O, 1] the HaarFourier expansion converges uniformly. {b) Construct a basis in C( �) .
8.
Let (cpn(t))�= l be the FaberSchauder system. Show that I = L:�= l an'Pn E LipQ [O, 1] , 0 < a < 1 if and only if l an l = O{n  Q ), so LipQ [O, 1] is isomorphic to £00 •
9.
Show the following isomorphisms: (a) Hp (T) "' Lp(T) for 1 < p < oo;
{b) Lipl [O, 1] "' Loo [0, 1] ; (c) C k [O, 1] "' C[O, 1] for k = 1, 2, 3, . . . ; {d) W;' [O, 1] "' Lv [O, 1] for k = 1, 2, 3, . . . and 1 � p < oo.
10. Suppose S C � is a closed subset. Prove the following.
(a) If S has continuum cardinality then C(S) "' C( � ) . (b) If S is countable then C(S) is not isomorphic to C{�). 11. Show the following isomorphisms. (a) C[O, 1] "' {EC [O , 1] )o ;
{b) C [O , 1 ] EB eo "' C[O , 1] ;
(c) Lv [O, 1] EB lv "' Lv [O, 1] .
48
II.B. Isomorphisms, Bases, Projections §Exercises
12. Prove the following. (a) Every isometric embedding of fp into Lp (J.L) (J.Larbitrary mea sure) , 1 � p < oo , p =F 2, maps unit vectors onto disjointly supported, normone functions. (b) Every subspace of Lp (J.L) isometric to ip , 1 � plemented.
p
<
oo
is com
13. Suppose X n �0 in X. Show that for every e > 0 and every N E N there exists a finite sequence such that = ei)..i xi ll < e for every choice of ei 1 >.i l = 1 and II with l ei l = 1.
"Lf=j.M
(>.j )f=+ff + "Lf=NM
"L.f=+,_M Aj
14. Let E be a aalgebra of subsets of n and let J.L be a probability measure on E. Let E1 be a subaalgebra of E. Prove the following. (a) For every I E £1 (E, J.L) there exists a unique function PI which is E 1 measurable, such that for every B E E 1 we have J8 l dJ.L = f8 P ldJ.L. (b) P is a linear map. (c) If I � 0 then PI > 0.
(d) P is a .normone projection in Lp(E, J.L) for 1 � p � (e) If g E L oo (Et . J.L) and I E Lt (E, J.L) then P(gf)
=
oo.
gPI .
15. Prove the following. (a) Every separable Banach space X is a quotient of £1 .
(b) If r is a set of continuum cardinality then i 1 (r) is isometric to a subspace of £00 •
(c) If r is a set of continuum cardinality then lp (r), 1 < and eo(r) are not isomorphic to any subspace of £00 •
p
<
oo
16. (The weak basis theorem) . Suppose that (x n )�= l C X is such that for every x E X there exists a unique sequence of scalars (tn )�= l such that the series tnXn converges weakly to x. Then (xn)�= l is a Schauder basis in X.
"L:'= 1
II. C . Weak Compactness
The sets compact in the u (X, X* ) topology are important in many ap plications. We study such sets in this section. The main result is the Eberlein S mulian theorem which says that weak compactness of a set is determined by properties of sequences, even when the u ( X, X* ) topology on this set is not metrizable. We apply this to study weakly compact operators, i.e. operators such that the image of any ball is contained in a weakly compact set. We show that each weakly compact operator fac torizes through a reflexive space, and use this to investigate properties of such operators. 1. This section is devoted to the study of weakly compact sets in Banach spaces, i.e. subsets A C X which are compact in the u (X, X* ) topology. We say that the set A C X is relatively weakly compact if its u ( X, X* ) closure in X is weakly compact. From Theorem II.A.14 we infer that every bounded subset of a reflexive space is relatively weakly compact. Also by Theorem II.A.4 and II.A.14 we get that every convex, bounded, normclosed subset of a reflexive space is weakly compact. Also if X is a reflexive space and if T: X + Y is a continuous linear operator, then T(Bx) is a weakly compact set. 2.
We have
A C X is relatively weakly compact if and only if it is bounded and the u ( X** , X* ) closure of i(A) in X** is contained in
Lemma. A subset
i( X )
.
Proof: Immediate from Alaoglu's theorem II.A.9 and the remarks after Proposition II.A. lO.
3. We have seen ( Exercises II.A. l and 2) that the weak topology ( even on a weakly compact set ) need not be metrizable. Thus it seems that we are not permitted to use sequential arguments. Actually it is not so. The following important and quite surprising theorem says essentially that sequential arguments about weakly compact sets are permissible.
Theorem. (EberleinSmulian).
compact if and only if every sequence subsequence.
A set
AA hasX aisweakly relatively weakly convergent
(an) c
C
50
II. C. Weak Compactness §3.
Assume A is relatively weakly compact in X and fix (an)�= l C A. Denote V = span{an}�= l · Clearly V is a separable subspace of X .
Proof:
Let us fix a sequence (x;)�=l C X * such that i f x E V and x; (x) = 0 for n = 1, 2, . . . then x = 0. (There are many ways to construct such a sequence, e.g. take (vn)�= l dense in V and use the HahnBanach theorem to get x; E X* such that ll x; ll = 1 and x; (vn) = ll vn ll for n = 1, 2, 3, . . . . ) Using a standard diagonal argument we find a subsequence (ank )�1 such that limk +oo x; (ank ) exists for every n = 1, 2, . . . . Let jj E X be any weak cluster point of the set {ank }� 1 C A; thus x; (y) = limk +oo x; (ank ) for n = 1, 2, . . . . Since V is a closed subspace, it is weakly closed, so jj E V. Thus jj is the unique cluster point of the sequence ( ank )� 1 • We have to show that ank �jj as k + oo, i.e. that for every x* E X* we have x* (ank ) + x* (y) as k + oo. But if for some x0 E X* we have x* (ank . ) + a =f. x0 (jj) , then there exists a cluster point of {ank . }� 1 , thus a cluster point of {ank }� 1 , different from jj. This contradicts the fact that jj was the unique cluster point of {ank }�1 . So in fact ank �jj. To prove the converse let us assume that A C X is not relatively weakly compact. We have to produce a sequence (an)�= l C A without a weakly convergent subsequence. Using Lemma 2 we find F E X** \i(X) such that F is in the a (X** , X*) closure of A. Let 9 = dist(F, i(X)) > 0. We will inductively construct (an , Yn)�= l C A x X* such that (1) Yn E X* and IIYn ll = 1, (2) an E A, (3) Re F(gn) > £9 for n = 1, 2, 3 . . . , (4) i Yn (a; ) i < a 9 for j < n, (5) Re Yn (a; ) > £9 for j � n. Such (an)�= l has no weakly convergent subsequence, because if (ank ) �a then by Corollary II.A.5 there is a convex combination such that From (4) we see that for n > nM we have IYn CL: ! N a k ank ) i < a 9 . Thus we infer that for n > nM we have iYn (a) i < �9 . On the other hand (5) implies Re Yn (a) � £9 for n = 1, 2, . . . , a contradiction. Digression. Let us consider the following simple example: A = Bco c Co · This is clearly not a weakly compact set. Put an = E;= l e; . Then F = (1, 1, . . . ) E £00 = cO* , 9 = 1 and Yn = e;. Conditions (1) (5) are clearly satisfied. Our construction is an attempt to imitate this example in every set not weakly compact. End of digression.
51
II. C. Weak Compactness §4. Inductive construction. Since II F II 2::: �< x · · ' x· )
() there exists 91 satisfying there exists a 1 E A such that
(1) and (3) . Since F E i(A) I F(91 )  91 (at ) l is small enough so that (5) holds. Suppose we have (ai , 9i) 'J= 1 satisfying (1)(5) . Use the Hahn Banach theorem to obtain
a
Now we would like to introduce and study the concept of a weakly compact operator. An operator T: X + Y is said to be weakly compact if the set T (Bx ) is relatively weakly compact. We infer from Theorem 3 that this is equivalent to the condition that for every bounded se quence (x n )�= 1 C X the sequence (T(x n ))�= 1 has a weakly convergent subsequence. From Theorem II.A. 14 we get easily that T: X + Y is weakly compact if X or Y is reflexive. On the other hand the operator T: Lt [O, 1] + C[O, 1 ] defined as Tf(x) = J; f(t)dt is not weakly compact since T(2n(X [ .!2  .!.n ' .!2 ]  X [ .!2 ' .!2 + .!.n ] )) has no weakly convergent subsequence. We should note that the requirement that T (Bx ) be weakly compact is far too strong. For example the identity id: C[O, 1] + L 2 [0, 1 ] is weakly compact but id(BC[o, 1 J ) is not normclosed in L 2 [0, 1] . We can also find 1 1 a functional on C[O, 1] , e.g.
5.
R.
Our investigation of weakly compact operators is based on the following:
Theorem. An operator T: X + Y is weakly compact if and only if T factors through a reflexive space, i.e. there exist a reflexive space R and operators a: X + R and /3 : R + Y such that T = /3 · a. Moreover such a factorization can be chosen such that ll a ll · ll/311 :::; 4 · II T II .
The 'if' part is obvious from 4, so we will discuss the 'only if' part. Proof: We can assume that T is 11. (If not consider T: X/ker T + Y
defined by T([x] ) = Tx.) Let us put T (Bx ) = W. The set W is convex and weakly compact. We define
Un = 2 n w + 2  n By = {yo E Y: Yo = 2n w + 2  n y with w E Wand
I I Y II :::; 1 } .
52 Each Un is convex and 2  n By C Un induces a norm II · lin on Y such that
II. C. Weak Compactness § 6. C
(2n ii T II + 2  n )By . Thus Un for all
y E Y.
We define R = { y E Y: I I IYI I I = CE�= l' IIYII �) ! < oo } . Obviously ( R, 1 1 1 · 1 1 1 ) is a closed subspace of (E� 1 (Y, II · lln )h so it is a Banach space. One checks ( see II.B.21 ) that R** can be canonically identified with {y** E Y** : (L:� 1 II y** ll �) ! < oo } . Since W is weakly compact the unit ball in ( Y** , I · lin ) equals 2 n w + 2  n By.. . Thus for y ** E Y** \ i ( Y ) we have ll y** lln � 2 n inf { ll y** y ll : y E Y } . This shows that for y ** E Y** \ i ( Y ) the series E�= 1 ll y** ll � diverges, in other words R** C (E�= l ( Y, I · ll n )h so R is reflexive. Now we define a : X  R as a(x) = T(x). Since Tx E ll x ll · W we have II Tx ll n :::; 2  n so Tx E R and a ll a ll :::; 2. We define f3(y ) = y for y E R. Clearly 11/311 :::; 2 II T II . 
We will now use Theorem 5 to deduce the fundamental properties of weakly compact operators from corresponding properties of reflexive spaces.
 Y is weakly compact and U: X1  X and V: Y  Y1 are continuous. Then VTU is weakly compact. (b) An operator T: X  Y is weakly compact if and only if T* : Y*  X* is weakly compact. (c) An operator T: X  Y is weakly compact if and only if T** (X**) c i( Y ) . (d) The set of all weakly compact operators from X into Y is a norm closed linear subspace of L(X, Y ) . 6 Theorem. (a) Suppose T: X
( a) follows immediately from the fact that normcontinuous linear operators are also weakly continuous and the well known fact that a continuous image of a compact set is compact. (b) Follows from Theorem 5 and Theorem II.A.14. (c ) If T is weakly compact then from Theorem 5 we get that T** (X**) C f3** ( R** ) = f3(R) C Y. Conversely, suppose T** (Bx·· ) C i(Y). Since T** (Bx·· ) is u ( Y** , Y* ) compact (see Theorem II.A.9 ) we infer ( cf. remarks after Proposition II.A. lO ) that i  1 (T** (Bx··) is u ( Y, Y* ) compact. Obviously T(Bx) C i  1 (T** (Bx··)) so T(Bx) is relatively weakly compact, so T is weakly compact. {d ) The fact that the set of weakly compact operators is linear can be seen easily from ( c ) . To show that it is complete it is enough to show
Proof:
II. C. Weak Compactness § 7.
53
that if T = L:: := l Tn with Tn weakly compact and II Tn ll :::; 4  n then T is weakly compact. For each Tn let us take a factorization as in Theorem 5 with Rn reflexive and llan ll :::; 2 · 2 n and II.Bn ll :::; 2 2  n . We define a: X  R by a (x) = (an(x))�= l and ,B (R)  Y R = (L:: := l Rn) 2 and 00 by ,B ((rn )�= 1 ) = L .Bn (rn ) · The space R is reflexive (II.B.21) and ·
n= l ,Ba (x) = T(x) so T is weakly compact.
a
7. A more elementary proof of the fact that the norm limit of weakly compact operators is weakly compact follows from the following
Lemma. A subset A C X is weakly compact provided for every e there exists a weakly compact set
>0 Ac c X such that A c Ac + eBx .
Since the ( X** , X*)closure of Ac + eBx in X** equals Ac + eBx·· (use Lemma 2) we infer that the a(X** , X*)closure of A is contained in n c>o Ac + eBx·· c X. Lemma 2 gives that A is weakly a
Proof:
a
compact.
The reader undoubtedly observed that the above arguments have already been used in the proof of Theorem 5. 8 Corollary.
If A
C
X is a relatively weakly compact set then conv A
is weakly compact.
Assume first that X is separable. Replacing A by its weak clo sure we can assume that A is actually weakly compact. Let f 1 (A) denote the space of functions x(a) defined on A such that L aE A l x(a) l < oo . This is clearly a Banach space. Define an operator T: £1 (A)  X by T((x(a)) aE A ) = L aE A x(a) · a. Since A is bounded (Lemma 2) the operator T is bounded. The adjoint operator T* : X*  foo (A) acts as T* (x* ) (a) = x* (a) ; thus actually T* (X*) C C(A, a ( X, X* )). Let denote the same map as T* but acting from X* into C(A). Let (x�) C Bx• . Since X is separable, passing to a subsequence we can as sume that x�  in the a ( X* , X)topology, so � point wise in C(A). Thus for every E C ( A ) * , the Riesz representation Theo rem I.B.ll and the Lebesgue theorem give fA � fA so (x�) converges weakly to Theorem 3 gives that S(Bx· ) is relatively weakly compact, so T* is weakly compact so by Theorem 6 the operator T is weakly compact. Since conv c T(Bt1 ( A ) ) we see that conv is weakly compact. Proof:
S
x0
S
A
J.L
S(x0).
S(x�) S(x0) S(x�)dJ.L S(x0)dJ.L, A
II. C. Weak Compactness §Notes.
54
For nonseparable X we argue as follows. If conv A is not weakly compact then by Theorem 3 there is a sequence in conv A that is not relatively weakly compact. Each element of such a sequence is the norm limit of some convex combinations of a countable subset of A. Thus we get a separable subspace Y C X such that conv{Y n A) is not weakly compact. This is impossible in view of the first part of our argument.a
Notes and remarks.
The fundamental Theorem 3 is, as usual, the result of much work by many mathematicians. The standard references to papers where various pieces appeared in full generality are Eberlein [1947] and Smulian [1940] . Our proof of sufficiency is taken from James [1981] . Other proofs can be found in DunfordSchwartz [1958] , PelczyD.ski [1964] , Whitley [1967] . Basic facts about weakly compact operators ( Theorem 6 (a, b,c)) have been proved in Gantmacher [1940] . Theorem 5 which is our basis for dis cussing weakly compact operators was proved by DavisFigielJohnson PelczyD.ski [1974] . This paper contains numerous applications of the technique of the proof to problems of the geometry of Banach spaces. Theorem 5 is the first instance of an application in this book of a factor ization idea. The reader will encounter many more examples later on. It should also be noted that the proof is a very elementary application of an interpolation argument. Corollary 7 was proved by M. KreinV. Smulian [1940] .
Exercises 1.
Let X be a Banach space. Show that i(X*) is complemented in
X*** .
( ) IYI
()
:5 !} is 0. Show that {g E Lt IL : weakly compact, and if IL i suppf is not purely atomic then it is not normcompact.
2. Let f E Lt IL and f � 3.
4.
£1 a((en)�= l ) = 0:::::=1 en );:': 1 .
Let X and Y be Banach spaces and let T: X  Y be an operator that is not weakly compact. Show that there exist S:  X and where U: Y  £00 such that UTS
=
=a
Let IL be a measure on T and let T,.. : Lt (T)  Lt (11') be given by *
T,.. { ! ) f IL· Show that the following conditions are equivalent: (a) T,.. is weakly compact; {b) T,.. is compact; (c)
1L
E Lt (11') .
II. C. Weak Compactness §Exercises
55
5.
Suppose that TK f(x) = J� K(x, y)f(y)dy where K(x, y ) is a measurable 1 function on [0, 1] x [0, 1] . Show that if supessx J0 jK ( x, y) j 2 dy < oo, then TK maps £00 [0, 1] into £00 [0, 1] and is weakly compact. Show that for K ( x, y) = x  1 X [O , xj (Y ) the operator TK0 maps C[O, 1] into itself but is not weakly compact.
o
6.
Let SN (IL) = l: �N [L(n)e i nB . Show that if IL is a measure on T such that sup N II SN (IL) Ih < oo then [l,(n)  0 as l n l  oo. Let X be the space of all measures IL such that l l l�tl l l = sup N II SN (IL) Ih < oo. Show that the Fourier transform X  co is a weakly compact operator. •
7.
Let IL be a measure on T such that [L(n) = 0 or 1 for n = 0, ±1, ±2, . . . . Show that {n E N: jJ,(n) = 1 } = (F1 U V)\F2 with F1 , F2 finite sets and V a periodic set, i.e. a finite sum of arithmetic progressions.
8.
:
=l
A basis ( x n) � in a Banach space X is called boundedly complete if
an = l =1 a a = l =o Show that if ( x n) �= l is a shrinking basis in X then ( x� ) �= l is a boundedly complete basis in X* . Let ( x n) �= l be a boundedly complete basis in X and let Y = span( x� ) �= l X* . Show that X is naturally isometric to Y* .
for every sequence of scalars ( ) � such that sup N 11 2:: n x n ll < oo the series l::' n X n converges. A basis ( x n) � is called shrinking if for every x* E X*, limNoo ll x* I span(xn)n� N II = 0. (a) (b)
C
(c) Show that every basis in a reflexive space is both shrinking and boundedly complete.
xn = l
(d) Show that if ( ) � is a shrinking basis in X then X** can be naturally identified with the set { ( ) � sup N 1 1 2:: ll < oo } . 9.
(James space) . Let
<ej ) � � =
ll < ei )II J = sup
us
an = l :
= l an xn
define the following norm on sequences
{ ( }; l ei·+'  ei. l 2 )
�:
}
1 � i1 < h < · · · < in ·
The James space J = { (ei )�1 e eo : ll (ei )IIJ < oo } . (a) Show that J is a Banach space.
Il. C. Weak Compactness §Exercises
56
(b) Show directly that the unit ball in J is not relatively weakly compact. (c) Show that the unit vectors form a basis in J and that this basis is shrinking. (d) Show that J** can be naturally identified with { (�i )� 1 E li (�i ) II J < oo}. Thus dim J** fi (J) = 1 .
c:
(e) Show that J is isomorphic to J** . (f) Show that J is not isomorphic to J EB J.
10. Apply the construction of Theorem 5 to the operator E: i 1 + c defined by E ( (�)� 1 ) = E;= 1 �i = . Show that the resulting 1 space X have the property that dim X** fi(X) = 1 .
(
)�
II. D . Convergence Of Series
We start this section with various characterizations of unconditionally and weakly unconditionally convergent series in Banach spaces. We prove the classical Orlicz theorem that every unconditionally convergent series in Lp , 1 � p � 2, has norms square summable. We also introduce the notion of an unconditional Schauder basis and show that C[O, 1] and £1 [0, 1] are not subspaces of any space with an unconditional basis. We conclude with the proof that the Haar system is an unconditional basis in Lp [O, 1] for 1 < p < oo . 1. This section deals with various types of convergence of series of elements of a Banach space X. The series E:=l X n is said to converge absolutely if E:=l ll xn I < oo. It is an obvious consequence of the triangle inequality that absolutely convergent series converge. The series E:=l Xn is said to converge unconditionally if the series E:=l c n Xn converges for all en with en = ±1 for n = 1, 2, 3, . . . . When X is finite dimensional then the classical Riemann theorem asserts that absolute and unconditional convergence coincide. This as sertion actually characterizes finite dimensional Banach spaces {Exercise III.F.8 ) . For the time being let us note that E:=l enen converges un conditionally in 'v • 1 < p < oo, whenever (en )�=l E lp but converges absolutely only when {en )�=l E '1 · The following characterize unconditional convergence. 2 Theorem. For a series E:= l X n in a Banach space X the following conditions are equivalent:
( a) the series
00 E=l Xn is unconditionally convergent; n 00 ( b ) the series E an Xn is convergent for every (an )�=l E i00; n =l ( c ) there exists a compact operator T: c0 + X such that T(en ) = X n for n = 1, 2, 3, . . . ; ( d ) for every permutation of the integers the series E:=l X u(n) con a
verges;
( e ) for every increasing sequence of integers
E:=l Xn k
converges.
(n k )�=l
the series
II.D. Convergence Of Series §3.
58
Proof:
We will prove the following implications: (c)

(b)  (e)  (a)
� (d) /

(c)
co.
(c)::::} ( b) . For every N we put VN = 2::= 1 anen E Since VN is weakly Cauchy T(vN ) is normCauchy, so 2:::"= 1 anXn converges. (b)::::} ( e) . Take appropriate sequence of zeros and ones. (e)::::} ( a) . Given a sequence en = ±1 we define n k in such a way that €nk = 1 for k = 1 , 2, 3, . . . and en =  1 if n =1 n k for k = 1 , 2, . . . . Then 2:::"= 1 €nXn = 2 2:�= 1 Xnk  2:::"= 1 Xn · Since both series on the right hand side converge the left hand side series also converges. (a)::::} ( c) . We will give the proof for real spaces. The changes for complex spaces are straightforward. Every vector in eo with a finite number of nonzero coefficients is a convex combination of vectors tak ing values 1,  1, 0. Thus for every N we have li T I span(en)n �N II :::; sup., n = ±1 ll l: ::"= N €nXn ll · Note that limN+oo sup., n = ±1 ll l: ::"= N €nXn ll = 0 since otherwise we can inductively produce a sequence of signs (cn)�= 1 such that 2:::"= 1 €nXn is not Cauchy. This shows that T is compact. (c)::::} ( d) . The permutation a induces an isometry of c0 defined as Iu (�n) = (�u (n) ) · The operator T o Iu: co + X is compact and sat isfies (T o Iu ) (en) = Xu(n) so the already proven implication (c)::::} (a) gives that 2:::"= 1 Xu(n) converges. (d)::::} ( e) . If (e) does not hold then we can find c > 0 and increasing se quences (n k )k'= 1 and (Ns )':: o with No = 0 such that 11 2: :,;,iJ. +1 Xnk I � c for k = 1, 2, 3, . . . . Let m8 be an increasing enumeration of N\{n k }f: 1 . One checks that for the permutation a defined by a
the series 2:::"= 1 Xu (n) diverges.
3. There is also a related notion of weakly unconditionally convergent series. The series 2:::"= 1 Xn is said to be weakly unconditionally conver gent if for every functional x* E X* the scalar series 2:� 1 x* (xn) is unconditionally convergent. Actually the name 'weakly unconditionally convergent' is a bit misleading, because such series need not converge (even weakly) . As an example take 2:::"= 1 en in More generally, the series l: ::"= o fn in the space C(K) , K compact, is weakly unconditionally convergent if and only if there is a such that
co.
c
Il.D. Convergence Of Series §4.
59
c < oo for every k E K. This can be easily checked using the Riesz representation theorem (I.B. l l ) .
E::'= o l/n (k) l :5
4 Proposition. For a series E;:'= 1 Xn in a Banach space X the follow ing conditions
are
equivalent:
(a) the series E ;:'= 1 X n is weakly unconditionally convergent; (b) there is a constant C such that
eo
L l x* (xn ) l :5 C ll x* ll for every x* E X*; n= l (c) there exists a constant C such that for every ( tn ) �= l E leo
(d) there exists an operator T: eo + X such that T(e n ) = X n · (a)::::} ( b) . We define S: X* + l 1 by S(x*) = (x* (x n )). The closed graph theorem implies that S is continuous, thus (b) holds. (b)::::} ( c) . For every N and every ( tn)�= l E leo we have Proof:
I nt= l tnXn I =
l ( nt= l tnxn ) I N I nL= l tnx* (xn ) l :5 ll (tn) lleo
S�p x •
ll x 11 9
= sup
ll x* ll 9
:5 C ll (t n ) lleo ·
sup
ll x * ll 9
eo
L l x* (xn) l
n= l
(c)::::} ( d) . Obvious. (d)::::} ( a) . For x• E X* we have 00
eo eo L l x * (xn) l = L lx * (Ten ) l = L I T* (x * ) (en ) l . n= l n =l n= l Since T* (x*) E l1 = c0 this is finite.
a
Il.D. Convergence Of Series §5.
60
5. Now we will investigate weakly unconditionally convergent series which are not convergent. It turns out that the example we gave in 3 is a canonical one. More precisely we have
I::'= t Xn P t r t P2 r2
Suppose that is a nonconvergent weakly un conditionally convergent series. Then there exist increasing sequences of integers with < < < < · · · such that the vectors
Proposition.
Pk , Tk
(1) form a basic sequence equivalent to the unit vector basis in CQ .
I::'= t Xn does not converge we can find sequences (Pk )k';. (rk ) k= l as above such that the vectors (uk ) k= l given by (1) satisfy1 l uk l � c > 0 for some c. From Proposition 4(d) we infer that uk �O. Corollary II.B. 18 yields a subsequence (which we will still denote u k ) which is basic. Since (u k ) k= l is basic, there exists a constant C such that for all finite sequences ( t k ) we have
Proof: Since
and
On the other hand Proposition 4(d) yields a constant
I for all
(tk ) E
Co ·
�::> k k ukl l :::; C l (tk ) l oo
This completes the proof.
C1 such that a
We would like now to investigate the unconditionally convergent series in Lp , 1 :::; p :::; 2. We have the following 6.
I::'= t Xn
converges uncondition Theorem. (Orlicz) . If the series ally in Lv (n, J..L ) with 1 :::; p :::; 2 and J..L is a probability measure then
I::'= t l xn l 2 < oo.
Let (rn(t))�=l denote the Rademacher functions. It fol lows from Propositkn 4( e) that there exists a constant such that
Proof:
II. D. Convergence Of Series § 7.
sup N
61
SUPte[o,1J I z:::= 1 rn (t)xn l v C. Thus we have 1 N 1 N p CP � I I n� 1 rn(t)xn l pdt = I I I n� 1 Tn(t)xn(w) I PdJL(w )dt 1 N = I I I nL= 1 rn (t)xn(w) I PdtdJL(w). �
0
0
n
n
{2)
0
The Khintchine inequality I.B.8 and the fact that p � 2 yield
CP � K: I ( nt= 1 l xn(w )1 2 ) P/2 dJL(w) N K = : I ( nL= 1 (l xn(w) IP)� ) 2 dJL(w ) n
n
E
{3)
� K: ( n� 1 ( I l xn(w) IPdJL(w) ) ) 2 N = K: ( � l xn l ; ) 2 �
N
P
n
E
E
•
Since N was arbitrary the theorem follows.
a
The important technical feature of the above proof is the use of the Rademacher functions to represent all possible choices of signs, each occurring with the same probability. This will be used extensively in the sequel.
Remark.
We will discuss this type of question in more detail in Let us note that what we actually proved is the following inequal ity: There exists a constant such that 7
liLA.
C
{4)
(xn)�= 1
C Lv (f!, J.L) , 1 � p � 2, and J.L a probability measure. for all finite A general Banach space X in which (4) holds is said to have the Orlicz property.
Il.D. Convergence Of Series §8.
62
(x�)�= l
(xn)�=l I:::'= t x�(x)xn
with biorthogonal functionals 8. A Schauder basis in a Banach space X (see II.B.5) is called an unconditional basis if for every E X the series converges unconditionally. A basis which is not unconditional is called conditional. Clearly the unit vectors in lp , 1 :5 p < oo, or in co form an unconditional basis. One easily checks that if is an unconditional basis in X then is an unconditional basic sequence in
x
(xn)�= l
9
Proposition.
tional basis in
Lp
(x�)�= l
x·.
()
The trigonometric system T only when p = 2.
(ein9 )�� oo is an uncondi
2( )
Proof: The trigonometric system, being complete and orthonormal is clearly an unconditional basis in L 1l' . By duality it is enough to consider only the case 1 :5 p < 2. Suppose that the series represents a function I E Lp T and converges unconditionally. Then Theorem 6 gives < oo, thus I E L 2 (1l') . •
() 2 I:�: l an l
I:�: aneinB
In view of Proposition 9 the question arises whether Lp , p =/:. 2, has an unconditional basis at all. The answer is positive for 1 < p < oo and negative for p = 1. We have the following. The space Lt [O, 1] does not embed into any space with an unconditional basis.
10
Theorem.
The proof of this theorem relies on the following general 11 Proposition. A block basic sequence of an unconditional basis is a an unconditional basic sequence.
Proof of Theorem 10.
(rn(t))�= l be the sequence of Radx E Lt [O, 1]
Let emacher functions. Note that for every
x · Tn�O
as n + oo
(5)
(6) l x + xrn l t l x l t as n + Assume that £1 [0, 1] embeds into a Banach space Y with an uncondi tional basis ( Yn)�= t· Starting with Xt = 1 we define
and
+
oo .
Il.D. Convergence Of Series § 1 2.
63
where kn increases so fast that
the sequence (x n )�= l considered in Y is equivalent to a blockbasic sequence of the basis ( Yn )�= l ·
{8)
Condition {8) can be ensured using {5) and arguments like those in II.B. 17. Condition {7) follows from (6) because x1 + · · · + Xn  1 = (x1 + · · · + Xn  2 ) + (x1 + · · · + Xn  2 )rkn  l . Conditions (7) , (8) and Proposition 1 1 show that for some constant C (depending on the norm of the embedding of £ 1 (0, 1] into Y) and for every choice of e n = ± 1 and for every N II
But this and (7) clearly contradict the inequality {4) .
12 Corollary. The space C(O , 1] does not embed into any space with an unconditional basis.
Proof: It is a special case of II.B.4 that £ 1 (0, 1] embeds into C(O , 1] , II so the proof follows from Theorem 10.
The existence of an unconditional basis in the spaces Lp[O, 1] , 1 < p < oo, is contained in the following theorem, which actually gives more precise information. 13 Theorem. numbers with
Let 1 < p < oo. If ( a k )k:,0 and (bk )k:, 0 l bk l ::5 lak l for k = 0, 1 , 2, then for all n � 0
are
complex
(9) where (h k ) k= O is the Haar system and p* =
ma:x.(p, pf(p  1)).
Let us recall that the Haar system was defined in II.B.9. Proof: One easily checks using duality that it is enough to consider only the case 2 < p < oo (when p* = p) . Let v: ([ x ([ + 1R. be defined as
v(x, y) = I Y I P  (p  1)P ix i P.
( 10)
Il.D. Convergence Of Series § 1 4.
64
The crucial part of the proof is the following 14 Lemma.
There exists a function then
a, b, x, y E CC and l b l � I a I
u: CC
x
CC
+
R.
such that if
v(x , y ) � u (x, y ), u(x, y ) u( x, y ), u (O, 0) 0, u(x + a , y + b) + u(x  a, y  b) � 2u(x, y ).
= =
= E �= O
{11) {12) {13) {14)
==E�=O =
Assuming this lemma for a moment we complete the proof as fol lows. Put fn bk h k . Assume also that ak h k and Yn (h k )k:, 0 is normalized so that ll hk ll oo 1 for k 0,1, . . . . This clearly does not influence the theorem but some formulas are a bit shorter. Then by (10) and {11)
= I v(fn (t), gn (t))dt � I u(fn (t), gn(t))dt. {15) Since both fn  1 and Yn  1 are constant on the In = supp hn we get from {14) 11 Yn ll �  {p  1) P II fn ll � "
1
I0 u(fn(t), gn(t))dt = (0,1I)\1.. uUn 1 (t), gn 1 (t))dt + I u{!n  1 + an , 9n  1 + bn)dt { h,. >O } + I uUn  1  an , Yn  1  bn)dt { h,. < O} � I u Un  9n  d dt [0 , 1 ] \1.. + � I [u{ !n  1 + an , Yn  1 + bn ) b
+ uUn  1  an , Yn  1  bn)]dt
::::::; J uUn 1 , 9n1 ) dt 1
0
(16)
II.D. Convergence Of Series § 14.
and thus inductively
llYn II �  (p  1) P II fn ll � �
65
J u(fo , 9o)dt = u(ao, bo)
= 21 (u(ao, bo) + u(ao, bo)) � u(O, O) = o. a
Proof of Lemma 14.
The desired function u (x, y ) is given by
( )
p 1 u(x, y ) = av( l x l + lyi ) P  1 ( IYI  (p 1) l x l ) where ap = p 1  p1  . (17) Conditions (12) and (13) are clearly satisfied. To prove (11), by homo geneity it is enough to assume l x l + I YI = 1. Letting l x l = s, (11) reduces then to the inequality
F(s) = ap( l ps)  (1  s) P + (p  1) P sP � 0 for 0 � s � 1 and p � 2. (18)
One checks by a direct computation that
F(O) > 0, F(1) > 0, F
G) = F' G) = 0, F" ( � ) > 0
and F" has only one zero in [0, 1] . This is enough to see that (18) holds. To prove (14) it is enough to assume that x and a are linearly indepen dent over IR and the same is true for y and b. Under this assumption the function G(t) = u(x + ta, y + tb) is infinitely differentiable. A routine but rather tedious computation gives
G"(O) = av {  p(p  1)( 1al 2  I W ) ( I x l + lyi ) P  2 (19)  p(p  2) [ l b l 2  Re < 1 1 ' b > 2 ] 1YI  1 ( 1xl + lyi ) P 1  p(p  1)(p  2) [Re < 1 1 ' a > +Re < I I ' b >] 2 1 x l ( l x l + lyi ) P 3 }
� :
�
Since l b l � l a l , the CauchySchwarz inequality gives G"(O) � 0. Since this holds for all x, y , a and b with l b l � Ia I we see that in general G"(t) � 0. This clearly implies (14). a Notes and remarks.
The interplay between conditional and unconditional, i.e. absolute con vergence for scalar series, was already the subject of research in the
II.D. Convergence Of Series §Exercises
66
nineteenth century, and probably even earlier. Thus, with the emer gence of a theory of Banach spaces, investigation of series in Banach spaces became an important topic. In this section we only scratch the surface of this vast area. We will present further development of these ideas in some of the subsequent chapters, most notably in liLA and III.F. The fundamental early paper on the subject is Orlicz [1933] . It contains our Theorem 2 (actually it contains the proof of (d) <* ( e ) and condition is used without comment in the proofs) and Theorem 6 with basically the same proofs. The notion of weakly unconditionally convergent series also appeared quite early. Already in Orlicz [1929] the following theorem is proved.
(a)
Theorem.
Every weakly unconditionally convergent series in a weakly sequentially complete space is unconditionally convergent.
Let us point out that a space X is weakly sequentially complete if every weakly Cauchy sequence is weakly convergent. This theorem is a forerunner of our Proposition 5 which is due to BessagaPelczynski [1958] . Unconditional bases (under the name of absolute bases) were already investigated in Karlin [1948] where among other results Propo sition 9 and a weak version of Corollary 12 were proved. Theorem 1 0 was shown in Pelczynski [1961] and the proof we present was given in Milman [1971] . Theorem 13 except for the constant p  1 is a classi cal result of Marcinkiewicz [1937] but the main work was done by Pa ley [1932] . This theorem was the starting point of much of the later work on orthogonal series and martingale inequalities. The proof we present is due to Burkholder [1988] . It has the advantage of being rather elementary and it gives the best constant in the inequality (9); see Exercise 9.
Exercises 1. (a) Show that if f E C[O, 1] then the series
"'£';:=0 ( , hn)hn, where (hn)�=O is the Haar system, converges uniformly! to f. (b) Show that if w (8) is the modulus of continuity of f E C[O, 1] then I L:= l (! , hn)hn  /lloo Cw( f;r ) . Show that "'£';:= 1 an'P n ( t ), where ('Pn)�= l is the FaberSchauder sys tem, converges unconditionally in C[O, 1] if L k sup2 k < n 9 k+1 ! ani < Show that this is not a necessary condition. :::;
2.
oo.
Il.D. Convergence Of Series §Exercises
67
3. {OrliczPettis theorem) . Suppose that for every subsequence (n s ) of the integers the series 2::, 1 xn. converges weakly to an element in X. Show that the series 2::"= 1 converges unconditionally. where { )�= O 4. For I E Lp [O, 1] , 1 < p � 2 with I = 2::"= 0 (/ , is the Haar system, define M{!) 2::"= 1 ( / , n f; Show that M: Lp [O, 1] L 2 [0, 1] . 5. Show that every linear operator T: C(K) l 1 is compact. 6. Show that lp , p > 2, does not embed isomorphically into any space Lq (/L) , 1 � q � 2. 7. Let X be a Banach space and let 2::"= 1 be a conditionally con vergent series in X with 2::"= 1 0. Let
=
+
Xn
hn)hn)hn,!  hn. hn
+
Xn = Xn
U(xn) = { x E X: there exists a permutation of such that X = � Xu(n) } (a) (Steinitz) . Let X be an ndimensional Banach space and let (xi )J'= 1 X be such that Ej= 1 Xj = 0 and ll xi ll � 1 for Show that there exists a permutation of j = 1, 2, u
N
C
u
. . . , m.
numbers {1, 2, . . . , m} such that
and the constant C(n) depends only on the dimension n.
Un
{b) {Steinitz). Show that if X is finite dimensional then ( x ) is a linear subspace.
Xn Un {d) Show that in L 2 [0, 1] there exists a conditionally convergent series 2::"= 1 Xn such that U ( x n ) consists of exactly two points.
{c) Show that in L 2 [0, 1] there exists a conditionally convergent series 2::"= 1 such that ( x ) is not a linear subspace.
8.
Suppose that X is a Banach space with an unconditional basis (xn)�= 1 {where N is finite or infinite) . We define the unconditional
II.D. Convergence Of Series §Exercises
68
basis constant of this basis ubc(xn) = sup
{I I t, cnanxn l = l en I = 1 for
n = 1, 2, . . . , N and
We define the unconditional basis constant of the space X as
ubc(X) = inf{ u bc { x n ) : ( xn ) �= l is an unconditional basis in X } .
9.
(a) Suppose that ( Yn) is a block basis of the basis ( xn ) �= l · Show that ubc ( yn ) $ ubc ( x n )· {b) Suppose that gn, k , n = 0, 1 , 2, . . . , k = 1, 2, . . . , 2 n , is a system of functions on [0,1] such that {1) supp gn, k = An , k and for every n, the sets An, k • k = 1, 2, . . . , 2 n , are disjoint and each has measure 2 n , {2) An+ 1 ,2 k  1 = {t: gn, k = 2 i } and An+I,2 k = {t: gn, k =  2 i } . Show that {gn, k } is an orthonormal system. Show that in Lp [O, 1] , 1 $ p $ oo it is a basic sequence equivalent to the Haar system in Lp [O, 1] . (c) Show that every basis in Lp [O, 1] , 1 $ p < oo, has a blockbasic sequence equivalent to the Haar system. {d) Show that for every complete orthonormal system (
II.E. Local Properties
In this chapter we present some results and notions concerning finite dimensional Banach spaces and the relation between an infinite dimen sional Banach space and its finite dimensional subspaces. We start with a discussion of the bounded approximation property and the 1r�spaces. We also prove the local reflexivity principle which connects the local properties of X and X** . We prove the Auerbach lemma which allows a good identification of an ndimensional Banach space with n or ccn . We also study the concept of BanachMazur distance.
R
1. By local properties of a Banach space we mean the properties which depend on the structure of finite dimensional subspaces of the space. Some examples of such properties will be pointed out in this chapter and many more will be encountered in the sequel. The basic aim of this chapter is to provide an elementary under standing of local phenomena. Even at this early stage it is apparent that one needs a clarification of two points:
(a) how the general Banach space is built up from finite dimensional subspaces; (b) what are the relevant properties of finite dimensional spaces. Let us start with some definitions and examples which explain point (a) a little . What we are really thinking about in (a) is the approxima tion problem: how well can we approximate the identity operator on the space X by finite dimensional operators? 2. A Banach space X is said to have the bounded approximation prop erty (b.a. p.) if there exists a constant C such that for every finite subset { . , Xn } C X and for every c > 0 there exists a finite dimensional operator T: X + X such that li T Xj I $ c, for j = 1, . . . , n and
x 1,
.
xi
.
II T II $ C.

A Banach space X is said to be a 1r�space if there exists a family (Xy )rer of finite dimensional subspaces of X such that
3.
/'I ./'2 f there exists
E for every and Xy2 c Xra •
1'3
E
f such that Xy1 C Xy3
(1)
70
II.E. Local Properties §4.
U "Y E r Xy is dense in X,
(2)
for every 'Y E r there is a projection P"Y : X � Xy with II P"Y I :5 A.
(3)
Many similar properties can be thought of and have been investi gated in the literature but we will discuss only these and the existence of a basis. We have 4
Proposition. A Banach space with a Schauder basis is a 71). space A. A 11>.space has the bounded approximation property.
for some
Proof: Let (xn)�= l be a Schauder basis for X. We let Xn span(x k )k= l and Pn be the partial sum projection. One obviously has (1)(3) . Now let X be a 7r>. space and x� , . . . , Xn be a fixed finite sub set of X. Given b > 0 we use (1) and (2) to find 'Y E r such that dist(x3 , Xy) < b for j = 1, . . . , n. Fix x3 E Xy such that l txj  x3 11 < b for j = 1 , . . . , n. Then we have ll x3  Pyxj I :5 ll x3  Xj II + ll x3  Pyxj II :5 b + II Poy (Xj  Xj ) ll :5 (A + 1) b. This shows that X has the b.a. p.
5
a
Examples. (a) Let (!1, tL) be a probability measure space. Then Lp ( n, 1L), 1 :5 p :5 oo is a 1r1space. We define r as the set of all finite partitions of n into sets of positive measure. For a parti tion 'Y = (!1 � , . . . , !1n ) we put Xy = span{xn }j= 1 and Py (f) = j L:7= 1 tL(nj )  1 J fXni dtL · Xni . The properties (1)(3) are clearly sat isfied (see the proof of II.B. lO(b) or Exercise II.B. 14) . Observe that Lp(n, tL), 1 :5 p :5 oo, may not be separable and then it does not have a Schauder basis. Note also that for each 'Y we have ImPy � e; . (b) The disc algebra A(JD) has the bounded approximation prop erty. To see this, it is enough to consider operators Fn (f) = f * Fn where Fn is the nth Fejer kernel (I.B. 16) . It also has a Schauder basis (see III.E.17) but is not a 1r1space (Exercise III.E.9) . (c) Every C(K), K compact, is a 7r i +e;space for every c > 0. (Ac tually it is a 1r1space, but this requires more care.) This can be proved directly. The alternative argument is to note that C(K)** = L00 (tL) for some (usually nonafinite) measure 1L · From this and the principle of local reflexivity (Theorem 15) we infer that there exists an increasing net (Xoy) of finite dimensional subspaces of C(K) such that U Xy is dense
71
II.E. Local Properties §6.
x
in C{K) and d(X..,. , ta:,m ., ) � 1 + c. {The notion d(X, Y) is defined in 6. ) It follows from the HahnBanach theorem that there is a projection P..,.: C(K) �X..,. with li P..,. II � 1 + c. 6. Since all ndimensional spaces are isomorphic, in order to investigate finite dimensional spaces more precisely we will use a more quantitative notion. Let X and Y be two isomorphic Banach spaces. The BanachMazur distance between X and Y denoted as d( X, Y) is defined as
d(X, Y) = inf{ II T II · II T  1 1 1 : T: X�Y is an isomorphism} .
{4)
If the spaces X and Y are not isomorphic we set d(X, Y) = oo . The BanachMazur distance has the following, almost obvious, properties: (5) d(X, Y) � d(X, Z) · d(Y, Z); if X and Y are isometric then d(X, Y)
=
1.
{6)
Thus the BanachMazur distance is not a metric in the geometrical sense, but its logarithm is. We follow, however, the long established custom, and use the definition {4) and the name 'BanachMazur distance'. We have the following converse to {6). 7
Proposition. If X is finite dimensional and d(X, Y) = 1
isometric to Y.
then
X is
Proof: Let us take Tn : X + Y such that II Tn ii · II T; 1 II + 1. Multiplying Tn by an appropriate scalar we can assume II Tn ll = 1 for n = 1, 2, . . . . Since L(X, Y) and L(Y, X) are finite dimensional Banach spaces, passing to a subsequence we can assume II Tn  T il 0 as n + oo and II T; 1 S ll 0 as n + oo for some T E L(X, Y) and S E L(Y, X). Obviously T is invertible and r 1 = s. Moreover II T II = II S II = 1 so T is the +
+
desired isometry.
II
8. As an easy example of a computation of the BanachMazur distance we offer the following
Proposition. For every n = 1, 2, 3, . . . and p such that 1 � p � oo have d(l;, .ey) = nl t  ! 1 .
we
Proof. Note that for reflexive spaces X and Y, d(X, Y) = d( X* , Y* ) ; thus it i s enough to consider 2 � p � oo . T he upper estimate i s obvious
72
ll.E. Local Properties §9.
T:id: l�l�. T I 1. n; = I t, ±ei 1 2 � I t, ±Tei 1 2 � j I t, rj (t)Tei 1 2dt = Jt l t, ri (t)Tei (k) l 2dt = t t, l rei (k) l 2 = l rej t, r Tei 2 n ;! I T111 � n !; . f: T n 1 p if we take £;: + operator £;: + II.D. { 4) we get
In order to check the lower estimate take any with l i Analogously as in the proof of �
s�p
s�p
Thus i�f ii 3
•
and
ll �
of trigonometric polynomials of order Let us consider the space with Lpnorm. More precisely the elements of Tf:, � � oo, are trigonometric polynomials of the form 9.
We have
Theorem.
There is a constant C such that
d(Tn•P lp2n+l )  pp2 1 · = e'· <2n+ l) k = 0, 1, n 1 2 (2n +1 L ) 2 )1 .! p 2� 1 ( 2n 1+ 1 Ln 1
.
21tk
Let fh for . . . , 2n. The theorem follows from the following two estimates valid for every h E Tf:.
Proof:
.!.
l h( lh W p � 3 l lhllp.
� �
oo ,
p ,
<
(7)
k=O
llh ii P
�C
p
lh(Ok ) I P
k=O
p
<
oo .
(8 )
73
II.E. Local Properties §9.
Proof of (7) : Let Vn be the de la VallePoussin kernel (cf. I.B. 17.). Clearly h Vn = h for h E Tf: . Thus *
so from the properties of the Fejer kernel i.B.16 we get
Summing these inequalities we get
2n
E�m ak eikB with
Let us observe that for every trigonometric polynomial cp( 0) m � we have
2 '��"
2n + 1 kL=O cp(Ok ) = a.o = 21!' I cp(O) dO. 1
2n
1
0
{10)
Interchanging the order of summation and integration in {9) and apply ing {10) for .1'2n  1 and Fn  1 we get {7) .
Proof of (8) . The condition {8) follows by duality from (7) . Let Pn be the natural projection from Lp(T) onto Tf: . We know {see II.B.ll) that II Pn ll � (;�tl) . Take g E Lq(T), � + � = 1 , IIYIIq = 1 such that
74
Il.E. Local Properties § 1 0.
(27r)  1 J�'lr g(9)h(9)d9 = ll h ll p · ll h llp = = =
Then, using
(10) we get
21<
� 1 g (9)h(9) d9
2
0
21<
� I Pn (g)(9)h(9)d9
2
0
2n
LO Pn (g)(Bk )h(Bk) · 2 n + 1 k= 1
Applying Holder's inequality and
(7) we get
(8) .
•
One of the most powerful tools available for investigating finite dimensional spaces which does not exist in the infinite dimensional sit uation is volume. Since we can treat an ndimensional space as Rn (or cr:n ) with some norm it is clear what the volume is. The normalization is arbitrary but in most cases it does not matter. As an illustration of it let us prove
10.
Proposition.

Let X be an ndimensional Banach space. If {xj }f=, 1 is an cnet in Bx , then N � g  n for real X and N � c 2 n for complex X. There exists an cnet in Bx , {xj }f=, 1 with N � ( �t for real X and
N�
{� ) 2n
for complex X.
Proof:
Let us consider real spaces first. If {xj }f= 1 is an cnet in Bx then Uf= 1 B(xj , c) ::> Bx so vol Bx � vol
�
N
N
U B(xi , c)
j=l
vol B(xj , c) = N · vol cBx = N · en · vol Bx . L j= l
Thus N � e  n . In order to find an cnet of a small cardinality let us fix a maximal 2cseparated set, i.e. a maximal set {xj }f= 1 C Bx such that ll xi  xi I > 2c for i =f. j . Such a set is obviously an cnet in Bx . Since B(xi , c) n B(xi, c) = 0 for i =/: j and B(xi , c) c (1 + c)Bx for j = 1, . . . , N we have N ·c n · vol Bx
=
vol
(ldN B(xi ,
c)
) � vol((1 +c)Bx ) = (1 +c) n vol Bx
75
II.E. Local Properties § 1 1 . n
so N :5 ( �) . Since an ndimensional complex Banach space is a 2n dimensional real space, the claim for the complex spaces also follows.a Another application of volume considerations is the following. 11 Lemma. {Auerbach) . Let X be an ndimensional Banach space. There exists a biorthogonal system (xj , xj ) j= 1 in X x X* with ll xi ll = 1 and ll xj II = 1 for j = 1 , . . . , n.
Let (yj , yj ) j= 1 be any biorthogonal system in X. The function V(zt , . . . , Zn ) = det(yj (zi )) i,j = l is a continuous function on the nfold product of Bx . Let V (x t , . . . , X n ) = max{V{zt , . . . , Zn ) : ll zi I :5 1 for j = 1 , . . . , n} . Clearly ll xi ll = 1 for j = 1 , . . . , n. We define Proof:
A moment's reflection gives that (xi , xj) j= 1 is the desired biorthogonal system. a As an application of Auerbach's lemma we show certain general perturbation results. 12 Proposition. (a) Let X be a Banach space and let (X,), e r be a fantily of subspaces satisfying (1) and (2). Let Xo be a finite dimensional subspace of X. For every c > 0 there exists a 'Y E r and XfJ C X, such that d(Xo, XfJ) < 1 + c. {b) Let X have the b.a.p. Then there exists a constant C such that for every finite dimensional subspace Xo C X there exists an operator T: X + X with II T II :5 C and T I Xo = idx0 • Proof: Let ( xj , xj ) j= 1 c X0 x X0 be given by Auerbach's lemma. In order to show (a) fix 'Y E r such that there are Xj E X, with I xi  Xj I :5 �  Define T: Xo + X, by T(xi ) = Xj and put XfJ = T(Xo). For x E Xo we have
I f: l I f: l t It ll x ll =
J=l
xj (x)xj =
J=l
xj (x) xj +
Since
xj (x)(xi  Xj ) :5
we get
lxj (x) l ll xi  Xj ll :5
:5 8 11 x ll
l
f: xj (x)(xi  Xj ) ·
J=l
t ll xj ll · ll x ll · ll xi  xi I
76
II.E. Local Properties § 1 3.
II Tx ll  6ll x ll :5 ll x ll :5 II Tx ll + 6ll x ll ·
If 6 is suitably chosen this gives ( a) . To show ( b ) we use the defini tion of b.a.p. and we fix an operator T1 : X + X such that II T1 (xi )  xi ll :5 � and II T II :5 cl . We define T(x) = Tl (x) + L:j= l xj (x)(xj  Txj) where xj is a HahnBanach extension of xj to a functional on X, j = 1, . . . , n. Clearly T(xj ) = Xj for j = 1, . . . , n so T I Xo = idx0 • Also
II Tx ll :5 II T1x ll +
n xj (x) x  Tx l l ll i i ll L j=l
Thus actually ( b ) holds with any constant C bigger than the constant given by (2) . a For every ndimensional Banach space X and for > 0 there exists an embedding i: X + t� with (1  c) ll x ll :5 ll i(x) ll :5 ll x ll where N :5 1 �e if X is a real space and N :5 1 �e if X is a complex space.
13 Proposition.
every c
( r
( fn
Let (xj )f= 1 be an cnet in Bx· given by Proposition 10. We define i: X + t� by i(x) = (xj (x))f= l · Given x E X let x* E Bx· be such that ll x ll = x* (x), and let j be such that ll xj  x* ll < c. Then
Proof:
l xj (x) l = l x* (x) + xj (x)  x* (x) l � l x* (x) l  l (xj  x*)(x) l � (1  c) ll x ll . a Thus ll i(x) ll � (1  c) ll x ll . Obviously ll i ll :5 1.
Remark: This result is a finite dimensional or 'local' version of Theo rem II.B.4. Actually without an estimate for N this result easily follows from Theorem II.B.4. 14. Now we will return to the interplay between local and global properties of Banach spaces. We start with the important Theorem. (Principle of local reflexivity.) Let X be a Banach space and let E c X** and F C X* be finite dimensional subspaces. Given c > 0 there exists an operator T: E + X such that
I I T II · II T  l I T(E) II :5 1 + c, T J E n X = id, f(Te) = e(f) for all f E F and e E E.
(11)
(12) ( 13)
77
Il.E. Local Properties § 1 5.
This theorem asserts that finite dimensional subspaces of X** are basically the same as finite dimensional subspaces of X. The proof relies on the following
and let X** .
A;
(A; )f= 1
be bounded, normopen convex subsets of X denote the norm interior of the a( X** , X * ) clos ure of A; in
15 Lemma. Let
N N If n;=1 A; =f. 0 then n;=1 A; =f. 0. (b) If we have a map T: X Y with Y space then T** C n_f= 1 A; ) = T(nf= 1 A;). 
(a)
+
Proof:
a finite dimensional Banach
(a) Let XN be the direct sum of N copies of X. The set
is a bounded, normopen, convex subset of XN . If n.f=, l A; = 0 then A n V = 0 where V = {(x;)f= 1 E XN : x; = x1 for j = 1 , 2, . . . , N} . Let
A  {( X;**) jN= l
and
{( X;** )jN= l
E X** N ·. X;** E
3,
A ·
X;**
J. 
1 , ... , N}
= x 1** £or J = 1 , . . . , N} . If A n V = 0 then there exists 0 for all a E A. Since A is a(X** , X* )dense in A (use Goldstine's theorem II.A. 13) we get 0 for all a** E A, N so A n V** = 0, i.e. n A; = 0 . j= l N N (b) Clearly T** C n;=1 A; ) and T(n;=l A;) are open, convex sets and N N T(ni = l A; ) c T** (n;=l A;). If they are not equal, then there exists a point p E T** C nf= 1 A; ) and a functional
(3 V**
such that
=
E X** N
· •
.
N
a > (3 >
x* E X* equal T* ( a}. Since ij = A; n {x** E X** : x** (X* ) > a} see that nf= 1 ij =f. 0 but nf= 1 Aj is empty. This contradicts (a) and Let
so proves (b) .
we
a
II.E. Local Properties § 1 5.
78
Let dim E = n and dim E n X = n  k. Fix a biorthogonal system (xj* , ej ) j=1 in E x E* such that span(xj* ) j= k+ l = E n X, and ll xj* ll = 1. The identity id: E � X** can be written as id ( e) = Ej=1 ej ( e )xj* . We want to find X t , . . . , Xk in X such that the map T: E � X defined as T (e) = E7=l ej (x)xi + E�+ l ej (e)xj* will have the desired properties. Property ( 12) is satisfied with this definition. Let Z be the direct sum of k copies of X and let 6 > 0 be a small number. Fix the following finite sets: { /i }� 1 is a basis in F, {xj }� 1 c Bx· is such that for every e E E ll e ll $; (1 + 6) sup{ l xj (e) l : j = 1, . . . , R } , { ei }_f= 1 i s a 6net i n BE. We have n "'""' A, jr X r** e3.  L..J
Proof of Theorem 14.
r= l
•
Now we form the following subsets of Z: Cj
= { (x s )==l : and ll x s ll
I t, A
� Xs +
< 1 + 6,
s
t
l
,\�x : • < I l ei II 1 8 = 1, . . . k}, j = 1, . . , N. .
Those subsets of Z are normopen, bounded and convex. Since (x: • )==l E n_f=, 1 Ci C Z** (where  has the same meaning as in Lemma 15) Lemma 15 gives (x s )== l E nf=1 Ci =f. 0 . Let us consider an operator s : z � RM · k $ R R · k (or into ccM · k $ (CR · k ) , defined as
From Lemma 15 we infer that there exists (x 8 )==l such that (x s ) != t E and
N
n=l Ci
j
S((x s)== l ) = S ** ((x : • )==l ) ·
With this choice of (x8 )== l we clearly get (13). Also we get ixj ( Te) i II Te ll ;::: j =sup l , ,R . . .
=
sup ixj (e) i ;::: ( 1 + 6 )  1 ll e ll . j = l , ,R ...
( 14)
79
II.E. Local Properties § 1 5.
Given e E BE let us fix have II Tei ll :5 ll e; ll · So
II Te ll
:5
ei
with
II Tei ll + II T(e  e;) ll
A very crude estimate for
:5
li e  ei ll
:5
6.
Since
(x s )!=1
E
C;
li e; II + II T II · 6 :5 ll e ll + 6 + 6II T II .
li T II is ( 1 + 6 ) :E7=1 ll ej II 6 is small enough ( 1 1 ) follows from ( 14 ) and (15 ) .
:5
2 :E 7= l
we
(15 )
ll ej II , so if a
15. We say that a Banach space X is finitely representable in a Banach space Y if there exists a constant C such that for every finite dimensional subspace X1 c X there exists Y1 c Y with d(X1 , YI ) :5 C. In other words X is finitely representable in Y if finite dimensional subspaces of X are subspaces of Y. Note that Proposition 12 yields that every Banach space X is finitely representable in c0 • From Proposition 12(a ) and Example 5(a) we get that Lp (O, ft ) is finitely representable in lp for 1 :5 p :5 oo. On the other hand from II.D. (4) we easily get that lp for p > 2 is not finitely representable in any Lq(O, ft ) for 1 :5 q :5 2. Theorem 14 shows that X * * is finitely representable in X . Notes and remarks. The general idea of approximation of a separable Banach space by finite dimensional ones is quite old and was around in Lw6w in the thirties. Banach ( 1932] asked the question if every separable Banach space has a basis. A notion of approximation property (a. p. ) , the concept even weaker than b.a.p. , was also invented then. We say that a Banach space X has the approximation property if for every normcompact set K c X and for every e > 0 there exists a finite dimensional operator T: X � X such that sup { II Tk  k ll : k E K} :5 e. The first deep study of a.p. and b.a.p. is contained in Grothendieck ( 1955] . The notion of 71'Aspace emerged in the sixties (see Lindenstrauss ( 1964] and MichaelPelczynski ( 1967] ) . The real breakthrough in the study of approximation proper ties come with Enflo's ( 1973] example of a Banach space without a.p. Many examples differentiating various approximation properties have been produced later. We refer the interested reader to Lindenstrauss Tzafriri ( 1977] and ( 1979] , Pisier ( 1986] and Szarek ( 1987] . All this is a very fascinating subject but beyond the scope of our book. As a first result of the local theory of Banach spaces one can consider the following well known fact proved in Jordanvon Neumann ( 1935] . Ac tually the analogous three dimensional characterisation was given earlier by Frechet [1935] .
80
II.E. Local Properties §Exercises
A Banach space X is isometric to a Hilbert space if and only if for all x, y E X . Note that this result implies that X is isometric to a Hilbert space if and only if every twodimensional subspace of X is isometric to a Hilbert space. For an isomorphic version of this see Exercise 9(b) . The local theory of Banach spaces gathered momentum in the sixties with the study of L1preduals (see Lacey [1974] ) and pabsolutely summing operators (III. F) . Today it is a vast subject having connections with operator theory, harmonic analysis, geometry of convex bodies etc. Some of it will be presented later. For a more detailed presentation of different aspects of the theory, the reader should consult MilmanSchechtman [1986] or TomczakJaegermann [1989] or Beauzamy [1985] . The notion of BanachMazur distance is in Banach [1932] . Theorem 9 is a classical result of Marcinkiewicz [1937a] (see also Zygmund [1968] chapter X §7) . Proposition 8 is an easy special case of a result of Gurarii KadecMacaev [1965] . The Auerbach lemma is mentioned without proof in Banach [1932] . The principle of local reflexivity ( Theorem 14) was proved in LindenstraussRosenthal [1969] . Exercises
1.
A reai Banach space X is called uniformly convex if there exists a function c,o (e) > 0 for e > 0 (called the modulus of convexity) such that if x, y E X, l l x l l = IIYII = 1 and l l x  Y ll > 2e then II (xty) I $ 1  c,o(e) (draw the picture) . (a) Let X be uniformly convex and let x• E X* , llx* ll that diam{x E X: l l x l l $ 1 and x* (x)
=
1. Show
> 1  e:} + 0 as e + 0.
(b) Show that uniformly convex spaces are reflexive. (c) Show that Lp [O, 1 ] , 1
< p < oo,
is uniformly convex.
(d) Show that if L:;::"=1 Xn converges unconditionally in X then n= l (e) Suppose that X is a uniformly convex space and that Y C X is a closed subspace. Show that for every x E X, there exists a unique y E Y such that ll x  Y l = dist(x, Y) .
i
81
Il.E. Local Properties §Exercises
2.
Let X be a complex Banach space. We say that X is complexly uniformly convex if there exists a function cp( e) > 0 for e > 0 (called the complex modulus of convexity) such that if y E X with IIYII � f: and llx + ei 6 y li :5 1 for all () then ll xll :5 1  cp(e).
x,
(a) Show that
L 1 [0, 1]
is complexly uniformly convex.
(b) Show that if cp is a complex modulus of convexity of X and the L: ::'= 1 Xn converges unconditionally in X then
L: ::'= 1 cp( ll xn ll ) <
00 .
,
=
3.
Find two Banach spaces X and Y such that d (X Y) Y are not isometric.
4.
Let T;:" be the space of trigonometric polynomials of the form L � n dk ei k(l with the supnorm. (a) Show that d (T;:"
, £�+ 1 ) :5 C log(n + 2) for n
=
1 , 2, 3, . . . .
(b) Show that T;:" contains a subspace isometric to
5.
1 but X and
£'�+ 1
Let T� 2 be the space of trigonometric polynomials of degree at most N in two variables (i.e. f(O , t) E T� 2 if and only if ' f(O, t) = L:: m = N an, m ei n(leimt ), equipped with the norm 11/llv =
; ( (411" 2 )  1 f02., fg.,. 1/(0, t) I P dOdt) t . cp2 � for 1 < p < oo .
Show that
,
2 d (T� 2 42N+ 1 ) ) < .
6.
Let X be an infinite dimensional Banach space. Show that there is no translation invariant Borel measure J.L on X such that J.L(U) > 0 for every open set U and such that J.L(UI ) < oo for some open set U1 . Translation invariant means that J.L(A + x ) = J.L(A) for every x e X.
7.
Suppose that T: X  Y. Let (Ya)aer be a net of finite dimensional subspaces of Y, ordered by inclusion and such that Uaer Ya = Y. Assume that there is a C such that for each a E r there is an operator Sa : Ya + X such that l i Sa I :5 and TSa = idy"' . Show that T * (Y * ) is complemented in X* .
onto
c
8.
(a) Show that £2 is isomorphic to a complemented subspace of (b)
( L: ::'= 1 £�) 00 . Let Mn denote the set of all ndimensional Banach spaces (up
to isometry, i.e. we identify isometric spaces). Show that for every n E N the set Mn with the BanachMazur distance (or rather, its logarithm) is a compact space.
82
Il.E. Local Properties §Exercises
(c) Let (Bk )� 1 be a sequence of finite dimensional spaces such that {Bk } � 1 n Mn is dense in Mn for all n E lN. Show that for every separable Banach space X the space (:�:::.: ;:'= 1 Bk) 00 contains a complemented copy of X* . 9.
(a) Show that if X is a Banach space finitely representable in a uniformly convex Banach space Y (Exercise 1 ) , then X has an equivalent uniformly convex norm. (b) Show that if a Banach space X is finitely representable in £2 then X is isomorphic to a Hilbert space.
10. A Banach space X has the uniform approximation property ( u.a.p.) if there exist a constant C and a function cp( e, n) , n E lN, e > 0, such that for all X 1 , . . . , Xn E X there exists an operator T: X + X such that IITx; II :::; e l l x; II for j = 1 , 2, . . . , n and liT II :::; C and dim T ( X ) :::; cp ( e , n) . Show that Lp [O, 1] , 1 :::; p :::; oo, have the
x2 , 
x;
u.a.p.
1 1 . Show that
£� is isometric to a subspace of £00 but is not isometric to any subspace of co .
Part III Selected Topics III . A Lp Spaces; Type And Cotype. In this chapter we investigate the Lp (JL)spaces, 1 < p < oo . We start by proving the isomorphisms of some natural spaces to spaces Lp (JL) . We show that the Sobolev space W� ("11'2 ) is isomorphic to Lp ("11'2 ) for 1 < p < oo and that the Bergman space Bp (D) is isomorphic to lp for 1 $ p < oo . Along the way we prove a useful criterion for the boundedness of integral operators on Lp(f!, JL) (Proposition 9). Later we borrow from probability theory and show the existence and basic properties of stable laws. These provide isometric embeddings of lp into Lq , q $ p $ 2. We continue the line of thought started in II.D.6 and introduce the general notion of type and cotype of a Banach space. In order to study these notions efficiently we prove the vector valued generalization of Khintchine inequality (Kahane's inequality) . A gener alization of a classical result of Carleman from the theory of orthonormal series is also presented. We conclude this chapter with the BanachSaks theorem and its generalizations to almost everywhere convergence. 1.
We start this chapter with some general observations.
A separable space Lp (f!, JL) , 1
to one of the following spaces: e; , n
Proposition.
t;)p, n = 1 , 2, . . . , (Lp (O, 1]
E9
lp)p o
=
$
p
<
oo,
is isometric
1, 2, . . . , lp , Lp [O, 1] , (Lp (O, 1]
E9
The proof is an immediate consequence of the characterization of nonatomic, separable measure spaces given in I.B. l. Let us also note that the above list contains at most two nonisomorphic infinite dimen sional spaces, namely ip and Lp (O, 1] (see II.B Exercise. 1 1 ) . That for P =I= 2, 1 $ p < oo these spaces are really nonisomorphic follows from Propositions 5 and 7. Thus there are rather few separable Lpspaces. Some questions about nonseparable Lpspaces can be reduced to the separable case using
84
III.A LpSpaces; Type And Cotype §2.
2 Proposition. Every separable subspace X c Lv(f2, f..L ) , 1 � p < oo, is contained in a separable Y c Lv(n, f..L ) isometric to some Lv(n, f..L I ) .
)� 1 A;,a ,b ·
Proof: Let us fix a countable dense subset ( x; in X, and consider sets = { w E n: a < x; ( w ) < b } where a, b are rational numbers. Let E be the aalgebra generated by all sets The space Lp(f2, E, f..L ) a of all Emeasurable, pintegrable functions is the desired Y.
A;,a ,b
3. One of the reasons why Lpspaces are important is that many other spaces common in analysis are isomorphic to Lpspaces (see II.B.Exercise 9). We want to present one more example of this type. Proposition. The space WJ (']['2) is isomorphic to Lp (1['2 ) , 1
< p < oo .
(an,m )�.:=  oo
For a doubly indexed sequence of numbers we say n i i that ( a ) is a multiplier on Lp(']['2) if the map L. } (n, m)e 6 l e m 62 1+ E a f ( n, m)e i n61 eim()2 extends to a continuous operator from Lp(Y2) into Lp(Y2 ) .
n,mn
,m
4 Lemma.
The doubly indexed sequences
( ( 1 + l nl+ l m l ) ) :.m=  oo
Lp(Y2) for 1 < p <
oo .
and
( ( l + l ni+ l ml ) ) :.m=  oo
( ( l +l ni+ l ml ) ) :.m=  oo '
are multipliers on
This Lemma is a special case of the multidimensional multiplier Theorem I.B.32. Proof of Proposition 3. Let T: Lv(']['2) + WJ (Y2) be defined by It follows easily from Lemma 4 the multiplier that T is continuous. We define E: WJ (Y2) + Lp(Y2) by E( f ) = f + 8d + 82 / where
( ( l + l ni+ l m l ) ) :,m=  oo .
1 ( n,Lm }(n, m)ein61 eim62 ) n,Lm l n i }(n, m)ein61 eim62
8 and
n,Lm
fh is defined analogously. } n (n, m)ein61 eim62 E Lp(']['2)
=
Since f E WJ (']['2) the function and using the Riesz projection (see
Theorem I.B.20 ) in the variable fh we get that 81 : WJ (Y2) + Lp(Y2 ) . Analogously for [h , so E is continuous. A routine calculation shows that E Tf = f for f E Lp(Y2). Since E is clearly 11 we get the desired isomorphism. a
85
Ill.A LpSpaces; Type And Cotype § 5.
Now we will investigate the spaces lp .
5.
Proposition. Let X be an infinite dimensional subspace of lp , 1
:5 p <
or of Co . Then X contains a subspace y such that y "' ep (or Co ) and is complemented in lp (or eo). oo,
(zn)�=1 =
be a blockbasic sequence of the unit vector basis Proof: Let in ep (or co , so Zn E Z:�11 with kn increasing to 00 and = 1. One checks that span{zn }�=1 is isometric to Let z� b e a functional and z� = E Z:�11 (or co such that z� (zn) on 1 by P n 1, 2, 3, . . . . We define P: = L:::'= 1 z� (x)zn . P is algebraically a projection onto span{zn}�=1 and
)
i =v
)
a ; e;
l zn l
iv . = = iz iv + iv l �(x)I
/3; ej ,
v )e x( I P(x) l v = ( n� 1 i z� (x) I P) ; :5 ( n� 1 � z�( k,.+ ; E1 j ) l ) ; :5 ( � I E x(j )e; I P ) t n  1 k,. + 1 :5 ( � l x (j ) I P ) p = l xl v so I P I = 1 . From II.B. 17 we infer that X contains a sequence (xn)�=1 very close to such a (zn)�=1 and II.B.15 gives that span(xn)�=1 is com 00
!.
plemented in lp (or Co) and isomorphic to span(zn)�=1 so to ep (or eo).a Using the above Proposition 5 and Theorem II.B.24 we get
ip , 1 :5
p
a
Every infinite dimensional complemented subspace of
6 Theorem.
< oo , or of Co is isomorphic to the whole space.
iv
7. This simple structure of complemented subspaces of is rather exceptional. For example Lp [O, 1] , 1 < p < oo, contains complemented subspaces isomorphic to Hilbert space as well as those isomorphic to or Lp [O, 1] itself (see II.B.2(b)) .
iv
Proposition. Let (rn)�= 1 b e Rademacher functions. Then the space span(rn)�= 1 Lp [O, 1] is isomorphic to £2 for 1 :5 < and is comple c
mented for 1
< p < oo .
p
oo
86
Ill.A LvSpaces; Type And Cotype §8.
Proof: The first claim is just the Khintchine inequality I.B.8. For the second let P: L2 [0, 1] + L2 [0, 1] be an orthonormal projection onto span(rn)�=l · For oo > p � 2 the Khintchine inequality gives
for By duality we get P: Lv [O, 1] + Lv [O, 1] for 1
f E Lv [O, 1] .
< p < oo .
a
Remark: The same property also holds for span(ei n k9)r=l • for any lacunary sequence (n k )� 1 , i.e. any n k such that inf k (n k+ l /n k ) > 1 . The proof i s exactly the same, only uses the analogue of the Khintchine inequality for lacunary sequences of characters (see I.B.8) .
lzl
{z
v
Let D = E ([: < 1} and be Lebesgue measure on D. Bv c Lv (D, denotes the Bergman space Bv (D) (see I.B.28 for definitions) . 8.
dv)
Theorem. For every 8
Ps f(z)
=
>0 8
the operator
+1 7r
f l wzl w)2t2f(+sw) dv(w)
D
(1 (1 
is a continuous projection from Lv (D,
dv) onto Bp , 1 :5
(1) p
< oo .
Note that this theorem is false for p = oo . Obviously Boo = H00 and there is no continuous linear projection from L00 onto H00 ; see re marks after Proposition III.E.15. In the proof we will need the following criterion for the boundedness of integral operators on Lv ( n , J.L ). 9 Proposition. Let (n, J.L ) b e a measure space an d let K(w t , w2 ) b e a measurable function on
n
X
n. Let us define
Then
(a) if I supessw 1 E n n
I K(wt , w2 ) 1 dJ.L (w2 ) < oo then
(b) if supessw2E n I I K(wt , W2 ) idJ.L(Wt ) n
< oo
then
87
III.A LpSpaces; Type And Ootype § 1 0
( c ) if 1
< p < oo and there exists a measurable positive function 0 and constants a, b such that for � + /J = 1 we have
g on
J I K(wt , w2 ) ig(wi )P' dJ.L(wi ) :::; [ag(w2 )]P' J.L  a.e. J I K(wt , w2 ) lg (w2 )PdJ.L(w2 ) :::; [bg(w1 )]P J.L  a.e.
!1
and
!1
Proof: The argument for ( a) and ( b ) is obvious, so we will prove ( c ) . We have
I T J (w 2 ) l I K(wl , w2 ) i l f(wl ) l dJ.L(w i )
:::; J J [I K(w1 , w2 ) l ? g(wi )] [I K(wl , w2 ) l � l f(wl ) l g(wl )  1 ]dJ.L(wi ) < ag(w2 ) { J I K(wl , w2 ) i ( l f l /g) P (w l )dJ.L (w i ) } ;; . !1
·
!1
1
·
Hence
!1
[! ( '�1 r(wl ) J gP (w2 ) I K(wt , W2 ) idJ.L(w2 )dJ.L(WI )] :::; ab [ j ( l f l fg ) P (wi ) gP (wi ) dJ.L (WI ) ] ab [ J if (wi ) I P dJ.L(wl ) ] ;; .
II T/IIv :::; a
!1
!1
=
10.
!1
1
p
1
v
1
!1
a
The other fact we will use is the following For every a and s such that  1 < a < 1 and cons tant Co:,s such that for l w l < 1 we have
Lemma. exists a
s
>
0, there
(2)
III.A LpSpaces; Type And Cotype § 1 0.
88
> 0 such that for all p with 0 p 1 1 1 'Y pei8 1 'Y (1  p + 1 0 1 ) . This gives � 2� d� 2 s < J 1 1  pe•8 1 2+s  'Y / (1  p + 1 0 1 )  2 sdO
Proof: There exists a constant and all with < 1r we have
0
101
::£
�
0
5
::£
�
Cs J o  2  s dO (3) 1p 1 Cs (1  p)  s . Since the integral in (2) depends only on p l w l , passing to polar coordinates and using (3) we get  l z� (1,..:1' 22)"+:s dv(z) 1 2� 1 r (1  r•2. ) a2+s drdO (4) 211" J J 1 1  pre 8 1 J 1 1  zw l 1 /1 (1  r2 ) a (!2� dO + ) dr 211" 11  prei8 1 2 s 1 C (1  r)"(1  pr)  1  8 dr. ::£
::£
=
=
D
0
::£
0
0
0
J
::£
0
Integrating the last integral by parts we see that it equctl.s
1
c1 + c2 J (1  r) 1 +"(1  pr)  2  8 dr 1 c1 + c2 J (1  pr) 1 +" (1  pr) 2  8 dr Ca,s (1  p) a  s , 0
::£
0
::£
so we get (2) .
( 1) 1 , v). (z) P*s (f) ( z ) s + 1 {1  l z l 2 ) s J {1g(z)d  zw)� 2+s
a
Proof of Theorem 8. First we show that defines a bounded operator on L ( D The adjoint operator is given by the formula =
7r
D
(5)
89
III.A LpSpaces; Type And Cotype § 1 1 .
()
P; is bounded on ( ) g (z) = (1  l z l 2 )  p\
so we infer from Proposition 9 b and Lemma 10 that L00 (D, so is bounded on L1 (D, For 1 < p < oo we apply Proposition 9 c for where � + � = 1 . Lemma 10 yields
v) P8
v).
$
[Cg(zW
and analogously
P8
so is bounded. Clearly for f E L1 (D, 11 ) , Moreover for n = 0, 1 , 2, . . . we have
P8 (!) is analytic in D.
Ps (zn )(z) = s +1r 1 J ( 1( 1l wzwl 2))2s+wsn dv(w) 2 + ;) r s 1 (zw) k dv(w) = ; [ ( 1  l w l 2 ) s w n � ��72 +s s + 1 r(� 7 2 + ; ) zn / ( 1  w 2 ) s w 2n dv(w). = ll ll 1r n. r 2 + s ][)
][)
P8 (zn ) zn
Evaluating the last integral in polar coordinates and using the well for known properties of Euler's beta function we get that = n = 0, 1 , 2, . . . . This shows that a is a projection onto Bv (D) .
P8
As an application of Theorem 8 we show 11 Theorem.
The space Bp is isomorphic to fp , 1 $ p <
oo .
The proof of this theorem will be based upon the following 12 Lermna. Every compact operator T: X
X
T
Lp
Q� f � p
+
Lp admits a factorization
90
III.A LvSpaces; Type And Cotype § 1 2.
with
ll a ll · ll fi ll
:::; 8 II T II ·
Proof: Since T is compact and Lp is a 1r1space (II.E.5(a)) there exists a sequence of normone projections Pn : Lp + Lp such that II T  PnT II :::; 4 n and d(ImPn , f�im ( ImPn ) ) = 1, for n = 1, 2, . . . . We identify fp with o = �=l ImPn )p and we define
and ,B ( fn ) = 2:: �=1 2 n + 1 fn · One checks the desired properties.
a
Proof of Theorem 11. Let us fix an increasing sequence of numbers (rn ) ;:,o= l tending to 1 with r1 > 0. Let us put
IDo =
{z E ID: l z l
::=;
r1 } and IDn =
{z E ID:
rn <
lzl
::=;
Tn + l } ·
Let In : Bp + Lp (IDn , dv) be the natural restriction operator. From I.B.28 and a standard normal family argument we infer that each In is compact. Let (an , fin ) be a factorization of In given by Lemma 12, with II an II = 1, II fin II :::; 8. Thus we have the commutative diagram
where a( f) = (an (f i1Dn ) ) ;:,o= 1 , I denotes the identity embedding, ,B ((xn ) ;:,o= 1 ) = ( ,Bn (xn ) ) ;:,o= l and � ((fn) ;:,o= l ) = L�=l fn and P is any projection onto Bp (see Theorem 8) . Since PI = idBv we get that a is an isomorphic embedding of Bp into (�fp)p � fp and P�,B is a projection a onto a( Bv) . Theorem 6 gives the claim. Remark: It is easy to see from (5) that Im(P; ) C L 00 (1D, v) is exactly
{f(z): f(z) = (1  l z 2 l ) s · g(z) with g(z) analytic} so we infer from Theorem 1 1 that the space Xs of all analytic functions f( z ) such that suplzl < l ( 1  l z l ) s lf (z) l < oo is isomorphic to foo , for s > 0.
III.A LpSpaces; Type And Cotype § 1 3.
91
13. Our aim now is to introduce the socalled stable laws. These are well known probability distributions, but because of their importance in Banach space theory we will discuss them here in some detail. Let us recall some general notions. To each real valued random variable f on a probability space ( 0, P) there corresponds a probability measure Ill on JR, called the distribution of J, determined by the relations P { w : f (w) < .>.} = Ill ((  oo , .>.)), >. E JR. Conversely for each probability measure p, on JR there exist random variables f such that p, 1 = p, . It is clear and well known that the integrability properties of f are reflected in properties of p, 1 . More precisely we have the following formula:
I
n
F (J(w) ) dP(w)
=
+co
I

oo
F ( x ) dp, 1 ( x ) ,
(6)
valid for every bounded continuous or positive continuous function F : JR + JR.
14 Theorem. For every 0 < p � 2 there exists a distribution P,p such that
Ico eio:xdp,p(a) = e lxiP . 00
(7)
Every function ( random variable ) whose distribution equals P,p is called pstable. Those variables do not exhaust the class of all pstable variables considered in probability theory. This is a simple special case, but sufficient for our purposes. Proof: Note that (7) defines what in probability theory is called the characteristic function and in harmonic analysis the Fourier transform of the measure p, . Its basic properties are well known and can be found in many books, e.g. Katznelson [1968] Chapter VI. Let B denote the class of functions on JR which are Fourier transforms of positive measures on JR. This class satisfies the following properties:
if JI , h E B, a1 , a2 ;::: 0 then ad1 + a2 !2 E B; (8) if ft , h E B then fi · h E B; (9) if Un )�=l C B and fn converges almost uniformly on JR to f then f E B. ( 10)
92
III.A LvSpaces; Type And Cotype § 1 4.
Our aim is to show that e l x l �' E B, 0 < p $ 2, because then we get from (7) that JLp(1R) 1 , so JLp is a probability measure on 1R . The case p 2 is the classical Gaussian ( normal ) distribution so dJL 2 (x) ., 2 (211") 21 e T dx. From now on we assume 0 < p < 2 . From the formula
=
=
=
(11)
which is easy to check using the substitution �
=
u
we get
Approximating this integral we see that e  lx i P is an almost uniform limit of the functions
So it is enough to check ( see (9) and (10)) that exp
(  1 +{! ) 2 ) E
B.
But
Since the convergence is almost uniform, from (8) , (9) and ( 10) we see that it is enough to check that (x 2 + b2 )  1 E B. This follows from the formula
00co = � cos ax e ba da
J 0
( 1 2)
III.A LpSpaces; Type And Cotype § 1 5.
93
which is easily verified using integration by parts twice.
a
15 Proposition. Let I be a pstable function on a probability measure space
(f!, JL).
( a) if p
Then q
= 2 then I E Lq (f!, JL) for 0 <
<
oo,
( b) if p < 2 then I E Lq (f!, JL) for 0 < q < p. The case ( a) easily follows from (6) , since we know that .,2 dJL2 (x) = (27r) 21 e  T dx. Let p < 2 and I be a pstable variable with the distribution JLp satisfying (7) . We have to estimate J�:; l x l q dJLp(x) ( see (6) ) . Since l x l q = Cq J000 (1  cos xt)r 1  q dt ( substitute xt = u here the condition q < 2 is important ) we get using (7) Proof:
oo
Joo
Joo 10 �lcz: xt dtdJLp(x) oo + oo Cq I t l�q I cos xt) dJLp(x)dt 0 oo oo + oo Cq I t l�q I Re eixt )dJLp (x)dt 0 oo 00 e  tP ] dt. Cq I t l +q 0
ixl q dJLp(x) Cq
=

=
=
=
oo
Substituting in the last integral
00
I0 which is finite for 0 <
1  e  tP
1
(1 (1 
1 [ 1
tP = u we get 1
00
I
t l +q dt = p 0
q
1  e u !±.i
up
du, a
< p.
16. Suppose now that (/n)':'= 1 is a sequence of independent pstable functions on a probability measure space ( f!, JL). Let ( an )':'= 1 be a finite sequence of real numbers with E:= l ian i P 1 and put E:=l an ln ·
=
I=
94
If
III.A LpSpaces; Type And Cotype § 1 7.
J.Lf is the distribution of f then we have + oo Ioo eiaxdJ.LJ (a.) nI eixf(w) dJ.L(w) =

=
=
In n:fi= l eixanfn (w) dJ.L(w ) :fi I eiXanfn (w) dJ.L(w) n= l n II e  l an xi P 00
=
n= l =
Thus
f is also pstable.
exp 
00
(L:: l an i P ) I x i P
=
n= l
e lxiP .
In particular ( see Proposition 15) we have
( )
J.L)
Let Un)':= 1 be a sequence of independent pstable func tions, 0 < p � 2. The span fn ':= 1 in real Lq (S1, is isometric to f.p if q < p and p < 2. For p 2 it is isometric to £2 for 0 < q < oo. D
Corollary.
=
17. We now wish to return to the circle of ideas connected with un conditional convergence of series in Banach spaces which were discussed in II.D.7. Motivated by Orlicz's theorem and in particular by II.D. ( 4 ) we introduce the following definitions. Definition. A Banach space X is said to have cotype p, 2 � p � there exists a constant C such that for all finite sets
(xi ) J= l
oo,
if
CX
( 13)
A Banach space is said to have type p, 1 � p � 2, if there exists a constant C such that for all finite sets ( xi ) j=1 C X ( 14)
III.A LpSpaces; Type And Cotype § 1 8.
95
Recall that (rj)� 1 are the Rademacher functions. A few comments about these definitions are in order. ( a) Since scalars satisfy neither ( 13 ) for p < 2 nor (14) for p > 2 ( see Khintchine's inequality I.B.8 ) we see that the above restrictions for p are essential if we hope to get nontrivial concepts. ( b ) If a Banach space X has type p and cotype q and Y is finitely representable in X then Y also has type p and cotype q. In particular X and X** have the same type and cotype; cf. principle of local reflexivity II.E. 14. ( c ) Every Banach space X has type 1 and cotype oo . Also if X has cotype p it has also cotype q for q > p and if X has type p it has also type q for q < p. (d) The smallest constant for which ( 13 ) holds for a given space X is called the cotype p constant of X and is denoted Cp(X). Similarly we define the type p constant of X, denoted by Tp(X). The following vector valued generalization of the Khintchine in equality is a fundamental tool for investigating types and cotypes. 18 Theorem. (Kahane's inequality) There exist constants Cp , 1 � p < oo such that for every Banach space X the inequality
holds for every finite sequence
(xi ) J= l
C
X.
The proof follows from the following distributional inequality. Let V(t) = II E.i=o ri (t)xi ll · {t: V(t) > 2a} l l � 4 l {t: V(t) > aW .
19 Proposition. we have
Proof: For k � n let us put the following sets:
Then for every
Vk (t) = II E7=o ri (t)xi ll
n
0
and let us define
Am = {t: Vk (t) � a , k = 0, . . . , m  1 , Vm (t) > a}, n A = U Am = {t: sup Vk (t) > a}, k m= l B = {t: V(t) > a}, C = {t: V(t) > 2a }, Cm = { t : I I L rj (t) xj l l > a } . J =m
a>
III.A LpSpaces; Type And Cotype § 1 9.
96
Let us write Ej=o Tj (t)xi = E;: o Tj (t)xj + Ej= m + l rj (t)xj = a(t) + b(t). It follows from properties of the Rademacher functions that b(t) is symmetric on every set where a(t) is constant. Since ll x ll � max( ll x + Yll , ll x  Yll ) for every x, y E X, we see that at least on half of the set Am we have ll a ll � ll a + b ll · Thus
(16) Analogously
(17) We put Am = (A m n{t: rm (t) = 1})U (A m n{t: rm (t) =  1}) = A;\:. uA;;;, and Cm = (Cm n {t: Tm (t) = 1}) u (Cm n {t: rm (t) =  1}) = c� u c;;;, . The independence of the Rademacher functions gives
and Since
l A� n C� I = 2 I A� I · I C� I l A;;;, n C;;;, l = 2 I A;;;, I · I C;;;, I . l A;\:. I = l A;;;, I = ! I Am l and I C� I = I C;;;, I = ! I Cm l we have (18)
Since obviously
C C B C A,
from
(18), (17)
and
(16) we get
I C I � L I Am n C I � L I Am n Cm l = L I Am i iCm l � sup i Cm l · L I Am l � 2 I B I · I A I � 4 I B I 2 m
•
Proof of Theorem 18. The left hand side inequality is obvious, while the right hand side inequality is a standard passage from a distribu 1 tional inequality to an integral one. We can assume J0 V(t)dt = 1, so l {t: V(t) > 8} 1 � � Applying Proposition 19 inductively for k = 1, 2, 3, . . we obtain .
This gives
1
k . 8) P · l {t: 2 k  l . 8 < V(t) � 2 k . 8} 1 J0 V(t)Pdt � 8 + kL(2 =l k � 8 + L (2 k . 8) P . T 2 = c:. lc = l 00
00
a
97
III.A LpSpaces; Type And Cotype §20.
20 Remarks. ( a) An obvious and immediate consequence of Theorem 18 is that in (13) and ( 14) instead of J II Erj Xj I we can use (J II Erj Xj ll q ) � for every q, 1 � q < oo. (b) For some applications the magnitude of the constant Cp is im portant. Our proof, as can be easily verified, gives Cp � C p for p ;::::: 2. The correct order of magnitude is Cp � C.,jP (see e.g. Milman Schechtman [1986] ) .
·
The following repeats arguments from II.D.6.
21.
Proposition.
If X has cotype p and E:: 1 X n is an unconditionally
convergent series in X, then
Proof.
E:: 1 ll xn li P < oo.
We have for every N
(� ) 1 I I n� 1 rn (t)xn l dt N
.!
ll xn ll p
P
N
� Cp (X)
0
� Cp (X) s�p
I n� 1 rn (t)xn l N
� C. a
22. A part of the relation between type and cotype is explained by the following. Proposition. If X has type p then X* has cotype q , where � + �
x1 , . . . , X n E X and xi , . . . , x� E X* 1 n n n xi (x i ) = ri (t)xi ri (t)x i dt •= 1 •= 1 •= 1 0
Proof:
For arbitrary
1.
we have
I ( ?:
?:
�
�
Since
=
) ( ?: ) ] I t•1 ri (t)xi 1 · 1 t•1 ri (t)xi l dt 2 ! 2 ! r (t)x r dt (t)x i l ) ( } I t• 1 i ( } I t• 1 i i l ) . 0
0
0
( 19)
98
III.A
LpSpaces; Type Cotype And
§23.
from (19) we get
1
!
( ?:t=1 l xill q) ; (Ji i ?=t=n1 ri(t)xi11 2) 2 { ( } I tt= 1 ri(t)xill 2) ! : tt=1 l xiiiP } n
�
0
· sup
� 1 ·
0
a
Using Theorem 18 we get the claim. 23.
The types and cotypes of Lpspaces are as follows.
The space Lp O, J.L , ) ( cotype L00(0, J.L)
Theorem. max(2, p) .
1 �p�
oo,
is of type
min(2, p) and
(x;) Lp (O, J.L)
Proof: Clearly has type 1 and cotype exactly like in II.D.6 we obtain for every c
For 1 � p <
oo.
( } I � r;(t)x; I pPdt) ; 1 = ( / � � � r;( t )x;(w ) I P dtdJ.L (w ) ) ; ( / ( � l x;(w W ) dJ.L (w ) ) ; ( / [ � (l x;(w) IP) � ] 2 dJ.L(w )) 0
J
n
""
=
n
n
1
o
of oo
(20)
J
�
J
.!
1?.
P
J
(x;) .
where '""' indicate, that there are inequalities in both directions with constants independent of the set If 1 � p � 2 then Theorem 18 and (20) give
1
x; l pdt J II �r;(t) J o
$
c (j L.J l x;(w )I PdJ.L (w )) n
1
p
$
C(� J
l x;ll�);
99
Ill.A LpSpaces; Type And Cotype §24.
and
! II � Tj (t)xj ll pdt � c ( � (! l xj (w) IPdJL(w) ) ) 2 2
1
0
For 2
3
=c
3
(� ) 3
0
ll xi ll �
p
!
.
:::; p < oo Theorem 18 and (20) give 1 ri (t)xi dt :::; l xi (wW p 3 3 0 0
JI �
l c(J (� ) ) )) :::; c ( � (! 3
0
1
£ 2
dJL(w)
l xj (w ) I P dJL(w)
A
p
� !
= C ( L ll xi ll � ) ! j
and
24. As we said earlier ( 17) our Definition 17 was motivated by the Orlicz property ( II.D. ( 4 ) ) . There is however a more formal connection between these two notions. Suppose the Banach space X has the Orlicz property and suppose that X "' ( E:: 1 X ) for some p, 1 :::; p :::; 2. Then X has P cotype 2.
Proposition.
Proof:
Let us take a finite set ( xj)j; 1 C X. Let rj (k), k = 'Rademacher type' functions on the
1, 2, . . . , 2 m , j 1 , . . . , m be the set {1, 2, . . . , 2 m }. Let us define
=
Xj
= 2 � kL2m= 1 fJ (k)xj
100
Ill.A LpSpaces; Type And Cotype §25.
where
0::: �
xj denotes the vector x; considered in the kth summand of 1 X) P . Using the isomorphism between X and 0::::= 1 X) P and
the Orlicz property we get
! (f: l x; l 2 ) ! :::; c sup l f r; (t) x; l 2 (f: l x ) l ; t E[O,l ] j = l j =l j=l 2"' ( )xj l = C tEsup(O,l ] 1 2  W" jL= l r; (t) L f; k k= l 2"' I L (t) ( )x l p ) = C tE[supO,l] 2  W" (I.: r; f; k ; k= l j = l 1 = C tE[supO,l ] ( / l jf= l r; (t) r; (u)x; I Pdu) ; = O " c ( I t, r; (u)x; l du) • . =
m
=
m
.l P
J
�
We see from Remark 20 that this completes the proof.
•
25. As a simple application of these ideas we will present the gener alization of a classical result of Carleman.
(cpn)� L [0, 1] l , ct'nW 1 for 2every c
Theorem. For every complete orthonormal system there exists an f E 1] such that E�= l ( f
p<
C [O ,
2.
= oo
2
Proof: Suppose it is not so. A standard category argument or the closed graph theorem applied to the space yields p < SUCh that < oo for all f E 1] . Thus we have a commutative E�= l I (!, diagram
ct'n W
C [O , 1]
C [O ,
�
"d
Up< 2 £p
L2 [0, 1]
�'\ fp �
where operators and � are defined by ( f = (! , and 1] is clearly a non 1] t Since � ) = E�= l compact operator the following lemma gives the contradiction.
(�n
26 Lemma.
cp �nct'n ·
cp ) ( n )� id: C [O , L2 [0, C,O ) 1 Every operator from C ( K) into fp, 1 < < 2, is compact. p
III.A LpSpaces; Type And Cotype §27.
101
Let T: C(K) + fp Take T* : fq + M(K), � + * = 1 . If T is not compact, nor is T* , thus there exists a sequence (x n )�= 1 C fq , ll xn I � 1 such that II T*xn  T*xm ll � 8 for n =1 m. Since f.q is reflexive we can pass to a subsequence such that X n k �x00 and to another subsequence such that Yk = X n k  X00 is equivalent to the block basis of the unit vector basis in f.q (apply II.B . 1 7) and thus to the unit vector basis in f.q . Since II T* yk I � 8 and M(K) has cotype 2, for N = 1 , 2, . . . we get Proof:

vN · 8 �
( k�1 ) 2 C J I �k1 rn(t)T* yk l dt 1 C II T II J I � rn (t) yk l dt C II T II ( � IIYk ll q ) k 1 k 1 N
II T * yk il 2
!
1
�
N
0
N
� Since
q
!
N
9
�
0
�
CN Jq . •
> 2 this is a contradiction.
This lemma is also true for £ 1 (Exercise II.D.5) .
27. We know from II.A.5 that for every weakly convergent sequence there exists a sequence of convex combinations convergent in norm. For Lp spaces, 1 < p < oo , this can be improved. Theorem. (BanachSaks) .
Every bounded sequence of functions such
(x n )�= 1 C Lp (O, J.L) , 1 < p < oo, contains a subsequence (xn k )� 1 that N  1 E�= 1 X n k converges in norm. This theorem is an obvious consequence of the following.
Every bounded sequence (x n )�= 1 c Lp (O, J.L) , 1 < contains a subsequence (x n k ) f= 1 such that for some x E Lp (O, J.L)
28 Proposition. p<
oo,
(21) for every finite subset of integers
A and for
s
= min(2, p) .
Propositions 1 and 2 show that it suffices to consider Lp [O, 1] . First we use reflexivity to choose the subsequence (x n k ) such that Xn k � x . If for some further subsequence (still call it Xn k ) we have llxnk  xll + 0 we take once more a subsequence such that llxn k  xll � 2  k so (21 ) holds. Otherwise llxn k  xll � 8 for k = 1 , 2, . . so (xnk  x)
Proof:
.
III.A LpSpaces; Type And Cotype §29.
102
has a further subsequence equivalent to the block basis of the Haar sys tem (II.B. 17), thus unconditional (see II.D . l l and II.D. 13) . Theorem 23 easily gives 21 in this case. a
( )
29. We also wish to present a similar result for almost everywhere convergence. Because of the application in III.C.8. we formulate it for countable family of sequences.
Z(x�)� 1
Theorem Suppose that for every m E is a bounded sequence in Lp ( O, J..L ) , 1 < p < oo. Then there exists an increasing sequence of integers such that for every m E Z there exists E Lp ( O, J..L ) such that 1 '"' J..L a.e. (22)
(nk ) f= 1
N m m N kLJ= 1 xu(k) � X
for every m E Z and every permutation
a
xm
of natural numbers.
(nk )� 1 ( ) (x�)� 1 ( ) (x�k )� 1 Xn,.< k l  x hk N  1 2::= 1 hk HN . 2:�= 1 (HN+ 1  HN ) (xn) I E i xn l l i P El l xHn l , l  H N+ N l: N ( N( + ) ) I HN I ( N(N�1 ) ) HN l:N ( N (N� 1 ) ) HN converges J..La.e. It remains to show that 00 h converges J..L  a.e. (23) L n= 2 n
A standard diagonal procedure and Proposition 28 shows that there exists an increasing sequence of integers such that for each m E Z the sequence satisfies 21 with some constant depending on m. We will show that any sequence satisfying 21 satisfies separately, (22). Clearly it suffices to consider each sequence so in the rest of the proof we will omit the superscript m. and = Denote = We have to show that converges J.La.e. we see that Since for C Lp ( O, J..L ) we have � absolutely convergent series converge J.La.e. We write = < oo so + h};;; > . From (21) we get � l
Proof:
�
Fix numbers 0<
a and (3 such that
a < 1, (3 > 1, a(3 < 1 , as > 1 and s(3 + 1 > s + (3.
K we have from (21) [( K + l ) /3 ] h < C ( [( K'"'+ l ) l3 ]  f' s < C ( f'  1 ) • . K (K ) I n=LJ[Ki3] I Kf's n=LJ[Ki3] )
For each integer '"'
� n
�
l
103
III.A LpSpaces; Type And Cotype §Notes. Since (3
> 1 we infer that 1.
[ K I3] hn ex1sts . 1m ""' K. oo nL..J n =l

J.l  a . e .
(24)
Let us write hn = h�+h� where h� = hn · X {w: l hni:Sn <> } · Since ll hn ll p � C we get J..L (supp h�) � nap so L n J..L (supp h�) < oo. This easily implies
h" !. L .! n= l n
00
converges
J.l  a . e .
(25)
From (24) and (25) we infer that
h' lim ""' .!!. exists K. oo nL..J =l n [ K I3 ]
J.l
(26)
a.e.
But for [K.B ] � N � [(K + 1 ) .8 ] we have
I t �I
� ( (K
[K/3 ]
+ 1 ) .8  K.B ) (K
;; )a.B
� CK a .B  1 .
The choice of a and (3 with (26) and (27) gives that lim K J.La.e. and this together with (25) yields (23) .
(27)
� ( �) exists
n l
a
There is an alternative argument for (23) . From the proof of In(n�l )hn s � C ! ns ln(n + Proposition 28 we infer that
Remark:
E:=2
li
I E�= N
E N
In(:+ l ) hn converges unconditionally in Lp (O.,
J.L) ,
1 )8 so the series so it unconditionally converges in measure. Corollary III.H.25. shows that (23) holds.
(J.L
Notes and remarks. The Lp )spaces are among the most important and widely used spaces in analysis. It is probably useless to trace back their first appearance in the literature but already in Banach [1932] they are the prime examples of Banach spaces. Proposition 3 is, as very often in this book, only a sample result. The same holds for spaces w; (M) for s 2: 0, 1 < p < oo , and M a sufficiently regular set in JR. or a differentiable manifold. It also holds for spaces analogously defined by different sets of derivatives. The reader should consult PelczynskiSenator [1986] for generalizations.
n
III.A
104
LpSpac
es ;
Type And Cotype §Notes.
Proposition 5 and Theorem 6 are taken from Pelczynski [1960] . Much work has been done on complemented subspaces of 1] , 1 < p < oo . Many local and global characterizations have been given. It was shown in BourgainRosenthalSchechtman [1981] that there are uncountably many nonisomorphic such subspaces. We will not discuss this subject in our book. The interested reader should consult the above mentioned paper and references quoted there. For p = 1 the situation is different. All known complemented subspaces of 1] are isomorphic either to £1 or to 1] . It is unknown if this is true for all complemented subspaces of 1] . Theorem 8 is a special case of results proved by ShieldsWilliams [1971] . Our proof of this result follows the presentation of ForelliRudin [1976] where these results are extended to the unit ball in ccn . The same approach is given in Axler [1988] . Proposition 9 is well known. Parts (a) and (b) are almost obvious and (c) is usually called Schur's lemma. Actually Schur proved only a very special case of it and the result evolved gradually. Theorem 1 1 is taken from LindenstraussPelczynski [1971] . It is also true that is isomorphic to for 0 < p < 1 (see Kalton Trautman [1982] ) but the proof has to be different since Theorem 8 is clearly false for p < 1 . Those results lead naturally to the following problem: Find a system of functions which is a basis in equivalent to the unit vector basis in £P " Wojtaszczyk [1984] has shown that for p :S: 1 analytic versions of spline systems analogous to the Franklin system have this property. In particular the Bockariov basis for A constructed in III.E. 16 and 17 is also a basis in � < p :S: 1 , and after suitable normalization is equivalent to the unit vector basis in It is unknown how those systems behave for p > 1. In the case p > 1 different bases have been constructed in MatelievicPavlovic [1984] . Proposition 7 and Corollary 1 6 address special cases of the follow ing question: for what p, q is the space 1] or isomorphic to a subspace of 1] ? Under the name of linear dimension this was stud ied already in Banach [1932] and BanachMazur [1933] . Today the full answer is known. It is summarized in the Table 1 . From the results given in this section the interested reader can easily deduce all these facts except the case 2 < p < q < oo, which is due to KadecPelczyD.ski [1962] (see Exercises 3 and 4) . The answers are the same if we replace by 1] . This follows from Proposition 2 and the following
Lp[O ,
£1 [0,
L1[0 , LI[O ,
Bp(D)
fp
Bp(D)
Bp(D),
fp
Lp[O ,
Lq[O ,
lp
fp Lp[O ,
:S:
Then there exists a measure subspace of oo.
Lp (J.t) .
:S:
J.t such that X is isomorphicfp, to1 a
Proposition. Let X be a Banach space finitely representable in
p
105
III.A LpSpaces; Type And Cotype §Notes.
Thble 1.
fp\Lq
q=1
1
q=2
2 < q < oo
q = 00
p= 1
Yes
No
No
No
Yes
1
Yes
If p � q yes if p < q no
No
No
Yes
p=2
Yes
Yes
Yes
Yes
Yes
2 < p < oo
No
No
No
If p = q yes p =/: q no
Yes
p = 00
No
No
No
No
Yes
The proof can be found in LindenstraussPelczynski [1968] ; it is basically a compactness argument. Corollary 1 6 in the context of Banach spaces was observed by Kadec [1958] , but all the probabilistic background as given in Theorem 14 and Proposition 15 was known much earlier. It can be found in Levy [1925] and our proof of Theorem 14 follows Bochner [1937] . We presented the existence of stable laws in detail not only to show the very useful Corollary 1 6 but also because they are important in Banach space theory (e.g. the notion of pstable type explained in Notes and remarks to III.H) . As remarked already the class of pstable variables investigated in probability theory contains many more variables than we discuss in this book. Note that our pstable variables are symmetric (this follows immediately from (7)). The general treatment can be found in many books on probability theory, e.g. Feller [1971] or Lukacs [1970] . The notions of type and cotype were in the air in the early 70's. Type 2 under the name of 'subquadratic Rademacher average' ap peared in DubinskyPelczynskiRosenthal [1972] and the general notion of type and cotype was introduced by Maurey in the Seminaire Maurey Schwartz 72/73 and HoffmannJ!Ilrgensen in the Aarhus University 72/73 preprint 'Sums of independent Banach space valued random variables'. It was the French group around L. Schwartz, B. Maurey and G. Pisier who showed the importance of the concept in Banach space theory and
106
III.A
Lp Spaces;
Type And Cotype §Exercises
in operator theory. We will discuss some important applications of the notions of type and cotype in III.F, III.H and III.I. Let us also note that it is not known in general if a space with the Orlicz property has cotype 2. Our Proposition 24 (which is due to FigielPisier [1974] ) represents all that is known about this problem. We will use it in III.I. Theorem 18 (Kahane's inequality) , which is basic to the theory of type and cotype, can be found in Kahane [1985] . It was first published in the first edition of this book which appeared in 1968. Theorem 25 for the trigonometric system was shown in Carleman [1918] . Actually the following was shown in KahaneKatznelsonde Leeuw [1977] . Theorem.
Given a sequence of numbers (a n)��= ' an < oo there exists f E C(1I') such that lf(n) l
L:!:' oo a; n = 0, ±1, ±2, . . . .
> 0 and > an for
As was shown by Kislyakov [1981] one can even get the above f with uniformly convergent Fourier series. Some generalizations of Theorem 25 are presented in Wojtaszczyk [1988] . A classical but more complicated proof of this theorem and more precise results can be found in Olevskii [1975] , p. 77 and Chapter III sec. 4. It is interesting that our proof of this Theorem is only a small modification of the arguments in Orlicz [1933] . Theorem 27 was proved in BanachSaks [1930] . In this formulation it clearly fails for L1 and L00 • On the other hand if we consider only weakly convergent sequences the theorem still holds in L1 . This was shown by Szlenk [1965] . Theorem 29 (also for p = 1) was proved by Berkes [1986] . It extends earlier results (without permutations being allowed) of Komlos [1967] and Aldous [1977] . Our proof is a modification of the arguments given in Lyons [1985] . Exercises
1.
1 !, 1
Let { f, },H c Lp (p,) , 1 � p � oo , be a subset such that �g for some g E Lp (f.L ) and all 1 E r. Show that there exists f E Lp (p,) , denoted f = sup{!, such that � f for all 1 E r and if h is such that h =1 f and h � f then there exists 11 E r such that we do not have J,, � h . All the inequalities between functions are understood to be p,a.e.
hEr
!,
Warning. Note that elements of Lp( J..t ) are really classes of functions equal p, a.e. , so if r is uncountable we have trouble with pointwise supremum.
107
III.A LpSpaces; Type And Ootype §Exercises 2.
3. 4.
Suppose X c Lp(J.L) is a closed subspace such that for some q < p 2:: we have for all E X. Show that 2:: for all s , 0 < s < p and E X.
l xl q cl xl p x
Suppose X C Lp(J.L) , 1 < p < oo and X "' lp  Show that there exists Y c X such that Y "' lp and Y is complemented in Lp (J.L) . 2:: 2. Show that every infinite dimensional subspace Lp [O, 1] contains an infinite dimensional subspace Y such that Y is complemented in Lp [O, 1] and either Y "' lp or Y
Suppose p
X
5. 6. 7.
l x l s Cs l x i P
x
c
"' £2 .
Let P: Lp (J.L) + Lp (J.L) , 1 � p � oo , be a projection of norm 1 with dim lm P = n. Show that Im P � £; .
0::::� 1 i�) P "' lp , 1 < p < Construct an unconditionally convergent series L: := l Xn in lp , 1 � p � 2 such that L': � 1 l x n 1 2 e = for every c > 0. Show that
oo .
oo
8.
A family of projections 'P = {Pi h EJ on a Banach space X is called a boolean algebra of projections if (a) for
j1,j2 E J we have Ph Ph = Ph Pil E
'P ,
(b) if ii > h E J and Ph Ph = 0 then Ph + Ph E 'P.
I lj
A boolean algebra of projections is bounded if sup { Pi : E J} < oo. Assume that X is a subspace of Lp (fl., J.L) , 1 � p < oo. Show that if we are given two commuting, bounded boolean algebras of projections 'P = {Pj };EJ and Q = {Q E s (i.e. we assume Pj Qs = Q8Pj for all s E E J ) , then the boolean algebra of projections generated by 'P U Q is bounded.
S,j
9.
s ls
(/n)::': 1 (z2n l z2n l _ 1 ):'= l (t2n l t2n l ; 1 ):'= l
C Bp (D) , 0 < = Show that the sequence p < oo , n = 1 , 2, . . . is a basic sequence equivalent to the unit vector C Lp [O, 1] , 0 < p < basis in lp . Show also that oo, n = 1 , 2, . . . is equivalent to the unit vector basis in lp 
10. For
f defined on D its Bergman projection is defined as 1 f(w) d (w ) . P( f ) ( z ) =  i (1 1r )2
Show that
10
 zw
v
(a) P is the orthogonal projection from L2 (D, v ) onto B (D) , (b) P is a bounded projection from Lp (D, v ) onto Bp (D) for 1 < p<
oo ,
2
108 11.
III.A LpSpaces; Type And Cotype §Exercises
(c) P is not bounded on
L1 (D, v) .
Show that if g is an analytic function on D such that the operator Tg(f) f · g maps Bp(D) into Bq (D) for some p, q, 0 p q then g 0. 12. Let i: W� ('1l'n ) + 0::::��/ Lp (r ) ) P be the natural 1embedding, and let P be the orthogonal projection from 0:::: ��1 L 2 (r)) onto i(WJ(r)). Show that P is continuous on ( 2::��11 Lp (r ) ) P2 , 1 p and of weak type (11). 13. Show that the operator Ta f(x) J; (x  y)a f(y)dy , 0 1, is bounded on Lp[O , 1] for 1 :5 p :5 14. Show that the operator Tf(x) x 1 J; j(t)dt is a bounded opera tor on Lp (O , ) for 1 p 15. Show that every linear operator from fp into lq , 0 q p is compact and also every operator from co into fq , 0 q is compact and that the same is true for T: X + Y where X is any subspace of lp and Y any subspace of fq . 16. Let (cpn)� 1 be a complete orthonormal system in L2 [0, 1]. (a) Show that for every U C [0, 1] with l U I > 0 there exists f E £00[0 , 1], supp f C U such that E� 1 l(f, cpn} I P for all 2. (b) Show that there exists an f E C [O , 1] such that f is linear on every interval of [0, 1]\ � (where � is a Cantor set) and E� 1 I (/, IPnW for all p 2. (c) Show that there exists a set U C [0, 1] such that for all p 2. E� 1 I (xu , ct'n} I P (d) Suppose that ('l/ln )�= l is an orthonormal system in L 2 [ 0 , 1] (not necessarily complete) such that infn J I t/In I > 0. Show that for there exists an f E C [O , 1] such that 2:: � 1 I ( ! , 'l/ln } I P all p 2. (e) Find an example of a noncomplete orthonormal system for all ('l/ln )� in L [0 , 1] such that E:'= 1 l(f,'¢'n }l f E C [O1, 1] . 2 <
=
=
< oo ,
<
<
< oo
< a <
=
oo .
oo
<
=
< oo .
<
<
<
< oo ,
< oo ,
= oo
p <
= oo
<
<
= oo
= oo
<
<
oo
17. Suppose that X is a Banach space of type p and that Y C X. Show
that the quotient space X/ Y has type p.
III.A LpSpaces; Type And Cotype §Exercises
109
18. Suppose that (Xn );;c'= 1 is a sequence of Banach spaces such that Tp(Xn) � C for n = 1 , 2, . . . and some p, 1 � p � 2. Show that 0:::: � 1 Xn) q has type min {p, If Cp (Xn ) � C for n = 1 , 2, . . . and some p, 2 � p � oo then 0:::: := 1 Xn) q has cotype max {p,
q).
q) .
Un )'::: 1
C
C[O, 1] such that ll fn ll = 1 , fn�O as n + oo but for every subsequence Un k ) �1 we have lim sup Noo N  1 11 E�=l fn k ll oo > 0.
19. { a) Find a sequence
{ b ) Show that a sequence like in ( a) can be found in a reflexive
space.
III.B. Projection Constants
We discuss projection and extension constants of Banach spaces. The Lewis estimate for the norm of a projection onto a finite dimensional subspace of Lp (JL) is given as well as the KadecSnobar estimate for the projection constant of ndimensional Banach space. We compute the projection constant of the space of trigonometric polynomials (Lozinski Kharshiladze theorem) and apply it to show the divergence of the general interpolating processes in C(1l') and to estimates of degrees of polynomial bases for C(T) . We also show that spaces of nhomogenous polynomials on lBd have projection constants bounded uniformly in n. This is ap plied to some questions in function theory, in particular a construction of a nonconstant inner function in lBd is presented. We conclude this chapter with the dual presentation of extension of operators; we intro duce the notion of the projective tensor product and present some basic observations about it.
an
1. In this paragraph we discuss quantitative notions connected with projections and extensions of linear operators. We say that a Banach space X is injective if for every Banach space Y and every subspace Z of Y and every operator T: Z + X there exists an extension T: Y + X. We define the extension constant e(X) by
e(X) = inf{c: for every Y :::> Z and T: Z + X there exists T : Y + X : T I Z T and II T II � c ii T II }.
=
Every injective space X has e(X) < oo , because if not then take Yn :::> Zn and Tn : Zn + X with II Tn ll = n  2 such that every Tn : Yn + X such that Tn i Zn = Tn has II T il n. Put Y = ( L �1 Yn) 2 :::> Z = ( z= :=l Zn) 2 and define T: Z + X by T((zn )) = (Tn (zn )). One sees that T is a continuous operator from Z into X without a continuous extension.
;::::
2. The HahnBanach theorem applied coordinatewise gives e(loo ) = 1 . The following generalizes this observation. Theorem. For every measure p, we have e(L00 (p,) )
w = (wj)
= 1.
Proof. Let n be a space on which p, i s a measure. Let r{f!) denote the set of all partitions of n into sets of finite, positive measure.
112
Ill.B. Projection Constants §3.
Note that r(n) is a directed set, if we put w < w if and only if for every E w there exist wi E w such that w; c Wi · Given a pair Y :::> Z and T: Z + L00 (p,) we define Tw : Z + L00 (p,) by
w;
Tw (z)
=
L j
1 ) 1 T(z)dp,] [(w, w, Jl
·
x
wr
Since the mapping z t+ p,(w; )  1 fwJ. T( z )dp, defines a linear functional on Z of norm not greater than II T II we extend each such functional to Y�; E Y* with IIY�; I � II T II . We define Tw: Y + L oo (JL) by Tw(Y) = L; Y�/ Y)Xwr Clearly, for every w E r(n) , Tw is an extension of Tw with II Tw ll � II T II . Let Bt be the closed ball in L00 (JL) with centre 0 and radius t, equipped with the a(L00 (p,) , Lt (JL) )topology. By II.A.9 it is a compact set. Put B = n y E Y Burn II Y II · Then B is a compact set with the usual product topology. We consider each Tw as an element of B; Tw at the coordinate y has the value Tw (y) . Let T be the limit point of the set {Tw}w Er ( O ) C B. One easily checks that T can be interpreted as a linear operator from Y into L00 (p,) with II T II � II T II · Also, since for every z E Z, Tw (z) + T(z) along the directed set r(f!) a in a(L00 (p,) , Lt (JL) )topology we get T I Z = T. Now we will define some other constants. For a pair (X, Y) con sisting of a Banach space Y and a subspace X of Y we define
3.
=
inf{ c : for every operator T: X + Z there exists an extension T: Y + Z : II T II � ciiTII }, A(X, Y) = inf{ II P II : P is a projection from Y onto X}. e(X, Y)
For a fixed space X we define e(X) A(X)
=
sup{e(Xt . Y) : X1 is a subspace of Y isometric to X}, = sup{A(Xt . Y) : X1 is a subspace of Y isometric to X}, 'Yoo (X) = inf{ ll aii · II.BII : X � L oo (JL) _!__. X with p, an arbitrary measure and ,Ba = idx }. The constant A(X, Y ) is called the relative projection constant while A(X) is called the projection constant. 4 Lermna.
A(X, Y) .
For every Y and its subspace X we have e(X, Y)
III.B. Projection Constants §5.
1 13
Proof: Since a projection from Y onto X is an extension of id: X
+ X to an operator from Y into X we see that A( X, Y) :5 e(X, Y). On the other hand every projection P: Y � X gives rise to the extension • T = TP with II T II :5 II P II · II T II . This shows e(X, Y) :5 A(X, Y).
5 Theorem. Let X be a Banach space and let X1 be any subspace of
Leo (p,) isometric to X. Then
Moreover if X is finite dimensional and X1 is a subspace of some C(K) space isometric to X then A( X) = A(Xt , C(K) ) .
Proof: We infer from Lemma 4 that e(X) = A(X) and e(Xt , Leo (P,)) =
A(Xl , Leo (P,)) . Obviously 'Yeo (X) :5 A(Xt , Leo (P,)) :5 A( X). If we have X � Leo � X with f3a = idx and ll a. ll · ll /3 11 :5 'Yeo (X) + c and are given Y :J Z and T: Z + X then Theorem 2 gives an operator 8: Y + Leo with 8 I Z = aT and 11 8 11 = ll a.T II . Since for z E Z we have {38(x) = f3aT(z) = T(z) the operator {38 is the extension of T with ll /38 11 :5 ll a ll · 11/311 II T II , so e(X) :5 'Yeo (X) . If X is a subspace of Y we can extend id: X + X to a projection P: Y + X with II P II :5 e(X), so A(X) :5 ·e(X) . This gives (1). In order to see the 'moreover' part note that from the definition A(Xt , C(K)) :5 A(X). Conversely if P is a projection from C(K) onto X1 then P** : C(K) ** � Xl is a projection of the same norm, so A(Xt , C(K)) ;::: A(Xt , C(K))** ) . Since C(K)** � a Leo (!L) for some measure p,, (1) gives the claim. 6 Corollary.
A(Y) · d(X, Y) .
For any two Banach spaces Xand Y we have A(X)
:5
From the definitions 'Yeo (X) :5 'Yeo (Y) · d(X, Y) so the claim follows from (1). a
Proof:
7. Before we proceed to more concrete questions, we want to address the general question how big A(X, Lp) , 1 :5 p :5 oo, can be for n dimensional spaces X. The answer is given in Theorem 10 below. We start with Proposition. Let X be an ndimensional subspace of the space Lp(p, ) , 1 < p < oo . Then there exist a basis (hj )J=I c X and a system of functions (gj ) J=I C Lp• (p,) such that
III. B.
114 (a) (b)
� I ( L:7= 1 I hji 2 ) 1 P � n t , � I ( L:;= 1 193 1 2 ) 1 p, � n t ,
Projection Constants § 7.
1
H = ( L:7= 1 l hj l 2 ) 2 then g3 = hj · HP 2 (we take § = 0) . Proof: The basis (h3 )' 'J= 1 is found as a solution of the following ex tremal problem: given ¢1 , . . . ,
1
1
1
X
1
:2::
Since both sides are polynomials in equal for 0 we have
t=
1
t of degree n and both sides are
III.B.
Projection Constants §8.
1 15
that is 1
so 1 1 1 '¢ 1 1 1 � nPT . Now we use the HahnBanach theorem and extend '¢ 1 to an element (g t , . . . , 9n) E ELp' with I l l (91 , . . . , 9n) I l i P ' � n P' . Since
n .P ( h, , · · · , h.. ) � �
t, I h; g; dp $ I ( t, l h; l 2 ) ( t, ) l
I Y; I ' l dp
� l l l (hi ) .i=t i i i P · l l l (gi ) .i= t l l l p' = n both inequalities are in fact equalities so the conditions for equality in a the Holder inequality (I.B.3) give (d) . 8 Lemma. With the notation of Proposition 7 we define a Hilbertian norm N(f ) N(f ) = (J l f l 2 wdJL) ! where w = HP 2 The following inequalities hold: (a) if 1 < < 2, then I I ! I I P � n "P  2 N(f) and for x E X we have N(x) � ll x ll p; (b) if 2 � < then N( f ) � n 2  "P I I J II P and for x E X we have as
•
1
p
p
1
1
oo,
1
ll x ii P � N(x) .
Proof: It follows directly from Proposition 7(d) that (hj ) J= l is an orthonormal basis in the norm N(·). For 1 < p < and x = "'£ o:i hi E X we have
2
� ll x ll�
( � lo:i l 2 )
� 2 H2 p · w l l oo = II
ll x ll� N(x) 2  p
so N(x) � ll x ll w For 1 < p < and arbitrary f we have
2
1 1 ! 11�
= = =
J ) � ( / lfl2 ) ( / N (f)P ( J ) N (f)P I I I (hi )J= t l l lf;P> ) P.
w dJL 2
l f i P w ! w  ! dJL HPdJL
N (f ) P n ( 2 2 p)
( 2 ) 2p
=
•
w < =f >< c 2 � p J ) dJL
(2 ) 2p
1 16
III.B. 1
1
Projection Constants §9
so II ! II P ::; n "P  2 N(f). This proves part (a) . For p :2:: 2 and arbitrary f we have
N(f) 2 :S
(/
= ll f ll; = 11 ! 11 ;
ll x ll:
�
I lx l'
I t.
= N(x) P  2 so ll x i i P
:S
o; h;
)
"P2 l f i P dJL
(/
( J HPdJL)
)
(p )
P2 w � dJL 
(p  2 )
P
n (p ; 2J
r'd� I l x l ' ( L I a; I ') '" ;" w'd� ,;
J lx l 2 wdJL = N(x)P
N(x). This gives (b) .
a
Let X be an ndimensional subspace of Lp(JL) , 1 ::; p ::; Then d(X,£�) ::; nl! tl. In particular for any ndimensional space X we have d(X, £�) ::; y'n. 9 Corollary.
oo .
For 1 < p < oo the corollary follows directly from Lemma 8(a) and (b) . Let X be any ndimensional Banach space. By the BanachMazur theorem II.B.4 X is isometric to a subspace X of C[O, 1] . Let Xp denote X with the norm from Lp [O, 1] . One easily sees that d(Xp, X) . 1 as p + oo, so Proof:
The estimates given in this corollary are in general best possible; see II.B.8.
Let X be an ndimensional subspace of Lp(JL) , 1 < p < Then there exists a projection P onto X with II P I I ::; nl!tl. If Y is any Banach space and X is an ndimensional subspace of Y, then there exists a projection P: Y �X with IIP II ::; y'n. In particular for any Banach space X we have A(X) ::; v'dim X. 10 Theorem. oo.
1 17
III.B. Projection Constants § 1 1 .
For 1 < p < oo we take as P the orthogonal projection with respect to the Hilbertian norm N(·) considered in Lemma 8. For 2 � p < oo Lemma 8 gives
Proof:
li P/lip � N(Pf) � N(f) � n !  t 11/ll w Given any Banach space Y we can assume that Y is a subspace of some
L00(p,). Let Xs be the space X with the norm Ls (f..L ) , s � 2. Let Ps be a projection from Ls (f..L ) onto Xs with li Ps II � n !  � . Let us fix any basis (xj )'J= 1 in X and write Ps( f ) = L:j= 1 cpj(f)xi . If cpj , j = 1, 2, . . . , n, is any cluster point of cpj as s + oo in L� , then P(f) = L:j= 1 cpi (f)xi is a projection from L00 (f..L) , and so from Y, onto X with II P II � ..fii . To show that the orthogonal projection P works also for 1 < p < 2
requires some additional duality arguments. The adjoint projection P* : Lp' + Lp' is given by P* (f) = L:j= 1 J fhi df..L Yi · This is a projec tion orthogonal in the scalar product (cp, t/J ) = J cp1jjw  1 df..L , and (gj )'J= 1 is an orthonormal basis with respect to this product. Analogously as in Lemma 8(b) we get
Thus for any f E Lp' we have
li P* /li P' � 1
1
(/
1
)
l/l 2 w  1 df..L 2 � II/Il P' n ! ; . a
So II P II = II P* II � n 2  ; . 11 Corollary.
Let X
.A( X, Y) � ..fii + 1.
C
Y be a subspace of codimension n. Then
Let P be a projection from Y* onto X .L with li P II � ..fii . We write P( y* ) = L:j= 1 yj* (y* ) xf with xf E X .L and yj* (xt) = Oj , k · Using the principle of local reflexivity, II.E.14, for span ( yj* )'J= 1 we find ( yj ) 'J= 1 C Y such that xf (yk) = Oj k and such that P defined as P (y) = L:j=1 xf (y)yj is a projection with norm � ..fii + c. Obviously ker P = X so I  P is a projection onto X with II I  P ll � ..fii + 1 + c. Since c was arbitrary we have the claim. a Proof:
,
118
III.B. Projection Constants § 1 2.
Suppose Z and V are finite dimensional subspaces of m, dim V = n. Then
12 Corollary.
X with Z
C
V, dim Z =
A(Z, X) � Jn  m + 1 , A(V, X) A(V, X) + 1 � Jn  m + 1. A(Z, X) + 1
(3) (4)
Clearly A(Z, X) � A(Z, V) · A(V, X) and Corollary 11 gives A(Z, V) � ..jn  m + 1. This yields (3) . Given a projection R from X onto Z and Q a projection from kerR onto kerR n V we put P = R + Q(I  R) . One checks that P is a projection from X onto V and II P II � II R II + II Q II ( II R II + 1). Thus
Proof:
A(V, X) + 1 � (A(Z, X) + 1)(A(kerR n V, kerR) + 1). Since dim(kerR n V)
=
n  m Theorem 10 gives (4) .
a
The following theorem is used quite often for computation of 13. projection constants of concrete spaces. Theorem. Let X be a Banach space and Y a complemented subspace. Let G be a compact group and let g 1+ Tg be a representation of G into L(X) such that
(a) Tg (x) is a continuous function of g, for every x E X,
(b) Tg (Y) c Y for all g E G. Then for every c > 0 there exists a projection P: X + Y such that TgP Tg1 = P for all g E G and II P II � (A(Y, X) + c) supg E G 11 Tg ll 2 • Observe first that the BanachSteinhaus theorem I.A.7 implies supg E G II Tg ll < oo. Let us fix a projection Q from X onto Y such that II Q II � A(Y, X) + c and for x E X define
Proof:
P(x)
=
i TgQTg1 (x)dg
(5)
where dg denotes the normalized Haar measure on G. Condition (a) ensures that there is no problem with the definition of the integral. For every x E X we see that TgQTg1 (x) E Y. Also for y E Y we see that TgQTg 1 (y) = y so P is a projection onto Y. For h E G Th PTh  1
=
L ThTg QTg 1 Th  1 dg L ThgQT( hg )  1 dg =
= P.
III.B. Projection Constants §14.
119
Also
II Px ll
:::; L II Tu ll II Q II II T9 1 II ll x ll dg
a
:::; II Q II ll x ll sup 11 Tu ll 2 
In many concrete application T9 are isometries and the invariant projec tion is unique. Then the theorem implies that the norm of the invariant projection equals .X(Y, X). 14.
Let us recall that
and that u is the normalized rotation invariant measure on Sd. By Wn (Sd ) we will denote the space of all homogenous polynomials of degree n on ([d . Equipped with the norm II P IIoo = sup < E Sd I P( ( ) I it will be denoted by W:' (Sd ) and equipped with the norm
it will be denoted by W�(Sd ) · Clearly every polynomial p(z1 , . . . , Zd ) can be uniquely written as p = L:Z =o Pk with Pk E Wk (Sd ) · The compact group of all unitary operators on ([d is denoted by U(d) . It acts on C(Sd ) by the formula U(d) : g 1+ T9 where T9 ( ! ) = f o g. Obviously T9 is an isometry on C(Sd) · 15 Theorem. Proof:
defined by
+ n)r( l + j) .X(Wn00 (Sd)) = r(d r( l + n) r(d+ 2) "
Let us consider the operator Pd,n from C(Sd) into W:' (Sd ) Pd , n ( /) ( ( ) =
(d  1 + n) ! { f(z) ( ( , z) n du(z). (d l ) !n! Jsd _
The binomial expansion of the Cauchy kernel 1
_
( 1  (( , z))d 
'"""'
n f='o (d  1) !n! (( , z) (d  1 + n) !
'
120
III.B. Projection Constants § 1 6.
together with the Cauchy formula I.B.21 and the fact that homogenous polynomials of different degrees are orthogonal, show that Pd,n is actu ally a projection onto W:O (§d) · The operator Pd,n commutes with the action T9 of the unitary group U(d) and Theorem 12.3.8 of Rudin [1980] , i.e. the fact that U(d) acts irreducibly on Wn ( §d ), shows that it is the unique such projection. Theorems 5 and 13 give
In order to estimate this integral we use the formula (see Rudin [1980] 1.4.5)
which reduces it to
Passing to polar coordinates and substituting l z l 2 = r we get 1 (d  1) J0 (1  r) d  2 r� dr. Known values of the Euler integrals of the first kind yield
�
r(d)f(1 + � ) l (z, () l n da( ( ) = f(d + ) sd 2
.
(7)
Substituting (7) into (6) we get the desired estimate.
a
!!
The integrals computed in (7) are clearly well known. An alternative argument can be found e.g. in Garling [1970] . Remark:
We want to note two special cases of Theorem 15. For n = 1 one easily checks that Wl' ( §d ) � cg. So using the properties of the r function we get 16 Corollary.
..fii � A(£�) = f(1 .5) n::!>>
An easy computation also yields
�
h/1r..fii .
121
III.B. Projection Constants §18.
18. Now we will present some applications to function theory on the unit ball of ([d . Our main goal is to present the construction of a non constant inner function in IBd . A function f(z) E H00 1Bd ) is called inner if 1/ (z) l = aa.e. on Sd {by f(z) for z E Sd we mean the radial boundary value; see I.B.21). Clearly inner functions are characterized by = ll/ ll2 = 11/lloo so the following proposition can be considered as a 'first approximation'.
{
1
1
n 1, 2, . . . , such that Proposition. =
For every
d 1 there exist polynomials Pn E Wn (Sd ), �
{8)
Proof: Let I denote the identity map from W:' (Sd ) into w;(Sd ) · We have to show II l ii �  d .,;rr . From Corollary we get
2
6
{9)
A (W�(Sd)) :5 A (W:' (Sd )) · d(W:' (Sd ), W�(Sd )) :5 A (W:' (Sd )) · II I II · II I  1 11 · The space w; (Sd) is obviously a Hilbert space. Since Pd,n is an orthog onal projection from L 2 (Sd , a) onto w;(Sd ) the dimension of w; (Sd ) equals the square of the HilbertSchmidt norm of Pd,n {III.G.12 and so { 2n d1. m Wn2 (Sd) = Jsd l (z, () l da( ( ) .
{7)
[ {d 1 +! n)! ! J2 {d 1) n
13)
From we infer that dim w;(sd ) = <�=._11�,:r . The same conclusion can also be reached counting monomials. Thus Corollaries and applied to give
{9)
p p{1, 0, . . . , 0) U(d) {1, 0, . . . , 0). 0, . . . , 0) c
16
17
1
{10)
In order to compute II I  1 11 take E Wn (Sd ) such that II P II 2 = and = II P IIoo · Let us aver II P IIoo = II I  1 11 · We can assume age over the subgroup of fixing For the averaged 1 polynomial Po we have Po(1, = II I  11 and II Po ll2 :5 II P II 2 , so Po(zb . . . , Zd ) = czf. The II Po ll = II P II 2 = But one easily sees that 1 2 normalization gives = ( fsd l z1 l da(z)r2 . Since = II I  1 11 , using and substituting into we get the claim. a
p
1.
c
(10)
{7)
19. Our next result is a technical device to construct the inner function.
122
III.B. Projection Constants § 1 9.
Proposition. For every cp E C (Sd), cp > 0 there exists a natural num ber N = N ( cp) such that for every n > N there exists a polynomial R = R
(a) I R I < cp on §d, (b) fsd I R I 2 da > 4  d fsd cp 2 da,
(c) R = '2:�!: Rk where Rk E Wk (§d) ·
It is enough to show the Proposition for cp polynomial in Z I , . . . , Zd and .Z1 . , Zd since such polynomials are dense in C(Sd) · For arbitrary f E C(Sd) we have
Proof:
•
•
.
Applying this for f = Pn , where the Pn satisfy (8) and using the fact that U(d) is connected we infer that there are polynomials Pn = Pn o 9n E Wn (Sd ), n = 1 , 2, . . . such that
II Pn ll oo = 1 , 11Pn ll 2 � 2  d ..;:rr and l cp( ( )Pn ( ( ) l 2 da( ( ) � 4  d 1r [sd cp2 da. [ s d l l
(11)
The desired polynomial R will be equal to �C(cp · Pn ) where C is the Cauchy projection (I.B.21) and n is big enough. Let a, (3, y, 6 denote dtuples of nonnegative integers. Then
One sees that infer that if
fs4 z.8+6 zr+a da( z)
=
0 if (3 + 6 =f.
'Y
+
a.
From this we
k  K�I aai �k + K This shows that for n big enough �C(cp · pn ) satisfies (c) with N(cp) = 2K. That � C(cp Pn) for large n also satisfies (a) and (b) follows from ·
limoo II'PPn  C(cppn ) l ! c(S4) n+
=
0.
(12)
123
III.B. Projection Constants §20.
This in turn is an immediate consequence of the fact that Pn tends weakly to zero in Hp (md), 1 :::; p < oo and the following lemma which will be also used later. 20 Lemma. Let cp E C(Sd ) satisfy the Lipschitz condition. Then Tcp ( / ) = cp · f  C(cpf) is a compact operator from Hq (md ) into C(Sd ) for q > 2d. Proof:
Using the Cauchy formula we can write
cp(z)  cp(() ( )da( ) = [ rz ( ) ( )da( ) . Tcp ( / ) (z) = J[sd (1 ( (/( (  (z, () ) d / ( Jsd
Since l z  (I :::; C.J I 1  (z, () I for ( E sd and z E md u sd, we have
so l rz ( ( ) l q is a uniformly integrable family of functions for q < ( 2J� 1 ) (see III.C. l l for definitions) . Since for (zn );:."= 1 c Sd u md such that Zn + ZO We have rZn + rzo pointwise 0"a.e. , the Uniform integrability and the Egorov theorem imply that z 1+ rz is a continuous map from Sd U md into Lq , for q < ( 2J� 1 ) . This yields the compactness of T"'. D 21.
Now we are ready to prove There exists a nonconstant inner function in
Theorem.
1, 2, 3, . . .
Proof:
md, d
=
Inductively using Proposition 19 we construct a sequence
(Rn );:_"= 1 of polynomials such that (a) Rn (o) = 0, n = 1, 2, . . . , (b) fsd Rn R k da = 0 if n =f. k, (c) I Rn + l l < 1  I E 7= 1 Ri l on sd, (d) fsd I Rn +I I 2 da > 4  d fsJ 1  I E 7= 1 Ri l ) 2 da. The series E :'= t Rn converges in H2 (md) since it is an orthogonal series (see (b) ) and its partial sums are uniformly bounded by 1 (see (c)). This implies that fsd 1 Rn l 2 da + 0 n + oo so by (d) I E7=l Ri l + 1 aa.e. , so E:'= t Rn is an inner function. Condition (a) gives that it is as
nonconstant.
'
a
124
III.B. Projection Constants §22.
22. Now we will discuss the projection constant of the space T� of trigonometric polynomials of degree at most n equipped with the sup norm. When naturally embedded into C(1l') , T� is a rotation invariant subspace and the unique rotation invariant projection is the Dirichlet projection Vn . Its norm equals (I.B. 15) 47r 2 log(n + 1) + o(1). Using Theorem 13 we obtain Theorem. (LozinskiKharshiladze) .
o(1).
.X(T�)
(�) log(n + 1)
a
+
23. This theorem implies some classical results about the divergence of trigonometric interpolation. For example we have Theorem. For every system of points x0, xf, . . . , x2n E 11', n = 1, 2, . . . there exists a function f E C(1l') such that the sequence of interpolat ing polynomials In ( / ) , i.e. polynomials in T� such that In ( / ) (xj ) = f(xj ), j = 0, . . . , 2n, is not uniformly bounded.
Observe that In is a projection from C(1l') onto T�, so II In il � .X(T�). Theorem 22 and the uniform boundedness principle (see I.A.7) give the claim. a Proof:
24.
We also have
Suppose ( /n )':= 1 is a Schauder basis for C(1l') such that with k(n) an increasing sequence of integers. Then k(n) � I + c log2 n for some c > 0.
Theorem.
fn E Tk(n)
Proof: Let Fn = span ( /k ) k= l · Since ( /n ) ';'= 1 is a basis for 0(11') , II.B.6 and Theorem 5 show that .X(Fn ) :5 C for some constant C, n = 1, 2, . . . . Using (4) we obtain �::
.X(Tk(n) ' C(1l') ) + 1 2k(n) + 1  n + 1. .X(Fn , C(1l') ) + 1 :5 vf Thus Theorem 22 yields log n :5 log 2k(n) + 1 :5 C J2k(n) + 1  n so k(n) �
i + c log2 n.
a
We would like to conclude this chapter with a more abstract 25. treatment of the extension of operators. This is connected with the
125
III.B. Projection Constants §26.
concept of tensor product. Actually we will consider only one tensor product. We assume that the reader is familiar with the algebraic notion of tensor product. Definition. If X and Y are Banach spaces then their projective tensor product X ® Y is a completion of the algebraic tensor product X ® Y with respect to the norm
l � xi ® Yi l N
J =l
II
= inf
{ I) xj ii iiYj ll : � xi ® Yi = � xj ® yj } . (13) N
3
3
J =l
26 Lemma. The dual space of X ® Y can be isometrically identified with L(X, Y* ) . Proof: The duality which we use is basically the trace duality ( compare with III.F. 16) . Explicitly it is given as
j
j
for T: X + Y* and I:j Xj ® Yi E X ® Y. One checks that this definition is independent of the representation of the tensor. Since
I (T, � xi ® Yi ) l � L I T(xi )(Yi ) l � II T II · � ll xi ii · IIYi ll 3
3
3
we infer from (13) that every T E L(X, Y* ) induces a linear functional on X ® Y of norm � II T II . Considering the elementary tensors we get
II T II = sup{ I (Tx)( y) l : IIYII = ll x ll = 1} = sup{ I (T, x ® y) l : ll x ii · IIYII = 1 } � sup (T, � xi ® Yj) : � Xj ® Yj
{I
3
l �
3
L � 1}.
This shows that each T E L(X, Y* ) gives a functional in (X ® Y)* of norm l iT II so L(X, Y* ) is isometrically a subspace of (X ® Y) * . Now suppose that we are given cp E (X ® Y)*. Note that for a fixed x E X, cpx (Y ) = cp (x ® y ) is a linear functional on Y with norm � ll cp ll  One easily checks that the map x ...... cpx is a linear operator from X into Y* . a This shows that L(X, Y*) = (X®Y ) * .
126
III.B. Projection Constants §27.
27 Corollary. Suppose Y conditions are equivalent:
C
X and Z* are given. The following
(a) every operator T: Y + Z* extends to an operator T: X + Z* ; (b) the map r: L(X, Z* ) + L(Y, Z*) given by r(T)
=
T I Y is onto;
(c) the identity map i: Y ® Z + x ® z is an isomorphic embedding. Proof: This is a routine duality (see I.A. 13 and 14) and the observation
that i*
=
a
r.
28. Now we will formulate the proposition which is the dual version of Theorem 2. Before we do so we will define the space £1 (S1, f.£, Y) of Bochner integrable Yvalued functions (Y is a Banach space) . First we consider the step functions of the form f(t) = Ej= 1 YiXA; (t) where Yi E Y, j = 1, . . . , N and (Aj ) f= 1 are disjoint subsets of S1 with f.£(Aj ) < oo for j = 1, 2, . . . , N. For such f we put
11/11
=
I: 11Yi ll f.£(Aj ) = In 11/(t) ll df.£(t) . i= l N
A Yvalued function f(t) is Bochner integrable if there exists a sequence (/n)':'= l of step functions, as above, such that Jn 1 1 /(t)  fn (t) ll df.£(t) + 0 as n + oo . The space of all Yvalued, Bochner integrable functions (when we iden tify functions equal f.£a.e.) is denoted by £1 (S1, f.£, Y) . One easily checks that it is a Banach space. This description makes the following proposition easy to believe. Proposition. If (S1, f.£) is a ufinite measure space and Y is a Banach space then
We define a map cp: Y ® £1 (f.£) + £1 (S1, f.£, Y) by the formula cp( Ef= l Yi ® /i ) = L:f= 1 /j (t) · Yi · One notes that this map is well defined and
Proof:
127
III.B. Projection Constants §Notes.
so ll
( � Yi ® XA; ) N
=
f and II I II =
� N
This shows that
I� N
IIYi ii iiXA; I �
Yi ® X A;
t·
a
Notes and remarks.
The question when a linear operator can be extended from a subspace to the whole space is quite natural and important. It was asked in Banach [1932] , remark to Chapter IV, and the first example showing that it is not always possible was given by Banach and Mazur [1933] in the form of an uncomplemented subspace of C[O , 1] . Thus the connection between extensions of operators and projections was noted very early. Theorem 2 is due to Kantorovic [1935] with the proof that directly mimics the proof of the HahnBanach theorem. Some extensions in terms of Banach lattice theory were given by Akilov [1947] and the following theorem was proved by Kelley [1952] building on earlier work of Nachbin [1950] and Goodner [1950] . Theorem. e(X) = 1 if and only if X is isometric to C(K) for some extremaly disconnected compact space K.
The notion of the projection constant and its properties summa rized in Theorem 5 evolved from those works in the 50's (except for the constant 'Yoo which is the child of the theory of operator ideals and appeared later) . The beautiful estimate >.(X) :5 v'dim X was proved by Kadec Snobar [1971] . The original proof used the important result of John [1948] , a special case of which is our Corollary 9 for p = oo . Theorem 10 and Corollary 9 were proved by Lewis [1978] . Our presentation fol lows LorentzTomczakJaegerman [1984] . Corollary 1 1 was observed by GarlingGordon [1971] . The KadecSnobar theorem stimulated further research in the local theory of Banach spaces, which is summarized in TomczakJaegerman [1989] . The KadecSnobar theorem is also quite useful in various questions of approximation theory. It may be of some interest that there exist finite dimensional spaces Yn , dim Yn = n such that >.(Yn) � .fii  Jn · Also an improvement in the KadecSnobar estimate is possible. There exists > 0 such that for any finite dimen sional Banach space X we have >. (X) :5 .fii  Jn · All this is beyond
c
128
III.B. Projection Constants §Notes.
the scope of this book (see Konig [1985] , KonigLewis [P] and Tomczak Jaegerman [1989] ) . It should be pointed out that methods used in those more sophisticated studies use extensively various Banach ideal norms, in particular those discussed in III.F. Corollary 12 and Theorem 24 are taken from Kadec [1974] . Theo rem 24 for k(n) = n and Theorem 23 are classical results of Faber [1914] . The estimate from below for k( n) in Theorem 24 was the best known for some time. Only recently Privalov [1987] has shown that k(n) � (1 + e)n (where e > 0 depends on the basis) . On the other hand Bockarev [1985] has constructed a basis Un )':'= ! for C(ll') such that fn E T4 n · Theorem 13 was proved by Rudin [1962] but its main idea, the averaging of projections, was already used in Faber [1914] in the case of the circle group and interpolating projections. Theorem 1 5, Corollary 1 7 and Proposition 1 8 are taken from RyllWojtaszczyk [1983] ; Corollary 16 however is quite old. With basically the same proof the upper bound was given in Griinbaum [1960] and this bound was shown to be exact by Rutowitz [1965] . The nonconstant inner function in 1Bd , d � 2 was constructed independently in Aleksandrov [1982] and L0w [1982] . This was an unexpected solution to the long standing problem. The proof presented here is due to Aleksandrov [1984] . Further results stemming from the solution of the inner function problem are discussed in detail in Rudin [1986] and Aleksandrov [1987] . Proposition 18 seems to be quite useful in function theory. Besides many applications presented in Rudin [1986] some are presented in RyllWojtaszczyk [1983] and Wojtaszczyk [1982] . Alexander [1982] used Proposition 1 8 to construct a function f E A(1Bd ) such that /  1 (0) has infinite (2d  2)dimensional volume. Also, the papers Ullrich [1988] and [P] should be consulted for further applications and improvements. Theorem 22 is due to S.M. Lozinski and F .I. Kharshiladze but the first published proof we are aware of is in Natanson [1949] . It is however one in the long line of averaging arguments starting from Faber [1914] or even earlier. The general theory of tensor products of Banach spaces was created in the 1940's by R. Schatten in a series of papers. The presentation of this work is contained in Schatten [1950] which even today seems to be the most complete exposition. Later Grothendieck [1955] and [1956] generalized the concept to more general linear topological spaces and applied it to Banach space theory in the most profound way. What we have presented here is mostly folklore and can be found in the works mentioned earlier.
III.B. Projection Constants §Exercises
129
The reader should be informed however that there are many impor tant norms on X ® Y besides the projective tensor norm. We have only scratched the surface.
Exercises
L00 [0, 1] . Find a onecomplemented subspace in A (11'2 ) isometric to T::O . 3. Let Wn = span{1, t, t 2 , , t n } C C[0, 1] . Show that .X(Wn ) > c log n for some positive c. 4. Show the analogue of Theorem 24 for £1 (11') and A(11'). 1 1 5. (a) Show that if 2 $ p $ oo then cnii $ .X(t;) $ nii for some c > 0. (b) Show that lp, 1 < p $ 2, is not isomorphic to a complemented 1. 2.
Show that
£00
is isomorphic to
•
•
•
subspace of any £1 (p,).
.X(£1). 7. Show that, if X is separable and Y C X is isomorphic to co, then Y is complemented in X. 8. (a) Show that ioo /co contains a subspace isometric to co(r) , where r has continuum cardinality. (b) Show that co C £00 is not complemented. (c) Show that if X C £00 and X rv c0 then X is not complemented 6.
Estimate
in i00 •
9.
10.
Suppose that p (z) is a polynomial of degree N on CCd , d 2:: 1 such that < 1. Show that there exists an inner function cp on md such that if cp = E:=o 'Pn with 'Pn a polynomial homogeneous of degree n, then p = E �= l 'Pn· Show also that for every r < 1 and every e > 0 this inner function cp can be constructed in such a way that sup{ l cp(z)  p(z) l : l z l < r} < e. For d > 1 let v c md be the set { (zb . . . ' Xd): l zi l $ d  ! j = ' 1 , . . , d}. Show that there exists an inner function cp on md such that l cp(z) l < � for z E V.
II P IIHoo(IBd)
.
11.
If f(z) is holomorphic on md , then its radial derivative Rf (z) is defined as Rf ( ) = E�= l Show that if the dth radial d derivative of J , R f , is in H1 (md) then f E A(md) .
z
(zi 8£t> ) .
130
III.B. Projection Constants §Exercises
£2 defined T( f ) = T: A ( JBd ) , where (Pk ) are the polynomials con =l structed in Proposition 18, is onto. 13. A function f(z) holomorphic in ID is called a Bloch function if sup 1/ '(z) l · (1  l z l 2 ) < oo .
12 .
Show that the operator
( fsd f( ( )P2n ( ( )da( ( ) )
�
+
as
zEID
( a) Show that L::= l anz 2 n is a Bloch function if and only if ( an ) ;::='= l E ioo . (b) Find a Bloch function which does not have radial limits a. e. on the circle 11'.
14.
For a function analytic in 1Bd and ( E Sd we define a slice function fc; (z) analytic in ID by fc; (z) = f(z · ( ) . We say that a function f analytic in 1Bd is Bloch if
Find a Bloch function in 1Bd which does not have radial limits aa.e. on Sd.
III. C . L1 (JL)Spaces This chapter discusses some topics connected with £1 ( JL ) spaces. We start with the general notion of semiembedding and investigate semi embeddings of £1 ( JL ) into various Banach spaces. This is applied to the class M0 (11' ) of all measures such that p,(n) + 0 as n + oo. We prove the classical Menchoff theorem that there are singular such measures and a theorem of Lyons characterising zero sets for the class Mo (ll' ) . Next we describe relatively weakly compact sets in £1 ( Schur's theorem ) and in general £1 ( JL ) spaces for a probability measure JL ( the Dunford Pettis theorem ) . We also discuss the connection between type and finite representability of £ 1 . Some characterizations of reflexive subspaces of £1 are given. We conclude with some results connected with the classical result of Nevanlinna about cosets F + H00 (11') C £00 (11') . We have already seen that properties of £1 spaces differ from properties of Lpspaces for 1 < p < oo. In particular, £1 being non reflexive, its unit ball is not weakly compact. Actually more is true.
1.
Proposition. If JL is a nonatomic measure and T: £1 ( JL ) + X is a 11, weakly compact linear operator then T(BL1 (p.) ) is not normclosed.
Since BL1 (p.) has no extreme points and T is 1 1 T(BL 1 (p.) ) also has no extreme points. The KreinMilman theorem I.A.22 implies that T(BL1 (p.) ) is not weakly compact, thus it cannot be normclosed.a
Proof:
2.
Let us introduce the following
Definition. A 11 linear operator T: embedding if T(Bx ) is closed in Y.
X
+
Y is called a semi
Clearly every isomorphic embedding is a semiembedding. Also if X is reflexive then every 11 map T: X + Y is a semiembedding. Proposition 1 clearly says that there is no semiembedding from £ 1 [0 , 1] into any reflexive space.
3. The following is a general, topological observation showing that r1 I T(Bx ) has at least one point of continuity. We formulate it for Banach spaces only, because this is the way we will use it.
132
III. C.
L 1 (p,)Spaces §4
Proposition. Let X and Y be separable Banach spaces and let be a semiembedding. Then there exists an x E X, ll x ll = 1 such that if (x n )�= I C X, ll xn I ::; 1, n = 1, 2, . . . and Tx n . Tx then
T: X . Y Xn ___. X.
Since T is a semiembedding, the image of every closed ball in X is closed in Y. Let us fix c: > 0 and a relatively open set V C T(Bx). Let us fix a sequence (vj ) � 1 C Bx such that we can write Bx = U;: 1 B(vj , c:) n Bx . Applying the Baire category theorem to the covering of V by closed sets T ( B ( vj , c:) n Bx) we get the following statement:
Proof:
> 0 and for every relatively open set V C T(Bx) there exists a nonempty relatively open set U C V such (1) 1 that diam T (U) < c:. for every c:
Applying (1) inductively we find a sequence of relatively open sets (Un )�= I in T(Bx) such that Un :J Un + I :J Un +I and diam T  1 (Un ) < � and 0 � ul . Clearly n�=l T  1 Un consists of exactly one point Xo . The a desired X equals .
I �� I
4.
Applying Proposition 3 for
X = L 1 [0, 1]
we get
Lemma. If T: L 1 [0, 1] . X is a semiembedding, then there exists an f E L 1 [0, 1] with II f II = 1 and a number 8 > 0 such that for all 1 realvalued functions r.p with J0 r.p(t)dt = 0 and I'PI = 1, p,a.e. we have
II T( rp f) ll
�
8.
From Proposition 3 we infer that there exists an f E L 1 [0 , 1] , 11 !11 = 1 and 8 > 0 such that II !  gi l < � whenever IIYII ::; 1 and li T f  Tg ll < 8. Given r.p above we take '1/J = r.p or 'ljJ = r.p so that we have J(1 + '1/J) I f l dp, ::; 1. For g = (1 + '1/J)f we have IIYII ::; 1 and IIY  ! II = 11 '1/Jf ll = 1, thus 8 < li T/  Tg ll · But II T( rpf ) ll = li T/  Tg ll
Proof:
as
so the claim follows.
5.
a
Now we are ready to prove
Theorem. Let p, be an atomfree, separable measure. There is no semiembedding from L 1 (p,) into Co · We see from III.A. 1 that it is enough to consider LI [O, 1] only. Assume to the contrary that T: £ 1 [0, 1 J . eo is a semiembedding, and
Proof:
III. C.
L 1 (p,)Spaces §6.
133
II T II = 1. Take f E L 1 [0 , 1] and 8 > 0 as in Lemma 4. Fix an integer N > 2 /8 , and let A� . . . . , A N be disjoint sets such that fA  1/1 = j;; . 1 For each n , n = 1 , 2, . . . , N let rj denote a sequence of Rademacherlike functions on An , i.e.
l r'J I = X An ' rj = 0,
J
= 1 , . . . , N, j = 1 , 2, . . . , n = 1, . . . , N, j = 1, 2, . . . , n
rj f �O as j + oo ,
for every
n, n
= 1, 2, . . . , N.
(2)
(3) (4)
Using the standard 'gliding hump' argument (cf. proof of II.B. 17) con dition (4) yields numbers j (n) , n = 1 , . . . , N, such that
But (2) , (3) and Lemma 4 give
which is impossible. This contradiction proves the theorem.
a
6. The spirit of the above results is that it is rather difficult to put a weaker, linear topology on the unit ball of L 1 [0, 1] and make it compact or even complete. The intuition is that somehow we always have to add some singular measures in the completion. The following classical theorem of Menchoff is only an example of this. Theorem. (Menchoff) . Let G b e a compact, infinite, metrizable abelian group with dual group r. Let M0 (G) c M(G) denote the set of all measures v with C'(y) E eo(r). Then Mo is a nonseparable closed band in M(G) . Since the Fourier transform A: M(G) + l00 (r) is continous and since Mo = (A  1 )(c0(r)), we see that M0 is a closed linear subspace. Also for v E Mo and p = :L: ..,. eA a..,. 'Y with finite A c r one easily checks that p · v E Mo. Since Mo is closed this gives that Mo is a band. If M0 is separable, then there exists a positive measure p, E M (G) such that Mo = L1 (p,) . One easily checks that p, has no atoms. By Theorem 5 (B M0 )A is not closed in co ( r} , i.e. there exist o!(y) E c0(r)\(BM0 )A
Proof:
134
III. C.
and fLn E such that a in of Since 1 11LII � 1 and contradicts the choice of a.
BMo {!Ln} �=1·
P,n +
P,
L1(IL)Spaces § 7.
eo(r).
= a
Let fL be an w*cluster point we infer that fL E This
BMo·
a
Note that the group structure plays almost no role in the above argument.
7. Given a class of measures it is natural to seek its zero sets, i.e. sets of measure zero with respect to every measure in the class. We will discuss zero sets for M0 (11'). This is a small part of the classical branch of the theory of Fourier series or harmonic analysis on more gen eral groups. In order to proceed we need some definitions. We say that a sequence C 11' has an asymptotic distribution if and only if N 8x n converges a(M(11') , C(T)) to some measure v E M(11'). Let
(xn )�=1 1 N n=L:1 us recall that 8x denote the Dirac measure concentrated at the point x. 1 Since for every m E we have zl(m) lim N +oo ( N L::= 1 8x n ) " (m) 1 lim N +oo ( N L::=1 exp (  im x n ) ) we see that the sequence (x n )�= 1 N 11' has an asymptotic distribution if and only if Nlim k L: exp(  im xn)) ( +oo =1 n exists for every m E Z. A Borel subset E C 11' is called a Weyl set if there exists an increasing sequence of integers (nk)k:, 1 such that for ev ery x E E the sequence (nk · x)f= 1 has asymptotic distribution with 7l
=
the corresponding measure measure.
8.
Vx
=
C
an
different from the normalized Lebesgue
The following theorem characterizes
M0 (11').
The measure fL is in Mo(11') if an d only if fL(E) every Weyl set E C 11'.
Theorem.
=
0
for
Throughout the whole proof it is sufficient to consider only positive measures (see Theorem 6). =>. Let fL be a nonzero measure in Mo 11' and let E be a Weyl set with the corresponding sequence For m E Z, m =/= 0 we put
Proof:
c
m
(t) =
{
(nk)k:, 1 .
1
lim N+ oo N
. 0,
()
t k =1 exp (  imnk t)
,
t E E, t (/_ E .
(5)
III. C.
£1 (p,)Spaces §9.
For a Borel subset
135
F c E we have (6)
Mo(ll') is a band (Theorem 6) p, I F E Mo(ll') so (6) gives JF Cm (t)dp,(t) = 0. Since F was an arbitrary Borel subset of E we infer that cm (t) = 0, p,a.e. for m =/= 0. But E is a Weyl set, thus for every t E E there is an m =I= 0 such that em (t) =I= 0. This shows that p,(E) = 0. ¢=: . Let us take p, fJ. M0(11') and let us fix a sequence of integers n k � oo as k  oo (or n k � oo as k  oo ) such that jL(n k ) � Since
a
=I= 0 as k 
oo.
Applying Theorem III.A.29 to the family of sequences
{exp(imn k t) }f= 1 in L2 (11', dp,) we get a further subsequence (n�)� 1 such that JJV (t) = N  1 L�= 1 exp(  imn�t) converges p,a.e for each m E 7l. Let us put E = {t E 11' : N+oo lim JJV(t) exists for each m E 7l and lim j].{t) is not zero}. N+oo
This is a Weyl set. We have
Thus
a
p,(E) > 0.
9. Our goal now is to characterize weakly compact sets in L 1 (p,) spaces. This characterization has many further applications (see III. C.19 or III.H. l O among others) and generalizations (see e.g. III.D.31 ) . It also nicely connects the general functional analytic notions with measure theoretical concepts. We start with the distinctive special case of the space £ 1 . Theorem. (Schur) . For a bounded subset H ditions are equivalent:
c
£1
the following con
III. C. L 1 ( !l ) Spaces § 1 0.
136 (a) (b) (c)
H is relatively compact; H is relatively weakly compact; there is no sequence (an )�= 1 C H which is a basic sequence equiv alent to the unit vector basis of € 1 . The proof clearly follows from the following.
10 Lemma. If H C € 1 is a bounded subset, not relatively compact, then there exists a basic sequence (an)�= 1 C H equivalent to the unit vector basis of € 1 . We find {bn}�= 1 C H such that I I bn II :::; C, n = 1 , 2, . . . for some C and ll bn  bm ll 2: 8, for m # n and {j > 0. A standard diagonal procedure gives a subsequence { bnJ � 1 such that bn1 (k) ___, b(k) as j + oo, for every k = 1 , 2, . . . . Clearly b E € 1 and l l bn1  b ll 2: 8, j = 1 , 2, . . . . II.B.17 gives a further subsequence (call it also bnJ such that (bnj  b)� 1 is equivalent to a blockbasic sequence, thus to the unit vector basis in € 1 . Let Y = span{(bnj  b) }� 1 . Omitting if necessary a finite number of j's we can assume that b � Y. Then
Proof:
I :L:O �j bnj ll = II �:::j:a� (bnj  b) + < :L:O �j )b ll 2: K L iaJ I , thus
{bnJ � 1
2:
K l l :L:O �j (bnj  b) ll II
is the desired sequence.
11. Now we will discuss relatively weakly compact sets in L 1 ( !l ) for a general probability measure 1l· Our main tool will be the notion of uniform integrability. Definition. A subset H C L 1 (1l) is called uniformly integrable if for every c: > 0 there exists an TJ > 0 such that sup
{ i l f l d!L : !L(A)
:::;
TJ,
fEH
}
:::;
c:.
(7)
If 1l is an atomfree probability measure then every uniformly in tegrable set in L1 (!l) is normbounded. This follows from (7) and the observation that for every f E L1 (!l) there exists a set A with !l(A) = * such that fA l f ld!L 2: n  1 J l f ldll. In the other direction let us observe
III. C. Ll (IL)Spaces §12.
137
that every oneelement set, and thus every finite set, is uniformly in tegrable. To see this put A n = {t : l f(t) l > n } . Since IL( A n ) + 0, the Lebesgue dominated convergence theorem gives l f l diL + 0 as
JAn
n + oo .
12. The next theorem gives the promised characterization of relatively weakly compact sets in £ 1 (IL) . This is the main result of this chapter. It says that basically there is only one reason for a bounded set not to be relatively weakly compact. In this sense it is similar to Proposition II.D.5 and also to Theorem 9. Note also the equivalence of finite and infinite conditions. This will be investigated later. Theorem. Let IL be a probability measure and let H be a bounded subset of Ll (IL) · The following conditions are equivalent: ( a) H is not relatively weakly compact in Ll (IL) i ( b) H is not uniformly integrable; ( c ) there exists an that
e > 0 and a sequence of disjoint sets (An) ;:::>= 1 n =
( d ) · there exists a basic sequence vector basis in e 1 ;
( Jn );:::>= 1
such
1 , 2, . . .
C H equivalent to the unit
( e ) there exists an e > 0 such that for every integer N there exist N disjoint sets A1 , . . . , A N such that sup
{ in l f l diL : f E } H
�
e,
n =
1 , 2 . . . , N;
( f ) there exists a constant K such that for every integer N there exist it , . . . , fN C H, K equivalent to the unit vector basis in if . Proof:
The proof will consist of the following implications:
(b)
/ ( a) � (e)
(d ) � ( f)
�(e) /
III. C.
138
Lt (Jt)Spaces § 1 2.
with the implications marked * being obvious. (a) + (b) . Suppose H C Lt (Jt) is uniformly integrable, thus bounded, i.e. for f E H we have 11!11 � M. Given an integer n we write every function f E H as f = r + fn = f · X{ l f l �n } + f · X{ l f l
}Bk }Bkf 1 1J I � 8j, k ,
i = 1 , 2, . . . , k
L
(c) + (d) . Fix disjoint sets ( A n )�=t and functions fn E H, n = 1 , 2, . . . such that JAn l fn l dJt � e > 0. Put hn = (JAn l fn l dJt )  t fn . X An and cpn = sgn fn · XAn · Clearly Y = span{hn }�= t is isometric to it and P(f) = I: :'= t J f cpndw hn is a projection from Lt (Jt) onto Y. One easily sees that P ({ fn }�=t ) is not relatively compact in norm. From Theorem 9 we get a subsequence Un; } � t such that { P C fn; H � t is equivalent to the unit vector basis in it . but this implies that Un; } � t itself is equivalent to the unit vector basis in it . (f) + (e) . We can assume II !J II � 1 , j = 1 , . . . , N and thus K  t I:f=t l aj l � J I I:f= t a3 f3 l for all sequences of scalars ( aj ) f=t · Let r3 (t) be, as usual, the Rademacher functions. We have
III. C. L t ( J.L )Spaces §13.
139
l
( �N l aj fi l ) dJ.L j N :::; ( J mr l aj fi i dJ.L) ( J � iaj fi l dJ.L) :::; (! mr l aj fi l dJ.L) ( �N l aj l )
:::; c mr l aj fi l ) �
·
2
l
2
l
(8) l
2
l
2
2
•
Thus (9) and in particular ( 10) Let B8, s = 1 , 2, . . . , N, be disjoint sets such that (some of them can be empty) . From (10) we get
11/s ll :::; 1 we JB. l fs l dJ.L � 2 k2 ·
Since
(maxj lfi i ) I Bs = l fs l
infer that for at least ( 2Kllf t ) indices
s
we have a
13. Remarks. (a) If we keep track of the constants in the proof of (f) + (e) we get the following statement: If (JJ ) f= 1 C L1 (J.L) are such that K
1
� N
l aj l :::;
JI� N
l � l aj l
aj fi dJ.L :::;
N
for all scalars(aJ ) f= 1
then for every 8 < 1 there exists a subset A C { 1 , . . . , N } and disjoint sets {Aj } jE A such that
where for every 8, r.p0 (N)
+
oo
as N
.......
oo .
III. C.
140
Lt (11)Spaces § 1 4.
(b) If 11 is an arbitrary measure on n and H c Lt (0, 11) is weakly relatively compact then there exists a set nl c n of afinite measure 11 such that all functions from H are supported on 0 1 . Thus when dealing with relatively weakly compact sets in £ 1 (11) we can restrict our attention to afinite 11 · But this case, as we know (II.B.2(c) ) , easily reduces to the probability measure 11 ·
14 Corollary. (Steinhaus) . The space Lt (l1) , 11 arbitrary, is weakly sequentially complete, i.e., weakly Cauchy sequences are weakly conver gent. Proof: Since every sequence is supported on a afinite set it is enough to consider a probability measure 11 (Remark 13(b) ) . A weakly Cauchy sequence which is not weakly convergent is not relatively weakly com pact, so from Theorem 12 it has a subsequence equivalent to the unit vector basis in ft , thus not weakly Cauchy. This contradiction proves the corollary. II 15. We have seen in Theorem 12 the interesting interplay between global notions like weak compactness and the local concept of finite representability of £ 1 . We want now to discuss the finite representability of £ 1 in a general Banach space X. We start with an interesting lemma about finite dimensional isomorphs of if . Lemma. Let X be an Ndimensional Banach space with d ( X , if ) = a. Then there exists a subspace X1 C X with dim X1 = [ VN] and
d(Xt , € 1[ v'NJ ) :::; yr;:.a.
Proof: scalars
Let us fix a basis we have
( a3) f= 1
(x3) f= 1
in
X such that for every sequence of
Let us also fix [ VN] disjoint subsets A s c { 1 , 2, . . . , N} each with car dinality [ VN] . For each s = 1 , 2, . . . , [ VN] we define
:
III. C. L1 (11)Spaces § 1 6.
141
If for some s we have ds � )a than xl = span{Xj j E A s is a good choice. On the other hand if for all s we have d8 < )a then we fix Ys = LjeA . fr.jXj such that IIYs ll < )a and L jeA. l ai l = 1 for s =
1 , 2, . . . ,
so for X1
=
[vr.:r] . 1v
For every sequence of scalars
span { Ys
}
[v'NJ we have
(f3s ) s = l
•
r;;.
} s[v'N= l ] we get d(Xt , i1[v'N] ) � y a .
16 Theorem. Let X be an infinite dimensional Banach space. The following conditions are equivalent: (a) X does not have type p for any p
> 1; and every e > 0 there exist normone vectors
(b) for every n = 1 , 2, . . . Xt , . . . , X n in X such that
(c) i 1 is finitely representable in X ; (d) for every n = 1 , 2, . . . and every e Xn , e C X with d ( Xn , e , ii' ) � 1 + e.
>
0 there exists a subspace
This is a remarkable theorem. We will use it in III.I to study some questions about the disc algebra. Note that it contains the passage from a probabilistic context of the definition of type to the purely determin istic situation described in (c) and (d) . It is also nice because it tells us that certain abstract things (like (a) ) happen only due to the presence of a very 'concrete' subspaces, namely il 's. For the proof of this theorem we introduce constants 'Yn (X) , n = 1, 2, . . . , defined by the formula 'Yn (X)
{ ')' : ( / I t
2 ri (t)x i dt
=
inf
�
'/' Vn ( t llxi ll 2) ' for all
( x i )� 1 C
X} ·
l )
�
(11)
III. C. L 1 (Jt)Spaces § 1 7.
142
Note that 'Yn (X) :5 1 for n = 1 , 2, . . . . The following lemma really explains some consequences of condition (a) of Theorem 16. 17 Lemma. (a) The constants 'Yn (X) are submultiplicative, 'Yn· k (X) :5 'Yn (X) · 'Yk (X) for all n, k E N. (b) If 'Yn(X) < 1 for some n then X has type p for some p > 1 .
(xj ) j;:1
(a) Fix integers such that
For
0, . . . , (k
Proof:
s =
 1)
n
and
k,
a number
e>
i.e.
0 and a sequence
define
( s+ l ) n r/Js (O) = L Tj (O)xj . j = s· n + l For every (} we have
and integrating over (} we get
k l
'Y� ( X ) k L
J
ll ¢s (O) II 2 d0 s=O (s+ l ) n kl ( X ) 'Y k (X)n :5 'Y� L ll xi ll 2 L � j = s· n + l s=O k ·n 2 = 'Y�(X) · 'Y� (X) · k · n L ll xi ll • j=l :5
( 13)
e was arbitrary, comparing (12) and (13) we get 'Yn· k (X) < 'Yk (X)'Yn (X). ( b ) Fix q > 1 such that 'Yn(X) = n  l / q' where * + f, = 1 . Observe also that it follows directly from ( 1 1 ) that ('Yn (X) Jn)�= l is an increasing Since
III. C. L ( !1 ) Spaces § 1 7.
1
143
(xj )j� 1 2::7= 1 I xi l i P = s+l ) /p :::; l xJ I :::; n  s/P } . ( A = n s 1 I As l :::; n8+ . 2 ! I ( ( J I t l rj (t)xj l 2 dt) ! ::; � L rj ( t )xj l dt) / s0 J E As J :::; L I'I A.I ( X ) JiA:T ( _L l xi l 1 2 ) 2 s=O J EAs 1 :::; L l'n • +l (X) vns +l (n8 + n  2 s iP )! j=O  � ) s+l ns+l n  s/p (n :::; L s=O :::; n 1  .!. s=O ns ( l  .!._1 ) This shows that X is of type for every
sequence. Take 1
< p < q.
For an arbitrary finite sequence with 1 (we can put in some additional zeros to have the right length) we define sets of indices {j : Clearly We have k
!
00
00
·
00
"\"' � 00
•'
p
•'
< oo .
v
a
p < q.
From Lemma 15 we see that (c){:}( d) . Also obviously both (c) and (d) imply (a) . (a)=>(b) . From Lemma 17 we see that l'n 1 for = 1, 2, 3 . . . . This means that for every Xn in such that = 1 , 2, . . . and every e > 0 there are vectors
Proof of Theorem 16.
(X) =
n
n
x1, , •
n)
•
•
X
l xi l , j =
If e is very small (depending on we see from (14) that 1 , 2, has to be practically constant so has the property for every and every e > 0 there exist vectors such that 1 for j 1, XI , , Xn E with
. . . , n,
n
X I xi I = (1  e)n :::; (J I j=lI:n rj (t)xJ I 2 ) 2 . •
.
.
1
X = . . . ,n
Since obviously
2 dt) ! 1 ( )x t j j J r ( I� :::; (T n
"�2�ll t€jXj l 2
+ (1  2  n )
( t 11 x1 11 r ) !
(15)
144
III. C. Lt (!L)Spaces § 1 8.
we infer from (15) (make c: very small) that X satisfies (b) . (b)=>(c). For each sequence T/ = (c:i )' J= 1 with C:j = ±1 there exists a functional x; E X* such that ll x; ll = 1 and E;= l x; (c:i xi ) > n  c: . Since I xi I = 1, an elementary computation shows that for every T/ and j, we have l x; (c:i xi )  1 1 < J&. Using this, for any sequence of numbers (ai ) J= 1 with Ej= 1 l ai l = 1 we obtain
a
18. As an application of our previous considerations we have the following useful Corollary. Let X be a closed subspace of Lt (!L) · The following con ditions are equivalent:
(a) X is reflexive;
(b) X has type p for some p > 1;
(c) X does not contain a subspace isomorphic t o f 1 ; (d) f 1 is not finitely representable in X. Since each of the conditions holds for X if and only if it holds for every separable subspace of X we can assume that 1L is a probability measure (see III.A.2). Now the corollary immediately follows a from Theorem 12 and Theorem 16 (see also II.A.14) . Proof:
19.
We wish to conclude this section with the proof of the following.
Theorem.
H00 (1I') } < 1. 0 such that
Suppose Fo E L00 (1I') is such that inf { I I Fo + h ll oo : h E Then there exists an F E Fo + Hoo (1I') and h E Ht (1I') , h =/=
F · h = lhl
a. e . on
1I' .
(16)
The proof of this theorem is a nice application of Theorem 12. More over the following lemma is relevant to some questions which will be discussed in Chapter III . E .
I45
Ill. C. £1 (JL)Spaces §20.
20 Lemma. If {En } is a sequence of measurable subsets of '][' such that I En l + 0, then there is a sequence {gn } of functions in H00 such that
(a) supess{ l9n (t) l : t E En } + 0 as n + oo , (b) 9n (O) = (211")  1 fv 9n (t)dt + I as n + oo , (c) IYn l + I I  9n l :5 I + en where lim en = 0. Fix numbers An such that An + oo as n + oo and An iEn l + 0 as n + oo. Let In be the Poisson integral of AnXEn + iAnXEn (XEn is the harmonic conjugate of XEn , I.B.22) . Clearly In is an analytic function on D taking values in the right half plane. Since the map z 1+ ( l ! z ) maps the right half plane onto the disc Proof:
( I 7) we get that hn (z) = ( 1 + /n ( z ) ) maps D into the disc given by (I7) . Since ln (O) = A n iEn l we get hn (O) + I as n + oo . Also supess{ l hn (t) l : t E En } :5 supess{ =
I : t E En } (I + Re ln (t))
I + 0 (I + An)
as
n + oo .
Now observe that the map z + z6 compresses the disc (I7) into the ellipse l w l + I I  w l :5 I + e(6) where e(6) + 0 as 6 + 0. All the above yields that for some sequence On + 0 slowly enough the functions a 9n = h�n satisfy (a) , (b) and (c) . Proof of Theorem 19. Put
Since the unit ball in H00 is a(L00 , £1 )compact this supremum is at tained at some F E F0 + H00 • Clearly I � dist(F, H�,) = inf{ II F  h ll : h E Hoo , h(O) = 0}. If II F  h lloo < I for some h E H! we see that for small e's the function F  h + c: is an admissible I in (IS) giving a larger mean than F. This shows that dist(F, H! ) = 1 . Since Hi = Leo/ H! we get by duality
146
III. C. L1 (Jt)Spaces §20.
Fix a sequence (hn)�= l in H1 with ll hn ll � 1 and
(19) If (hn)�= l has a weakly convergent subsequence (hnk )k:: 1 we put h w lim hnk E H1 (Y) . Now (19) gives
=
so in particular h =i 0. Since II F IIoo � 1 and ll h lh � 1 we get (16). We complete the proof by showing that the assumption that (hn)�= l has no weakly convergent subsequence leads to a contradic tion. If (hn )�= l has no weakly convergent subsequence Theorem II.C.3 (EberleinSmulian) and Theorem 12 give sets (En) C 1I' with I En l + 0 such that (20) hn (e i8)d0 > {3 > 0 2
1
� � Ln
at least for a subsequence of (hn) · Now let 9n and C:n be given by Lemma 20 and put (1  9n)hn . Hn  9nhn and Kn = 1 + C:n 1 + C:n We infer from Lemma 20(c) that II Hn l l 1 + II Kn l l 1 � ll hn ll < 1. Also since C:n + 0, (19) gives _
1 1'1m 1  n+oo 2 71'
1 1211" 0
F
hn 1 + C:n

I
(2 1 )
Thus the limit in the middle exists and equals 1. From (20) and Lemma 20 (a) we get limn II Kn l h � {3 > 0 so (21) yields (2 2 ) Note that Lemma 20 (b) gives Kn (O) + 0 as n + oo. The dual ity relation (HP)* = L 00 / Hoo and ( 22 ) give dist(F, H00) > 1. But
III. C. L1 (tL)Spaces §21 .
147
dist(F, Hoo ) = dist(Fo , H00 ) < 1. This contradiction shows that a (h n );:"= 1 actually has a weakly convergent subsequence.
21 Corollary. Let (z3 ) �1 C 1D be such that 2:::3 (1  l zJ I ) < oo . For every f E H00 with ll f ll oo < 1 there exists an inner function r.p with r.p ( zj ) = f(zj), j = 1, 2, 3, . . . . Let B be the Blaschke product with zeros (z3 ) �1 . We de fine Fo E L00 (1r) by Fo = B f and apply Theorem 19. We obtain a unimodular F = F0 + g for some g E H00 • Since on the circle '][' we have BF = f + Bg, we see that BF is a boundary value of an analytic function. This function is clearly inner and satisfies BF(z3 ) = f (z3 ) for a j = 1, 2, . . . . Thus BF is the desired r.p.
Proof:
Notes and remarks. The fact that there is no weaker topology on L1 (f..L ) , f..Latomfree, with some compactness properties is well established. Our Proposition 1 only states the easiest fact of this type. A more detailed study of semi embeddings is in BourgainRosenthal [1983] . This paper contains our Theorem 5 and the proof of Theorem 6. The first proof of Theorem 6 was given in Menchoff [1916] . Much more detailed information on supports of measures in Mo(G) can be found in Varopoulos [1966] . The following fact was shown in PignoSaeki [1973] .
Theorem A. lE f..L is a measure in M(G) such that then f..L E L 1 ( G) .
f..L * Mo(G) c L1 ( G )
Theorem 8 is taken from Lyons [1985] . It completes a long line of investigations in the theory of Fourier series and solves problems going back to Rajchman in the 20 ' s. The nontrivial implication (b)::::} (a) in Theorem 9 is due to Schur [1921] in the language of summability methods. Banach spaces where this implication holds are nowadays said to satisfy the Schur property. The fundamental Theorem 12 is usually called the DunfordPettis theorem. They established the equivalence between (a) and (b) and successfully used it in their papers Dunford [1939] and DunfordPettis [1940] . The relevance of condition (d) was realized in KadecPelczynski [1962] . This theorem is by now classical and various versions of it with many different applications are presented in DunfordSchwartz [1958] , DiestelUhl [1977] , Kopp [1984] . This last book shows the fundamental importance of this theorem in probability
III. C. L1 ( J.L )Spaces §Exercises
148
theory. Corollary 14 is a classical theorem of Steinhaus [1919] . We will discuss important generalizations of these facts in the next chapter. Lemma 1 5 is a well known finite dimensional version of a result of James [1964] (see also Exercise 9) . The important Theorem 1 6 was proved by Pisier [1974] . We basically reproduce his proof here with the changes necessary to obtain the result for complex spaces as well. This theorem was the beginning of the study of connections between type and cotype on one side and geometry on the other side. By now the subject has grown enormously. A presentation of this is contained in the beautiful monograph MilmanSchechtman [1986] . Let us only note the direct generalization of Theorem 1 6 proved by MaureyPisier [1976] .
Theorem B. Let X be an infinite dimensional Banach space and let Px = sup{p: X has type p} , qx = inf{q: X has cotype q}.
Then fpx and lqx are finitely representable in X .
The connection between the reflexivity of subspaces of L1 ( J..L ) and conditions like (b) in Corollary 1 8 was recognized in the important paper Rosenthal [1973] . Theorem 1 9 and Lemma 20 is due to Garnett [1977] (see also Gar nett [1981] ) . The first version of Lemma 20, with a more complicated proof, was discovered by Ravin [1973] . Various variants, usually referred to as the Ravin lemma, are known. The main idea is to show that on small sets there are analytic functions almost peaking on them. Theorem 111.1.9 presents a very elaborate version of this idea. Corollary 21 is an easy special case of a classical theorem of Nevanlinna. For more details on such matters the reader should consult Garnett [1981] , in particular chapter IV.4.
Exercises
1.
Suppose that T : c0 + X is a semiembedding. Show that T is an embedding.
2.
Suppose X is a Banach space and X* is separable. Show that X* does not contain a subspace isomorphic to L1 [0, 1] .
3.
Construct a sequence which does not have the asymptotic distribu tion.
III. C.
4.
L 1 (J.L)Spaces §Exercises
149
Suppose that J.L is an arbitrary measure on f! and that H c £ 1 (J.L) is a relatively weakly compact subset. Show that there exists a subset V C f! of afinite measure such that for every f E H we have suppf C V.
5.
Suppose (f!, J.L) is a probability measure space and K is a compact space. Show that, if T: £ 1 (J.L)  M ( K) is a continuous linear opera tor, then there exists 11 E M(K) , 11 � 0 such that T(L 1 (J.L)) C £ 1 ( 11 ) .
6.
Show that H C £1 (J.L) is relatively normcompact if and only if it is relatively weakly compact and relatively compact with respect to convergence in measure.
7.
Find a weakly compact set H c £ 1 [0, 1] such that there is no func tion cp � 1 , cp finite almost everywhere, such that {!: f · cp E H} C Lp [O, 1] for some p > 1.
8.
Suppose that Un)':'= 1 is a sequence in H1 (1l') and that fn � f and 11/n ll  II/II as n  oo for some f E H1 (1l'). Show that 11/n  /II  0 as n  oo .
9.
Suppose X is isomorphic to £ 1 . Show that for every c > 0 there exists an infinite dimensional subspace X 1 c X with d (X 1 . £ 1 ) �
1 + c.
10. Show that if £ 1 is not finitely representable in finitely representable in X* .
X,
then
£1
is not
1 1 . Suppose that the sets (En ) in Lemma 20 are closed. Show that the functions (gn ) can be chosen in the disc algebra A.
12. Show that if H c £1 [0, 1] is not uniformly integrable then there exists a sequence (cp n)':'= 1 C C[O, 1] such that E:'= 1 I 'Pn(t) l � 1 1 and lim sup n+oo sup hEH I f0 'Pn(t)h (t)dt l > 0.
13. Suppose that T: X  £ 1 is a noncompact operator. Show that there exists a complemented subspace X1 and T I X1 is an isomorphic embedding.
c
X such that X1 "' £ 1
III.D. C(K) S paces We start this chapter with the general notion of an Mideal and show that for every element one can find a best approximation to it in any Mideal. We discuss the space Hoo + C and show that every function f E L00(1l') has a best approximation in H00 + C. We prove the linear extension theorem of Michael and Pelczyfiski and the Milutin theorem that all spaces C(K) for K compact, metric and uncountable are iso morphic. We present the construction of the periodic Franklin system and prove its basic properties. We investigate its behaviour in Lp (1r) , C(1l') and Lip0 (1l'). We show that Lip0(1r) "' £00 • Then we investigate weakly compact sets in duals of C k (1r8 ) and A(1Bd ) and show that they have properties similar to weakly compact sets in £ 1 The unifying framework for this study is provided by the concept of a rich subspace of C(K, E). We also introduce and study the concepts of the Dunford Pettis property and the Pelczyfiski property.
(J.L) .
1. Our subject in this chapter is the spaces of the form C(K) where K is a compact space. This includes also spaces L00 ( J.L ) since they can be realized as C(K) (I.B.lO) . This fact is a standard result in the theory of Banach algebras. It shows the importance of the multiplicative structure existing in C(K)spaces. Given a closed subset S C K put C(K; S) = {! E C(K) : / I S = 0 } . This is clearly a closed subspace and actually an ideal in the algebra C(K) . Proposition. Every closed ideal in C(K) is of the form C(K; S) for some closed S C K. Given a nontrivial closed ideal I C C(K) we define 81 = Since every I E I is noninvertible we get ,  1 (0) =F 0. Given It and h in I we see that 1/1 1 2 + 1 12 1 2 = !d1 + !d2 E I which gives that the family {f 1 (0) } J E I is a family with the finite intersection property so its intersection 81 is not empty. Thus C(K; 81) is a proper ideal containing I. Note that if f E I then f E I. To see this let g E C(K) be such that g = f/1 on the set {k E K: 1/1 2:: c } and IIYII � 1. Such a g exists by the Tietze extension theorem. Then g · f E I and IIY · f  fl loo � 2c. Since c was arbitrary and I was closed we get f E I . Note also that for any two points k1 =F k2 in K\81 there is an
Proof:
n! E l , 1 (0).
152
III.D. C(K)Spaces §2.
f E I with f (kl ) =F f(k2 ) =F 0. Thus the StoneWeierstrass theorem a I.B.12 yields that I = C(K; SI ) . The notion of an ideal is not a linear concept but the following concept generalizes the above considerations.
2.
Definition. A subspace M of a Banach space X is called an Mideal if there exists a projection E from X* onto Ml. = {x* E X*: x* I M = 0 } such that for every x* E X* we have
ll x * ll = II Ex * ll + ll x *  Ex * II · One checks that every subspace of C(K) of the form C(K; S) is an Mideal. The projection E of a measure p, is given by E(p,) = p, I S. Other examples of Mideals are given in Exercises 1 and 2 and also in Theorem 8.
3 Proposition. Let M be an Mideal in X and let open balls B(xt , rt ) = B1 and B(x 2 , r2 ) = B2 be such that B 1 nB2 =F 0, B 1 nM =F 0 and B2 n M =F 0. Then B 1 n B2 n M =F 0. Proof: Consider B 1 x B2 C X E9 X and let a = {(m, m) : m E M} C X E9X. If the conclusion does not hold then (B 1 x B2 ) na = 0. Since B 1 and B2 are open we can diminish r1 and r2 a bit in such a way that the assumptions are still satisfied and there exists cp = ('Pt . cp2 ) E X* E9 X*
such that
for some positive c. Passing to biduals we get from Goldstine's theorem II.A.13 that a** = { ( m ** ' m **) : m ** E M** } is disjoint from Bi* X B2* = B(x1 , rt ) x B(x 2 , r2 ) where the balls are in X** this time. This translates back to the statement that there are two open balls Bi* , B2* in X** such that Bi* n M** =F 0, B2* n M** =F 0, Bi* n B2* =F 0 and Bi* n B2* n M** = 0. But X** = (M** E9 Z) oo (since M is an Mideal) a so this is clearly impossible. Using this proposition we will obtain the following useful
4.
If M is an M ideal in X, then the quotient map X/M maps the closed unit ball in X onto the closed unit
Theorem. q:
X
+
153
III.D. C(K)Spaces §5. ball in X/M, or equivalently, for every M : ll x  m il = dist(x, M) } =1 0.
xE
X the set
PM(x)
=
{m E
Let us note that the open mapping theorem yields that every quo tient map maps the open unit ball onto the open unit ball. For the closed unit ball this is generally false. As an example take map T: £ 1 + £2 de fined by T(e n) = Xn where ( xn) �= 1 is a sequence dense in the open unit ball of £2 . Clearly T maps the closed unit ball of £ 1 onto the open unit ball of £2 . Fix x E X\M and let d = dist(x, M) . Given c > 0 there exists m 1 E M such that ll x  m 1 ll < d + c. The open balls B(mt . �c) and B(x, d + !c) satisfy the assumptions of Proposition 3 so there exists m2 E M such that ll m 1  m2 ll :::; �c and ll x  m2 ll < d + !c. Now the balls B(m2 , ( i} 2 c) and B(x, d + (!) 2 c) satisfy the assumptions of Proposition 3, so we get m3 E M such that ll m2  m3 ll < ( � ) 2 c and ll x  m3 11 < d + (!) 3 c. Continuing in this man ner we find a sequence (m k )k"= 1 in M such that ll m k  m k + l ll < ( � ) k c and ll x  m k + l ll < d + ( ! ) k c. This sequence clearly converges to a limit m E M and ll x  m il :::; d. a
Proof of the theorem.
Remark.
Actually the above argument gives the following statement.
For every m 1 cM with
m E PM ( x) such that
5.
ll m 1  x ll < d + c there exists li m  m 1 ll :::; 3c.
(1)
Using (1) we obtain the following improvement of Theorem 4.
Proposition. If M is an Mideal in X, then for every x set PM (x) algebraically spans M.
E X\M the
Proof: For every m E M with ll m ll < d = dist(x, M) we will show that m = z  v for some z, v E PM (x) . Since PM (x + m) = m + PM (x) we have to show that PM (x) n PM (x + m) =1 0. Assume to the contrary that PM (x) n PM (X + m) = 0. Then
dist(PM(x), PM(x + 2m)) � l i m II ·
From Proposition 3 we get that for every
c > 0 there exists a point
me E B(x, d + c) n B(x + 2m, d + c) n M.
(2)
154
III.D. C(K)Spaces §6.
From (1) we get m 1 E PM (x) such that ll m 1  me ll $ 3c and m2 E PM (x+2m) such that ll m2  me ll ::5 3c, so dist(PM (x), PM (x+2m) ) $ 6c. a This contradicts (2) if we choose c < lo II m il · This proposition shows that the best approximation by elements of an Mideal is always possible and in a very nonunique way. A concrete application of this fact will be given in Corollary 9.
6. Let us introduce now the space H00 + C which has many important applications in operator theory and function theory. To be more precise Hoo + C is the subspace of L00 {1r) algebraically spanned by C(1l') and boundary values of H00 • Our study of this space is based on the following general Lemma. Suppose Y and Z are closed subspaces of a Banach space X and suppose that there is a family � of uniformly bounded operators from X into X such that (a) every A in � maps X into Y, (b) every A in � maps Z into Z, (c) Eo� every y E Y and c
> 0 there exists A E � such that ll y Ay ll < c.
Then the algebraic sum Y + Z is closed. Let x E Y + Z. We can find a sequence X n E Y + Z such that E�= l Xn = x and ll xn ll $ 2  n for n > 1. Every Xn, n = 1, 2, . . . can be written as X n = Yn + Zn with Yn E Y and Zn E Z. Using (c) we can find An, n = 1, 2, . . . , in � such that llYn  A n Yn I < 2  n . Let us put Yn = Yn  An Yn + AnXn and Zn = Zn  An Zn · Properties (a) and (b) show that Yn E Y and Zn E Z. Since � is uniformly bounded we see that II:Yn ll ::5 c 2  n . Since X n = Yn + Zn we have ll zn ll ::5 c2  n . This gives x = E�=l Xn = E�=l Yn + E�=l Zn · But Y and Z are complete a SO X is really in Y + Z.
Proof:
7 Corollary. The algebraic sum H00 + C is closed. It is also a Banach algebra. Proof: We apply Lemma 6 with X = L00 (1l') , Y = C(Y) , Z = H00 (Y) and � the family of Fejer operators. This shows that H00 + C is closed.
155
III.D. C(K)Spaces §8.
In order to show that H00 + C is a Banach algebra it is enough to show that h · I E H00 + C for h E H00 and I E C(11') . Since H00 + C is closed it is enough to consider only trigonometric polynomials I, but then co N h . I = L anei n6 . L bnein6
n= O
N
( nt0 anein() ) . ( f.N bn ein() ) + ( f an e in() ) ( f. bn ein () ) . N+l N =
The first summand is in C(1r) and the second is in H00 , so H00 + C is an algebra. a 8 Theorem.
L oo/Hoo .
The quotient space (Hoo + C)/Hoo is an Mideal in
:
Let us identify L00 (11') with C(M) where M = M (L00 (1r)) is the space of all nonzero linear and multiplicative functionals on L00 (11') (I.B.lO) . Then (L oo /Hoo )* = H� = {JL E M (M) f ldJL = 0 for all I E H00 } . The annihilator of (Hoo + C)/Hoo in (L00 /H00)* can be identified with
Proof:
(Hoo + C)l. = {JL E M (M)
:J
ldJL = 0 for all I E H00 + C} .
Let m denote the Lebesgue measure considered as a measure on M . From the generalized F.M. Riesz theorem (see Garnett [1981] V.4.4 or Gamelin [1969] 1I.7.9) we get (3) where Msing denotes the space of all measures on M singular with re spect to m. Suppose that JL E (Msing n H� ) . Let a = J e  i 6 dJL. Clearly e ie JL  am E H� and from (3) we get e  ie JL  am = lm + v. But both v and e  i e JL are singular with respect to m so a = I and since I E H� we get a = 0. This means that e i e JL E H� . Repeating this we get that J ldJL = 0 for I E u:=l e  i n() . Hoo which is dense in Hoo + c, so JL E (Hco + C)l. . Since no I E L1 (m) n H! can annihilate H00 + C we get that (Hoo + C).L = Msi ng n H! c H! = (Leo / H00) * . From(3) a we see that (Hoo + C)/Hoo is an Mideal in L00 /H00 •
156
III.D. C(K)Spaces §9.
For every f E Leo there exists h E Heo + C such that II J  h ll eo = inf{ II J  Yll : g E Heo + C}.
9 Corollary.
From Theorems 4 and 8 and the definition of the quotient norm we infer that given f E Leo there exists a g E Heo + C such that dist(f  g, Heo ) = dist(f, Heo + C) . Since balls from Heo are w* compact in Leo we obtain that there exists an h1 E Heo such that ll fg ht ll = dist(f g, Heo ) = dist(f, Heo +C) . The function h = g +h t is in Heo + C, so it is the desired best approximant. Proof:
10. Now we want to address the problem of linear extensions. If K is a compact space and S C K a closed subset then for every f E C(S) there exists a g E C(K) such that IIYII = II I II and gi S = f. This is easy to see directly but can also be derived from Theorem 4 since the map g + gi S is a quotient map from C(K) onto C(S) = C(K)/C(K; S) . The question is when there exists a linear map u: C(S) + C(K) such that u(f) I S = f. In some cases such a map obviously exists (see II.B.4) and in some cases it does not exist (see Exercise III.B. 8). Because of applications to the disc algebra (III.E.3) we will study this question in some generality. Let T be a topological space and let S be a closed subset. Suppose we are given linear subspaces E c C(S) and H c C ( T ) . We say that u: E + H is a linear extension operator if u(f) I S = f for all f E E. Clearly, in order to be able to talk about such operators we need H I S :::> E. Actually we will always assume that H I S = E. We will denote the set of such extension operators A(E, H) . If a linear extension u E A(E, H) exists, then the operator P(h) = h  u(h i S) defines a projection in H with ImP = {h E H: hi S = 0} and kerF = u(E) � E. 11 Definition. The pair ( E, H) as above has the bounded extension property if there exists a constant C such that for every c > 0 and every open set W :::> S and for every f E E there exists g E H such that IIYII ::::; C II J II , g i S = f and l g(t) l ::::; c ll f ll for t E T\W.
Let us start with two easy observations. 12 Lenuna. If G c C(S) is a finite dimensional subspace and u E A (G, C(T)) then for every c > 0 there exists an open set W :::> S such that lu(g) (t) l ::::; ( 1 + c) II Y II for g E G and t E W.
157
III.D. C(K)Spaces §13.
Proof: Since the closed unit ball of G is compact and u is continuous one checks that cp(t) = sup{ l u(g) (t) l : g E G, IIYII :5 1 } is a continuous function. Since cp(t) :5 1 for t E S the lemma follows. a 13 Lemma. If ( E, H) ha.s the bounded extension property and G C E is a finite dimensional subspace, then there exists a constant C such that for every open set W :::) S and every c > 0 there exists u E A(G, H) with ll u ll < C and l u(g) (t) l :5 c iiYII for t E T\W. Proof: We choose an algebraic ba.sis in G and extend each function separately using Definition 1 1 . This yields the desired operator u. a 14 Proposition. Let F C G be finite dimensional subspaces of E. Assume that 11': G�F is a projection with 11 11' 11 :5 1 . Given c > 0 and u E A(F, H) there exists u E A(G, H) such that u i F u and
=
ll u ll :5 max(1, ll u ll ) + c.
Remark. The mysterious expression max(1 , ll u ll ) is justified by the fact that we allow F = {0} and u = 0.
VI
Let us put F1 = ker 11'. We start with an arbitrary open set wl :::) s and from Lemma 13 we get E A(Fl , H) with ll vl ll :5 c and l v1 (f) (t) l :5 � c ll f ll for t E T\W1 . We define u1 E A(G, H) a.s u1 (g) = u(7r(g)) + v1 (9  11' (g) ) . Now Lemma 12 gives an open set W2 , S c W2 c W1 such that l ul (g) (t) l :5 (1 + i ) IIYII for g E G and t E w2 . Using Lemma 13 we get v2 E A(F1 , H) such that ll v2 ll :5 C and l v2 (f) (t) l :5 ! c ll f ll for t E T\W2 . We define u2 E A(G, H) a.s u2 (g) = u(7rg) + v2 (9  7r(g)) . Repeating this procedure N times we obtain a decreasing sequence of open sets W1 :::) W2 :::) :::) WN + 1 :::) 8 and a sequence of extensions ui = u o 11' + vi o (id  11') E A(G, H) , j = 1, 2, . . . , N, such that ui i F = u, j = 1, 2, . . . , N and Proof:
·
·
·
for
t E Wi+ l •
for t E T\Wi , otherwise.
}
(4)
The desired extension u is defined a.s u = N 1 L: f= 1 Uj Obviously u i F = u. From (4) we see that for any given t E T, we have ·
III.D.
158
C(K)Spaces § 1 5.
l ui (g)(t) l > max(1, ll u ll ) + � for at most one index j, so we obtain that a ll u ll � max(1, ll u ll ) + c, provided N was big enough. The same argument gives for every e E E and every c > 0 and every open set W :J S an extension h E H such that ll h ll � 2 ll e ll and l h(t) l � c for t E T\W. Remark.
15 Corollary. If ( E, H) has the bounded extension property and E is a separable 1r1 space then there exists a linear extension operator
u : E + H.
In E we have an increasing sequence of finite dimensional subspaces En and a sequence of projections ?rn : E � En with ll 1rn ll = 1. Using Proposition 14 with en such that Ec n < oo we get a sequence of extensions Un E A(En , H) with un i Ek = Uk for k < n and supn ll un ll < oo. Since UEn = E we infer that u(f) = limn+ oo Un ( f ) extends to a a well defined linear extension operator on E. Proof:
16. Being a 1r1space is a rather restrictive condition. It is difficult however to modify the above proof using the weaker approximation con dition. Nevertheless the following theorem is true, albeit with a rather roundabout proof. Theorem. Let T be a topological space and let S C T be a closed subset. Let E C C (S ) and H C C(T) be closed linear subspaces and let (E, H) have the bounded extension property. Assume that E is separable and has the bounded approximation property. Then there exists a linear extension operator u: E + H.
We will deduce Theorem 16 from Corollary 15 applied to a properly enlarged space. Let w = { 1, 2, 3, . . . } U { oo } be the onepoint compactification of the natural numbers, and let S = S x w C T = T x w. Let Tn : E + E, n = 1 , 2, . . . , be a sequence of finite dimensional operators with Tn (e) + e for e E E. Denote En = span U�= l Tk (E) . We define E C C ( S) by Proof:
E
We also define
!oo E E, fn E En , n = 1, 2, . . . , and fn + foo as n + oo } .
= { ( f')'hEw : H=
{( f"Y ) "YEw : f"Y E H for 'Y E w
and fn + foo as n + 00 }.
(5)
159
III.D. C(K)Spaces § 1 7.
One easily sees that E is a closed subspace of C(S) and ii is a closed subspace of C( T ) . Claim.
The pair (E, H) has the bounded extension property.
Proof of the claim. If W C T is an open set containing S then there exists an open set W00 C T, such that W00 :J S and W00 x w :J S. Given ( /y)yEw E E and c > 0 we can find 9oo E H such that 9oo i S = foo and l 9oo (t) l < � for t E T\W00 • Using the remark after Proposition 14 we can find hn E H with ll hn ll :5 2 ll foo  fn ll , hn i S = foo  fn and l hn(t) l < � for t E T\Woo. Since ll foo  fn ll + 0 as n + oo we also have ll hn ll + 0 as n + oo . The desired extension (gy)yEw is defined by 9n = 9oo  hn·
Returning to the proof of the theorem let us observe that E is a 1r1space. To see this we define En = {( /y)yEw E E: fk = fn for  onto k 2: n } . The projections Pn: E+En are defined by Pn (( /y)yEw) = ( JI , h , . . . , fn, fn, . . . , fn)· Obviously II Pn ll = 1 and UEn is dense in E. From Corollary 15 we get a linear extension operator u: E + ii. We define
u:
E + H by
u ( f ) = u(Tl (f), T2 (f), . . . ' !) I T
X
{ 00 }.
a
17 Corollary. If S c T and S is compact metric and T is normal then there exists a linear extension operator C(S) + C(T) .
u:
Proof: Since S is compact metrizable the space C(S) is separable. The
Tietze extension theorem implies that (C(S) , C(T)) has the bounded extension property. Since C(S) has the bounded approximation property a (see II.E.5(c)) the corollary follows. 18. The above corollary exhibits many complemented subspaces of C(K)spaces. One more, different and very important example is pro vided by the following. Proposition. {Milutin) Let yN denote the countable product of circles. The space C(A) contians a 1complemented copy of C(1l'N ) .
Let us identify A with {  1 , 1}1N . By p we mean the classical Cantor map from A onto 1l', i.e. p((c3)� 1 ) = 21r E� 1 ! ( 1 + c3 ) 2  j . It is an easy and well known fact that there exists a map 7': '][' + A such
Proof:
160
III.D. C(K)Spaces § 1 9.
that jYf = and f is measurable and continuous on Y\D where D is a countable set of dyadic points. Since � is homeomorphic to � N we infer (take products) that there exists a continuous map p: ��yiN and a measurable map yN + � such that p = and T is continuous on YIN\Doc where Doc has measure 0. Note that both � and yiN have a ·natural group structure, so we can perform algebraic operations. We define the isometric embedding C ( � x �) � C � as a {3 = f(p( a) + p(f3 )). We define a norm1 map C ( � x �) + C(YN ) by
idy
,
r idyN
r:
I(f)( , )
I : C (YN ) +
/1fN
Q:
()
Q (g)(t) = g(r(s) , r(t  s))ds . To see that Q (g) E C (YIN ) let us take a sequence tn E YIN, tn + t . Then g(r(s) , r(tn  s)) + g(r(s) , r(t  s)) as n + oo for all s E YN \(U := l (t n  Doc ) (t  Doc ) Doc ) , i.e. for almost all s E By the Lebesgue dominated convergence theorem Q (g)(tn) = JTN g(r(s) , r(tns))ds + JTN g(r(s), r(ts))ds = Q (g)(t) as n + oo, so Q (g) is continuous. Since QI(f)(t) = lT{ N f(p( r(s)) + p(r(t  s)))ds = }yN{ f(t)ds = f(t) yiN .
U
U
a
the proposition follows.
This proposition should be compared with Exercise 4, which indi cates that some ingenious embedding is needed. 19.
result.
The above proposition allows us to prove the following surprising
Theorem. {Milutin) For every compact, metric, uncountable space K, the space C(K) is isomorphic to C � .
()
Proof: As is well known, every uncountable, compact metric space K contains a subset K1 homeomorphic to � (see Kuratowski [1968] or Semadeni [1971] ) . So by Corollary 17 the space C(K) contains a com plemented subspace isomorphic to C(�) . It is also elementary and well
known (cf. Kuratowski [1968] ch.4§4l.VI.) that every compact metric
161
Ill.D. C(K)Spaces §20.
space is homeomorphic to a subset of yiN so in particular we obtain from Corollary 17 that C(K) is isomorphic to a complemented subspace of C(YN ) . This and Proposition 18 yield that C(K) is isomorphic to a complemented subspace of C(�) . Since C(�) ,...., (EC(�))o Theorem a II.B.24 gives the claim. This theorem in particular implies that every C(K)space for K compact, metric, and uncountable has a Schauder basis. Also, such a C(K)space is isomorphic to (EC(K)) 0 • 20. Our aim now is to present in some detail the orthonormal Franklin system. Usually it is constructed on [0, 1] . We will present the detailed construction on the circle (i.e. we will construct the periodic Franklin system). This will be useful for some of the future applications, in particular Theorem III.E. 17. The reader interested in [� ,) ]Should be able to repeat the construction in this case witho¢ any difficulty. We will identify the circle Y with the interval [0,1). For an integer n = 2 k + j, with k = 0, 1, 2, . . . and 0 � j < 2 k we_"'define tn = <;it:> and we put to = 0. The Franklin system Un) ':'= o is an orthonormal system of real valued, continuous, piecewise linear functions such that In has nodes at points t3 , j = 0, 1, . . . , n, for n = 0, 1, 2, . . . . This definition specifies In up to the sign. Let Fn = span{IJ }J::;n · Clearly Fn is the space of all continuous, piecewise linear functions with nodes at { t3 }J::;; n · For a fixed n we will denote by (s3) 'J=o the increasing renumbering of (tj ) 'J=o • i.e. 0 = so < s1 < · · · < Sn = 1. Let Zn + l denote the group of integers 0, 1 , . . . , n with addition mod (n + 1). The natural group invariant distance p( · , · ) on Zn + l is defined as p(k, l) = min( l k  l l , in + 1 + l  ki, I n + 1 + k  l i ) , for k, l = 0, 1, . . . , n. We define also (for fixed n) the 'tent' functions Aj , j E Zn + l by the conditions A3 E Fn , A3 (s3 ) = 1 and A3 (s k ) = 0 for k =f. j. Let us also note that 1 (6) � dist(s3 , s3+ 1 ) � � for j E Zn + l · n 2n 21. Our main goal now is to establish the following technical proposi tion. It describes the behaviour of an individual Franklin function and of the integral kernel of the partial sum projection. This proposition will allow us to investigate the properties of the FranklinFourier series E ':= 0 (/, In) In for I in various classes of functions. Proposition. There exist constants C e very n = 0, 1, 2 . . . , we have
<
oo and
q <
1 such that for
1 62
III.D. C (K) Spaces §22.
(a) l/n(t) l � C ../n + 1 qn ·dist ( t , tn ) ' (b) I L� = O /j (x) /j (y) i � C(n + 1)qn · dist ( x , y) . Before we start the proof we need some preliminary remarks. Let us write
n
Kn (x, y) = L Jj (x) /j (y) = L O! ijAi (x)Aj (y). j =O
(7)
Since Pn f(x) = fr f(y)Kn(x, y)dy is an orthogonal projection onto the space Fn we see that Pn (Aj ) = Aj , for j E Zn+ l · From this we infer that ( a ij ) i ,jE Zn +l is the inverse matrix to the matrix (fy Ai (x)A3 (x)dx) t,. J· ezn +l . The key to the proof of Proposition 21 is the analysis of the matrix ( a ij) i , jE Zn +l " This is what we will do now. 22 Lemma. the matrix
There exists a constant C (independent of n) such that
(8) defines .an operator
II A� 1 11 � c.
Proof:
An : f2(Zn+1)
+
f2(Zn+1)
with
II An ll < C
and
One checks that for any affine function f ( t) on the interval
[a , b] we have
� (b  a)[ lf (a) l 2 + 1/ (bW J � 1b l f(x) l 2 dx � � (b  a)[ lf (a) l 2 + l/ (b) l 2 ] .
Define an operator S: f2(Zn+1) + L2(Y) by S(e3 ) j E Zn+l · From (6) and (9) we obtain
� II S(x) ll � ll x ll � 2 v'3II S(x) ll
Since 23.
=
(9) Jn + 1Aj , for
for x E f2(Zn+1)·
a
An = S* S (10) implies the claim.
The important feature of the matrix
(10)
A
given by (8) is that
Jy Ai (x)A3 (x)dx = 0 if p ( i, j) > 1. We say that the matrix (aij ) i ,jEZn + I is mbanded if a ij = 0 for p ( i , j) > m. One easily sees that the inverse
of an mbanded matrix (if it exists at all) need not be banded. The
163
III.D. C(K)Spaces §24.
following proposition shows that something remains. The entries of the inverse of an mbanded matrix are exponentially small far away from the diagonal. Proposition. Let A = (aij ) i , j EZn + l be an mbanded invertible matrix with II A II � 1 and II A 1 11 � C where the norm is understood as the operator norm on There exist numbers K = K(C, m) and q = q(C, m) , 0 < q < 1 such that
i2(Zn+l) ·
l bij l
�
KqP (i ,j )
where A  1 =
(bij ) i ,jE Zn + l '
(11)
The proof of this proposition uses the following easy approximation fact.
f(x)
n
< a < b and let = � · Then a is an algebraic polynomial of degree l < n } Kqn + 1 for some K = K(a, b) and q = q(a, b) < 1 . 24 Lemma.
inf{ ll /

Let 0
P ll c [a , b] :
p
Let = ( a� b) . Then � converges in C(a, b] . We have
� 00
<
Lk=O ( (e x) ) k and this series 1 00 l ( c  x ) k l 1 00 ( b  a ) k � = n L a � k=n+1 c C [a, b] � k=Ln+1 bn+1a b  a)  . < K ( b+a c
Proof:
c
II
First note that if A 1 is m1banded and A2 is m2 banded then A 1 A 2 is (m 1 + m2 )banded. In particular AA* is a positive definite 2mbanded invertible matrix with (AA* )  1 = ( uij) i ,jEZn + l ' Obviously a(AA* ) C (C  2 , 1] so from the spectral map ping theorem and Lemma 24 there exists a sequence of algebraic poly nomials with deg p � k such that Proof of Proposition 23.
(Pk ) �0
k
For a given pair (i, j) E Zn+l x Zn+l we take the largest integer k such that 2k m < p(i, j) . Since P ( AA* ) is 2kmbanded it has a zero entry at the place (i, j) so ·
k
Since the product of an m1banded matrix and a matrix satisfying (11) also satisfies ( 11) (for different K and q < 1) the formula A  1 = A* (AA* )  1 gives the claim. a
164
III.D. C(K)Spaces §25.
Proof of Proposition 21. Part (b) follows immediately from Lemma 22, Proposition 23 and (7) . Given n let A(t) E Fn + l be such that
A(tn + l ) = 1 and A(tj ) = O for j = 0, 1, . . . , n. Clearly fn + l = II A  Pn A II 2 1 (A  Pn (A)). From (7) and Proposition 23 we get I Pn A(x) l � Cq ( n + l ) dist (x , t n + d so also I A(x)  Pn A(x) l < Cq ( n + l ) dist ( x , t n + d for some C < oo and q < 1. Since II A  Pn A II 2 ;:::: inf ab
(
1
I A(x)  ax  b l 2 dx , uppA s 1 = C v l suppA I ;:::: c v'n"+T n+1
the proposition follows. 25 Theorem. The Franklin system space C(Y) .
)
1
2
•
Un)�= O is a Schauder basis in the
Clearly span{ /n }�= O = C(Y) . In order to estimate the norm of Pn (see Proposition II.B.8) we have from Proposition 21 {b)
Proof:
I Pnf(x) l
�
�
I fv t, fv I t,
IJ (x) /j (y)f(y)dy
11/lloo
IJ (x) /j (y) dy
C ll/ll oo · ( n + 1) � C ll/ll oo · �
26 Corollary.
1�p<
00 .
l
l
100
q x
n dx
•
The Franklin system is a Schauder basis in Lp (Y) for,
The case p = 1 follows by duality from Theorem 25, because the Franklin system is orthonormal. It also follows from Theorem 25 that the partial sum projections Pn are uniformly bounded on L00 ('U') . Thus the RieszThorin theorem I.B.6 gives that they are uniformly bounded on Lp (1I') for 1 < p < oo. This gives the claim; see Proposition II.B.8. Proof:
a
165
III.D. C(K)Spaces §27.
27. Our next application of the Franklin system is to the spaces Lipa (Y) , 0 < a < 1. We have Theorem. The space Lipa (Y) , 0 < a < 1 , is isomorphic to £00 • The isomorphism is given explicitly a.s f �+ ( (n + 1)"' + ! JT f(x)fn(x)dx )' :;=o· 28 Lemma. There exists a constant C such that for n = 0 , 1, 2, . . . , the circle Y can be partitioned into intervals (Ij) with l lj l < (n� 1 ) and such that Jl':o fn(t)dt = 0 for all j . 3
Since fn is orthogonal to Fn  1 (20) the function cp(x) = J; fn(t)dt is orthogonal to all step functions with discontinuities at t0, t1 , . . . , tn_1 . This implies that cp(x) has a zero in each interval (sj , Bj + I ) · The zeros of cp(x) give the endpoints of the desired inter a vals (Ij). Proof:
Proof of Theorem 27. For a given integer n, let (Ij)j = 1 be the intervals given by Lemma 28. Fix points Uj E Ij , j = 1 , . . . , s. For f E Lipa (Y) we have from Lemma 28 and Proposition 21 ( a)
I h J(t)fn(t)dt l = I t ji J (t)fn (t)dt l j l f(t)  f(uj ) l l fn (t) l dt :::; t j= 1 :::; t �BJ: l f(t)  f(uj ) l ji l fn (t) l dt :::; 11/II Lipa z1 "' h l fn (t) l dt I;
(n
)
:::; CII JIIL ip0 ( n + 1 )  a  ! ·
In order to show the other estimate let us fix a sequence of scalars (an )�= O such that l an l ( n + 1 )  a  ! . Let us write 00 00 2 k + l _ 1 00
:::;
f = L anfn = aofo + L L anfn = aofo + L Fk . n=O k =O 2 k k =O
From Proposition 21 ( a) we infer
2k + 1 _ 1
I Fk (t) l :::; L lan l lfn (t) l :::; C2  k "' . 2k
( 12 )
166
III.D. C(K)Spaces §29.
Each Fk (t) is a continuous, piecewise linear function with nodes at least 2  k  1 apart, so
For t1 . t 2 E Y let N be such that 2  N  1 < dist(tt , t 2 ) :5 2  N . Using ( 1 2) and (13) we have oo
N + t ) :5 (t dist(tb t ) ) ao 1 f( 1 Fk (t1 )  Fk (t 2 ) l + 2 L II Fk ll oo L / l l 2 2 l l N+ 1 k=O N :5 c dist(tb t2 ) 0 + c L: T k o2 k + l dist(h , t2 ) + c L T k o k =O k=N+ 1 N a :5 c dist(tb t2 ) 0 + CT o :5 c dist(tb t 2 ) 0 • oo
Now we will introduce a special class of subspaces of C(S, E), 29. the space of continuous Evalued functions on S where E is a finite dimensional Banach space. Before we proceed a few comments about notation are in order. The norm in E will be denoted by 1 · 1 · Actually, since all finite dimensional spaces of a given dimension are isomorphic different choices of norm in E give isomorphic spaces C(S, E) . Any linear functional on C(S, E) can be identified with an E*valued mea sure. Really, we can identify E with IR.n or ern using the Auerbach lemma (see II.E. l l ) , so then each functional on C(S, E) can be iden tified with a sequence ( J.L b . . . , J.Ln ) of measures. For every E* valued measure J.L = ( J.L b . . . , J.Ln ) there exists a positive measure IJ.LI such that for f E C(S, E) we have (14) and IIJ.LII c{S,E)• = I IJ.LI II · This requires some work, but the measure E7= 1 IJ.Li I clearly satisfies (14) and also I E7= 1 IJ.Li I II :5 C IIJ.LIIc {S,E)• . This is all we will need. Definition. Let E be a finite dimensional Banach space and let S be a compact space. A subspace X c C(S, E) is called rich if there exists a probability measure a on S such that for every cp E C(S) and every bounded sequence (xn)�= 1 C X such that fs lxn lda + 0 as n + oo we have dist(cp X n, X) + 0 as n + oo . ·
III.D. C(K)Spaces §30.
167
As we will see later, rich subspaces of C(S, E) and their duals share with C(K)and £ 1 (J.L)spaces many properties related to weak compact ness. Also some important spaces can be realized as rich subspaces of C(S, E) . (a) C(S, E) itself is obviously rich. (b) Let us consider the space C k (1l'8 ) , k ;::: 0, s ;::: 1 (see I.B.30) . If D is the set of all multiindices i = (i i . . . . , i s ) such that ij is a nonnegative integer and L: j= 1 l ii l � k then we can embed C k (1l'8 ) into C(1l's , .e)£> 1 ) . The embedding cP simply assigns to f the collection ( 8, 1 ., 1��:� �• x. ) i e D = • «P (f). The subspace «P(C k (1l'8 ) ) c C(']['s , .e)£> 1 ) is rich. To see this let us take a trigonometric polynomial cp E C(1l'8 ) and a function f E C k (1l'8 ) . It is easy to check that «P ( cp · f ) = cp · «P (f) + G where G is the sum of derivatives of f of order < k multiplied by some derivatives of cp. This shows that 30. Examples
ll cp«P (f)  «P (cpf) ll· � Ccp · 11/llc k 1 (11'• ) · Take now a bounded sequence ( /n )::'= l C Ck (1l'8 ) such that f..rs I«P (fn ) l dm + 0 as n + oo where m is the normalized Lebesgue measure on 1l'8 • Since i d: C k (1l'8 ) + ck  1 (1l'8 ) is a compact operator (see Exercise 23) , we get ll/n llc k 1 (T• ) + 0 as n + oo. This shows that for every trigonometric polynomial cp we have
dist(cp · «P (fn ), «P (C k (1l'8 ))) � ll cp«P ( /n )  «P(cp · fn) ll � Ccp ll/n llc k  1 (11'• ) + 0. Since the trigonometric polynomials are dense in C(1l'8 ) the above re lation holds for all cp E C(1l'8 ) . This shows that «P(C k (1l'8 )) is a rich subspace. (c) Let us consider A (md) c C(Sd) where md is the ball in ([d and Sd is the unit sphere. Let C denote the Cauchy projection (I.B.21 ) . Given cp , a polynomial i n z 1 . . . . , Zd , z1 , . . . , Zd, and a bounded sequence Un )::'= 1 C A (md) such that fsd l fn l da + 0 (a is a normalized rotation invariant measure on Sd) we get from III.B.20
ll cp · fn  C(cp · /n ) lloo + 0 as n +
00 .
This gives that A(md ) is a rich subspace of C(Sd) · 31. The following theorem characterizes weakly compact subsets of X* , where X is a rich subspace of C(S, E). It is a generalization of Theorem III.C.12.
168
III.D. C(K)Spaces §31 .
Theorem. Suppose that X C C(S, E) is a rich subspace and SUJ:r pose that K C X* is a bounded subset. The following conditions are equivalent.
(a) K is not relatively weakly compact. (b) There exists a sequence (kn)�= t
c
K equivalent to the unit vector
basis of it .
(c) There exists a constant C such that for every N there exists a subset in K, Cequivalent to the unit vector basis in
if .
(d) There exists a weakly unconditionally convergent series L �= t (/)n in X such that lim sup sup{ l k(cpn) l : k E K} > 0. n  oo (e) There exists a sequence (xn)�= t
C
X, xn�O such that
lim sup sup{ l k(xn) l : k E K} > 0. n  oo
We will prove the following implications (the implications marked ( * ) are obvious) (d)
(!j
(b)
*
(c)  (d)
� (a) �
(d) ::::} ( b) . This is still easy. The w.u.c series (cpn)�= t C X defines an operator T: X* + it as T(x* ) = (x* (cpn))�= t · It follows from (d) that T(K) is not normcompact in it so (b) follows from Theorem III.C.9 (see Exercise III.C.13) . (e)::::} (d) . Using (e) let us fix a 6 > 0 , a sequence (x�)�= t C K and a sequence (xn)�= t C X such that Xn �O and x� (xn) > 6 and iixn I � 1 for n = 1, 2, . . . . Let us also fix a sequence of positive num bers (cn )�= t such that L�= t en < too . Let J.Ln be the HahnBanach extension of x� , n = 1, 2, . . . and let 'Y = sup{ ll a + I J.Ln l ll : n = 1 , 2, . . . } . Note also that since Xn�O we get (see Excercise II.A.3) that l xn i �O in C(S) . Thus there exists a finite convex combination such that I z= := t Aj l xi l lloo < � (see II.A.5). From this we infer that there exist n t , 1 � n t � N and an infinite set N t C N with min N t > N such that
J lxn, l d( IJ.Ln l + a) < ct
for n E Nt .
(15)
III.D.
169
C(K)Spaces §31 .
Repeating the same argument we find inductively a sequence ( nk )k= 1 such that
J
(16) l xn k I da is so small that dist
k 1
( j=1 II (1  l xn1 l ) xn k , X ) < €k . (17)
Let 'P k E X be such that II n;:; (1  l xnj l )x n k  'P k II < C k o The series 2::: � 1 'Pk is w.u.c. because from (17) we get 00 00 00 k  1 L I 'P k l � L €k + L II (1  l xn1 l ) l xn k I k = 1 j= 1 k= 1 k= 1
From ( 17) we also have
l x� k (r.p k ) l
= I J 'Pkd/Lnk l � I J ll, ( 1  I Xn1 l )xnk dJLnk l  €k � I f Xnk dJLnk 1  1 J ( 1 ll. (1  l xn1 I ) ) xnk dJLnk 1  €k ·
Since for 0 � a1 � 1 we have (18) and (16) we infer that
1  IJ;= 1 (1  a1 ) � 2:::;= 1 a1 ,
(18)
from
(c)=>(d) and (a)=>( d) . The proof of both implications is practically the same. Both (a) and (c) imply the following. ( * ) There exist elements (x]', . . . , x�) C X, and (k]', . . . , k� ) C K, n 1, 2, . . . and numbers C and {; > 0 such that
=
ll kj ll � c l l xj ll � 1 l kj (x J: ) I � kj (x J: ) ?:: {;
�
= =
for n 1 , 2, . . . and j for n 1, 2, . . . , and j
= 1, = 1, 2,
if k > j for n = 1, 2, . . . , if k S j for n = 1 , 2, . . . .
. . . , n, . . . , n,
(19) (20)
(21) (22)
170
III.D. C(K)Spaces §32.
When we assume (a) then (*) follows from the proof of the Eberlein Smulian theorem II.C.3. When we assume (c) , then we take (kj) j= 1 to be vectors in K equivalent to the unit vector basis in /!f and take xj_* such that xt,* (kj) 0 if k > j and x t,* (kj) 1 if k :S j. Clearly such x r with norms uniformly bounded can be found, so the local reflexivity principle II.E.14 (and some renormalization) gives (*) . To show ( *) =} ( d) we will need the following fact.
=
=
H IJ :S 1 j = 1 , . . . , n. I I A, E { 1 , . . . , n} max A < min E
be a Hilbert space and let /I , . . , fn E H be Then there exist nonempty sets for with such that
Let
32 Lemma. vectors with C
.
(23) Proof:
Let us put
ak
=
inf
{ l l �l L /j l : I A I = 2k ,A
C
{ 1 , 2, . . . , n}
JEA
}
and
{ 1 1 �1 � 1�1 � /j l : A, E { 1, . . . , n} , I A I = l E I = 2k and max A < minE} for k = 0, 1, . . . , [log 2 n] 1 . (y �z ) we Elementary geometry shows that for x, z E H with x z 2 2 ( have max( l l z l l , I Y I 2 ) l x l l + I y ; ) 1 2 • This observation implies that a2 so a 2k  a 2k+l + 4J'k fJ 
f3k = inf
C

>
1
2:
y,
=
a
Since a0 :S 1 we get the claim.
=
=
(*) *(d). Let us start with the observation (to be repeatedly used later) that if we divide the set of pairs ( n, j), j 1 , 2, , n, n .
.
.
171
III.D. C(K)Spaces §33.
1, 2, . . . , into a finite number of sets then at least one of those sets will contain a subset of the form
{(n k , i!): nk is an increasing sequence of integers and s = 1, . . . , k and jf < j� < · · · < j� }.
(24)
Let us fix a sequence of positive numbers (c k )k'= 1 such that E%"= 1 c k < 160 • Let (as previously) J.Lj be the HahnBanach extension of kj . Using Lemma 32 we find n 1 such that for every (n, j) with n > n 1 there are subsets A, B C { 1 , . . . , n } such that
Since there is a only finite number of subsets in { 1 , . . . , n 1 } we find sets A and B such that (25) holds for this pair of subsets A, B on the set of
the form (24) . Thus (after renumbering) we can assume that there are vectors (xi, . . . , x�) C X and (ki, . . . , k�) C K for n > n 1 , satisfying (19)(22) . We put xi = k:1 where s = max A. In the second step we analogously find n 2 > n 1 . and z2 = I A I  1 Lj e A xj2 I B I  1 L j eB xj2 such that J i z2 id ( iJ.Lj i + a) < c 2 for all (after passing to appropriate subset as before) n, j with n > n 2 , j = 1, . . . , n, and such that J i z2 ida is so small that dist((1 l z 1 i )z2 , X) < c 2 . We put x2 = k: 2 where s = max A. Continuing in this manner we find sequences ( zj ) � 1 C X and ( xj ) � 1 C K such that


�6 
(this follows from (21) and (22)) , i xj ( zj ) i > 1 kdist ( IJ (1 i zi i )zk , X < c k , j=1 i zi i idJ.Lk i < C"j for k > j where J.Lk is the HahnBanach extension of xA; .
J
)
(26) (27) (28)
Using (26) , (27) and (28) we construct the desired weakly uncondition a ally convergent series exactly like in the proof of (e)::::} ( d) . 33. Now we wish to cast the above considerations into a more general context. Let us introduce the following concepts. We say that a Banach space X has the DunfordPettis property (for short DP) if for every
III.D. C ( K ) Spaces §34.
172
Xn�O in X and x� �O in X* we have x� (xn) + 0. Clearly if X* has DP then also X has DP. We say that a Banach space X has the Pelczynski property (for short P) if for every subset K c X* that is not relatively weaklycompact there exists a weakly unconditionally convergent series E �= l Xn in X such that infn supx* E K x* (xn) > 0. Clearly Theorem 31 shows that every rich subspace of C(S, E) has DP and P. Also any L 1 (J.L) space has DP. 34.
We have the following, rather routine
Proposition. equivalent:
The following conditions on the Banach space X are
(a) X has the DunfordPettis property; (b) every weakly compact operator T: X
__.
Y transforms weakly
Cauchy sequences into norm Cauchy sequences; __. c0 transforms weakly Cauchy sequences into norm Cauchy sequences.
(c) every weakly compact operator T: X
(a)=> (b) Passing to differences it is enough to show that II Txn ll __. 0 for every Xn�O. But if Xn�O and II Txn l l ;::: 8 > 0 for n = 1, 2, . . . , then we can take y� E Y* with IIY� II = 1 such that w* y�T(xn) ;::: 8. Passing to a subsequence we can assume that y� +y* for some y* E Y* . But y* (Txn) __. 0 so we can replace y� by y�  y* and additionally assume that y� � O. But T* is weakly compact (see II.C.6(b)) so T* (y� ) �O. This contradicts (a) since T* (y�) (xn) = y� (Txn) ;::: 8. (b)=>(c) . Obvious. (c)=>(a). Let us take x� E X* such that x� �O and define an operator T: X __. co by T(x) = ( x� (x) ) :'= l · Clearly T** (x**) = (x** (x�))�= l E eo so T is weakly compact by II.C.6(c) . Applying (c) we get that for Xn �O in X, II Txn ll __. 0 so in particular x� (xn) __. 0. a Proof:
35 Proposition.
Suppose X has the Pelczyriski property. Then
(a) X* is weakly sequentially complete, (b) for every operator T: X __. Y that is not weakly compact there exists a subspace xl c X such that xl Co and T I Xl is an "'
isomorphic embedding.
If K is a subset of X* that is not relatively weakly com pact then there exists a sequence {x� }�=l C K which is not relatively
Proof:
III.D. C(K)Spaces §Notes.
173
weakly compact (see II.C.3) . Thus there exists a weakly unconditionally convergent series :L:;:: 1 X k in X and a subsequence ( x � k )� 1 such that l x� k (x k ) l > 6 > 0 for k = 1 , 2, . . . . From II.D.5 we see that we can additionally assume that ( x k )� 1 is equivalent to the unit vector basis of Co · In order to prove (a) we take K = {x � } �= l where x � is weakly Cauchy but not weakly convergent. Let T: c0 + X be defined by T(e k ) = X k . Then T* : X* + i 1 and one easily sees that T* ( x� k ) has no norm Cauchy subsequence, so by III.C.9 T* (x� k ) has no weak Cauchy subsequence. This contradicts the fact that (x�)�= l was weakly Cauchy. In order to prove (b) let us put K = T* (By. ) . Then X1 = span (x n )�= l is clearly isomorphic to Co and for x = L�= l anXn we have
so T I X1 is an isomorphic embedding.
•
Notes and remarks.
The C(K)spaces are among the most widely used Banach spaces. They are also the easiest examples of Banach algebras. As usual in this book we discuss the multiplicative structure only so far as it relates to the linear structure. Thus the well known Proposition 1 serves as an in troduction to the concept of an Mideal. This concept was introduced by AlfsenEffros [1972] where Theorem 4 is also proved. Our proof is a modification of proofs given in Behrends [1979] and Lima [1982] . A similar proof and many applications of the theorem can be found in GamelinMarshalYounisZame [1985] . We will present some other ap plications of the concept of an Mideal in III.E. The space Hoo + C was introduced into analysis in the sixties by D. Sarason and A. Devinatz. Corollary 7 is due to Sarason but the simple proof presented here is from Rudin [1975] ; a similar proof is given in Garnett [1980] . The analysis of the particular example H00 + C led to the general theory of Douglas algebras, i.e. closed algebras X such that H00 C X C L00 • H00 + C is the smallest such algebra. It is a deep theorem of Marshall and Chang that each such X is the smallest algebra generated by H00 and complex conjugates of certain inner functions. For detailed information about all this we refer to Garnett [1980] or Sarason [1979] . Corollary 9 was proved by complicated operatortheoretical argu ments by AxlerBergJewellShields [1979] . The simple proof given here
174
III.D. C(K)Spaces §Notes.
is due to Luecking [1980] . This started the investigation of Mideals in Douglas algebras and other spaces connected with function theory. We refer the interested reader to GamelinMarshalYounisZame [1985] and to the references quoted there. There are also important applications of the concept of Mideal to the theory of C* algebras; see ChoiEffros [1977] or AlfsenEffros [1972] . It seems that the first linear extension theorem was proved by Bor suk [1933] where he established a version of our Corollary 1 7 and used it to show that C [O, 1] "' ( I: C [O, 1] )0. Borsuk's argument was different and together with later improvements by Dugundji [1951] it gives the following Theorem A. (BorsukDugundji) . Let S be a closed nonempty sub set of a metric space T and let X be a normed vector space. Then there exists a linear extension operator C(S, X) + C(T, X) such that = 1 and for every g in C(S, X) the values of the function (g) belong to the convex hull of the set g(S) . Our Corollary 15 was proved by MichaelPelczynski [1967] . Actually
llull
u:
u
it was proved with the additional information that llull = 1. We decided not to present this improvement because we are mainly interested in isomorphic theory and our goal is Theorem 1 6. This theorem was proved by RyllNardzewski (unpublished) and the proof we follow here was later given in PelczynskiWojtaszczyk [1971] . Davie [1976] used Proposition 18 in his discussion of classification of operators on Hilbert space. It is his version of the proof that we present. It should be stressed, however, that questions of linear extensions are not limited to spaces with the sup norm. There is an extensive literature on the existence and nonexistence of linear extensions when other norms are involved, in particular Sobolev or Besov norms; see Stein [1970] or Triebel [1978] and the references quoted there. Proposition 18 and Theorem 1 9 are due to A.A. Milutin. He proved them in his Candidate of Sciences dissertation presented to the Moscow State University in 1952. Those results were only published in Milutin [1966] . These are important results. Some reasons for this opinion are as follows. (a) Results of an isomorphic nature, once established for one 'sim ple' space K, like K = � or K = 11' are valid for C(K) with more complicated compact, uncountable metric spaces K. One modest exam ple of this is the comment made after Theorem 1 9. (b) This is an important step in the programme of isomorphic clas sification of C(K)spaces. For separable spaces C(K) , i.e. Kcompact,
III.D. C(K)Spaces §Notes.
175
metric such classification is known. For countable, metric compact spaces this was done in BessagaPelczynski [1960] . The situation for nonseparable spaces is more complicated and only partial results are known. (c ) The Milutin theorem shows that for most important compact sets K the multiplicative structure of C(K) has nothing to do with its lineartopological structure. This contrasts sharply with the isometric situation. Namely we have the following. Theorem B. If d(C(S) , C(K)) < 2 then S is homeomorphic to K, so in fact there exists a linear isometry i: C(K) � C(S) preserving the multiplication.
This theorem under the assumption that C(K) and C(S) are ac tually isometric ( with the a description of the isometries ) was proved for metric K and S in Banach [1932] and for general K and S in Stone [1937] . This version was given independently by Amir [1965] and Cam bern [1967] . The Franklin system was introduced in Franklin [1928] , where The orem 25 was proved. We follow an approach developed by Ciesielski and Domsta in order to deal with systems of more general spline functions; see CiesielskiDomsta [1972] . The Franklin system itself was earlier in vestigated in detail in Ciesielski [1963] , [1966] where the fundamental Proposition 21 was proved. Proposition 23 was proved by Demko [1977] . The very ingenious proof presented here was given in DemkoMossSmith
[1984].
Theorem 2 7 was first proved in Ciesielski
[1960] using the Faber
Schauder system and later in Ciesielski [1963] using the Franklin system. The Franklin system is one of the most important orthonormal systems (see KashinSaakian [1984]). The DunfordPettis property as defined in 3 3 was explicitly defined by Grothendieck [1953] who undertook an extensive study of this and related properties. He was directly influenced by the important paper DunfordPettis [1940] where among other things it was proved that ev ery weakly compact operator T: £1 [0, 1]  X maps weakly compact sets into normcompact sets. Our Proposition 34 and much more can be found in Grothendieck [1953] . The PelczyD.ski property (obviously under a different name, property V) appeared first in Pelczynski [1962] , where he showed that C(K) has P. The class of rich subspaces of C(K, E) ap peared in Bourgain [ 1984b] . In Bourgain [ 1983] and [ 1984b] our Theorem 31 and Examples 30 were proved. Theorem 31 for the particular case
176
III.D. C(K)Spaces §Exercises
of the disc algebra was shown earlier by Delbaen [1977] and Kislyakov [1975] (independently and almost simultaneously) . Clearly Theorem 31 when applied to C(K) gives information about subsets of L 1 ( ll) . This information is akin to that given in Theo rem III. C. 12. One can derive Theorem III.C.12 from Theorem 31 but even then many of the measuretheoretical arguments from the proof of III.C.12 have to remain. We have chosen to present a separate proof of Theorem III. C. 12 because it is an important theorem and the direct argument is relatively simple. It stresses the important notion of uni form integrability which cannot appear explicitly in the more general Theorem 31 .
The paper Diestel [1980] contains a nice survey and exposition of the DunfordPettis property and related topics. It does not however, discuss Theorem 31 . Exercises
1.
Show that the space of compact operators on lp , 1 < p < Mideal in L(lp ) ·
2.
Show that, if (E, H) has the bounded extension property then H0 = {! E H: ! I S = 0} is an Mideal in H. (The notation agrees with
oo ,
is an
10.)
3.
Show that every C(K)space, K compact, is a 1r1space.
4.
Let cp: � [0, 1] be the classical Cantor map, i.e. if � = {  1, 1} N then cp(ei ) = E� 1 (ei + 1)  l  j . Let I"': C[O, 1]  C(�) be given by Icp (f) = f o cp . Show that Icp (C[O, 1] ) is uncomplemented in C(�) .
5.
Find two nonhomeomorphic, compact metric spaces K1 and K2 such that d( C(KI ), C(K2 ) ) = 2.
6.
A matrix (aj k )j, k � o is called a Toeplitz matrix if aj, k
onto
=
cp(j  k).
(a) Show that a Toeplitz matrix is a matrix of an operator on l2 if and only if cp(s) = j ( s ) , s = 0, ± 1, ±2, . . . for some f E Loo (Y) . (b) A matrix (bj k )j, k �O is a Schur multiplier if for every matrix (mj k )j, k �o of a linear operator on l2 the matrix (bi k · mj k )j, k �O is a matrix of a linear operator on l2 . Show that the Toeplitz matrix is a Schur multiplier if and only if cp(s) = [l,( s ) , s = 0, ± 1 , ±2, . . for some measure ll on 1r.
177
III.D. C(K)Spaces §Exercises
(c) Show that the main triangle projection, i.e. the map (aj k ) j, k ?. O  (bj k ) j, k ?_ O where b.
{
if j � k, aj J k  0k otherwise,
is unbounded on L (£2 ) . 7.
Show that if (n k )%"= 1 is a lacunary sequence of integers and the Fourier series 2: �= 1 a k ein k B represents a bounded function, then 2::'= 1 J a k I < oo.
8.
(Korovkin theorem) . Suppose that Tn: C [O, 1]  C[O, 1] is a se quence of linear operators such that J I Tn ll  1 as n  oo and Tn (P)  p as n  oo for every quadratic polynomial p. Show that Tn (f)  f in norm for every function f E C[O, 1] .
9.
For f E C[O, 1] we put Bn (f)(x) = I: �= O f(�)( ! )x k (1  x) n  k . The operators Bn are called Bernstein operators. (a) Show that each Bn is a linear, norm1 operator. (b) Show that for f E C[O, 1] we have Bn (f)  f uniformly.
10. For s > 0 we define
X8 = {f(z): f is analytic in ID and J / (z) J (1  Jz l ) 8 E L00 (ID) } and X� = {! E X8 : J / (z) J = o (1  J z l )  8 }. Show that Xs rv £00 , X� rv C() and (X�)** = X8 • 11. Show that there exists a function f E A(IBd ), d � 1, such that fmd I Rf(z) Jdv(z) = oo where R is the radial derivative (see Exercise III.B. ll) and v is the Lebesgue measure on IBd .
12. A sequence of finite dimensional Banach spaces (Xn )�= 1 is called a sequence of big subspaces of £! if there exist constants C and a such that for each n there exists a subspace Xn C f!n such that d(Xn , Xn) :S: C and Nn :S: a dim Xn . (a) Let T;;" C C(Y) be the space of trigonometric polynomials of degree :S: n, n = 1, 2, . . . . Show that (T;;" )�= 1 is a sequence of big subspaces of £! . (b) Let W;;" (IBd ) C C(IBd ) be the space of all polynomials homoge nous of degree n. Show that for every d = 2, 3, . . the sequence (W;;" ( 1Bd ) ) �= 1 is a sequence of big subspaces of £! . .
178
Ill.D. C(K)Spaces §Exercises
(c) Show that, if E C £! , dim E = n and 0 < 8 < 1, then there exist an integer k > 2 � and a subspace G c E, dim G = k, such that d(G, £� ) :::; �i��� ·
13. Construct the system of quadratic splines analogous to the Franklin system. More precisely, construct an orthonormal system of func tions ( gn ) �= O c L 2 (Y) such that each g� is a continuous, piece wise linear function with nodes at points t3 , j = 0, 1, . . . , n for n = 0, 1, 2, . . . , . (The points t3 are defined in 20.)
14. Let I c '][' be an interval. The function a 1 ( t ) is defined as a1 (t) =
{
0 if t ¢ I, I I I  1 if t is in the left hand half of I,  I I I  1 if t is in the right hand half of I.
We define the space B as the space of ail functions f(t) which admit a representation f(t) = >.o + I:: :'= 1 >.na ln (t) for some se quence of intervals ( Jn ) �= 1 and some sequence of scalars (>. n ) �= O with I:: :'= o l >.n l < oo. Then inf :L:: :'= o l >.n l over all representations of f is the norm denoted by II/Il B · (a) Show that B is a Banach space. (b) Show that f E B if and only if
� (n + 1)  ! l l f(t)fn(t)dt l
< oo
where Un ) �= O is the Franklin system.
15. The space A. (the Zygmund class) is defined as the space of all functions in C (Y) such that 11 /11
* 
_
sup
{I f(x  h) + f(x + h)  2f(x) .· x E '][' h > o } I h ,
< oo .
Show that f E A. if and only if I J1f f(x)fn (x)dx l = O(n  � ) , where ( /n) �= O is the Franklin system.
16. Show that the Franklin system is not an unconditional basic se quence in Lip1 (Y). 17. Let (!n ) ':'= be the Franklin system and let f E L 1 ('][') . Show that 1
the series "2.:'=0 (.f, fn ) fn converges almost everywhere.
179
III.D. C(K)Spaces §Exercises
R
18. Let A = 7L U �'ll  and let us consider the subspace V c L 2 ( ) consisting of all continuous, piecewise linear functions on with nodes at the points of A. Let r be a function which is continuous, piecewise linear with nodes at Ao = A U { � } and orthogonal to V. Assume also that ll r ll2 = 1 .
R
( a) Show that
l r(x) l
::; Cq l x l for some C
> 0 and q < 1. ( b ) Show that the family of functions {2i 1 2 r(2i x  k)}(j, k )EZxZ is a complete orthonormal system in L 2 (R). 19. Suppose that X is a separable Banach space with the Dunford Pettis property. Assume also that X "' Y* for some Banach space Y. Show that X has the Schur property, i.e. weakly convergent sequences converge in norm. It follows that Ld H1 and L1 [0, 1] are not isomorphic to a dual space. 20. Show that, spaces L1 (J.L) and C(K) do not have complemented, infinite dimensional, reflexive subspaces. 21. Find a Banach space X with the DunfordPettis property, such that X* does not have it.
22. Show that if X is a rich subspace of C(K) , then X* has the Dunford Pettis property.
23. Show that every operator 24.
space, is weakly compact.
T: £00 + X for X a separable Banach
Show that the identity operator id: C k ('l£'8 ) + c k  1 (Y8 ) , k � 1, s � 1 is compact. 25 . Show that C 1 (Y2 ) * = M EB F where M is isomorphic to M(Y) and F is separable.
26. Let us consider H00 (S) + C(S) where S is the unit sphere in ([d , d > 1 . Show that H00 (S) + C(S) is a closed subalgebra.
27. Let us consider the algebraic sum H00 (1l"" ) + C(11"" ) . Show that it is closed in L00 (1l"" ) but is not a subalgebra if n > 1.
III.E. The Disc Algebra
First we study some interpolation problems in the disc algebra A. We describe those subsets V c ID for which we have A I V = C(V) . We also describe the sets V c 11' such that every f E C(V) can be extended to a function F E A with finite Dirichlet integral. We also show that every £2 sequence is a sequence of lacunary Fourier coefficients of a function in A. Next we show that A rv (EA)o. We show that A is not isomorphic to any C(K)space. We present the construction of a Schauder basis in A and give different isomorphic representations of H00 • 1. Let us recall that the disc algebra A is the space of all functions continuous for l z l $ 1 and analytic for l z l < 1, equipped with the norm II/II = sup l z l 9 1 /(z) l. The maximal modulus principlen easily implies 1 1 / 11 = sup l z l = l 1/( z) l so A can be identified with span{e i B }n�o C C(T) . The disc algebra appears prominently in many parts of analysis; it is the canonical example of a uniform algebra, the von Neumann inequality (see III.F. (25)) gives the functional calculus on A for every contraction on a Hilbert space, etc. It is also a very interesting Banach space. From the F.M. Riesz theorem I.B.26 we get immediately that
A* = C(11') * /AJ.. = LI/Jtt EB1 Ms
(1)
where M8 denote the space of measures on 11' which are singular with respect to the Lebesgue measure. 2. We will discuss some interpolation results for the disc algebra and other related spaces. More precisely, given a space X of analytic functions on ID we look for sets V c ID (or even V c ID if elements of X extend naturally to 11') such that the restrictions X IV fill the space C(V) or £00 (V). We want to impose minimal, sensible conditions. The above description (1) of A* easily yields the following result. Proposition. (RudinCarleson) Let Ll = Ll c 11' be a set of Lebesgue measure 0 and let cp E C(11') be a strictly positive function with cpi Ll = 1 . Then for every f E C(Ll) there exists a g E A such that giLl = f and lg(O) I $ 11 /11 cp( O) for all 0 E 11'. ·
III.E.
182
The Disc Algebra §3.
Let us introduce an equivalent norm on A by ll g ll "' = Proof: sup{ l g( O ) I cp 1 (0) : 9 E 1I'} and let r: (A, I · II "') + C(Ll) be the re striction operator r(g ) = gi Ll. Since Ll has Lebesgue measure 0 and cpl .!l = 1 we infer that r* is an isometric embedding, so r is a quotient map. Note that r* assigns to the measure JL on Ll the same measure treated as a measure on 1I' and considered as a functional on A. Since (ker r) .L = r* (C(Ll) * ) we infer that ker r is an Mideal. Thus from The orem III.D.4 we get that there exists g E (A, I · II "') with g i Ll = I and a ll gii 'P � II I II , i.e. l g( O ) I � cp( O ) II I II · Note that I Ll l = 0 is also a necessary condition for A I Ll = C(Ll) for a closed set Ll c 1I'. If I Ll l > 0 then the restriction r: A + C(Ll) is 11 (It is known that I E Hp (1I') cannot vanish on a set of positive measure. This fact is hidden in the canonical factorisation and the form of an outer function; see I.B.23) and it easily follows from Proposition 2 that it is not an isomorphism. Thus it cannot be onto. 3.
From this proposition and Theorem III.D. 16 we get
Corollary. If Ll c 1I' has Lebesgue measure 0 then there exists a linear extension operator u: C(Ll) + A. a
Interpolating sequences in the open disc are also of considerable in terest. First we consider them for the space H00 • A sequence (>.n)�=O C D is called interpolating if the map I t+ (f(>.n))�=O transforms H00 onto l00 • Obviously (see I.B.23) any interpolating sequence has to sat isfy the Blaschke condition :E:= 0 (1  l >.n l ) < oo. Thus we can form the Blaschke products B(z) = TI:= o Mn (z) and Bn (z) = B(z)/Mn (z) where Mn (z) = l >.n l >.; 1 (>.n  z) ( 1  Anz)  1 • The following gives some information about interpolating sequences. 4.
Theorem. The following conditions on the sequence (>.n)�=O C D are equivalent:
(a) (>.n)�=O is an interpolating sequence; (b) infn ;::: o I Bn (>.n) l
=
8 > 0;
(c) there exists a bounded linear map T: loo + Hoo such that T(e) (>. k ) = ek for k = 0, 1 , 2, and any e = (en ) E loo Note that condition (c) gives a linear lifting to the m ap I t+ (!(>.n))�=O ·
1 83
III.E. The Disc Algebra §4.
(a)=*(b) . The open mapping theorem yields a constant C such that for each n = 0, 1, 2, . . . there exists In E Hoo with Illn il ::; C and ln (>.j ) = 6ni · From the canonical factorization I.B.23 we get that In = cpn · Bn with cpn E Hoo and ll cpn ll = llln ll · This gives condition (b) . (b) =*(c) . Let us assume 0 < l >.o l ::; l >. 1 l ::; . . . . We define Proof:
(2) where an(z) = L k > n (1 + X k z) (1  X k z)  1 (1  l >. k l 2 ). Clearly ¢n (>. k f= 6n,k so the operator T( en ) = L n > O en¢n satisfies (c) provided it is continuous. This will follow from
L: l¢n(z) l ::; c{j for l z l n�O
<
1.
(3)
Since (4) we get
k �n ::; 2  4 log
k >n
I kfmII Mk (>.n) l ::; 2  4 log 6 = Kl) .
Using (4) , (b) and (5) we get
L l¢n (z) l ::; L [Rean (z)  Rean+l (z)] 6  1 exp Kli exp(  Rean(z )) n�O n�O ::; c{j L exp(Rean (z)  Rean+ l (z))  1 . exp  Re an(z) n�O = c{j L [exp (  Rean+ l (z))  exp( Rean(z)) ::; c{j . n �O
[
(c)=*(a) . is obvious .
]
]
a
184
III.E. The Disc Algebra §5.
5. If (.Xn)�= O C ][) is an interpolating sequence then it is quite difficult to describe the set {(f(.Xn))�=o= f E A } C C00 • This set clearly depends on the topological structure of the closure of (.X n)�= O in 1D. One can easily construct examples (see Exercise 6) where this set is not closed in C00 • The following theorem characterizes sets of 'free' interpolation for the disc algebra. Theorem. Let V c 1D be a closed subset. Then A I V = C(V) if and only if I V n lr l = 0 and V n 1D is an interpolating sequence.
If A I V = C(V) then A I (V n lr) = C(V n lr) so by the obser vation made after Proposition 2 we have I V n lr l = 0. Clearly V n 1D is countable and there exists a constant C such that for every finite subset {.A 1 , . . . , .An} C V n 1D and any numbers (a 1 , . . . , an ) there exists f E A such that 1 1 /11 � C max 1 :::; j :::; n l ni l and f(.Xj ) = Ctj . By the standard normal family argument we get that V n ][) is an interpolating sequence. Conversely let I v n lr l = 0 and v n ][) be an interpolating sequence. Let us write A = A0 EB A 1 where A0 = {! E A: !I V n 1[' = 0 } and A 1 = u (C(V n lr)) where u is a linear extension operator from C(V nlr) into A (cf. Corollary 3). Let us also split C(V) = C0 EB C 1 where C0 = {! E C(V): f i V n lr = 0 } and C 1 = A 1 1 V. In order to prove the theorem it is enough to show that A0 1 V = C0 • Let V n 1D = {.An}n;:: o with I.Ao l � I .A1 I � I .X 2 I � . . . . Let (rf>n)n;:: o be given by (2) . It is known (and easily follows from the standard proof that the Blaschke product converges) that each Bn(z), n = 0, 1, 2, . . . is continuous on 1D\(V n lr). Also the an(z) , n = 0, 1 , 2, are continuous on 1D\(V n lr) so all the (rf>n)n;:: o are continuous on 1D\(V n lr) . Let 'Pn E A be such that II 'Pn ll � 2, 'Pn i V n 1[' = 0 and 'Pn(.Xn) = 1 for n = 0, 1, 2, . . . . The existence of such 'Pn 's easily follows from Proposition 2. It follows from (3) that for 1/Jn = 'Pn · r/>n , n = 0, 1, 2, . . . we have Proof:
L 1 1/Jn (z) l n;:: o
�
C for l z l
<
1
and 1/Jn (.X k ) = 8n ,k · For f E C0 we define F = L n > O f(.Xn)'I/Jn · Since a F E A and F I V = f the theorem is proved. 6. Let us consider now a more specialized interpolation result. Before we proceed we have to recall the notion of a Dirichlet integral (Dirichlet norm) which is instrumental in proving the Dirichlet principle by vari ational methods. We restrict our attention to analytic functions only.
III.E. The Disc Algebra §6.
185
We define the Dirichlet space
{
D = f(z): f(z) is analytic for l z l < 1 and
(L
l f'(z) l 2 dv(z)
)
1
2
<
<x}
An easy computation shows that f(z) = L n >o ](n)z n E D if and only if L n >o n l ](n) i 2 < oo. We are interested in D n A or, more precisely the sets E C T for which we have D n A l E = C ( E ) . Clearly E has to have measure zero but this is not enough. The proper condition involves the notion of capacity which we will now recall. For a closed set E c 1I' and a Borel measure f..L on E we define the energy of f..L as
2 e(f..L ) = r r log  df..L ( x)df..L ( y), jEjE 1 X  y 1
(6)
where this integral is understood as the Lebesgue integral on ExE with respect to the measure f..LXf..L . Since for x = e i6 and y = ei"', l x  Yl = 2 1 sin ! ( (J  cp) I we can write the integral (6) as
r r log . � df..L ( fJ)df..L ( cp) . J.11:J.1f I sm 7I Writing log I sin ! O I  l rv I: � : 'YneinB we get e(f..L ) = I: � : 'Yn i J1 (n) i 2 . Integration by parts yields 'Yn = 1 �1 + o ( 1 � 1 ) . Moreover one can show that 'Yn � 0. This gives e(f..L ) � 0 for every Borel measure f..L on 1I' and (7) e(f..L ) < oo if and only if :L: l n i  1 1M n) l 2 < oo. n ;<\0 Note also that if £(f..L ) < oo and
e(9 . JL ) < oo.
g
is a bounded function on 1I' then
The capacity of E , denoted by 'Y(E) is defined as
'Y( E ) = exp  inf { e (f..L ) : f..L is a probability measure on E } . If l E I > 0 then 'Y ( E ) > 0 because the Lebesgue measure restricted to E has finite energy. Now we are read to state: Theorem. Let E only if 7(E) = 0.
c
1I' be a closed set. Then ( D n A ) I E
For the proof we will need two lemmas.
=
C ( E ) if and
Ill. E.
186 7 Lermna. Let Jt E M(Y) and £(Jt) < E C '][' with 'Y(E) = 0 we have Jt(E) = 0.
oo.
The Disc Algebra § 7.
Then for every closed set
Proof: It is enough to check for positive Jt only. Now if Jt(E) > 0, then, since 'Y(E) = 0 we have oo
2 = £( �t i E) = }[ }[ log 1 d�t(x)d�t( Y ) E E 1x  y 2 $ [ [ log  d�t(x)dJt( Y ) = £(Jt) < oo j.lf j'Jf 1 X  Y 1
a
which is absurd. 8 Lermna. Let Jt be a measure in M(T) . � < "'+ oo
Lm=  oo , n ;o!'O
Proof:
(
lnl
)
IE L:�= 1 ( I P.<:W)
<
oo
then
00 ·
We will actually show the inequality (8)
It is enough to establish (8) , which obviously implies the lemma, for absolutely continuous 1" · If Jt = f d>. we put f* ( 0 ) = ! (  0 ) and g = f * f * · Clearly g(n) = 27rl f (nW = 27rljl (nW. It is known and easily checked that the function h(O) = � (1r  101) sgnO defined for 1r $ (} $ 1r has the Fourier series L:� n  1 sin nO. We have 1l"
2 II Jt ll
1l"
2 � 2 I IY II1 = ll h lloo ii Y ih �
I /_: h(O)g(O)d(} l = 1 211"2 � � ( IMnW
so (8) follows.

1 11 (

nW )
I
a
We identify D n A with the subspace X of D E9 A consisting of pairs ( !, f). This allows us to describe (D n A)* as D * E9 A* / X 1.. . Let r denote the restriction operator from D n A into C(E). It is enough to consider E with l E I = 0. For Jt E M(E) = C(E) * we get r * ( Jt ) = (O, Jt) + X1.. . Thus, by duality, r is onto if and only if there exists a c > 0 such that Proof of the Theorem.
inf{ II (O, Jt)  (S, v) ! l : (S, v) E X l.. } ;::: CII JL II for all Jt E M(E) . (9)
Ill.E. The Disc Algebra §9.
187
Assume now that 7(E) = 0. Since (e inB , einB ) E X for n = 0, 1, 2, . . . we get that for (S, v) E XJ. we have S(ei n6) + v(n) = 0 for n = 0, 1 , 2, . . . . For S E D* we get easily that L::= l n 1 I S(e inB ) j 2 < oo so for {S, v) e Xl. we have L::= l n 1 l v{n) l 2 < oo. From Lemma 8 and {7) we get e(v) < oo. From Lemma 7 we get v(E) = 0. Thus for (S, v) E XJ. and p, E M(E) we have 11 {0, p,)  (S, v) ll = II S IIv • + lip,  vii � ll llll so by {9) r is onto. Assume now 7(E) > 0. From {7) we get that for every p, E M(ll') with £(p,) < oo the functional Sp. defined on D as Sp. { / ) = L n > o f(n)[l,(n) is continuous and II Sp. l l v · � Ce(p,) ! . In particular if e(p,) < oo, then ( Sp. , p,) E X 1. . Since 7(E) > 0 there exists a probability measure p, on E with £(p,) < oo. Let cpn E L00 {p,) be a sequence of functions such that lcpn l = 1 p,a.e. and cpn + 0 as n + oo in the a(L00 (p,) , £ 1 (p,))topology. From {6) we infer that e(cpn Jl ) + 0 as n + oo. So inf{ II {O, cpnP,))  {S, v) ll : (S, v) E XJ. } � II {O, cpn Jl )  (S'Pn l' • cpn Jl) ll = II S'Pn P. II D• � C£( cpn Jl ) ! + 0 as n + 00 . This shows that {9) is violated, so r is not onto.
a
9. In the previous sections we have been mostly interested in interpo lation taking into account values of the function. The other very natural and important problem is to impose some conditions on Fourier coeffi cients. The following result is interesting in itself and will be used in Chapter III.F. Theorem. Let (n k )k:: 1 be a sequence of positive integers such that n k+ l � 2nk for all k in N and let (vk )k:: 1 E £2 . Then there exists g E A with g(z) = L::= 0 g(n)z n and g(n k ) = Vk and IIY II oo � C ll (v k ) ll 2 ·
Since (g(n ) ) n�o E £2 for every g E A one cannot relax the condi tion on the sequence (v k )k:: 1 . Thus the theorem can be rephrased as { f (n k ) : f E A} = f2 . Let us consider only numbers z with l z l = 1 . Assume also = 1 . We define inductively two sequences of polynomials 9k (z) and h k (z) , k � 0 as follows : Proof:
L:%': 1 l vk l 2
9o (z)
=
vo z n o
and ho ( z )
=
1
(10)
188
Ill.E. The Disc Algebra § 1 0.
and for k > 0 we put
9k (z) = 9k  l (z) + Vk zn k h k  l (z), hk (z) = h k  l (z)  'ihzn k 9k  l (z).
{11)
Using the elementary identity
b, v we inductively obtain that k = {1 + l vi l 2 ) � C. (zW (zW + h IT I Yk l k j =O
valid for all complex numbers
a,
{12)
We also obtain inductively that
nk 0 9k = 'L, §k (j)zi and hk = 'L, h k (j)zi . j� �� Since n k + l � 2n k we infer that there is no cancellation of Fourier co efficients in {11). In particular we get h k (O) = 1 for all k and thus Yk (ns ) = V8 for s � k. Thus {12) and the open mapping theorem I.A.5 give the claim.
a
10. We have seen projections in A whose image is a C(K)space. Now we will investigate projections whose image is isomorphic to A. This will lead to the proof of Theorem 12. Given a positive number c and an interval I in 11' we say that a function I E A is csupported on I if ll (t) l � c lllll for t E 11'\f. We say that a subspace X C A is csupported on I if every I E X is csupported on I. We have the following. Proposition. For every c > 0 and an interval I C 11' there exists a subspace X C A and a projection P: A�X such that
(a) X is csupported on I, (b) d(A, X) � 1 + c,
( c ) P 1 = 0 and II P II � 1 + c,
(d) for every g E A, t5 supported on 11'\I we have II Pgll � (c + t5 ) II Y II ·
189
Ill.E. The Disc Algebra § 1 1 .
This is an interplay between averaging projections and con formal mappings. Every conformal map cp: [) + [) induces an isometry I'P : A + A defined by I'P ( f ) = f o cp. Suppose the proposition holds for some e > 0 and some interval I. Then it holds for the same e > 0 and any other interval I1 · To see this let us take a conformal map cp such that cp(ft ) = I. Then I'P (X) is esupported on I and I'PPI'P 1 is a projection onto I'P (X) and (a)(d) hold. Let Proof:
(13) This is a norm1 projection and ImQ n � A. For a positive number � < ! we find a function F E A, with II F II � 1 such that
27r I F(e '"(J )  1 1 < � for n

< (J < 21r   and F(1)
27r n
= 0.
Such a function is easy to construct using conformal mappings as before. Observe that for f E ImQn we have and
11/11 ;;::: II F . /II ;;::: (1  � ) II/II II Q n (F · f)  /II = II Qn C f · (F  1)) 11
(14)
� 2 � 11 / 11 ·
(15)
Let I be such that I F(t) l < � for Y\f. From (14) we infer that X = F · ImQn is a closed subspace of A, ( � )supported on I, with d(X, A) � (1  �) 1 . The condition (15) shows that Q n i X is an isomor phism between X and ImQ n with II (Qn i X)  1 11 � (1  2 �) 1 . We define P = (Qn i X)  1 Q n . This is a projection onto X with II P II � (1  2�)  1 . Since Qn 1 = 0 P(1) = 0 as well. If g is a function in A which is 8supported on Y\f then 
If � was chosen small enough and
��
n
big enough we see that (a)(d) a
1 1 Proposition. The space A contains a complemented subspace iso morphic to (I: A)o .
190
Ill.E. The Disc Algebra § 1 2.
Proof: Let us take a sequence of disjoint closed intervals {In )�= l in '][' and a sequence of positive numbers (en ) �= l with L:: :'= l en = e < 0. 1. Using Proposition 10 we construct subspaces Xn C A, ensupported on In for n = 1 , 2, . . . , and projections Pn from A onto Xn satisfying (a)(d). It is routine to show that for Xn E Xn , n = 1, 2, . . .
(1  2e) sup ll xn ll :5
I�
l
Xn :5 (1 + e) sup ll xn ll
(16)
so X = span{Xn} �= l "' (EA)o . Let R(f) = L:: :'= l Pnf · In order to show that R is continuous it is enough to check that for every f E A we have II Pn /11 + 0. Fix tn E In. Since f is uniformly continuous in [) we get that f  f(tn) is Onsupported on ll'\In for some On + 0. From Proposition 10 (c) and (d) we get II Pn/11 = II Pn (f  f(tn)) ll :5 (on + en) II / II , and this yields the continuity of R. Clearly R: A + X. For x = L:: :'= l Xn E X with Xn E Xn and ll x ll = 1 we define hn = L:;;:: l ,n # X k · Since Rx  x = L:: :'= l Pnhn we get from (16) that II Rx  x ll :5 (1 + e) supn II Pnhn ll · Once more using (16) we get ll xn ll :5 (1  2e)  1 for 1 n = 1, 2, . . . so ll hn ll :5 (1 + e) (1  2c)  . Since each Xn is ensupported on In we get that for t E In we have l hn (t) l :::; e(1  2c)  1 . Proposition 10 (d) gives II Pn (hn) ll :5 [en + e(1  2e)  1 J II hn ll so we conclude that II Rx  x ll < 0.8. This shows that R I X is an isomorphism of X (see II.B. 14) and one checks that (R I X) 1 is a continuous projection from A onto X. a 12. The decomposition method (see II.B.21) and Proposition 11 yield immediately Theorem. The disc algebra A is isomorphic to its infinite eosum.D 1 3 Remark. Since the projections constructed in Proposition 10 have the property that P* (Lt fHt ) C Lt fH1 we easily see that (ELI/H1 ) 1 "' Ltf H1 and passing to the duals we get (EHoo)oo "' Hoo . 14. Most of the results in this chapter show the analogy between A and C(K)spaces. We would like to point out however that A is not a C(K)space. To see this we need to observe that every C(K)space has the fol lowing extension property:
Ill.E. The Disc Algebra § 15.
191
There exists a constant C such that for every Banach space X, every subspace Y of it and every finite rank operator T: Y + C(K) there exists an operator T: X + C(K) such that T I Y = T and II T II :5 C II T II · This follows {for any C > 1 ) directly from II.E.5{c) and III.B.2 (or the HahnBanach theorem) . Let I = 'E::N aneinB be a trigonometric polynomial. The operator Tt: A + A defined by Tt (g ) = I* g is a finite dimensional operator and II Tt ll :5 11 1 11 1 · The operator Tt has a unique rotation invariant extension Tt: C(11') + A which is given by Tt (g ) = 'RI * g where 'R is the Riesz projection (see I.B.20). The standard averaging argument shows that for any extension T we have II T II � II Tt ll · But II Tt ll = II 'RI II 1 · Since the Riesz projection is unbounded on £ 1 (11') (see I.B.20 and 25) we see that A does not have the above property (*). Since if Z has (*) its complemented subspaces also have (*) we get Theorem. The disc algebra A is not isomorphic to any complemented subspace of any C(K)space.
Despite this theorem and some other striking differences between and C(K)spaces, which we will exploit e.g. in III.F.7, the general impression is that A is quite similar to a C{K)space. Actually the idea of comparing A to C(K) underlies most of the results about the disc algebra A presented in this book. A
15. Another proof of the above theorem, which is different, although similar in spirit, follows from the LozinskiKharshiladze theorem III.B.22 and the following Proposition. Let T� denote the space of all trigonometric polyno mials of degree :5 n with the supnorm. There exists a sequence of projections (Pn)�= 1 in the disc algebra such that
{a) II Pn ll :5 C,
{b) d(Im Pn , T�) :5 C for some constant C and all n = 1,2,3, . . . . Proof: The proof depends on the properties of the Fejer kernels; see I.B. 16. Let An = span{ 1 , z, . . . , z 2n ) C A. We define i n : A n + A EBoo A
192
Ill.E. The Disc Algebra § 1 6.
) (
(
by in L:: �:o a k z k L:: �: o a k z k , L:: �: o a k z 2 n  k We also define ¢n: A ffi A + A n by
)
.
Clearly I l i n I :5 1.
Using the properties of Fejer operators once more we get II
The desired isomorphism from A onto A r is defined by
We intend to show that the disc algebra has a Schauder basis. Clearly the monomials (z n )�= O are not a Schauder basis in A. The pre vious proposition gives us a different representation of the disc algebra. In this representation we can forget for a while about analyticity. It also makes the proof of our next theorem more transparent. 17.
Theorem.
The Franklin system is a Schauder basis in
Ar .
The definition and properties of the Franklin system have been given in 111.0.2027. In the proof of Theorem 17 we will freely use notation from there.
193
III.E. The Disc Algebra § 1 7.
If Fn is the nth Fejer kernel (see I.B. 16) then for f E Ar we have II/  f * Fn ll + 0 as n + oo so the trigonometric polynomials are dense in Ar . For every trigonometric polynomial cp( O ) there exists a sequence 'Pn of finite sums of Franklin functions ( 'Pn can be taken to interpolate cp in to, t 1 , . . . , tn) such that 'Pn + cp in Lip 1 . Since, as is well known, ll tPn  c,OI I oo :5 C ll cpn  cpii Lip 1 we see that the Franklin system in linearly dense in Ar . Thus our theorem follows from Theorem III.D.25 and the following inequality which is the main point of the Proof: There exists a constant C such that for every N and every f E C (11') Proof:
(17) For a fixed x E 11' and f E C(11') we have
N SN ( / )(x)=df � � ( !, fn } fn (x) n=O cot .(KN (x  t, y )  KN (x + t, y))dtf(y )dy =
h 1"' �
(18 )
N
where KN (x, y ) = L fn(x)fn(y) . n=O I f we define
AN,x (t) =
{
O
if I t  x i < ..!..N •
cot t 2 x if
I t  x i > 1J
then we can write
1
h f(y) fo"N cot � · (KN (x  t, y )  KN (x + t, y) ) dtdy + h f(y) [ l AN,x (t)KN (t, y)dt  AN,x (y) ] dy ( 19 ) + h f(y)AN,x ( y )dy = + /2 +
SN ( / )(x) =
h
h
For every y , KN (x, y) is a piecewise linear function with nodes at least 2� apart so Proposition II.D.21 (b) yields the estimate for the slope
194
Ill.E. The Disc Algebra §18.
which gives
k cot � · I KN (x  t, y )  KN (X + t, y ) l dtdy 1 h N � 11 /lloo h 1 cot � · C(N + 1) 2 q Ndi s t ( x, y ) tdtdy � C ll/lloo (N + 1) h q Ndi s t ( x, y ) dy � C ll/ll oo ·
h �
11 /lloo ·
1
(20)
The estimate
(21) follows immediately from the known properties of the trigonometric con jugation operator; see Zygmund [1968] p. 92 or Katznelson [1968] Corol lary 111.2.6. In order to estimate 12 we write
where PN is the Nth partial sum projection with respect to the Franklin system. Let 'PN,x be a piecewise linear function in C(ll') with nodes at to, tt , . . . , t N such that 'P N,x (tj ) = AN,x (tj ) for j = 0, 1, . . . , N. One checks (draw the picture) that (23) Since the Franklin system is a basis in £1 (ll') (see III.D.26) from(23) we get
From (20), (21), (22) and (23) , (17) follows immediately.
a
18. Now we want to give different isomorphic representations of H00 • Having different isomorphic representations of an important space is generally useful because each representation carries with it different in tuitions, and even the possibility of using different analytical tools. Let Ui ) f'=. o be the basis in A which exists by Theorem 17 and let H� = span{ /j };::;n · Let us recall also that An = span(1, . . . , z 2 n ) C A (see 15) .
III.E. The Disc Algebra §Notes. Theorem. The spaces morphic.
( I:�= 1 H;;, ) 00 , ( I:�= 1 An) 00 and H00
195 are iso
First note that ( I:�= 1 H;;, ) 00 is isomorphic to its infinite £00sum. This follows from II.B.24. A standard perturbation argument (see II.E.12) shows that for certain A each H;;, is Aisomorphic to a A complemented subspace of A k ( n ) and analogously it follows from Propo sition 15 that each An is Aisomorphic to a Acomplemented subspace of H'j, n ) . This yields that ( 2:: �= 1 H;;, ) 00 is isomorphic to a comple mented subspace of ( 2:: �= 1 An) 00 and also ( 2:: �= 1 An) 00 is isomorphic to a complemented subspace of ( 2:: �= 1 H;;, ) 00 so our first observation and II.B.24 give ( 2:: �= 1 H;;, ) oo rv ( 2:: �= 1 An) 00 • Remark 13 yields 00 that ( L H;;, ) oo is isomorphic to a complemented subspace of H00 • Proof:
n= 1
To complete the proof we need to show that ( 2:: �= 1 An) 00 contains a complemented copy of H00 • Let :Fn be the Fejer kernel and define i: Hoo + ( 2:: �= 1 An) 00 by i(f) = (f * :Fn )�= 1 . The properties of the Fejer kernel (see I.B. 16) give that i is an isometry. To define the projec tion onto i(Hoo ) we use a compactness argument. Let B denote the unit ball of H00 with a(H00 , LI/ H00 )topology, and let Bn be the unit ball in An · We define maps 1fn: f1�= 1 Bn + B by 1fn(h1 , h2 , . . . ) = hn. Since the space of all maps from I1�= 1 Bn into B is compact we take 7f to be a cluster point of {7rn} �= 1 . One checks that 7f is homogenous and ad ditive so it extends to a continuous linear map 1r: ( 2:: �= 1 An ) 00 + H00 • Moreover
This shows that i1r is a normone projection onto i(H00 ) .
a
Notes and remarks. As noted in 1 the disc algebra is an important space. It is a prototypic
uniform algebra, so much information about it can be found in Gamelin [1969] , Hoffman [1962] or Garnett [1981] . The closely related space H00 is even more fascinating; the whole book Garnett [1981] deals with it. A more Banach space oriented exposition is in Pelczynski [1977] . The connection with operator theory hinted in 1 is presented in detail in the beautiful lectures of Nikolski [1980] . The theory of peak sets and peakinterpolation sets is a well developed topic in uniform algebra theory; see Gamelin [1969] 11. 12. Our Proposition 2 is a prototype of
196
Ill.E. The Disc Algebra §Notes.
this theory. It was proved by Rudin [1956] and Carleson [1957] . The appeal to Theorem III.D.4 can be avoided but it saves some calculations. The problem of characterizing interpolating sequences for H00 was an object of very intense study in the late 50's; Hoffman's book [1962] contains a nice presentation of these early results. By now it has grown into a vast area (see Garnett [1981] ) . The beautiful and simple proof of (b)=>(c) in Theorem 4 is due to Peter Jones (see GorinHruscov Vinogradov [1981] ) . Our Theorem 5 is a particular case of results in CasazzaPengra Sundberg [1980] where complemented ideals in A are fully described. The description of ideals in A is contained in Hoffman [1962] . This says that every closed ideal I in A is of the form AK · F where K C '][' is a closed subset of Lebesgue measure 0 and AK = {! E A: JI K = 0} and F is an inner function such that F  1 (0) n ll' C K and the measure determining the singular part of F is supported on K. The result of CasazzaPengraSundberg [1980] asserts that I is com plemented in A if and only if F is a Blaschke product whose zeros form an interpolating sequence. In paragraph 6 we gave a crash course in ele mentary potential theory for subsets of ll'. Chapter III of KahaneSalem [1963] contains everything we state and use. Theorem 6 is one of the re sults contained in HruscovPeller [1982] . Our presentation follows Koosis [1981] . The direct proof of Theorem 9 is taken from Fournier [1974] . Much more general theorems are proved in Vinogradov [1970] . In partic ular he has shown that given ( v k ) � 1 E £2 there exists f(z) = E�o anz n such that a2 k = Vk and f(z) is holomorphic in G and continuous in G where G is any region in cr with smooth boundary and aG n ll' contains an interval. Theorem 12 and its proof are taken from Wojtaszczyk [1979] . There are many proofs that A is not isomorphic to any C(K)space. We will see some more in 111.1. The fact was first observed by Pelczynski [1964a] with basically the same proof as the one given in 14. The argument was extended to the context of ordered groups by Rosenthal [1966] . Proposition 15 is an unpublished observation of J. Bourgain and A. Pelczynski. The question if A has a Schauder basis was asked by Banach [1932] . The answer was given by Bockariov [1974] . The use of the Franklin system in the construction was rather unexpected. Theorem 18 is a rather routine consequence of previous results. It was observed in Wojtaszczyk [1979a] . It shows in particular that H00 is isomorphic to the second dual of a Banach space. Note that with the natural duality we have H00 9:! (LI /H1 ) * . This is the unique isometric predual of H00 (see Ando [1978] and Wojtaszczyk [1979a] ) .
197
III.E. The Disc Algebra §Exercises.
The space Ltf H1 in its turn is not isomorphic to a dual Banach space; this was noted in Pelczyilski (1977] and Wojtaszczyk (1979a] ; see Exercise III.D. 19. Exercises
1.
Let � c T be a compact set of measure zero and let (nk)k:: 1 be a lacunary sequence of natural numbers. Given f E C(�) and a sequence (ak) E £2 show that there exists h E A ( D) such that h i � = f and h(nk) = ak for k = 1, 2, . . . .
Suppose that x * E A* . Show that there exists only one measure on T such that ll�tll = ll x * ll and �ti A = x * . 3. Let V c A* be a relatively weakly compact subset. For each E V let v E M(T) be its normpreserving extension (see Exercise 2). Show that V = { v : E V} is relatively weakly compact in M(T). 4. (a) Let f E D (the Dirichlet space) . Show that f induces a func tional on H1 (T), i.e. we have an inequality I Jy g(e i9 )f(e i9 )dO I ::;; cf . II Yil t . (b) Show that there are unbounded functions in D. 5. The matrix ( a ii ) i ,j�O is called a Hankel matrix if Cl! ij = cp( i + j) for 2.
v
v
some cp. An operator on £2 is called a Hankel operator if its matrix in the natural unit vector basis (ej ) � 0 is a Hankel matrix. (a) Show that L 00 (T)/ H! is isometric to the space of all Hankel operators.
(b) Show that C(T)/Ao is isometric to the space of all compact Hankel operators. (c) Show that for every Hankel operator T there exists a best ap proximation by a compact Hankel operator.
6.
(a) Suppose that (zk)k:: 1 C D is such that d�1"1:��)1 ) < c < 1 for k = 1, 2, 3, . . . . Show that (zk)k:: 1 is an interpolating sequence. Show that if (zk)i:'= t are positive real numbers then the above condition is also necessary for (zk)i:'= 1 to be interpolating. (b) Find an interpolating sequence (A n)�= l C D such that { (/(An))�= 1 : / E A } is not closed in £00 •
7.
Let IPr = {z E CI:: : r ::;; l z l ::;; r  1 } for 0 < r < 1 and let A(IPr ) denote the space of all functions continuous in IPr and analytic in the interior.
198
III.E. The Disc Algebra §Exercises.
(a) Show that A(1Pr) contains a !complemented isometric copy of
A(D) .
8.
(b) Show that A(D) contains a complemented copy of A(JPr) with the constants independent of r. (c) Every f E A( 1Pr ) can be written as L:: � : anz n . Show that the map Pr ( L:: � : anz n ) = L:: := o anzn is a projection from A(JPr) onto A(r  1 D). Show that sup l< r< l II Pr ll = oo. (a) Show that f E Hoo (D) is a Blaschke product if and only if 1/ (z) l ::; 1 and limr+1 f:_1r log lf (rei9 ) l d0 = 0. (b) Suppose f E H00 (D) is an inner function. Show that for every p with 0 < p < 1 the functions
 peicp wcp (z) = 1 f(z)  pe icp f(z)
9.
are Blaschke products for almost all cp, 0 < cp ::; 21r. (c) Show that every f E H00 (D) is a limit in the topology of uniform convergence on compact subsets of D, of a sequence of finite Blaschke products. (d) Show that the closed unit ball in A is the closed convex hull of finite Blaschke products. (e) (von Neumann inequality) . Let T: H + H (H a Hilbert space) be a contraction, i.e. II T II ::; 1. Show that for every polynomial p (z) we have ll p (T) II ::5 sup zEID lp (z) l . (a) Show that if P: A + A is a normone, finite dimensional projection with dim /m P > 1 then Im P* c {J.q.t .l m} where m is the Lebesgue measure on Y.
(b) Show that the disc algebra is not a 1r1space. 10. (a) Suppose f E L00 ('U') . Show that there exists 9 E H00 ( 'U') such that I I/  9 11 = inf{ l l /  h ll : h E Hoo (Y) }. (b) Suppose f E C(Y) . Show that dist(f, H00 ) = dist(f, A) . (c) Suppose that f E C(Y) . Show that there exists only one 9 E H00 ('U') such that I I/  9 11 = dist(f, Hoo ), and that 1 9  /I = const. (d) Show that there exists f E C (Y) such that its best approximation in H00 (Y) , i.e. 9 E H00 (Y) such that II/  9 11 dist(f, H00 ) , is not continuous.
III.F. Absolutely Summing And Related Operators .
We discuss in detail pabsolutely summing operators. The Pietsch fac torization theorem, which is basic to this theory, is proved. The funda mental Grothendieck theorem is proved in its three most useful forms. Later we improve it and show the GrothendieckMaurey theorem, that every operator from any £ 1 space into a Hilbert space is pabsolutely summing for all p > 0. We present the trace duality and show that the p' nuclear norm is dual to the pabsolutely summing norm. We also introduce and discuss pintegral operators. We show the connection be tween cotype 2 and the coincidence of classes of pabsolutely summing operators for various p's. The extrapolation result for pabsolutely sum ming operators is proved. We apply Grothendieck's theorem to exhibit examples of power bounded but not polynomially bounded operators on a Hilbert space and to give some estimates for the norm of a polynomial of a powerbounded operator. We also present many applications to harmonic analysis: we construct good local units in L 1 { G ) , we prove the classical OrliczPaleySidon theorem and give some characterizations of Sidon sets. 1. In this chapter we will discuss several important classes of operators, namely pabsolutely summing, pintegral and pnuclear operators. All these classes have some ideal properties so we will introduce the general concept of an operator ideal. We are given an operator ideal if for each pair of Banach spaces X, Y we have a class of operators J(X, Y) such that { 1 ) J(X, Y) is a linear subspace {not necessarily closed) of L(X, Y)
containing all finite rank operators,
E J(X, Y), A E L(Z, X) and B E L(Y, V) that BTA E I(Z, V) for all Banach spaces X, Y, Z, V and all operators A, B.
{2) if T
An operator ideal is a Banach ( quasiBanach) operator ideal if on each I(X, Y) we have a norm {quasinorm) i such that {3) (J{X, Y ) , i) is complete for each X, Y
(4) i(BTA) ::; II B I !i(T) II A II whenever the composition makes sense
III.F Absolutely Summing And Related Operators §2.
200
(5) for every rankone operator where = · y.
T(x) x*(x)
T: X + Y we have i(T) l x *l l I Y I , =
·
Actually the reader has already encountered some examples of Ba nach operator ideals. Compact operators (see I A.15) and weakly com pact operators (see II.C.6) form Banach operator ideals with the opera tor norm. .
T: X + xi X
2. An operator Y is pabsolutely summing, 0 < p � oo (we write lip ( Y)), if there exists a constant C < oo such that for all finite sequences ( ) J=l C we have
T E X,
n
.!
n
!.
( � 1 Txi 1 P) p � C sup { ( � i x* (xi ) I P) p= x* E X*, I x* l � 1 } . (6) We define the psumming norm of an operator T by 7rp(T) inf{C: (6) holds for all (xj )J=l X,n 1, 2, . . . } . (7) c
=
=
Let us observe that for p = 1 the condition (6) is equivalent to the fact that transforms weakly unconditionally convergent series into absolutely convergent series.
T
3.
We have the following
Theorem. For 0 < p � oo the pabsolutely summing operators form a quasiBanach (Banach if 1 � p � oo) operator ideal when considered with the pabsolutely summing norm 7rp ( · ) .
The proof of this theorem consists of routine verification of condia tions (1)(5) and is omitted.
X,
II00(X,
L(X,
Y) = One easily checks that for all Y we have Y) and = For p < oo the situation is less trivial. The iden tity id: £2 £2 is not pabsolutely summing for any p < oo. To see that condition (6) fails it is enough to consider finite orthonormal sets. The canonical example of a pabsolutely summing map is given as fol lows: Let /.L be a probability Borel measure on a compact space K. Let 4.
1r00 (T) I T I . +
lii.F Absolutely Summing And Related Operators §5.
idp: C(K) + Lp(K, f..L ) be the formal identity. Then for have Trp(i dp) = 1 . Simply we have
�
(:�� f; l /; (k)IP)p { ( t, I j l ) · n
= sup
201 1�
p < oo
.l
/;dv ' ; ' v
E M(K),
I
we
(8) v ii = 1
}
so Trp(idp) � 1 , but taking the one element family consisting of a con stant function we see that Trp(idp) = 1 . A slight but useful variation of this example is the map f �+ f g defined as a map from L oo (f..L ) into Lp(f..L ) (clearly g E Lp(f..L ) ). Here we do not assume that f..L is a proba bility measure; it can be arbitrary. The same calculation as in (8) gives 'lrp( / 1+ f . g ) =
I YI p·
5. The following proposition describes some formal but useful proper ties of pabsolutely summing operators.
I
I
Proposition. ( a) lE T E IIp(X , Y), 0 < p < oo and X1 c X is a closed subspace then T X 1 E Ilp(Xt . Y) and 7rp( T X t ) � 7rp(T) . (b ) lE T E IIp(X, Y), 0 < p < oo and Y1 C Y is a closed subspace and T (X) C Y1 then T E Ilp(X, Yt ) and the norms of T in both spaces are the same. ( c ) If ( X. J'YEr is a net of subspaces of X directed by inclusion such that U "Y H X"Y is dense in X and T: X + Y then Trp (T) = sup"Y Trp (TIX"Y ) for 0 < p � oo . ..
Parts ( a) and ( b ) are obvious from the definition. Part (c ) requires a simple approximation argument (see II.E. 12) and is omitted.a
Proof:
Now we will give some very important examples of psumming maps. Actually these are one operator acting between different spaces. Later on we will call this operator the Paley operator. 6.
Proposition. For f
E L 1 (Y)
let P(f) = ( / (2 n
) ':=t ·
( a) P: A + £2 is 1absolutely summing and onto. ( b ) P : C(Y) + £2 is pabsolutely summing for 1 < p < oo and onto.
202
III.F Absolutely Summing And Related Operators § 7.
(c) P: C(Y) + co is 1absolutely summing. It is clear that P is continuous in all the cases (a) , (b) and (c) . That P is onto in cases (a) and (b) follows directly from Theorem III.E.9. In order to see (a) let us factor
Proof:
where P1 is the Paley projection, i.e. the operator P acting on H1 (Y) . It follows from Paley's inequality I.B.24 that P1 is continuous. We see from Proposition 5 (a) , (b) and from 4 that id is !absolutely summing so P is also (see ( 4)). To see (b) we consider the factorization
where Pl ( L�: J(n)e inB ) = ( } (2 n ))':= l · The operator P1 is bounded by the remark after Proposition III. A. 7 so P is pabsolutely summing by (4) and 4. For (c) we use the factorization a
The above proposition easily yields the fundamental theorem of 7. Grothendieck.
Theorem. (Grothendieck) . Every operator T: L1 (/.L) + H, where H is a Hilbert space, is 1absolutely summing.
Remark. It follows from the closed graph theorem or from the proof given below that there exists a constant K such that 1r1 (T) � KIITII for all T: £1 (t.L ) + H. The smallest such constant is called the Grothendieck constant and denoted by Kc .
III.F Absolutely Summing And Related Operators §8.
Let us start with an operator T: 6 ( a) we have a commutative diagram Proof:
i1 i2. +
203
Using Proposition
where
Theorem. operator T:
Let 0
X + Y.
<
p <
oo .
The following are equivalent for an
( a) T is pabsolutely summing with 1rp(T)
C. (b) There exist a Borel probability measure JL on (Bx· , a(X* , X)) �
a constant C such that
I I Tx ll � C
( Lx. l x(x* ) I PdJL(x * ) )
Moreover if 1 � p <
oo
.! p
for
and
x E X.
(9)
conditions (a) and (b) are equivalent to
(c ) for every (equivalently for some) isometric embedding i: X + C(K) there exists a Borel probability measure JL on K and a constant C such that
II Txll
�
c ([ l i(x) I PdJL) �
for
X
E X.
( 10)
204
III.F Absolutely Summing And Related Operators §8.
Let us rephrase the condition {10) (or {9)) . Denote by Xp c Lp(K, f..L ) the closure of i { X) in Lp(K, f..L ) . Then {10) means that T induces a continuous linear operator T: Xp + Y. Thus we have the commutative diagram
X i! i{X) !
C(K)
T
+ +
id
+
id
y
lT Xp !
{10')
Lp (K, f..L )
The fact that {b) and (c) imply (a) follows directly from{10'). One simply has to check that 7rp{id) < oo . For p � 1 this was done in {8) . For K = Bx· the same argument works also for p < 1. We will prove that (a) implies (b) and (c) . Let us take any isometric embedding i: X + C(K). Let V c C(K) be the set of all functions �(k) of the form
Proof:
�(k) = CP L l i( x;) (k) I P  L II Tx; II P for some finite (x; ) C X. j
j
The set V is a convex cone in C(K) . Let c  be the open, convex cone of all negative functions in C(K). If K = Bx· then {6) yields V n c  = 0 for 0 < p < oo. For other K's and p � 1 we get
=
{ { IJ {
}
L l x * (x; ) I P: x * E X * , ll x * ll � 1 P sup L i(x; )df..L = f..L E M(K) , llfLII � 1
sup
I
= sup L l i {x; ){k) I P: k E K
}
}
since cp(f..L ) = L; I J i( x;)df..L I P is a a(M(K), C(K) )continuous, convex function on BM ( K ) • so it attains its maximum on an extreme point. Thus V n c  = 0 in this case also. By the HahnBanach (see I.A. l l) and the Riesz (see I.B. ll) theorems there exists a measure f..L on K with ll fL II = 1 such that
and
L �( k ) df..L � O
[ f(k)df..L � 0
for all � E V
{11)
for all f E c  .
{12)
205
III.F Absolutely Summing And Related Operators §8.
Since c  consists of all negative functions ( 12 ) implies that J..L is a prob ability measure. Taking ¢ corresponding to the oneelement family { x} we get from (11)
II Tx ii P � CP
I l i(x) I PdJ..L
for all x E X.
• This shows that (a) implies (b) and (c). In general we have almost no control on the measure J..L · Let us however consider the following special situation. Let G be a compact, abelian group with the Haar measure m. Assume that G acts as a group of homeomorphisms of a compact space K, i.e. for each g E G we have a homeomorphism i9 of K such that i9 ·ih = ig·h · For a function f E C(K) and g E G we define 19 / (k) = f(i9 (k)). Assume that X c C(K) is a closed invariant subspace, i.e. 19 (X) c X for all g E G. Assume that T: X Y is a pabsolutely summing operator, 1 � p � oo such that II TI9x ll = II Tx ll for all g E G and x E X. Then there exists a measure J..L on K such that

II Tx ll � Trp(T)
1
( I ) ;; and i9 (J..L) = J..L for all g E G. l xi PdJ..L
(13)
Let ji, be a probability measure on K given by Theorem 8. We have
Ia II TI9x ii Pdm(g) � Trp(T)P Ia L ix (i9(k)) I Pdji,(k)dm( g) = Trp(T)P L i x(k) I PdJ..L ( k)
II Tx ii P =
where for A c K we have J..L ( A) = fa ji,(i9(A))dm( g ), so (13) follows. Let us also note that (a) does not imply (c) if p < 1. To see this let us take the operator (really a functional) T: C(O, 1] R. defined by Tf = J; f(s)ds. Clearly it is psumming for every p > 0 with 7rp(T) = 1 . Suppose that there is a probability measure J..L on (0, 1 ] satisfying (10) , so we have

1 1 1 J(s)ds r � c J l f(s) I PdJ..L ( s)
for all 1 E c [o, 1] .
(14)
One checks that if ( 14) holds for some measure J..L it also holds for the part of it which is absolutely continuous with respect to the Lebesgue
206
III.F Absolutely Summing And Related Operators §9.
measure, f..Lc · But f..Lc is a nonzero, nonatomic measure and (14) says that T induces a nonzero linear functional on Lp( [O, 1] , f..Lc ) · Since p < 1 this is impossible; see I.B.4. From Pietsch's theorem we can derive some properties of 9. pabsolutely summing operators. Corollary. (a) If O < p < q < oo then llp(X, Y) c ll q ( X, Y) for all Banach spaces X, Y. Also 7rp(· ) 2: 7rq (·). (b) Every pabsolutely summing operator 0 < p < oo is weakly compact and maps relatively weakly compact sets onto normprecomp act sets. (c) If X is a subspace of X1 and T E flp (X, Y) with p 2: 2, then T extends to an operator T1 E fl 2 ( XI , Y).
(a) and (b) follow directly from (10') and the fact that Xp is qabsolutely summing and satisfies (b) . For (c) ob serve that by (a) we can assume p = 2. Then use (10') and the fact that every subspace in a Hilbert space is complemented. Proof:
id: i(x)
+
All this may seem rather trivial and quite abstract. In order to convince the reader that these are important and powerful concepts, before investigating them any further, let us give some applications of the results already obtained. 10.
Theorem. Let X be a complemented subspace of L 1 (J.L) . Assume also that ( xn, x�)�= l is a normalized unconditional basis in X. Then there exists a constant C such that L::= l l x�(x) l :::; C ll x ll for all x E X, so (xn)�= l is equivalent to the unit vector basis in £1 . Proof: Since X has cotype 2 (see III.A.2123) or by the Orlicz theorem (see II.D.6) we get that the map T: X + £2 defined by T(x) = (x� (x))�= l is continuous. Let P: L 1 (J.L)�X be a projection. By Theorem 7 the operator TP is !absolutely summing so T = TP I X is also !absolutely summing. Thus for every finite sequence of scalars (an);{= 1 we have
N N L l an l = L II T(anXn) ll n= l n= l = 1r 1 (T) �up
:S
N 1r 1 ( T ) sup L l x * (anxn) l ll x * ll:9 n = l
l
sup x *
ll x * ll 9 lc: n l = l
( nt= l Enanxn ) I
III.F Absolutely Summing And Related Operators § 1 1 .
207 a
Let us note some special instances of this theorem (a) Since £1 [0, 1] is not isomorphic to £ 1 (see III.A.5 and III.A.7) we get a new proof that £ 1 [0, 1] does not have an unconditional basis (see II.D.lO) . (b) The spaces Lp [O, 1] and fp, 1 < p :::; 2, are not isomorphic to complemented subspaces of L 1 [0, 1] , although they are isomorphic to subspaces of L1 [0, 1] (see Notes and remarks to III. A) . The same conclusion can be easily derived from III.D.34(b) (see Exercise III.D.20) . Note however the important difference between these two arguments. Using III.D.34(b) we get no information about the complementation of e; in £ 1 [0, 1] . A careful reading of the above proof gives that if P is a projection from L 1 (f..l ) onto a subspace isometric to e; then II P II :::: Cn 1  i , 1 :::; p :::; 2. This reflects the difference between the global, infinite dimensional approach and local, more quantitative one. (c) £ 1 has only one unconditional basis in the sense that every normalized unconditional basis in £ 1 is equivalent to the unit vector basis. Let Wf (lf2 ), l 1 , denote the Sobolev space (see I.B.30) . 2: By the Sobolev embedding Theorem I.B.30 the identity operator 1 2 2 id: Wf (1r ) + wJ (1f ) is continuous. Note that wJ  1 (1f2 ) is a Hilbert space. We want to show that it is not 2absolutely summing. Since WJ (1r2 ) c Wf (lf2 ) it suffices to show that id: WJ (1r2 ) + wJ  1 (1f2 ) is not 2absolutely summing. Let us put fn,m (O, r.p) = (m + n)  1 ei nll ei m'P, n, m > 0. We have 11.
L
O < m +n <::: N
ll id( fn ,m ) ll 2
1  1 j 1  1 j 2 2: O n m) l < n <:;N N 1 1 1 j 2 = L: r  21 L 'L ni m l  n+ m = r j =O r= O N ( 1 ?: Cl L r  21 . r . r 2 l  ) = O r
� � [� : J ] [ N
= Ct L 1
r =1
r
( 1 5)
208
III.F Absolutely Summing And Related Operators § 12. =
C1 log N .
On the other hand 11 /n,m ll wJ (P ) :5 C and Un,m ) is orthogonal in W4 ('1I'2 ) so we get (16) Clearly (15) and (16) show that i d is not 2absolutely summing. Comparing the above argument with Theorem 7 we infer that Wf ('1I'2 ) is not isomorphic to any L1 ( J.L) space. Averaging over k  2 variables we see that Wf ('1I'2 ) is complemented in Wf ('1I'k ) for l � 1 and k � 2. Thus we obtain Theorem. For l � 1 and k � 2 the Sobolev space W1 ('1I'k ) is not a isomorphic to a complemented subspace of any £1 ( J.L )space.
We also want to show how these general concepts apply to some concrete problems in harmonic analysis. We will see more of this later. Let G be a compact abelian group with Haar measure m. 12.
Proposition. An invariant operator T: C ( G )  C(G) is 1absolutely summing if and only if T f = f * h for some h E L00 (G) . We have 1T 1 ( T ) = ll h lloo ·
Suppose h E L00 (G) . As is well known and easy to check ( see I.B. 13) f * h E C(G) for all f E L 1 (G) . Thus every T: C(G)  C(G) which is of the form Tf = f * h for some h E L00 (G) is !absolutely sum ming with 7r 1 (T) :5 ll h ll oo · Conversely if T: C(G)  C(G) is translation invariant and 11"1 (T) = 1, from Theorem 8 and (13) we get
Proof:
II T /IIoo :5
L l f(g) l dm(g)
for every f E C(G) .
( 17 )
Since T is translation invariant, there exists a measure J.L such that T f = f * J.L · From ( 17) we infer that II / * J.L IIoo :5 11/lh for all f E L1 ( G ) . An easy limiting argument shows that ll v * J.L IIoo :5 ll v ll l for all v E M(G) . Taking as v the Dirac measure at the neutral element of G we get that J.L = hdm for some h E Leo ( G) , with ll h ll oo :5 1. a This proposition can yield some very concrete results.
III.F Absolutely Summing And Related Operators §13.
209
13 Theorem. Suppose A C N is a set of cardinality k. Given c there exists a trigonometric polynomial f such that
>0
}(f)
=
1 for all 11/ lh $ 1 + c,
.
f E A,
( �c ) 2k for some constant c.
l { f E N: / (f) =f. O} l $
(18) ( 19 ) (20)
In operatortheoretical terms we are looking for a rotation invariant operator T: C(1l') + C(1l') such that
Proof:
T I E = id where E = span{ einB } ne A, II T II $ 1 + c, rank T $
( � fk .
(18') (19') (20')
Let us fix a number 8 such that 0 < 8 < 1. From II.E.13 there exist a natural number N < ( �) 2 k and an embedding u: E + £! with !l u ll · ll u  1 11 $ 1 + 8. Let ii: C(1l') + £! be an extension of u with !l u ll = ll ii ll (see III.B.1) and let v : £! + C(T) be an extension of u 1 with llvll $ (1 + 8) ll u  1 11 . Observe that by II.E.5(c) there exists a subspace F with E c F c C (1l') such that d(F, e'!m F) $ 1 + 8, so such an extension exists (see also III.E.14) . Let us put T1 = vii: C(T) + C(1l') . Clearly T1 I E = id and
11'1 (TI) $ llvll · ll ii ll11'1 (id: e! + e! ) $ llv ll · ll ii ll · N $ (1 + 8) 2 N. We now define
1 T2 = 2 71'
1271" l9 T1 L 8 d(} 0
where !9 is the rotation by the angle invariant and satisfies
0.
The operator T2 is rotation
T2 I E = id, II T2 11 $ II T1 11 $ (1 + 8) 2 , 11'1 (T2 ) $ 11'1 (T1 ) $ (1 + 8) 2 N. From Proposition 12 we know that T2 is convolution with a function h which satisfies
h(i) = 1 for f E A, lihll l $ ( 1 + 8) 2 , ll hlloo $ (1 + 8)2 N.
(18") ( 19" ) (20") certain (18"') (19"') (20"' )
210
III.F Absolutely Summing And Related Operators § 14.
Observe that (19"') and (20"') give ll h ll 2 :5 (1 + 8 ) 2 ./N . We define g = h * h * h and put
f (O ) = s: l § (s) I > N  4
We have
11/lh :5 II Y II l + II Y  /lh :5 (1 + 8)6 +
s : l § ( s) I < N4
l .§ (s) l
:5 (1 + 8)6 + N � L l h(s) l 2 = (1 + 8)6 + N  � ll h ll � :5 (1 + 8)6 + (1 + 8 )4 N  i and also
l {k E N: j ( k ) =/: O} l = l {k: l h(k) l > N  � }1 :5 ll h ii � N i :5 ( 1 + 8 )4 N4• Since j ( k) = 1 for k E A we see that for appropriately chosen polynomial f satisfies (18)(20) .
8 the a
Clearly the same argument works for any compact abelian group. 14. Now we will present a different form of Grothendieck's theorem, the socalled Grothendieck inequality. We will derive it from Theorem 7. Actually it is also easy to derive Theorem 7 from the Grothendieck inequality. Theorem. (Grothendieck's inequality) . Let ( an,m)� m = l be a fi nite or infinite matrix such that for every two sequences of scalars (an)�= l and (.Bm );:;= l we have
I n,mt= l an,mGn.Bn l :5 s�p l an l · s�p I.Bn l ·
(21)
Then for any two sequences (hn) �= l and (km )�= l in an arbitrary Hilbert space H we have
I n,mt= l an,m ( hn, km) i = Ka s�p ll hn ll · s�p llkm ll
(22)
III.F Absolutely Summing And Related Operators § 1 5.
21 1
where Kc is the Grothendieck constant.
Let us define an operator T: if + H by T(en ) = hn for 1 , 2, . . . , N. Clearly II T II = supn ll hn ll · For every n = 1 , 2, . . . , N let us put Zn = ( an,m )�= l E £f . From Theorem 7 and ( 21 ) we get Proof:
n =
n= l
{ t, l (v , Zn ) l : v £! , llvlloo :5 1 } = Kc li T I sup { t. l t, f3man,m I : lf3m l :5 1 for m = 1 , 2, . . . , N } :5 Kc ii T II sup { t O. nf3man,m : lo. n l :5 1, lf3m l :5 1 } n,m = l :5 Kc ii T II sup
E
(23)
= Kc sup ll hn ll ·
n
On t�e other hand
t, II Tzn ll t, I t, anmhm l h kn, � sup � kn ll l t, ( t, anm m) l = an,m ( kn, hm) l · sup � kn ll l n � l =
Comparing (23) and (24) we get (22) .
(24)
a
Note that (21) means simply that the matrix ( an,m ) defines a norm one operator from £! into if. The Hilbert space figuring so prominently in Grothendieck's the orem, it is not surprising that it has important applications to Hilbert space operators. We will discuss some of them presently. An operator T: H + H is said to be power bounded if SUPn;::: o II Tn ll < 00. Clearly every operator of the form T = VTt v l 15.
212
III. F Absolutely Summing And Related Operators § 1 5.
where II Tt ll � 1 and V is an isomorphism of H, is power bounded. Actually the classical von Neumann inequality proved in von Neumann [1951] (see Exercise III.E.8 or Exercise III.H.19 or NagyFoias [1967] ) gives that for every such T and every polynomial cp
2
ll cp(T) II � C sup l cp(z) l .
( 5)
lzl9
2
An operator satisfying ( 5 ) is said to be polynomially bounded. We will discuss those notions a little. Let us start with some examples.
For every sequence c = (cn)� with = ± 1 if n = 2k and Cn = 0 otherwise, there exists an operator=OTc : H Cn H such that sup sup ur; I < ( 26) c n there are e , E H such that (T;e, "'} = Cn for n = 0, 1, 2, . . . . ( 27)
Proposition.
oo ,
'f/
Let us �bserve that not every operator Tc is polynomially bounded. For a polynomial cp we have
� � c,O(n) (r;e, .,.,} l � � c,O(n)cnl � ��>2kc,0 (2k)l·
ll cp(Tc ) ll ll e ll ll"'ll � l (cp(Tc)e, .,.,} l = =
(28)
=
Observe that Theorem III.E.9 gives polynomials cp with ll cp lloo � 1 but as big as we please.
I Lk C2kc,0 (2k) I
Proof of the proposition.
{ cno m b(n, m) = { cno m
a(n, m) = and
Let us put +
if 0 <_n <_m, if n > m � 0,
+
if n > m � 0, if m � n � 0. One easily checks that a(n, m) is a matrix of a bounded operator A on £1 while b(n, m) is a matrix of a bounded operator B on £00 with norms bounded by a constant independent of On sequences (en )�= O we formally define operators
c.
Sk ((�n) �= o ) = (�k , �k+b · · · ) and SA; ((�n )�=o) = (O, . . . , O , �o. � l , · · · ) · ........_.. k t i mes
213
III.F Absolutely Summing And Related Operators § 1 6.
One checks that for finite sequences (en)�= O one has ASZ  Sk A = Sk B  BS;'�fVk . From these relations we infer that Vk acts on both £ 1 and €00 with norms uniformly bounded in k. Thus by interpolation (see I.B.6) Vk acts also on f2 and II Vk ll :5 C. We now define H as a completion of £ 1 EB £ 1 with respect to the scalar product
where ( ·, · ) is the usual scalar product. We define Tc: H  H by Tc(f, g) = (Sif, S1g ). Clearly r: (f, g) = (Sk f, Sk g ). Since
li T: ( !, g) II � = II Sk f ll� + II ASk f + Sk g ll� :5 II/II � + ( II Vk/ 112 + Il Sk A! + SkYII 2 ) 2 :5 11/11� + (C II /11 2 + II A J + Yll2 ) 2 :5 C ll (f, g) ll � we get ( 26 ) With e = (eo, 0) and TJ = (0, eo) we get for n = 0, 1 , 2, . . . .
((T0 e , TJ)) = (( (en, O), (O, eo) )) = ( A(en), eo ) = a(n, O ) = cn . a
So (27) also holds.
Since the estimate (25) does not hold for every powerbounded operator one has to seek other estimates for ll cp(T) II · For a polynomial cp(z) = L n <:: o cp(n)z n we put 16.
a (cp) = inf{ l l l (an ,m ) n,m<:: o l l l :
L anm = cp(k), k = 0, 1, 2, . . . } n+m=k
where
l l l (an, m ) n,m <:: o l l l = sup
{I n,m L cnTJmanm J : l cn l = 1 , ITJm l = 1 } ·
Theorem. Let T: H  H be such that every polynomial cp ( z) we have
supn > O II Tn ll = C.
ll cp (T) II :5 C2 Ka a ( cp ) .
Then for
(29)
214
III.F Absolutely Summing And Related Operators § 1 7.
For arbitrary x, y E H with ll x ll = IIYII ( an,m )n ,m2: 0 with l: n + m= k anm = cp(k) we have ( cp(T)x, y ) = (L cp(k)Tk x, y )
Proof:
=
= 1 and arbitrtary
L cp(k) ( Tk x, y) = L anm ( Tn Tm x, y ) n,m k
(30)
n,m Since II Tm x ll � C and II (T n )* yll � C for all m and n and all x, y as above the Grothendieck inequality (Theorem 14) applied to (30) yields
l ( cp(T) x, y) i
�
C2 Kc l l i (an,m) l l l ·
(31)
Since x, y are arbitrary vectors with ll x ll = IIYII = 1 and ( an,m ) is also arbitrary, (31) implies (29). a 17. In order to apply Theorem 16 we need some method of computing or estimating o:(cp). This is generally a nontrivial matter. The following lemma will nevertheless give some interesting information. Lemma. Let wn(t) for n = 1 , 2, . . . be the piecewise linear function determined by the conditions wn (t) = 0 for t � 2 n  l and t 2: 2 n + l and Wn ( 2 n ) = 1 . Let us define a sequence of polynomials (Wn (z))n;2: 0 as follows. We put W0 (z) = 1 + z and for n 2: 1 we put Wn (z) l: k2: o wn (k)z k . Then for every polynomial cp(z) we have
o:(cp)
�
L II 'P * Wn lloo · n2: 0
(32)
For every polynomial cp we have cp = l: n > O cp * Wn  Since o:( · ) is a rotation invariant norm on polynomials one gets o:(f * cp) � II J II Io:(cp). As is well known, II Wn lh � 2 for n = 0, 1 , 2, . . . (this can be seen by writing Wn as a sum of shifted Fejer kernels; see I.B. 16) , so it is enough to show that for cp such that supp cp C [2 n  l , 2n + 1 ] one has o:(cp) � C II 'P IIoo· Given such a cp let us define 'ljJ = 2: k  1 cp(k)z k . We can write 'ljJ = z 2 "1 · [ (cp · z  2 "  1 ) *gn] where gn = I: t:'_ 00 (2 n  l + l k l )  l e i kll. Comparing Fourier coefficients we see that Proof:
III.F Absolutely Summing And Related Operators §18.
215
where :Fk is a Fej er kernel. This shows that 9n � 0 and
L00 (
1 1_ 2 1 ll Yn Il l < k 2n  l + k 1 + 2n  l + k + 1  2n  l + k )  _ n 2 l k= l so 11 1/J II oo � II �P II oo . IIYn ll l � 2 n + l ll cp lloo· Let ( an,m ) n,m�O be the ma trix given by an,m = �( n + m ) . It is known ( see Exercise III.E.5 or Duren [1970] ) that this matrix acts on l2 with norm � 11 1/J II oo � 2  n+ l ll cp ll oo · Observe also that a (cp) � ll (a n,m ): loo + l d . Since actually ( an,m ): .e� +1 + .er +1 we have the factorization _
so
a (cp) � ll (an,m): loo + l1 ll +1 2" +1 2" +1 2" +1 � ll id: l2" oo + l2 ll · ll (an ,m ): l2 + l2 ll · ll id: l2 + ll II a � 2 II �P II oo · The estimate for IIYn ll1 is a repetition of a well known theorem in the theory of Fourier series ( see Katznelson [1968] Th. 1.4. 1 ) . Remark.
1 8 Corollary. If cp is a polynomial of degree n and T is a power bounded operator on a Hilbert space then ll cp(T) II � C·log ( n+2 ) · 11 �P II oo · Proof:
This follows directly from ( 29 ) and (32).
•
It seems to be unknown if this is the best estimate. The opera tors constructed in Proposition 15 can be used to show that at least yllog ( n + 2) is needed. There is also a direct proof of Corollary 18. Using the canonical factorization I.B.23 we write the Dirichlet kernel E�= O ei k B as a product of two functions f = E � o a k ei k B and g = E�= O f3k e i k B such that
Then we can write
216
III.F Absolutely Summing And Related Operators § 1 9.
so for every x, y E H with ll x ll
=
IIYII = 1 we have
Representing the Hilbert space H as £ 2 (!1, J.t ) and using the orthonor mality of the sequence (ei k t)t:=_ oo in L 2 (1l') we get
and analogously
Putting this together we get
i (p (T)x, Y } l
(
00
1
1
( I: 00
:5 IIPIIoo L l a k l 2 r . I ,Bk l 2 r . m:x II Tk x ll m:x ii (Tk ) * y ll = = O O k k :5 C IIPIIoo log(n + 2) . 19. We hope that the previous discussion justifies the opinion that a more detailed study of pabsolutely summing operators is interesting and important. Before we resume such a study we would like to introduce two new families of Banach operator ideals. An operator T: X  Y (X, Y arbitrary Banach spaces) is pnuclear, 1 :5 p :5 oo, if it can be written in the form
T (x)
00
=
L: xj (x)y3
j =l
(33)
217
III.F Absolutely Summing And Related Operators §20.
with 00
( jL= l
)
�
" ll xj ii P
00
sup
y* E Y* , IIY" II 9
1 1 where  + ' = 1. p
(I: j= l
)
�
' iy * ( yj ) I P' "
< oo
(34)
p
Obvious modifications are made in (34) for p = 1 and p = oo . We define the pnuclear norm np (T) as the infimum of the quantities (34) over all representations (33). The class of all pnuclear operators from X into Y is denoted by Np(X, Y) . An equivalent definition of pnuclear operators runs as follows: an operator T: X + Y is pnuclear if it admits a factorization
X u!
�
y
i
(35)
v
where u, v are continuous operators and A is a diagonal operator, i.e. A(ej ) = (83e3 ) for some (83 ) E lp. Given a representation (33) satisfying (34) we define u(x) = 1 and (83) j;:1 = ( ll xj ll ) j;:1 and v(e3) = L.3 e3 y3 . The dia gram (35) then commutes. Conversely given (35) the above formulas easily yield (33) and (34) . Note also that
(j1��iD :
np(T) = inf { ll u ll · llvii · II A II : u, v , A satisfy (35) } .
We can interpret A: £00 + lp as id: £00 + Lp ( N, J.L) with the measure J.L defined by J.L(S) = Lj e S 1 83 I P. This shows that Np ( X, Y) C Ilp (X, Y) for 1 :5 p :5 oo with 7rp(·) :5 np (·). It is also pretty clear from (35) that Np (X, Y) is an operator ideal. Actually the proof of the following is routine. 20 Theorem. The pnuclear operators with pnuclear norm np ( · ) :5 oo form a Banach operator ideal.
p
,
1 :5 a
The most important case is that of the 1nuclear operators, simply called nuclear or trace class operators. Note also that N00 (X, Y) is simply the class of operators admitting a factorization through £00 • Thus if E is a (finite dimensional) Banach space and idE denotes the identity
218
Ill.F Absolutely Summing And Related Operators §21.
operator on E, then n00(i dE) = 'Yoo (E) , where 'Yoo (E) is as defined and studied in III.B.35. 21. We say that an operator T: X + Y is pintegral, 1 :::; p :::; write T E Ip(X, Y) if it admits a factorization
and
oo ,
(36) where f..L is a probability measure on a compact space K and a, (3 are continuous linear operators. We define the pintegral norm
ip ( T ) = inf{ ll a ll · ll f3 11 : there exists factorization (36) } . We have the following routine theorem whose proof is omitted. Theorem. The pintegral operators p :::; oo, form a Banach operator ideal. 22.
Ip(X, Y)
with norm
ip (T ) , 1 :::; a
A comparison between (36) , (19) and (10') gives
Proposition. For all Banach spaces X , Y and numbers p with 1 :::; p :::; oo we have
with the corresponding inequality for norms, i.e. a
Let us also note that the Pietsch theorem (Theorem 8) implies that for every Y and every p with 1 :::; p :::; oo we have Ip(C(K) , Y) = ITp(C(K), Y) with equality of norms. Note also that it follows from the Pietsch theorem (see 10' ) and the fact that every subspace of a Hilbert space is 1complemented, that for all Banach spaces X, Y we have J2 (X, Y) = ll 2 ( X, Y). With equality of norms. 23 Theorem. Let T: Assume also that i: X
X +
+
Y be a pintegral operator, 1 :::; p :::;
C(S)
oo .
is an isomorphic embedding. Then
T
III.F Absolutely Summing And Related Operators §24.
219
admits a factorization
where v is a probability measure on S and B is a continuous linear operator with II BII � ip(T) · IIi II · ll i  1 11 + E .
The proof is best illustrated by the following diagram where (3 have the same meaning as in (36) . K, J..L , a,
Proof:
i(X) ' X
!
id
1
T
___.
____
 . � /
C(S)
a
a
L oo (K, J..L )
id
1
Y
fi
Lp(K, J..L ) ·
Lp(S, v )
On' i(X) C C(S) we consider the operator ai  1 : i(X) + L00 (K, J..L ) . Using III.B.2 we find an extension ii : C(S) L00 (K, p,) with ll ii ll = ll a · i 1 11 � ll a ll ll i  1 11 · Since i da is pabsolutely summing with 11'p (i dii) � ll ii ll the Pietsch factorization theorem (Theorem 8) gives a probability measure v on S and an operator A, with IIAII � ll ii ll · The desired +
a
factorization is given by B = (3A.
Let X be a finite dimensional Banach space and let Then ip(T) = np(T) for 1 � p � oo .
24 Corollary.
T: X + Y.
It is enough to show that np(T) � ip(T) (see Proposition 22) . Given e > 0 we get from II.E. 13 an isomorphic embedding i: X + e;;, with II i ll . ll i 1 11 � 1 + E . Since e;;, = C(S) where s = { 1, 2, . . . ' n } we apply Theorem 23 to get a measure v. Since id: C(S) + Lp(S, v ) is the same as �: e;;, + e: with � = ( 6n ) ;;' 1 where 6n = v ( { n }) � we get the pnuclear factorization. a Proof:
=
25 Corollary.
If T: X + Y
with
X and Y finite dimensional,
then
220
III.F Absolutely Summing And Related Operators §26.
From remarks made after Proposition 22 we know that 11'2 ( · ) =
h ( · ) (always) so the claim follows from Corollary 24. Proof:
a
The reader should consult the exercises to find examples showing that the above classes are different in general. One of the reasons that � nuclear and �integral operators are important is that they are connected with �absolutely summing operators via duality. Before we proceed we have to discuss duality as applied to Banach operator ideals. 26. Given a Banach operator ideal I(X, Y) we want to describe the dual space I(X, Y) * . In general this subject is quite involved; the theory of general tensor products and the approximation property play impor tant roles. We will discuss it only for finite dimensional Banach spaces X and Y. Despite this restriction, the results can be applied in the study of �absolutely summing operators on infinitive dimensional spaces. This follows from Proposition 5. If T: X > X (X finite dimensional) is a linear operator then T has a representation (nonunique of course) in the form T(x) = Lj xj (x)xi . The trace of T is defined as tr(T) = Lj xj (xi ) · As is well known and easily checked this definition is correct, i.e. it does not depend on the particular representation of T. Obviously the trace is a linear functional on L(X, X). Given T: X > Y and S: Y > X we see that tr(ST) = tr(TS) . For every operator S: Y > X the formula r.ps (T) = tr(TS) , for T E L(X, Y ) , defines a linear functional on L ( X, Y). Counting dimensions we realize that we can identify L(X, Y)* with L(Y, X) if the duality is given by the trace, i.e. (S, T ) = tr(ST) . All this is elementary linear algebra and can be found in most textbooks on the subject. If we have a norm i on L(X, Y ) then the norm i* on L(Y, X ) is dual to i (with respect to trace duality) if for every T E L(X, Y ) we have
i(T) = sup{tr(ST) : i * (S) � 1}. 27.
(37)
The following important theorem identifies 71'; .
Theorem. Let X and Y be finite dimensional Banach spaces. Then ( IIp ( X, Y ) , 11'p )* = (Np' (Y, X), np' ) , 1 � p � oo , with the trace duality.
Let us fix T E L(X, Y ) . For S: Y > X with the representation S(y) = Lj yj (y) · xi we have
Proof:
j
j
221
III.F Absolutely Summing And Related Operators §28.
(38) ( � i!Yj ll p' ) ? ( � I! Txi ii P) ; :5 ( � IIYj li p' ) 11'p (T) sup { ( � l x * (xi W ) "P : x * E X* , ll x * ll :5 1 } · :5
J
J
1
pr
J
1
J
Since (38) holds for every representation of S we get l tr(TS) I :5 np' (S) ·
11'p(T).
On the other hand let us fix X1 , , Xn in X such that Ej l x* (xj) I P 1 :5 1 for every x* E X* with ll x* I :5 1 and such that ( Ei II Txi l i P ) "P � (1  c:)11'p(T) . Let us choose yj E Y* such that IIYj ll = 1 and yj (Txi ) = li Txi II for j = 1, 2, . . . , n. Let us also fix numbers ai � 0, j1 = 1, 2, . . . , n such that Ei a� = 1 and Ei ai II Txi I = ( E i II Txi li P ) "P . We define S: Y + X by the formula S( y) = E;=l ai yj (y)xi . Since •
•
•
I
np' (S) :5 and
(� ) J
a�'
1
pr
·sup
{ ( � l x* (xi ) I P) J
;;1
}
: x * E X* , l! x * ll :5 1 :5 1
tr(TS) = L ai yj (Txi ) � (1  c:)11'p(T) j
we get the claim.
a
28. Our aim now is to establish the dual version of Grothendieck's theorem. Because of later applications in III.I we give a more abstract presentation than is really necessary.
Proposition. Let X be a finite dimensional Banach space. The fol lowing conditions are equivalent:
(a) for every T: X + £1 we have 11'2 (T) :5 C II T II ; (b) for every T: X* + £2 we have 11' 1 ( T) :5 CI !I T II · More precisely if (a) holds then C1 :5 Ka · C. If (b) holds then C :5 C1 . (a) =?(b) . Clearly we can restrict our attention to T: X* + i!f . From Theorem 27 we see that we have to estimate tr(T S) for every
Proof:
222
III.F Absolutely Summing And Related Operators §28.
00
i!f + X* with n (S) = 1. Using the standard approximation (see Proposition II.E. 12) we get the diagram
S:
s
�iM00 /fi
i!f
X*
T
X
T*
i!f
with II a ll · 11 .8 11 ::; 1 + c. Dualizing this diagram we get
iN2 a
s•
\iM/
iN2
�
1
From Corollary 25 and Theorem 7 we get using ( a )
l tr(TS) I = l tr(S*T* ) I ::; 1r2 ( a * )1r2 (,8 *T* ) ::; 7r2 (a * )7r2 (,8 * ) 11 T* II ::; Ka ll a * II C II .B* II · II T* I I ::; Ka C(1 + c) II T II ·
Since c was arbitrary we get ( b ) . ( b ) => ( a) . As before we use Theorem 27 so we have to look for tr(TS) where we have the following diagram:
with ll a ll · 11 .8 11 · I A ll ::; 1 + c. Dualizing this diagram we get
III.F Absolutely Summing And Related Operators §29.
223
Using ( b ) we get 1r1 (S*T*) :5 ll a* II II� * II II T* II 7r i ( .B * ) :5 C1 (1 + c) II T II · But S*T* is an operator on .e� so by remarks made after Proposition 22 we have 1r1 (S*T* ) = i 1 (S*T* ). Corollary 24 gives i 1 (S*T*) = n 1 (S*T*) so we have
l tr(TS) I = l tr(S*T* ) I :5 n 1 (S*T*) :5 C1 ( 1 + c) II T II · This gives ( a) .
•
29 Theorem. For any C(K)space (in particular for L oo (!£) ) and 1 :5
p
:5 2 we have
L(C(K) , Lp) = Ih (C(K) , Lp) with 1r2 (·) :5 Ka ll · II · From Proposition 5 and II.E.5 we infer that it is enough to show 1r2 (T) :5 Ka li T II for every T: .e� + lif , 1 :5 N, M < oo . The Grothendieck theorem ( Theorem 7) says that X = if satisfies ( b ) of Proposition 28 so we get
Proof:
1r2 (T) :5 Ka i i T II for every T: .e� + .efl.
(39)
If 1 < p :5 2 then lif is isometric to a subspace of LI [O, 1] ( see III.A. 16) . From II.E.12 and an easy approximation we see that we can assume a lif c .efl' so (39) and Proposition 5 give the theorem. 30. This dual version of the Grothendieck Theorem also has some nice applications. We will discuss some of these, connected with harmonic analysis. Let G be a compact, abelian group with dual group r. The following is a classical result of Orlicz, Paley and Sidon.
Let A = (>. r ) r er be a function on r. The map A(!) = b)) (>.r f r er maps C (G) into £1 (r) if and only if A E £2 (r) .
Theorem.
If A E £2 (r) then A acts not only from C (G) into £ 1 (r) but also from L 2 (G) into £1 (r) , simply because characters form a complete orthonormal system in L 2 (G). Conversely if A: C(G) + £ 1 (r) , then by Theorem 29 1r2 ( A ) :5 Ka ll A l . The Pietsch factorization theorem ( Theorem 8) and (13) give Proof:
!
II A /11 :5 Ka ll A l! ·
( i if(g) l2dm(g) )
1
2•
224
III.F Absolutely Summing And Related Operators §31 .
But this means that
(
L 1 X'Y I I ib ) l :5 Ka llAl! L 1] ( 7 ) 12 'YEr "Y Er
!
)2
a
31. Let us recall that a subset S C r is called a Sidon set if { 'Y h es C C (G) is equivalent to the unit vector basis in i 1 ( S ) . For S c r the symbol Cs (G) will denote span {'Y : 'Y E S} C C(G). The following
theorem shows that the Banach space structure of Cs (G) determines if S is Sidon or not. Theorem. If Cs(G)* is isomorphic to some Sidon set.
C( K )
space than S is a
Proof: Let us fix f E Cs (G) and let us consider the operator T1 : M(G) + C(G) given by the formula Tt (J.t) = f * J.t · We can factorize
this operator as follows:
M(G)
Tr
_.
Cs (G) � C(G)
•\ jr ___
M(G)/Is
where q is the quotient map and Is = {J.L E M(G): [l,('y) = 0 for 'Y E S} . Clearly M(G)/Is � Cs(G) * rv C( K ) and also the space Cs(G) can be considered as a subspace of Cs (G) ** rv M (K) so by Theorem 29 we get '11'2 (T) < oo. This implies '11'2 (Tt) < oo. But Grothendieck's theorem implies that also '11' 1 (Tt) < oo. The proof is completed by an application of the following. If an operator T: M(G) + C(G) given by T(J.t) = L: 'Y e r a'Y[l,('y)'Y is 1absolutely summing then L: 'Y e r l a'Y I < oo, and con
32 Lemma.
versely.
Let L c r be a finite set and let E c C(G. ) be such that d( E, i�) :5 1 + c and 'Y E E for every 'Y E L. Let P be a projection from C(G) onto E such that II P II :5 1 + c . Let Lf = span{'Y : 'Y E L} C L1 (G) c M(G) . For the operator S: Lf + E defined by S(J.t) = PT ( J.t ) Proof:
225
III.F Absolutely Summing And Related Operators §33.
we have 11" 1 (8) :5 (1 + c)11"1 (T) . For every sequence TJ = ( TJ he L with ,. I TJ,. I = 1 we define a map u71: E > L f by u11 (!) = ( 'L. ,. e L TJ,. f ('y ) · r} Since I 'E e L TJ,. J('y ) · 'Y II 1 :5 I 'E e L TJ,. J('y ) · 'Y II2 :5 ll/ll2 :5 11 /ll oo we get ,. ,. ll u11 ll :5 1 and thus also noo (u 11 ) :5 1 + c. From Theorem 27 we get
I I:>,.a,. l = 1 tr(u11 8) 1 :5 11"1 (8) · n00(u11 ) :5 (1
+
')'E L
Since this holds for every c > 0, every finite set L get 'E ,. e r l a,. l :5 11" 1 (T). The converse is obvious.
C
c) 2 7r1 (T).
r and every 17 we a
Now we return to the study of pabsolutely summing operators. One lesson to be learnt from our previous discussions is that it is very useful to have equalities of the form Ilp ( X, Y) = L ( X, Y). This is only rarely true so we try to look for equalities of the form Ilp( X, Y) = II q (X, Y). These are also useful ( see III.H. 12). Before we proceed let us state some easy but useful facts. 33.
Proposition.
( a) Let 1 :5 p :5
oo
and let T: X >
Y.
Then
11"p(T) = sup{7rp(T8): 8: I!� > X, 11 811 :5 1 , m = 1 , 2, . . . } .
(40)
( b ) If T: X > Y is an operator and 0 < p :5 oo and (0, J.L) is any measure space and f(w) is an Xvalued Bochner integrable function (see III.B.28) then
( J II Tf(w ) II PdJ.L(w) )
1
'P
:5 11"p(T)
sup
x• e x • , Ux* l l 9
( j l x* (f(w )) I PdJ.L(w) )
1
'P .
(41)
Proof: ( a) Clearly the quantity on the right hand side of (40) does not exceed 11"p(T) ( see (4) and Theorem 3). On the other hand let us take 1 XI . , Xm C X such that sup { ( 'L:;'= 1 lx* (xJ ) I P) 'P : llx* II :5 1 } = 1 and 1 ( 'L:;'= 1 II T(xj ) li P) 'P 2: 11"p(T)  c. Let us define S: I!� + X by S(e1 ) = x1 .
•
.
226
III.F Absolutely Summing And Related Operators §34.
for j = 1, 2, . . . , m. We have
li S I = sup = sup = sup
{I � I { t, { (�
ai xi :
t, l ai lp' 1 } �
ai x * (xi ) : ll x * ll
m
l x * (xi ) I P
)
.!
P :
� 1 and
ll x * ll
� 1
�
l ai l p'
�1
} = 1.
}
Also
(b) By a standard approximation argument it is enough to check ( 41) for step functions of the form :E;'!: 1 Xj �A; for disjoint sets Aj , j = 1 , 2, . . . , m. If for some j we have JL(Ai ) = oo then both sides of (34) are 1 infinite. If for all j we have JL( Ai ) < oo then we put Yi = JL( Ai ) "P xi for j = 1, 2, . . . , m and we see that (41) takes the form n
( j=l L II TYi ii P)
l. P
�
trp(T )
sup
llx II ::::;I ,x . E X •
n
•
(L: l x * (xi ) I P ) j=l
!. P
which is clearly true. 34.
II
We now return to the investigation of the identity Ilp (X , Y) =
IIq (X, Y). We will prove an important extrapolation type result. It is
analogous to Exercise III.A.2. We will use it in III.H and it also permits some improvements of our previous results. It is also the first result, besides the Pietsch factorization theorem, which deals with pabsolutely summing operators for p < 1. Theorem. Let X b e a Banach space and let 1 � for some number p, with 0 < p < q we have IIq (X, Y) =
Ilp (X, Y)
q
� 2. Suppose that
for all Banach spaces Y.
Then for all Banach spaces Y and for all numbers p such that 0 have IIp (X, Y) = Ilq (X, Y ) .
we
(42) < p < q
227
III.F Absolutely Summing And Related Operators §34.
(42)
First note that implies that there exists a constant C such that for every and for every operator we have Proof:
Y
T: X + Y
(43) (43) does not hold then there are Tn: X + Yn with 7rq(Tn) = 1 but 1rp(Tn) > 4n , n = 1,2,3,n . . . . Then the operator T: X + ( E� 1 Yn) 2 defined by T(x) = (2  Tn(x))� 1 is qabsolutely summing but not absolutely summing. Let IC denote ( Bx· , ( X * , X)) and let P denote the set of all prob ability measures on We will identify X with its canonical image in C(IC). Using Theorem 8 and (43) we see that (42) is equivalent to If
rr
a
/C.
for every .>. E P there exists .>.1 E P such that
( l l x (x* Wd>.(x*))
1
9
5 C
1
( l l x (x*) I Pd>.1 (x*)) ;; for x E X. (44)
Analogously we see that in order to show the theorem it is enough to show that for every r < r < p and for every .>. E P there exists E P and Cr > 0 such that
1,
( l l x (x* Wd>.(x* ))
1
9
5
J.L
1
Cr
( l l x (x* )r dJ.L (x * )) ;: for all x E X. (45)
(44)
Note that if .>. = .>.1 in then the Holder inequality ( see Exercise for all r, O < r < q with = .>.. In general however III.A.2 ) gives .>. =I= >.1 . We put .>. = and inductively applying we get a sequence .>. 2 , . . . in P such that for all E
(45)
>.o, >.I.
We fix
8,
0<
8
< 1
>.o
x X,
J.L
(44)
such that � = � + (l � O) and put a n = 2 n  l . For
228
III.F Absolutely Summing And Related Operators §35.
x E X we have n 2': 0
< C L an ll x ii L ( >,n +I ) · ll x ll t1> n +tl n 2': 0 l  (J (J < L an ll x ii Lr ( A n +t l L an ll xii L .(>. n+ t l n 2': 0 n 2': 0
c(
)(
)
Thus Since
r
:::;
1 we get
From (46) and (47) we obtain
where /l (45) .
L n ;::: o 2  r (n+ l ) An. Normalizing
/l appropriately we get a
35 Corollary. {GrothendieckMaurey) .
Every operator from L1 (!l) into a Hilbert space H is pabsolutely summing for every p > 0. Proof: From Theorem 7 and Corollary 9 we get IT 2 (L1 (/l) , H) = IT1 (L1 (!l) , H) = L(L1 (/l) , H). Theorem 34 gives the claim. II
Note that this is a strengthening of Theorem 7 which does not follow from our proof of it, simply because the operator P: A + £2 ( see
229
III.F Absolutely Summing And Related Operators §36.
Proposition 6 ( a)) is pabsolutely summing for p < 1, ( see Exercise III.I.2 ) . Note also that for the above argument to work we do not need the full power of Theorem 7. It is enough to know that (J.L) , H) = (J.L) , H) for some p < 2. This fact can be derived from Proposition 6 ( b) exactly like Theorem 7 was derived from Proposition 6 ( a) . If we avoid the use of Proposition 6 we can obtain some equalities of the form = without knowing that = The following theorem is a useful example of such situation.
not
L(L 1
llv(L 1
Tip( X, Y) llq (X, Y)
llv(X, Y) L(X, Y).
X is a Banach space of cotype 2 then for any Banach llv(X, Y) = TI2 (X, Y) for all :5 2.
36 Theorem. If space Y
p
Let us start with the following. 37 Lemma. If Y is a Banach space of cotype 2, then for any Banach
X and any C2 (Y) · Cv · 1rv(T).
p
space
� 2 we have
Tip(X, Y) = TI2 (X, Y), and 1r2 (T) :5
T Tip(X, Y)
,
X.
Let us take E and X I . Xn E From the definition of cotype ( see III.A. 17 ) and Kahane's inequality ( see III.A.20 ) we get
Proof:
•
•
•
p (48) ( � 1 Txi 1 2 ) 2 ::; c( J I � ri (t)T(xi ) l dt) p " ,; C (/ l r ( t, r; (t ) x; ) I dt) l , with C = C2 (Y) · Cv · Applying Proposition 33 ( b ) and Khintchine's inequality to ( 48) we n
l
n
.!.
get
:5 C1rp (T) , and the constant is of the right form. The remaining inclusion is always true. a
so 1r2 (T)
230
III.F Absolutely Summing And Related Operators §Notes.
We know from Theorem 34 that it is enough to show the theorem for a fixed p, 1 < p < 2. Also it is enough to consider X, Y finite dimensional and to keep track of the constants (see Proposition 5) . Under these assumptions, for T: X + Y Theorem 27 and Lemma 37 give
Proof of Theorem 36.
1rp (T)
sup{tr(ST) : S: Y + X, np' (S) ::5 1 } ::5 sup{tr(ST) : S: Y + X, 1rp' (S) ::5 1 } ::5 sup{ tr(ST) : S: Y + X, 1r2 (S) ::5 C2 (Y) · Cp} = C2 (Y) · Cp · sup{tr(ST) : S: Y + X, 1r2 (S) ::5 1 } . =
Applying Corollary 25 we get a
Notes and Remarks.
Much of this chapter, as well as much of modern Banach space theory, is the outgrowth of Grothendieck [1956] . The work of Grothendieck was phrased, however, in the language of tensor products and bilinear forms. This language, although still used by some and known by many, seems to have been generally replaced by the language of operators. In our book we adhere to this usage and avoid tensor products almost entirely. The notion of Banach operator ideal emerged in the late 60's, mainly as a result of many attempts to understand Grothendieck [1956] . From this time on, A. Pietsch and his students and collaborators have stud ied many aspects of the abstract concept, contributing greatly to the creation of the theory of operator ideals as presented in Pietsch [1978] . The pabsolutely summing operators (probably the most important op erator ideal) were introduced by Pietsch [1967] as a generalization of Grothendieck's 'application semiintegrale a droite' which are now called !absolutely summing operators. In this paper A. Pietsch proved the basic properties of pabsolutely summing operators, in particular the fundamental Theorem 8 (the idea of using the separation argument in the proof is due to S. Kwapien) and Corollary 9. The Grothendieck theorem ( Theorem 7) was proved by Grothendieck [1956] who called it the fundamental theorem of the metric theory of tensor products. The proof was understood and presented in the language of 1summing op erators by LindenstraussPelczynski [1968] . These authors proved the Grothendieck inequality (our Theorem 1 4) directly and derived The orem 7 from it. The prmtf presented here was found by A. Pelczynski with some help from the present author and was published in Pelczynski
III.F Absolutely Summing And Related Operators §Notes.
231
[1977] . Numerous other proofs have been published for various versions of the Grothendieck theorem. We refer to Pisier [1986] , Haagerup [1987] and Jameson [1987] for references. Considerable effort has gone into eval uating the Grothendieck constant Ka. It is known that this constant is different for real and complex scalars. The most precise estimates for the complex case are 1.338 :5 Kg :5 1.40491 (see Haagerup [1987] for the proof and a discussion of the known results) . There exist also C* algebra versions of the Grothendieck theorem. They are quite involved, but useful in the theory of C* algebras. As an example let us quote the following Theorem A. If A is a C* algebra and Y is a Banach space of cotype 2, then every linear operator T: A + Y factors through a Hilbert space.
This was proved in Pisier [1986a] . This theorem is rather in the spirit of Chapter III.H but it is also the most Banach space theoretical statement. A more detailed presentation of even the most important results in this area requires, quite naturally, some familiarity with the theory of C* algebras. We refer the interested reader to Pisier [1986a] for the proof of this result and for a detailed description of and references to the earlier works. Theorem 1 0 was proved in LindenstraussPelczynski [1968] . The orem 1 1 is a special case of a result proved by Kislyakov [1976] . This theorem should be compared with Proposition III.A.3. It is a Banach space manifestation of a phenomenon common in harmonic analysis that certain continuity results for important operators which hold for 1 < p < oo fail for p = 1 or p = oo . Incidentally one gets a rather crazy proof that the multipliers considered in Lemma III.A.4 are not contin uous in L1 (T2 ). The fact (Exercise III.G. 13) that C 1 (11'2 ) is not iso morphic to any C(K)space was stated by Grothendieck [1956a] and the first proof was published by Henkin [1967] . Actually Henkin proved the much stronger result that C k (11'e) for k � 1 and i > 1 is not homeomor phic to any C(K)space with the homeomorphism and its inverse being uniformly continuous. The basic idea of the proof of Theorem 1 1 has been applied in a similar but much more general context in Pelczynski Senator [1986] . Theorem 13 is taken from Bourgain [1987] . Its main point is the estimate (20) . Apart from this estimate facts of this type are well known and much used in harmonic analysis (see Rudin [1962a] 2.6) . Problems centred around the von Neumann inequality (25) are among the most interesting in operator theory on Hilbert space. Even
232
III.F Absolutely Summing And Related Operators §Notes.
our small sample shows the remarkable variety of methods used. The first example of a powerbounded but not polynomially bounded opera tor was given by Lebow [1968] . He showed that an example constructed by Foguel [1964] of a power bounded operator which is not similar to any contraction has this property. We present here some results taken from Peller [1982] and Bozejko [1987] . More precisely the proof of Proposi tion 15 is taken from Bozejko [1987] while Theorem 16 and Lemma 1 7 and Corollary 1 8 are due to Peller. The direct proof of Corollary 18 was communicated to the author by G. Pisier. Observe that estimates of the type 11 (T) II � (3(4>) where (3 is some norm on polynomials lead to a functional calculus for T on some class of functions. This is a very important subject in operator theory. It has many connections with other branches of analysis. The reader may consult Nikolski [1980] for a more complete picture. This subject has also a branch in the theory of Banach algebras. A nice result (once more relying on the Grothendieck theorem) is a theorem of Varopoulos [1975] that any Banach algebra X isomorphic as a Banach space to a C(K)space is algebraically and topo logically isomorphic to some subalgebra of the algebra of all operators on a Hilbert space. It is an open problem asked by P. Halmos if every polynomially bounded operator T is similar to a contraction, i.e. is of the form T = VT1 v  1 with II T1 I I � 1 and V an isomorphism of an underlying Hilbert space. The best partial result in this direction seems to be contained in Bourgain [1986a] . He proved Theorem B. If T: H + H satisfies ll p(T) II � M II P II oo for every polynomial p and if dim H = N < oo then there exists S: H + H such that II STS  1 11 � 1 and II S II II S  1 11 � M log N.
4
The proof is quite complicated and uses, among other things, Grothendieck's inequality and Theorem 111.1. 10. For applications of the Grothendieck theorem in the theory of stochastic processes the reader may consult Rao [1982] . For applications to interpolation theory the pa pers of V.I. Ovchinnikov should be consulted (e.g. Ovchinnikov [1976] and [1985] ) . The notion of 1nuclear operator goes back to Grothendieck. It is a generalization of a1 operators on a Hilbert space (see Remarks af ter III.G. 18). The theory of pnuclear operators has been developed in Chevet [1969] and PerssonPietsch [1969] . The important and useful Theorem 27 can be found in PerssonPietsch [1969] . It is only a small sample of various duality results for other operator ideals. There is also
III.F Absolutely Summing And Related Operators §Exercises
233
a (more complicated) duality theory for operators on infinite dimen sional spaces. Proposition 28 is folklore. It is a formalization of the connection between Theorem 7 and its dual form, Theorem 29. This connection was already known to Grothendieck [1956] and was quite explicit in LindenstraussPelczynski [1968] . The notion of pintegral op erator and all our results about them can be traced to PerssonPietsch [1969] . Actually our pintegral operators are quite often called in the lit erature strictly pintegral, with the name 'pintegral operator' reserved for operators T: X + Y such that iT is strictly pintegral (i.e. integral according to the definition given in 21) where i is the canonical em bedding of Y into Y** . Such operators appear naturally in the duality theory for operator ideals when the spaces are infinite dimensional. Theorem 30 is a classical result of the theory of Fourier series. It is very similar in spirit to Theorem III.A.25. The theory of Sidon sets is an interesting part of commutative harmonic analysis. The standard, but a bit outdated, reference is LopezRoss [1975] . More recent advances in this area are connected with the use of Banach space methods. Our Theorem 31 is one such example. It is a variant of a result of Varopoulos [1976] . A stronger result proved in BourgainMilman [1985] is Theorem C. If G is a compact abelian group with dual group r and if for S C r the space Cs ( G ) has ootype p for some finite p, then S is a Sidon set.
Theorem 34 and 36 are due to Maurey [1974] . Our proof of Theorem 34 follows Pisier [1986] and is due to Maurey and Pisier. These theorems
will be useful later on in Chapters III.H and 111.1. There are many places where the theory of pabsolutely summing, pintegral and pnuclear operators is presented. We conclude these re marks by listing some of them: Pietsch [1978] , Pisier [1986] , Tomczak Jaegermann [1989] , Kislyakov [1977] , Jameson [1987] . Exercises
1. 2.
Let p , q, � 1 be such that � + � = � and let T E IIp (X, Y) and S E IIq (Y, Z) . Show that ST E IIr(X, Z) and 7rr(ST) :5 7rp(T)7rq (S). r
Show that id: £1 + £2 is not 1integral. Is it pintegral for some > 1? Show that id: £1 + £00 is 1integral.
p
3.
(a) Let T: L1 [0, 1] + C [O, 1] be given by Tf (x) = J; f ( t)dt. Show that T is not !absolutely summing.
234
III.F Absolutely Summing And Related Operators §Exercises
(b) Let a: if" � i� be defined as a((ei }.f= 1 ) = (}::: j= 1 ei ) := 1 · Show that 1r 1 (a) "' C log(N + 1 ) . n 4. For I = .E� 0 anz E H1 (D) we define T(f) = ( v'�� 1 ) := o · Show that T: H1 (D) � £2 is bounded but not !absolutely summing. 5. Let K(x, y) be a measurable function on [0, 1] x [0, 1] . Let T l (x ) = J; K(x, y)l(y)dy. Find necessary and sufficient conditions for K(x, y ) so that the operator T maps C[O, 1] into itself and is ! absolutely summing. 6. Show that id: C[O, 1] � Lp [O, 1] is not qabsolutely summing for any q < p. 7. (a) Show that every pnuclear operator, 1 $ p < oo, is compact. (b) Show that if 1 $ p < oo and X is reflexive then Ip(X, Y) = Np (X, Y). 8. Show that if id: X � X is pabsolutely summing for some p < oo, then X is finite dimensional. In particular in every infinite di mensional Banach space there exists an unconditionally convergent series which is not absolutely convergent. 9. (a) Let G be an infinite, compact, metrisable abelian group and let f..L E M(G) be such that jJ,('Y) � 0 as 'Y � oo, f..L � 0 and f..L is singular with respect to the Haar measure m. (Note that III.C.6 shows that such a f..L exists.) Let T�' : C(G) � L 1 (G, m ) be defined by T�'(f) = I * f..L · Show that T is a compact, 1integral but not 1nuclear operator. (b) Let r.p E L00 (T)\C(T) . Show that Tl = I * r.p considered as an operator on C(T) is compact and 1integral but not 1nuclear. 10. Show Theorem 7 assuming Theorem 14. 11. Suppose that I E L 1 (T) is such that / ( 2 k ) = 1 for k = 1, 2, . . . , N and fr I l l $ 1 + c. Show that for every a, 0 < a < 1 , there exists a constant C = C ( a , c ) > 0 such that 1 {£: /(£) =/: O} � CNo. tn N .
l
12. (a) Suppose that X is a Banach space with unconditional ba sis ( xn) ;;::'= l · Show that every !absolutely summing operator T: X � Y factors through £ 1 . (b) Show that C[O, 1] does not have an unconditional basis. This is a special case of 11.0. 12 but try to prove it using (a) .
III.F Absolutely Summing And Related Operators §Exercises
235
(c) Suppose that F C N is a A(2) set (see I.B. 14) and suppose that L: = span{ ein9 } n EF C Lp(T), p > 2, has an unconditional basis. Show that the characters are unconditional in L:. The same holds for Cp = span { ein 9 } n EF C C (T) .
13. Show that for every p =f. 2, there exists a subset F C N such that idp : Cp + L: (for notation compare Exercise 12 (c)) is p absolutely summing but not pintegral.
III. G . SchattenVon Neumann Classes
In this chapter we consider Schattenvon Neumann classes of operators on a Hilbert space and their applications in the theory of Banach spaces. We start with the notion of an approximation number of an operator between Banach spaces. We prove that the approximation numbers of an operator and its adjoint are the same. Then we study operators on Hilbert space. We prove the Weyl inequality and basic facts connecting eigenvalues, snumbers and approximation numbers. Various character izations of HilbertSchmidt operators are presented. We also show the classical FredholmBernsteinSzasz theorem about Fourier coefficients of Holder continuous functions. Next we give results about summability of eigenvalues of pabsolutely summing operators on a general Banach space and apply them to eigenvalues of HilleTamarkin integral opera tors. 1.
Given an operator T: X + Y we define its approximation numbers
an(T) = inf{ II T  Tn ll : Tn : X + Y, rank Tn < n } ,
n
= 1 , 2, . . . .
Clearly a1 {T) = II T II and the sequence { an (T))�= l is decreasing. If an (T) + 0 then T is a norm limit of finite dimensional operators, thus compact. One proves routinely that if Y has b.a.p. (see II.E.2) and T: X + Y is compact then an(T) + 0. 2 Proposition. The following inequalities hold for every n , m � all operators T, S:
1 and
an + m  l (T + S) $ an (T) + am(S); an + m  l (T S) $ an(T) · a m (S).
{1) {2)
o
us
Proof: The argument for {1) is obvious. To prove {2) let take any Tn with rank Tn < n and any Sm with rank Sm < m. Then
Since
rank {Tn S + TSm  TnSm ) � rank{Tn(S  Sm )) + rank(TSm ) < n+m 1
238
Ill. G. SchattenVon Neumann Classes §3.
we get
For 0 < p < oo we define AP(X, Y) to be the set of all operators T: X  Y such that 2:: :::'=1 an(T) P < oo. We denote (2:: :::'=1 an (T)P) ! as ap(T) . 3.
The quantity ap (T) is a quasinorm on AP(X, Y) for with this quasinorm is a quasiBanach operator
Proposition.
0
oo .
AP(X, Y)
ideal. Proof: Since an (T)
= 0 if rank T < n we get that finite rank operators are in AP(X, Y). From (1) we get
ap(T + S)
=
!.
( � an(T + S)P) ( � a n 1 ( � an + an(S)]P) ; 00
:5 2 ;
p :5
[
00
2
2

)
.1
(T + S) P p
(3)
:5 Cp [ap(T) + ap(S)] .
(T)
Since ap (AT) = IAiap(T) we get that ap (T) is a quasinorm and III.F (1) holds. The conditions III.F (2) , (4) and (5) are obvious. The proof of completeness follows the usual lines and is left to the reader. a We would like to point out that ap (T) is not a norm even for p � 1 . In general on AP(X, Y) there is no equivalent norm ( see Exercise 6) , so AP(X, Y) cannot be made into a Banach operator ideal. The detailed analysis of the convexity of AP(X, Y) can be found in Pietsch [1987] 2.3.7. 4 Proposition.
!
If T E = p! + !q .
AP(X, Y)
and S E
Aq (Y, Z)
A8 (X Z)
with
Proof:
Using (2) and Holder's inequality we get
'
s
:5 2 �
( � an (S) 8 00
•
an(T) 8
)•
!.
then ST E
:5 2 � ap(T)aq (S) .
a
Ill. G. SchattenVon Neumann Classes §5. 5 Proposition.
239
If T: X + Y is a compact operator then an (T) =
an(T*), n = 1 , 2, . . .
.
Since an(T*) � an (T) it is enough to check that an(T** ) � 1 , 2, . . . . For a fixed n and arbitrary c > 0 let us take V, a finite cnet in T ( Bx ) . Since T is compact T ( Bx ) is normdense in T** ( Bx·• ) so V is also an cnet in T** ( Bx·· ) . Fix also an operator Tn : X** + Y** with rank Tn < n and li T**  Tn ll � an (T**) + c. Put F = span { V U Tn (X**) } . From the principle of local reflexivity II.E. 14 we get an operator cp: F + X with ll cp ll � 1 + c and cp I F n X = i d. In particular cp j V = i d. For x E X with ll x ll � 1 let us fix v E V such that II Tx  vii � c. Then we have Proof:
an (T), n =
II Tx  cpTnxll � c + ll v  cpTnx ll � c + ( 1 + c) ll v  Tnxll � c + (1 + c) ( ll v  Tx ll + ( II T** x  Tnxll ) � c + ( 1 + c ) ( 2c + an (T**)). Since cpTn has rank less than n and c was arbitrary we obtain an(T) �
an(T**).
a
Let T: X + X be an operator such that Tk is compact for some k . Such operators are called powercompact. For a powercompact operator T we define the sequence (A n (T))�= l which consists of all eigenvalues of T counted with multiplicities ( cf. I. A. 18.) ordered in such a way that I A 1 (T) I � I A2 (T) I � I A 3 (T) I � · · · . Let us recall that A E ([ is an eigenvalue of an operator T if Tx = Ax for some x E X, x =/= 0. Since T is power compact, the Riesz theory holds for T ( see I.A. 1619.) so the multiplicity of each eigenvalue is finite. Let H and L be Hilbert spaces and let T: H + L be a compact oper ator. Then I T I = /T*T is a positive compact operator and there exists an isometry U: I T I (H) + L such that T = U I T I . The spectral theorem for compact, positive operators shows that there exists an orthonor mal system ( vn)�= l such that I T I (x) = E�= l An ( I T I ) (x, vn)Vn · Since T = U I T I we have that for an arbitrary compact operator T: H + L there exist orthonormal systems (vn )�= l and ( un )�= l such that 6.
00
(4) T(x) = L An( I T I ) (x , Vn }Un · n= l Clearly Un = Uvn, n = 1 , 2, . . This is called the Schmidt decomposi tion and the numbers A n( I T I ) are called the singular numbers of T and are denoted sn (T). .
.
III. G. SchattenVon Neumann Classes § 7.
240
If H and L are Hilbert spaces then AP (H, L) is called the pth Schattenvon Neumann class and denoted ap( H , L) and ap(T) is then denoted ap(T). This may seem confusing but the hilbertian theory is so rich and special that it has its own traditional language. 7 Theorem. Let H, L be Hilbert spaces and let T: H pact operator. Then s n (T) = an(T), n = 1 , 2, . . . .
Proof:
t
L be a com
Using (4) we get immediately
Conversely, given any operator Tk : H t L with rank Tk < k let us take x = E7= l /3j Vj (where (vj ) �1 is given by (4) ) such that ll x ll = 1 and Tk x = 0. From (4) we get
I! Tx !l = so
(� k
A n ( I T I ) 2 1 /3i l 2
)
.!
2
?
Bk(T) I! x ll •
8. Our basic tool for the study of Schattenvon Neumann classes is the following.
Theorem. (Weyl's inequality) . Let T: H t H be a compact oper ator on a Hilbert space H. Then for evezy n = 1 , 2, . . .
n
n
k= l
k l
II ! A k (T) I :5 II= ak (T).
(5)
Proof: Clearly we can assume A n (T) =/= 0 since otherwise there is nothing to prove. Using the spectral theorem for compact operators and the Jordan decomposition in finite dimensional spaces we infer that for every n there exists a subspace Hn C H such that
= n, T(Hn) Hn, T I Hn : Hn Hn dim Hn
c
t
has eigenvalues
At (T) , . . . , An (T).
(6) (7) (8)
III. G. SchattenVon Neumann Classes §9.
241
Let Tn denote T I Hn : Hn + Hn. Let us fix an orthonormal basis (xj ) j= 1 in Hn , and for any operator S: Hn + Hn let det S = det [ (Sxk , Xj) ] k,j =t · From well known properties of determinants of finite matrices we get det S = Il �= l .Ak (S). Thus (8) gives
n
n
k= l
k= l
II I.Ak (T) I = II I.Ak (Tn ) l = I det Tn l ·
(9)
Using the polar decomposition we get Tn = U I Tn l where U is a unitary operator. Obviously o:k (Tn ) :::; o:k (T) , k = 1 , . . . , n and Theorem 7 gives o:k (Tn ) = o:k ( I Tn l ) = A k ( I Tn l ), k = 1, 2, . . . , n. Thus we get
I det Tn l = det I Tn l =
n
n
II= l I.Ak ( I Tn l ) l ::=; II= l o:k (T) k
k
and comparing (9) and (10) we get the claim.
( 10)
•
9. To use all the information contained in (5) we need a lemma about sequences of positive numbers.
Lemma. Let ( o:k )f= 1 and ( f3k )f= 1 be decreasing sequences of positive numbers such that L:�= l O:k :::; L:�= l f3k for n = 1 , 2, . . . , N and let r.p: 1R + 1R be a convex function such that r.p(x) :::; r.p( l x l ) . Then
N
r.p( o:k ) 2::: k= l Proof:
N
r.p(f3k ) · :::; 2::: =l k
In 1RN we define a convex set V by
V = conv{ (c k f3u(k) )f= 1 : E" k = ± 1 and a is a permutation of the set { 1 , 2, . . . , N}}. If ( o:k )f= 1 f/. V then by the HahnBanach theorem I.A. lO there exists a functional on :JRN such that 4>/V :::; 1 and ¢((o:k )f= 1 ) > 1 . Since V
is invariant under permutations of coordinates and changes of signs and (o:k )� 1 is a positive, decreasing sequence we can assume that (( x1 Jf= 1 )
=
N
:�:::>1 x1
j=l
with
c
1 � c2 > · · · � eN � 0.
242
Ill. G. SchattenVon Neumann Classes § 1 0.
But 1<
j N N N1 ( ) Cj + O! l = CN a ak C j L: L L: L k k k c + j=1 k= 1 k= 1 k= 1 j N1 N N ( ) + C $ N L .Bk L cj  Cj + l L .Bk = L: ck,Bk $ 1 . j=1 k= 1 k= 1 k= 1
This contradiction shows that Ej= 1 Aj (e{.Bu; (k J )k'= 1 for some gives
N
N
� c,o (ak ) = � N
C,O
(
8
s
�
(a k )k'= 1 E V, so .>.j 's with E Aj = .>.je{.Bu; (k)
)
we have ( a k )k'= 1 = 1 and Aj � 0. This
8
N
j=1
k= 1
$ L L Aj c,o (e{ ,Bu; (k) ) $ L Aj L c,o ( .Bu; (k) )
k= 1 j = 1 N = L: c,o ( ,ak ). k= 1
10 Theorem. For every compact operator T: H < oo and every N = 1 , 2, . . . we have
p
• +
H and every p, 0 <
(11)
In particular
ll (.>.n (T)) II P $ ap(T) .
Without loss of generality we can assume .>. N (T) � 1 and O! N (T) � 1 . Applying Lemma 9 for O!n = p log .>.n (T) and .Bn = p log an (T) and c,o (t) = exp t we get the claim.
Proof:
II
11. Weyl's inequality also allows us to show that ap ( H ) , 1 $ p < oo , is actually a Banach space. As we know ( see 3) , this is not true for general Banach spaces. We have
Proposition. Let T and S be compact operators on a Hilbert space H. For every n = 1 , 2, . . . and every 1 $ p < oo we have
Ill. G. SchattenVon Neumann Classes §12.
243
Using the Schmidt decomposition (4) and Theorem 7 we can
Proof: write
(S + T)x = L a k (S + T) (x, vk )uk . k Let U be a partial isometry defined by U( vk ) = U k and orthogonal projection onto span{ v k }k= l · We have n
let
P
be the
n
L a k (S + T) = L( U* (S + T)vk , vk) = trPU* (S + T)P k= l k= l ::; i tr(PU*SP) I + i tr(PU*TP) I . Since for finite rank operators the trace equals the sum of eigenvalues the last expression is majorized by
Thus from ( 1 1 ) we get for n
n = 1 , 2, . . . n
n
L ak (S + T) ::; L ak (PV* SP) + L a k (PV*TP) k= l k= l k= l n
n
k= l
k= l
(12)
$ L a k (S) + L a k (T) so our proposition is proved for p = 1 . Applying Lemma 9 for a k = ak (S + T) and f3k = ak (S) + ak (T) and cp(t) = tP we get (use (12)) E�= l a k (S + T)P ::; E�= l (a k (S) + ak (T)) P . Now the Holder inequality gives the claim. a 12. The operators of class a2 are called HilbertSchmidt operators. Here are some equivalent characterizations.
Proposition. Let H and L be Hilbert spaces and let T: H > L. The following conditions are equivalent: (a)
T E a2 (H, L) ;
(b) for every orthonormal basis
LjEJ II Th1 ll 2 < oo ;
(h1 ) 1 0
(c) there exists an orthonormal basis
l:j EJ II Th1 112 < oo ;
in the space H we have
( h1 ) 1 EJ
in the space H such that
Ill. G. SchattenVon Neumann Classes § 1 2.
244
(d)
(e)
T admits a factorization
for some Lt ( JJ, ) space;
T admits a factorization
T
H
L
� � C(K)
for some C ( K ) space; (f)
(g)
T E Tip(H, L) for every p, T E Tip(H, L) for some p,
1�p< 1 �p<
oo;
oo;
Proof: ( a) => ( b ) . Let ( hj ) jE J be any orthonormal basis in H. Using the Schmidt decomposition (4) we get
I
L II ThJ II 2 = L L ak (T) (hj , vk )u k jEJ jEJ k = L L a k (T) 2 i (hj , Vk) l 2 jEJ k
l2
( 13)
jEJ k 2 = L a k (T) = a2 (T) 2 . k
( b ) => ( c ) . This is obvious. (c ) => ( d ) . We define v : H + ft (J) by and �: £1 (J) + L by
v( x ) = ((x , hJ ) II ThJ II ) JEJ with the convention
� = 0.
245
Ill. G. SchattenVon Neumann Classes §13.
Since T = E o v we get the desired factorization. ( d ) =? ( f ) . By Grothendieck's Theorem III.F.7 and Corollary III.F.9. ( f ) =? ( g ) . Obvious. ( g) =? ( a) . Since Hilbert space has cotype 2 ( see III.A.23 ) we get from Lemma III.F.37 that T E II2 (H, L) . From Corollary III.F.9 ( b ) we get that T is compact thus the Schmidt decomposition gives T ( x ) = 1:�1 A k ( x, vk) u k . The definition of 2absolutely summing map gives
L I.Xk l 2 = L 11 Tvk ll 2 $ 1r2 ( T ) 2 sup { L l (x, vk) l 2 : ll x ll $ 1} = 1r2 ( T ) 2 k k k
(14) so a2 (T) $ 1r2 (T). Since ( e ) is a dual condition to ( d ) and ( a) is a self dual condition ( see Proposition 5) we infer that also ( e ) is equivalent to all the others.• If an operator T: H  L satisfies any of the conditions of the Proposition then
Remark.
(
1r2 (T) = a2 (T) = � 11 ThJ II 2 J
)
1
2
( 15)
for any orthonormal basis ( hJ )je J is H. We see from ( 13) and (14) that only 1r2 (T) $ a2 (T) remains to be proved. If (lJ ) JeJ is any orthonormal basis in L and h E H then
11 Th ll 2 = L I (Th , lj) l 2 = L l (h, T*lj) l 2 . jEJ jE J
When we view this inequality as a special case of III.F. ( 9 ) we see that
1r2 (T) 2 $ LjEJ II T*lj ll 2 . From ( 13) we get 1r2 (T) $ a2 (T* ) = a2 (T) .
13. One of the reasons why the HilbertSchmidt operators are impor tant is that they admit a nice integral representation. Proposition. An operator T: L 2 (0, J.L)  L 2 ( E , v ) is HilbertSchmidt if and only if there exists a function K E £ 2 (0 x E, J.L x v ) such that
Tf( a) =
k K (w, a) f(w)dJ.L(w)
(16)
III. G. SchattenVon Neumann Classes § 14.
246 Proof:
From the Schmidt decomposition we get
Tf(a) = L >.. k r f(w)vk (w )df1 (w)uk(a) k ln = >.. k vk (w ) · u k (a) f (w )df1 (w ).
In [ �
]
and vk (w ) · u k (a) is an orthonormal system in £2 (!1 x I:, 11 x v ) we get the desired function. Conversely let ( fj (w ))j EJ and ( h 8 (a)) 8 E S be orthonormal bases in L2 (!1, 11) and L2 (I:, v) respectively. Then ( fj (w ) · h8 (a)) (j, 8 )EJ x S is an orthonormal basis in the space £ 2 (!1 x I:, 11 x v ) . We have Since I: l>.. n l 2 <
j
oo
j 8 = II K IILc n xE,Jtxv) ·
j 8
So Proposition 12 implies that T is HilbertSchmidt and a2 (T) = •
II K IILc n xE,Jtx v) ·
Example. There exists an operator T: £ 2 [0, 1] . £ 2 [0, 1] such that T E ap for every p > 2 and there is no function K (x, y ) on [0, 1] x [0, 1]
such that
Tf(x) =
1 1 K(x, y)f(y)dy
a.e.
where the integral is understood as a Lebesgue integral. Observe that if such a representation exists then the function n(x) = J I K (x, Y ) l dy is finite almost everywhere, so
I Tf(x) l S 11/l loo · D (x).
(17)
We define our example as n = 1, 2, . . . , n = 0,  1, 2, . . . ,
where hn is the Haar system ( see II.B.9) and An = n ! · log n. Obviously T E ap(L 2 [0, 1]) for p > 2. Since sup n I T(e 2,.in y) (x) l = sup n l >..n hn (x) l = oo a.e. we see that (17) does not hold. 14. Now we would like to apply these general notions to investigate the connection between the smoothness of functions and the size of Fourier coefficients.
247
III. G. SchattenVon Neumann Classes §15.
Theorem. The Fourier coefficients of every function I a $ 1 belong to f.p for every p > (2 a� l ) .
E Lipa (T), 0 <
Proof:
Given I E Lipa (1l') let us consider the operator H1: L 2 (1l') t L 2 (T) given by HJ (g) = I * g. Since the eigenvalues of H1 are (j(n))�� oo by Theorem 10 it is enough to show that H1 E uv  Since I E Lipa (1l') the operator n, actually maps L 1 (1l') into Lipa (T). Let Un )<;;'= O be the Franklin system (see III.D.20) and let F(f) = ( (!, ln ) (n + 1) a + ! )<;:= o · From III.D.27 we get that F: Lipa (1l')  £00 • Let :E: £00 t L 2 (T) be given by :E (�n ) = L ::'=o �n (n + 1)  a  ! In · We have the commutative diagram
L2 (T) id
HI
+
l
H1
L2 (T)
Ei�
F
£2 f.oo Lipa (T) £1 (T) where ?J (�n ) = ((n + 1)  !  e �n ) <;;'=o and "Y (�n ) = L ::'= o �n (n + 1)  a + e In · Clearly ")' E with s > (a � e ) and 7JFH1 id E 2 by Proposition 12. Thus by Proposition 4 H1 E up with � = � + .; . Since c was arbitrary 0'
0'8
•
we get the theorem.
15. Now we want to discuss eigenvalues of operators on general Banach spaces. Estimating eigenvalues of an operator on a Banach space is more difficult than on a Hilbert space since the geometry is more complicated. The main idea is to reduce the problem to a related problem on a Hilbert space. One way to do it is to use the concept of related operators. Two operators 8: X t X and T: Y t Y are related if there exist operators A: X t Y and B: Y  X such that
S = BA and T = AB. The following simple lemma is very useful when estimating eigenvalues. Lemma. Let S and T be related operators. If one is powercompact then the other is and (.Xn(T))<;;'= 1 = (.X n (S)) �= l ·
l 1 Proof: Since T k + ·= AS k B and B k +
=
BTk A the first claim follows.
Let .X =1 0 b e an eigenvalue of S with multiplicity k and let V be the
248
III. G. SchattenVon Neumann Classes § 1 6.
V= v
8)8 = V for some A (v) = 0. Then
corresponding eigenspace, so dim k and ker(>. s. Suppose that for some E with =I= 0 we have
v V
is impossible, so for W = A(V) we have dim W = k. >.w==0.AvButE this W we have
so For
(>.  T)8w = jt= l ( 1 )j (�) >.s i (AB)i w = jt= l ( 1)j (�) >.s i (AB )i Av = A(t(  1 )i (;) >.s i (BA)i v) = A( >.  8)8v = 0. J
J
This shows that W is contained in the eigenspace of T corresponding to >.. Repeating the same argument with and T interchanged we get the a lemma.
8
16.
Directly from Lemma
15 we get
Theorem. IfT: X + X is a pabsolutely summing operator with p :::; 2 then (>.n (T) )�= l E i2 .
It is enough to assume 1r2 (T) Then we have a factorization
Proof:
X
C(K)
8:
T
= 1 (see Corollary III.F.9(a)). X
id
8 = o o B is
The operator L 2 (JL) + L2 (JL) defined as id i similar to T and from Proposition 12 we get a2 < oo. Theorem 10 completes the a proof. 17.
(8)
Surprisingly this is an optimal result. Namely we have
Proposition. There exists a nuclear operator T such that (>.n (T))�=l � lp for any p < 2.
249
III. G. SchattenVon Neumann Classes §18.
This follows directly from III.A.25 and the following. 18 Proposition. Let G be a compact abelian group with Haar measure *
m, and let T: C(G)  C(G) be given by Tf = f h for some h E C(G) . Then T is nuclear.
It is clearly sufficient to show that n1 (T) :5 ll h lloo for h a finite combination of characters. Such T being finite dimensional is nuclear. It follows from II.E.5(e) and the definition of the nuclear norm that T can be approximated in the nuclear norm by operators of the form P1 TP2 where P1 and P2 are projections onto finite dimensional subspaces in C(G) with d(ImP1 . £�) :5 1 + c and d(ImP2 , £::!,) :5 1 + c. From III.F.12 we infer that 1r1 (T) :5 ll hlloo so it is enough to show that for 8: £�  £::1, we have n1 (8) :5 1r1 (8). This follows directly from Corollary III.F.24 since 1r1 (8) = i1 (8) (see remarks after III.F.22) . a
Proof:
The above Proposition 17 contrasts with the situation in Hilbert spaces. For T: H  H we have n1 (T) = a1 ( T ) . If T has a nuclear representation T(x) = Lj ( x, Xj) Yi then from Proposition 1 1 we get a1 (T) :5 Lj a1 ( ( ·, xi ) Yi ) = Lj llxi ll II Yi ll so a1 (T) :5 n1 ( T ) . The converse follows directly from the Schmidt decomposition. Thus Theorem 10 shows that any nuclear operator on a Hilbert space has absolutely summable eigenvalues. Remark.
19.
The behaviour of eigenvalues of pabsolutely summing maps for
p > 2 is given in Theorem. then
If T: X  X is a pabsolutely summing operator, p � 2, (18)
The proof of this theorem follows from the following two facts. 20 Lemma. There exists a constant Cp such that for every operator T: X  X we have
21 Proposition. Let P be given. Suppose that for some q there exists a constant Cq such that for every operator T: X  X and any Banach
250 space
III. G. SchattenVon Neumann Classes §21.
X
Cq = 1 .
we have
(2:�= 1 1Xn (T) I q ) � ::; Cq rrp(T) .
Proof of Theorem 19.
1
Then we can take
It follows from Lemma 2 0 that for every
q > p we have (2: I .Xn (T) I q F ::; Cq rrp(T). Applying Proposition 21 and a
passing to the limit as q + p we get (18) .
From finite dimensional linear algebra we infer that for every n = 1 , 2, 3, . . . there exists a subspace Xn C X, dim Xn = n such that T(Xn ) C Xn and Aj (T I Xn ) = Aj (T), j = 1 , 2 , . . . , n. Clearly 1rp(T I Xn ) ::; 7rp(T) so applying the Pietsch factorization Theorem III.F.8 we have the factorization Proof of Lemma 20.
T IXn
Xn i
l
x=
Xn
I
id
a
XPn n where rrp(id) ::; 1 , ll a ll ::; 7rp(T) and X� is an ndimensional subspace of some Lp(/L) · From Corollary III.B.9 we get operators A and B such that X��f'2�X� and such that BA = i dx:;, and II A II · II B II ::; n !  � . The operator T I Xn : Xn + Xn is related to the following composition which we will call S: on B XnP "' Xn i xn= id XnP A {.on2 · {. 2 +
+
+
+
+ 1
1
Since rrp(S) ::; rrp(id) II A II · l l i l l · ll a ll · II B II ::; n 2  "P rrp(T) we get from Proposition 12 that there exists a constant Cp such that (20) From Theorem 10 and (20) we get
so (19) follows.
a
Proof of Proposition 21. Let us denote the smallest possible Cq by K. If K > 1 then there exists an operator T: X + X such that
251
Ill. G. SchattenVon Neumann Classes §21 .
1rp(T) = 1 and (�=:"= 1 1Xn (T) I q ) � > .[K. Without loss of generality we can treat X as a subspace of C(O) for some compact spac� n. From III.F .8 we get a probability measure f..L on n and an operator T: Xp + X with II T II = 1 where Xp is the closure of X in Lp(O, f..L) . Let Y c C(O ® O) be the closure of the set of functions of the form r;;=l x; (wl ) · z; (w2 ) where x; , z; E X, j = 1 , . . . , n and let Yp be the closure in Lp(O X n, f..L X f..L) of Y. We define an operator T ® T: y  y by the formula T ® T("[; x; (w l )y; (w2 )) = 'E T(x;)(w l ) · T(y; )(w2 ) · One easily
checks that T is continuous and that 00
00
00
L 1Xn (T ® TW � L 1Xn (TW · L 1Xn (T W . n =l n= l n= l
(21)
Let T ® T: Yp + Y be defined by
Since the formal identity from Y into Yp has norm at most 1 we get 11'p(T ® T) � li T ® T il· But for F = 'E;=l f; (wl ) g; (w2 ) E Yp we have
I ( t,
) l l ) (fn i � n p = II T 11 (fn i T ( � f; (wi ) g; ) (w2 ) 1 df..L (wi ) ) n p 2 T � II II ( i l � f; (w1 ) 9; L df..L (w l) ) n p 2 T II I1 (1 l ?: /; (w1 )9; (w2 ) 1 d(f..L X f..L ) (w b w2 ) ) J =l
T (g;)(w2 ) /; (wl ) I (T ® T)(F) (w1 , w2 ) l = T n T (g;)(w2 ) /; (wl ) df..L (wl ) P � II T 11 �
P
�
P
�
P
�
P
�
nxn
� II T II 2 II F II Yp · Thus 11'p(T ® T) � 1, so (21) together with the choice of T gives
This contradicts the definition of K.
a
252
III. G. SchattenVon Neumann Classes §22.
Remark. The reader familiar with tensor products will easily see that the above argument gives that an c:tensor product of psumming maps is psumming. 22. As an example of the applicability of previous results let us consider HilleTamarkin integral operators. Let (0, p.) be a probability measure space and let K(w1 , w2 ) be a function on n X n such that
2�p<
00 .
(22)
Then the formula TK ( f ) (w) = fn K(w , w2 ) J (w2 )dp.( w2 ) defines, as is easily seen, a linear operator TK : Lp( O, p.) > Lp( O, p.). For such an operator we have 2:�= 1 IAn (TK ) I P < oo. This follows from Theorem 19 and the fact that TK is pabsolutely summing. To see this put cp(w1 ) = (J0 I K(w1 , w2 ) 1 P dp.(w2 )) _!_v' . One checks that S(f) = TK (f) · cp  1 is a linear map from Lp (O, p.) into £00 (0, p.) so we have to check that for cp E Lp the map f �+ f · cp is pabsolutely summing from £00 (0) into Lp (n, p.) . This was observed in III.F.4. Note that HilleTamarkin integral operators are direct generaliza tions of HilbertSchmidt operators (see Proposition 13) . Let G be an abelian compact group with normalized Haar measure m and dual group For f E Lp' (G, m) , p � 2 we define a kernel K(g1 , 92 ) = !(91  92 ) · This kernel clearly satisfies (22). Since TK (9) = f * g we see that eigenvalues of TK coincide with j('y ) so we get the HausdorffYoung inequality I
f.
(
L l f('y ) I P
I'Er
)
1
p
� IIJII P1 > p � 2.
(23)
Notes and Remarks.
There are two excellent books which treat the matters explained in this chapter, and much more. They are Pietsch [1987] and Konig [1986] . The concept of approximation number is so natural that we have been unable to trace proper historical references. Nowadays it is an example of the general notion of snumbers (see Pietsch [1987] ) . Proposition 5 is due to Hutton [1974] . It is a quantitative version of the classical Schauder Theorem asserting that an operator is compact if and only if its adjoint is compact. Our material on Schattenvon Neumann classes is classical and can be found in many places. The above mentioned books contain nice
III. G. SchattenVon Neumann Classes §Exercises
253
presentations but also GohbergKrein [1969] and Simon [1979] should be mentioned. Proposition 12 showing the connection between HilbertSchmidt and pabsolutely summing operators on a Hilbert space was proved by Pelczyiiski [1967] . Theorem 14 is the classical result of Fredholm [1903] but in the theory of Fourier series it is usually associated with Bernstein [1914] and Szasz [1922] . Our proof is taken from Wojtaszczyk [1988] . Theorem1 6 and Proposition 1 7 and 18 are basically due to Grothen dieck [1955] . In the present generality Theorem 16 was proved by Pietsch [1963] . The fact that the eigenvalues of a nuclear operator on Hilbert space are absolutely summable actually characterizes spaces isomorphic to Hilbert space among all Banach spaces (see JohnsonKonigMaurey Retherford [1979] ) . This paper contains also the first proof of Theorem 19. Our proof of Theorem 19 is a modification of a proof given in Pietsch [1986] . The application of Corollary III.B.9 allows us to avoid the use of the general theory of Weyl's numbers. One should be aware that the subject of eigenvalue estimates of operators on X is related to best projections on finite dimensional subspaces of X. This is made clear in Konig [1986] 4.b where the estimates for eigenvalues are used to prove Corollary III.B.9. Our discussion of HilleTamar kin integral operators is taken from JohnsonKonigMaureyRetherford [1979] . Exercises 1.
Show that up (£2 )* = uq (£2 ) for 1 < p < oo and � + � = 1 , and also O'oo (£2 )* = 0'1 (£2 ) and 0'1 (£2 )* = L(£2 ). The duality is given by
(T, S)
2. 3.
=
trTS.
Show that the space u1 (£2 ) has cotype 2.
For A C N let PA : £2 t £2 denote the natural coordinate projection defined by PA (E� 1 a;e;) = L; E A a;e; . (a) Show that maps T �+ TPA and T �+ PAT are contractions on O'p(£2), 1 � p � 00 . (b) Show that if p =1 2 then the operators Ti; , i, j = 1, 2, . . . defined by Ti; (E;;: 1 a k e k ) = a i ei do not form an unconditional basis in O'p ( £2 ) ·
(c) Show that O'p, P =1 2 and p =1subspace of Lp(JL) .
oo,
is not isomorphic to any
III. G. SchattenVon Neumann Classes §Exercises
254 4.
If X is an infinite dimensional subspace of a00 (i2 ) , then X contains an infinite dimensional subspace x1 complemented in O'oo (i2 ) such that either X1 rv Co Or X1 rv £2 .
5.
Show that
6.
Show that there is no norm on A 1 (i00, il ) which is equivalent to the quasinorm a1 (·).
7.
Show that for every (.X n )�= 1 E i2 there exists a nuclear operator i 1 EB 00 + i1 EB 00 such that the eigenvalues of T are precisely
O'p
rv
CL:
O'p)p for 1 :5 p <
00
and O'oo
9.
(E O'oo ) o.
T: l
l Suppose that K( x , y) is integrable on [0, 1] [0, 1] and that 1 1 I K (xi. y)  K(x2 , y) i 2 dy + 0 as l x 1  x2 l + 0. 1 Show that the operator Tf( x ) = f0 K( x , y)f(y) dy acts from C [O, 1] into C [O , 1] and has square summable eigenvalues. Suppose (f!, f..L ) is a probability measure space and K(e , 11 ) is a func tion on f! f! such that pq ( , "l) ( K ( e df..L i ) fn fn .,W 1 df..L (e) < ± .\1 , ± .\2 , ± .\3 , . . . .
8.
rv
x
X
oo ,
where p and q are positive numbers such that � + � :5 1. Show that given as = fn the operator maps where � + � = 1, and (.Xn (Tk ))�= 1 E into q + = max q' 2).
TK TK(f)(e ) K(e , "l)f("l) df..L ("l) lq + Lq' (!"!, f..L) , (, Let In : a00 (i�) + a2 (i�) be the formal identity. Show that there exist constants 0 < < C such that 1r 1 ( ln ) :5 C
Lq' (!"! , f..L)
10. (a)
c
for n = 1 , 2, . . . .
en
:5
n
(b) Let Jn : a1 (�) + a2 (i�) be the formal identity. Show that 1r 1 Jn :5 c..fii for some c > 0.
( )
(c) Show that 'Yl (Jn ) ;:::: en for some constant c > 0, where i 1 and {3: £1 + a2 (i�) 'Yl (Jn ) = inf { ll a L B : a : a1 � and {3a = Jn} ·
ii I
( ) +
(d) Use the above to show that there exist a 1absolutely sum ming operator which does not factor through any £1 )space. Another such example is given in Exercise 111.1.2.
(f..L
255
III. G. SchattenVon Neumann Classes §Exercises
(e ) Show that, if X is a complemented subspace of a Banach space
with an unconditional basis, then every !absolutely summing operator from X into any Banach space factors through £ 1 .
( f ) Show that
a 1 (£2 ) and a00 (£2 ) are not isomorphic to any com plemented subspace of a space with an unconditional basis.
11. Suppose that T: ey + £� is such that li Te; II Show that rank T ;:::: 1 ;1 •
2
2:::
1 for j
=
1, 2, . . . , n.
12. Let u: £� + ey be a linear operator given by the matrix (u( i, j) ) i,;= 1 . Show that L l u(i, j) l � na1 (u).
i ,j
13. ( a) Show that every !absolutely summing operator T: C(K) + £2 is 1nuclear. Note that A(D) does not have this property ( see III.F.6 ( a)) . (b ) Show that the space C 1 ('1'2 ) is not isomorphic to a quotient space of any C ( S ) space.
( c ) Show that every linear operator
T: L 1 (J..L ) + L2 ( v ) maps order bounded sets into order bounded sets, i.e. sets of the form V9 = { ! E L 1 (J..L) : 1/1 � g for g E L1 (J..L) } are mapped into sets { ! E L 2 (v): 1 /1 � g, g E L 2 (v)}.
14. ( a) Let E be any ndimensional Banach space and let i dE be the identity operator on E. Show that 1r2 (idE) = yn. (b ) Let (0, J..L ) be a probability measure space and let E
C
L00 (0, J..L ) be an ndimensional subspace. Show that there exists e E E such that ll e ll oo = 1 and ll e ll2 � n ! . ( c ) Let E be an ndimensional Banach space. Show using ( a) that d ( E, £�) � yn. This gives an alternative proof of an important special case of III.B.9.
( d ) Let
E be an ndimensional Banach space. Show using ( a) that A(E) � yn. This gives an alternative proof of an important
special case of III.B. lO.
15. Suppose that I · 11 1 and II · 11 2 are two Hilbertian norms on an n dimensional space X. Show that there exists a subspace X1 c X, dim X1 � � and a constant a such that
256
III. G. Schatten Von Neumann Classes §Exercises
16. Let E be a subspace of f� . (a) Show that there exists an x E E such that llxll l {i: lx(i) l = 1} 1 � dim E.
1 and
(b) Show that d(E, £'2) � � JN where n = dim E.
17. (a) Let T:;::' be the space of trigonometric polynomials on 1I', of degree at most n, with the supnorm. Suppose that E is a subspace of T;::' and dim E � a(2n + 1 ) . Show that there exists a polynomial p(O) = L �= n a k ei k O E E such that II P II oo = 1 but (2: �= n l a k l 2 ) ! � C(a)y'n where C(a) > 0, does not depend on n. (b) Let W;:" (§d) be the space of polynomials homogeneous of degree n restricted to §d and equipped with the sup norm. Suppose that E is a subspace of W:;::' (§d) with dim E � a dim W;:" (§d) · Show that there exists p E E such that II P II oo � C(a, d) II P II 2 where C(a, d) > 0 does not depend on n (for more information about the spaces Wrf(§d) see III.B. 14) .
III.H. Factorization Theorems
This chapter discusses some very important and powerful theorems about factorization of operators with values in £,spaces, 0 :::; p :::; oo . These theorems express the fact that sometimes the map ( not necessar ily linear ) from a Banach space X into L,(O, JJ. ) is 'essentially' a map into some smaller function space, notably Lq or L q , oo for some q > p. Contrary to our general custom we will discuss operators more general than linear. We start with the Nikishin theorems which discuss a factor ization through weak L,. As a preparation we discuss some properties of operators from a Banach space X into Lo(O, JJ. ) . Then we discuss factor ization through L, spaces. For linear operators we show the connection between factorization and pabsolutely summing operators and prove that every linear operator from L,, 2 :::; p :::; oo, into Lq , 0 < q :::; 2, fac tors strongly through £ 2 . We apply these results to reflexive subspaces of £1 (JJ. ) , to the structure of series unconditionally convergent in mea sure, to the MenchoffRademacher theorem about almost everywhere convergence of orthogonal series or series unconditionally convergent in measure, to Fourier coefficients of Holder functions and to multipliers of some spaces of analytic functions. Let us start with the precise definition of sublinear operators, for which we will prove our basic factorization results. 1.
Definition. An operator T: X + Banach space is called sublinear if
L,(O, JJ.) , 0 :::; p :::; oo ,
where X is a
( a) I T(x + y) i :::; I T(x) i + I T(y) i for x, y E X,
( b ) IT(Ax) i = IAI · I T(x) i for x E X and all scalars A, where the above inequalities are understood pointwise JJ.almost every where.
Clearly every linear operator is sublinear. If 8: X + L0 is a linear operator then Tx(w) = I S(x) (w) i is a sublinear operator. More generally if S: X + Lo (Y) is a linear operator, where X, Y are Banach spaces and L0(Y) is the space of strongly measurable Yvalued functions, then
258
III.H. Factorization Theorems §2.
= I I S(x) ll v is a sublinear operator. In particular if we have a sequence Sn : X � L0 of linear operators then
T(x)
M(x) (w) = max n I Sn(x)(w) l
and
Q(x) (w)
=
00
L I Sn(x) (w) l 2
n= l
are sublinear operators ( provided they do exist ) , because we can consider the operator S: X � Lo (foo) or S: X � Lo (£2)
defined by S(x) = (Sn (x));;:"= l · This covers a wide variety of maximal operators and square function type operators. It is such examples that convinced us to consider sublinear operators. 2 Proposition. Let T: X � £0 (0, J.L), with J.L a finite measure, be a sublinear operator. The following conditions are equivalent:
( a) T is continuous at 0; (b) T(Bx ) is a bounded set in L0 (0, J.L); (c ) there exists a function C ( .A ) defined for .A � 0 such that J.L{W
E n: I Tx(w) l > A · llxl l } :::; C ( .A )
and lim C ( .A ) = 0 as .A � 0. a =? b and c =? a are obvious from =? a. Since T(Bx ) is absorbed by every
the definition. Let us prove b neighbourhood of zero in £0 (0, J.L) we get that for every .A > 0 there exists an N>. such that Proof:
T(Bx )
s;;;
N>. · { f E Lo ( O, J.L): J.L{w E 0: 1/ (w) l > .A } < 1 / .A } .
This gives J.L{w: ITx(w) l � .A · N>. · llxll } :::; 1 / .A, so the desired C ( .A ) a exists. Let Tn: X � L0 (0, J.L) , n = 1 , 2, . . . , be a sequence of continuous linear operators. Assume that T• (x) = supn I Tn (x) (w) l < oo, J.Lalmost everywhere for every x E X . Then T• is a continuous sublinear operator from X into Lo (O, J.L) . 3 Proposition.
Let us put Lo(foo) = { ( /n );;:"= 1 : 11/n (w) ll oo E Lo(O, J.L)}. With the natural structure it is an Fspace. It follows from the closed
Proof:
259
III.H. Factorization Theorems §4.
graph theorem I.A.6 that the operator U: X + Lo (ioo ) defined by U(x) = (Tn (x))�= l is continuous. Thus T• (x) (w) = II U(x) (w) lloo is a also continuous. A word about notation. It is customary in harmonic analysis to denote the maximal operators such as our T• = supn I Tn(x) (w) l as T* . This, however contradicts the functional analysis usage of T* as an ad joint operator. Let X be a Banach space and let T: X + Lo (f! , J.L) be a sublinear operator. We say that T factors strongly through Lp ,oo if there exists a function g on n such that
4 Definition.
In more operatortheoretical terms we can express this mutative diagram
as
a com
T
X + Lo (f!, J.L)
� � Lp, oo
where M9 (f) = g · f.
5 Proposition. Let X be a Banach space and let T: X + L0 (f!, J.L) be a sublinear operator continuous at 0. Let us assume that J.L(f!) = 1 and 0 < p < oo. The following conditions are equivalent:
(a) T factors strongly through Lp ,oo i (b) there exists a function C( .X ) with lim>.+ oo C( .X ) = 0 such that J.L { w E f!: sup I T(xi ) (w) l
;?:
.X ( L I xi I IP ) 1 1P} j
::;
C( .X )
for all finite sequences (xi ) in X ;
(c) for every positive c there exists a constant Ce > 0 and a set EE with J.L(f!\Ee) < c such that
J.L{ W E Ee : I T(x) (w) l > .X} :5 ce ( 11 � 11 r
c
n
Ill.H. Factorization Theorems §5.
260 for all x in X and all positive >.. Remark.
The condition (b) means that the operator T : (EX)p + L0(ioo ) defined by T (xi ) = (T(xi ) ) is continuous. (a)=>(b) . Let us take (xn)�= l with :E ll xn ii P = 1. Using the notation from 4 we have
Proof:
J.t{w: sup I T(xi ) l � >. } = J.t{w: g(w) · sup I To (xj ) l � >. } $ J.t{w: g(w) � JA } + J.t{w: sup i To (xi ) l � JA } $ J.t{w: g(w) � JA } + L J.t{w: I To (xj ) l � JA } j $ J.t{w: g(w) � JA} + � J
( '1�1 r
$ J.t{w: g(w) � JA } + >. P / 2 = C(>.) . One checks that this function C(>.) satisfies the desired conditions so we have (b) . (b)=>(c) . Let us fix a function C(>.) as in (b) and for a given c > 0 let us fix a number R such that C(R) < c. Let us consider the following condition on a subset F c !l 3x E X, ll x ll $ 1 such that J.t(F) · I Tx(w) I P > RP for all w E F. If no subset F have
c
n satisfies (*) then for every x E X with ll x ll tt{w e n: I Tx(w) l > >.} $
( �Y ·
=
(*) 1 we
If there are sets in n satisfying ( *) let us fix a maximal family of disjoint sets (Fj ) satisfying (*) with corresponding Xj E X, ll xi ll $ 1. For ci = J.t(Fj ) l !P we have E ll cixi iiP $ 1 and supi I T(cjXj ) (w) l > R a.e. on F = Ui = l Fi . Condition (b) yields tt(F) $ C(R) $ c. We will show that E = 0\F satisfies (c) with Ce = RP. If not, there exist x E X with ll x ll $ 1 and a number >. > 0 such that J.t{w E E: I Tx(w) l � >.} > (Rj>.)P. Thus {w E E: I Tx(w) l � >.} satisfies (*) and is disjoint with F. Since ( Fj ) was a maximal family we get (c) . 1/n, n 1 , 2, . . . let En and Cn CE n be (c)=>(a) . For c n given by (c) . Let us fix a sequence of positive numbers O n such that I: Cna ;; P 1 and for Fn = En\ Ui
=
=
=
III.H. Factorization Theorems § 6.
261
g(w) = E:= l anXFn · One sees that u:= l En = u:= l Fn = 0 and Fn 's are disjont. For x E X and >. > 0 we have w I Tx(w) l > an >.} � f: cn ( M f an >. n= l = (� Cna;? ) · ( l x i / >. ) P = ( l x i / >. ) P . II 00
� L tt{ E En:
n= l
The really important implication in Proposition 5 is (b)=>(a) . It allows us to establish that some operators do factor strongly through Lp,oo · The main example of it is the following
T
6 Theorem. (Nikishin) . Let X be a Banach space of type p and let (0, tt) be a afinite measure space. Then every sublinear operator from X into L0 (0, tt) continuous at zero, factors strongly through Lp,oo ·
As is well known (see II.B.2) there exist a probability measure on 0 and a measurable function ¢ such that the map f �+ f · ¢ is a topological isomorphism between L0 (0, tt) and L0 (0, Jt) . Thus without loss of generality we can assume tt(O) = 1 . In order to check condition (b) of Proposition 5 let us fix . . . , Xn E X with E � 1 . Let are Rademacher functions; see E X (the us put = where = 1 and I.B.8) . For every k, 1 � k � n , let = + = + 1 for j "I k. Clearly = � €j For almost all E 0, the functions and have the same distribution as a function of so Proof: v
x1. x2 , ' I xi l i P rj s cjrj(t)xj c 2 I T(xk )gfl I TLj(gt gf) l I T(gt ) l k I T(gf ) l . w I T(g )(w ) l I T(gf)(w) l t, t l {t E [0, 1] : I Tgt (w) l � I Tx k (w) l}l � 1 / 2. Since I Tx k (w) l does not depend on t we get that for almost all w E 0 l {t E [0, 1 ] : ITgt (w) l m:X ITxk (w) l } l � 1 / 2. 9t Li ri (t)xi
�
III.H. Factorization Theorems § 7.
262
A>0 tt{w : s�p I Txk (w) l A} $ 2. I i {t E [0, 1] : I Tgt (w) l A}i dtt (w ) :::; 2 · l [ l {t E [o , 1] : l ut l ;:::: v'X}I (1) + l {t E [0, 1] : I Tut (w) l v'XI I ut l }l ] dtt(w) = 2 · l {t E [0, 1] : I Yt l ;:::: v'X}I + 2 . I JL{w E n: I Tgt (w) l ;:::: v'XI I ut l } dt. Since X has type p, the Markov inequality, sometimes also called Cheby
Thus for every
2::
2::
2::
shev's inequality, gives
T
l {t E [0, 1] : I Yt l ;:::: v'X}I :::; T, (X)P . A p/2 • A +
Since is continuous at zero, Proposition 2 shows that the last integral in (1) tends to zero as oo. This verifies condition ( b ) of Proposition a 5 and proves the theorem.
X g 19
7 Corollary. Let G be a compact abelian group with Haar measure m. Assume that is a Banach space of type p and that we have a represen
X. T9f(h) = f(h + g) T: X + T9TI91 = T g
tation �+ of G into the isomorphisms of Let for E L0 (G, m) . Assume that L0 (G, m) is a continuous sub linear operator such that for all E G . Then T is of weak type q with q min (p, 2 ) .
f
=
Using the Nikishin theorem we see that condition (c ) of Propo sition 5 gives a set E C G with � such that
m(E) > m( {g E G : I Tx(g) l > A} n E) :::; c ( I � I r for all E X. Since T9TI91 = T we get {g E G: I Th (x)(g) l > A} = h + {g E G: I Tx(g) l > A}. Comparing (2) and (3) we see that for every h E G
Proof:
X
(2)
(3)
263
III.H. Factorization Theorems §8.
Integrating (4) with respect to h and using the fact that h E G we get
l h l � C for
m {g E G: I Tx(g) l > .X } � m(E)  1 c ( " � " r·
a
8 Comment. The Nikishin theorem and Corollary 7 formalize several well known equivalences between existence almost everywhere and weak type (11) for certain operators. In particular we have
Let f E £1 (T) and let f(rei 9 ) denote its harmonic extension. Let f(rei9) denote the harmonic function conjugate to f(rei9 ). For f � 0 the function G(rei9 ) = (1 + f(rei9 ) + if(rei9 ))  1 is bounded and analytic. The Fatou theorem shows that lilllr + l G(re i 9 ) exists a.e. . This implies that limr+1 j(re i9 ) exists a.e. . This implies that such a limit exists a.e. for arbitrary f. For the background on this see I.B.19 and the references given there. So (see Proposition 3) Mf(O) = maxo< r
Example 2.
The following are equivalent for 1
� p � 2.
(a) For every f E Lp(1l') the series :E �oo ] (n) · e in 9 converges to f a.e. (b) For every f E Lp(1l') we have max N
L �N
I L�N ](n)ein9 1 <
oo
a.e.
] (n)ein9 1 is of weak type (p, p). (c) The map f �+ maxN I The only nontrivial (given Nikishin's theorem) implication is (c)=>(a) . This is however a standard approximation argument. We have almost everywhere convergence on the dense set of smooth functions and (c) allows us to get the same for all f E Lp(T). Actually the well known CarlesonHunt theorem (see Carleson [1966) and Hunt [1968) ) says that (a) ,(b),(c) hold for p 1 and the classical Kolmogorov [1926) example shows that they are false for p = 1.
>
9.
The following definition is clearly analogous to Definition 4.
Definition. Let X be a Banach space and let T: X + Lp(f!, 1L), 0 < oo , be a sublinear operator. We say that factors strongly through Lq , q p if there exists a positive function g E Lr ( f! , /l) , with 1/ q + 1/r = 1/p, and a cl!ntinuous sublinear operator u: X + Lq (f!, 1L) such that Tx = g u ( ) for x E X .
p<
>
T
· x
264
III.H. Factorization Theorems § 1 0.
Note that if T: X + Lp(f!, f.L) factors strongly through Lq , 0 $ p < then T { X ) c Lp(E, f.L) where E = supp g is a ufinite set. Also, there exists a function cp on E and a probability measure v on E such that the map I �+ I · cp is an isometry between Lp(E, f.L) and Lp(E, v ) ; see III.A.2. These remarks show that we can always assume that {f!, f.L) is a probability measure space. q$
oo,
10. Our next important result will be the following proposition which gives conditions on an individual operator to factor strongly through Lp(f!, f.L) . It is analogous to Proposition 5.
Proposition. Let T be a sublinear operator from X into Lp(f!, f.L) , 0 < oo. Then, the following conditions are equivalent:
<
p
(a) T factors strongly through L q , q > p; (b) there exists a constant K such that for every finite sequence (xi )'J= 1 c X
Proof: Observe that condition {b) implies that T(X) C Lp(E, f.L) for some ufinite set E c f! . If this is not the case then there exist a sequence (xi ) �1 c Bx and disjoint subsets of n, (Ai ) �1 such that JA3. IT(xi ) I P df.L 2::: c for some c > 0 and all j's. But then for all N we have
2:::
(�
i3 I T(xi ) IPdf.L) 1 /P
2:::
(cN) 1 /P
which is impossible for large N. Thus we can assume (see 9) that {f!, f.L) is a probability measure space. (a)=?{b) . This is easy. We have
(fo ( t I T(xi W) pfq df.L) 1 1p p/ q /p q (In gP ( t j u(Xj ) i ) df.Lr =
265
III.H. Factorization Theorems § 1 0.
j
j
g
J..L)
The implication (b):::} ( a) is more difficult. Let us observe that ac tually we are looking for a positive function in Lr(O, such that The case q = oo is rather simple. In this case (b) � reads
l g 1 T(x) l q K l x l ·
This implies that
g = sup{ I T(x)l:x E X, l x l � 1 } exists in Lp ( O, J..L ) and I 9 I � 1 (see Exercise III.A.1). This is the desired g (note that for q = weP get p = ) In the case q <
oo
oo
r .
let us define
(5)
xf, . . . ,x� 1 /n) p J0(Ej I Txjl q )pfqdJ..L (Ej I Txj l q ) fq. (Ak );;'= 1
E7= 1 l xjl l q K;;: q (1 n (nk );;'= 1
with We put K = lim Kn· Fix + � = 1 . Let us define functions In = such that Clearly In � 0 and J In = 1 for = 1, 2, . . . . For any sequence of disjoint subsets of n and any sequence of integers, we have using (5)
266
III.H. Factorization Theorems § 1 1 .
This shows (see III.C. 12) that the sequence Un )':= t is uniformly inte grable in Lt (O, f.L) . Let h be any u(Lt (O, f.L) , L00(0, f.L)) cluster point of this sequence. Clearly h is nonnegative and J hdf.L = 1. For any element E X with = 1 and any number 2::: 0 and any n = 1 , 2, . . . we have from (5)
x
t
l xl
1n (f�fp + tq i Tx l q )pfq df.L = Jrn ( t I Txjl q + I T(txw) pfq df.L pf q :5 K� + l ( t l xjl l q + tq ) :5 K� + l (K_; q (1 + 1/n) + tq ) pf q . 3=1
Passing to the limit with n we get
t
Since for = 0 this inequality becomes an equality we see that for small we have
s
Routine differentiation gives
a
Thus we can put g = h�  � 
11. It is possible to present a result analogous to Nikishin's theorem 6 for strong factorization through L (O , f.L) of linear operators. It requires, however, a different notion of type (see Notes and remarks) . Only the case of type 2, the most important for applications, can be easily handled with our concept of type.
p
T
Assume that X is a Banach space of type 2 and that 2, be a continuous linear operator. Then factors strongly through L2 (n, f.L) . Corollary.
T : X + Lp (O, f.L) , 0 < p <
267
III.H. Factorization Theorems §12.
We will check condition ( b) of Proposition 10 assuming that
J.L) is a probability measure space. As we know (see 9 and the begin ning of the proof of 10) this is enough. For a finite sequence ( xi ) j= 1 X Proof:
(0,
C
we have from the Khintchine inequality ( see I.B.8 )
(j ci; I Txii 2 )PI2 dJ.LY IP cp ( j j 1 � ri (t)Txi (w) I PdtdJ.L(w) y 1P 1 = cp ( J J I T ( � rj (t ) xj )(w) I P dJ.L(w)dt ) /P 1 Cp i T I (/ I � ri (t) xii i Pdt) /P Cp i T I T2 (X)(Lj l xil l 2 ) 1 12 • '5:
3
3
3
'5:
3
'5:
II
12. For spaces of type 8, with 8 < 2 we have the following result which is weaker than one might expect looking at Corollary 1 1 . Unfortunately it cannot be improved.
X is a Banach space of type 1 < < 2 and T: X Lp(O, J.L) is a linear operator, 0 < p < then for every q, with p < q < T factorizes strongly through L q (O, J.L ) .
Theorem. +
8,
If
8,
8,
8
We will check that condition (b ) of Proposition 10 holds. Let be a finite sequence in Let be a sequence of qstable independent random variables (see III.A. 13 ) . We have ( see III.A.16 )
X.
(xj )j=1
Proof:
n
l!.
("lj (r ))j= 1
.l
( j ( � I T(xj )(w W) dJ.L(w) ) c ( J 1 1 t, �; (T)T(x; )(w) l .drdp (w) ) l = c ( j J I T (t 1Ji (r)xi) (w ) I PdJ.L(w)dr) ; C II T II ( j I 1Ji ( r )xj l i P dr) ; . q
P
�
'5:
t
(6 )
268
III.H. Factorization Theorems §13.
T/j ( ) are symmetric variables and X has type s we have vd vd d r ) ) ( ( ) (7) ( = xj xj t j j j r TJ r r rt j J l l l l � f � "' :::; Ts (X)P j ( � I TJj (rW I xi l i s ) ; dr. Now we take a sequence (''1j( O )) j= 1 of sstable, independent random variables. Like previously we get T
Because
j ( � I TJi (rWI I xil l s ) ; dr = j j I � 'Yj (O)TJj (r) l xil fdOdr (8) = j ( � I 'Yi (OW I xil l q ) � dO :::; ( j � I 'Yi (OW I xil l qdo) � = (! i'Y1(0 W dO) � ( � l xil l q ) � Since q < s we get from III.A. 15 that J iy1 (0) I q d(} < so putting together (6) , (7) , (8) we see that condition ( b ) of Proposition 10 holds.a ·
oo
X
L1 (J.L) , J.L X L1 (J.L) = l gl r =
13. Now we would like to discuss an application of the previous the orem. Let be a reflexive subspace of a probability measure space. From III.C.18 we see that has type s for some s > 1. Applying Theorem 12 to the identity operator i: � we get a factorization through for any 1 < q < s. Explicitly for every 1 < q < s there exists a function 1, � + � 1 and a constant K such � 0, that
Lq (J.L)
X
g(w)
�(w) = gr (w) we have Corollary. If X is a reflexive subspace of L 1 ( J.L), J.L a probability mea 1 , and K and a positive function � with sure , then there exist If we write
q >
269
III.H. Factorization Theorems § 1 4.
l 6 l 1 = 1 such that for every x E X an p
L1 (p,) space is isomorphic to a subspace of some Lp (p,) space for
Note that this means in particular that every reflexive subspace of
> 1.
There i s a connection between absolutely summing operators and the factorization problems discussed here. We do not intend to present here a full picture, so we will limit ourselves to the following remarks. Suppose we know that for some and some p, = a factor This gives us for every operator ization 14.
X
llp (C(K), X).
C(K)
L(C(K), X) T: C(K) + X
X
T
�Lp (K,p,)�
Dualizing we get
X*  M(K) T*
�Lp'(K,p,)�
M(K) L1(v) id*: Lp'(K,p,) + L1(v) 'll (C(K), X) X* L1 (p,) p
Note that if we treat as then is a multiplication operator. Since both properties = and 'every operator from into factorizes strongly through are local ( for the last one see Proposition 10) there is no problem with taking duals. This is rather imprecise but for future reference we will prove
L(C (K), X)' Lp (p,)'
X* has type p for some p > 1 then L(C(K), X) = ' llq (C(K), X) for every p , where p p1, = 1 . Proof: It is enough to check that L ( c0, X) = llq ( c0, X). If T: eo + X then T*: X* + £1 . Theorem 12 gives a factorization Proposition.
If
q >
1 +
III.H. Factorization Theorems § 1 5.
270
e e
Thus T** : ioo + X** has a factorization M*Ti* . Observe that M ( n ) = mn n for some m n E lq , thus M* : i00 + lq is given by M* ( n ) = mn n · It is immediate (see III.F.4) that M* is qabsolutely sum ming. So T** is qabsolutely summing and we get that T is qabsolutely a summing.
( e) ( e)
( )
15. The following proposition, going in the other direction, is more involved. It gives another condition for a space X ensuring that every linear operator from X into Lp( O. , p,) factors strongly through L q (O., p,) . Proposition. Let X be a Banach space and let 0 < p < q, q � 1 . Let us assume that for some C < oo and for every linear operator u: X* + lq we have 11"p(u) � C1rq (u) . Let us ass ume moreover that if p < 1 then X has the bounded approximation property. Then every linear operator T: X + Lp(O., p,) factorizes strongly through L q (O., p,) .
We will check condition (b) of Proposition 10. Given a finite sequence (xi ) J= l in X we can find (using the b.a.p. of X if p < 1 or the b.a.p. of Lp( O. , p,) if p � 1) an operator X + Lp(O., p,) such that T (x) = 2:: � 1 x k (x) X A k for some disjoint sets Ak c 0. � and
Proof.
I TI KI TI ,
T:
(9)
Let us define u : X* + i� by the formula u(x* ) = (xi (x* ) ) j= l · One can easily check that 11"q (u) � ( 2:j= 1 ll xi ll q ) l f q so 11"p(u) � C(2:j= 1 xi ll q ) l / q . Let us define a function
I
271
III.H. Factorization Theorems § 1 6.
We have ( see III.F.33 and (9) )
n n 2 · c · (jL= l l xj l q ) l fq Wfl l � 2KC I T I (jL= l l xj l q ) 1 1q . In the above 8: X** + Lp (O., Jt) is given by S(x**) 2:: � 1 x**(xk)XA k · Clearly I B I = I T I · 16 Corollary. Every linear operator T: Lp + L q (O., Jt) , 0 < q � 2 � p� factors strongly through L 2 (0., Jt) . �
=
a
oo,
The case p < oo follows immediately from Corollary 11 and Theorem III.A.23. For the case p = oo let us note that by Corollary III.F.35 we have for 0 < p � so we can apply = Proposition 15.
Proof:
2,
B(L1, £2 ) IIp (£1, £2 )
Remark. Corollary 16 for q � 1 and p = from Corollary 11 and Theorem III.F.29.
oo
can be derived directly
2::
We say that the series In is unconditionally convergent in measure if L Cnln is convergent in measure for every (en ) E £00 • Let us discuss now the structure of such series. 17.
Theorem. (0rno) . ·I£ 2:: � 1 In is unconditionally convergent in mea sure on [0, 1}, then there exist a function g(t) on {0, 1}, a sequence
272
III.H. Factorization Theorems § 1 8.
(an);;:='= 1 E £2 and an orthonormal system (wn);;:='= 1 in L2 [0, 2] such that fn(t) = Gn · g (t) Wn(t) for t E [0, 1] and n = 1, 2, . . . . Proof: By the closed graph theorem I.A.6. the map T: £00 L 0 [0, 1] defined by T(cn) = 2::, 1 cnfn is continuous. We have the following commuting diagram: ·
+
co . foo
id
j
id
T
• Lo [O, 1]
ul /
2 ( JL)  L2 [0, 1] Operators and M9 ( recall that M9(f) = g· f) are obtained by applying first Theorem 6 and next Corollary 16 to the operator T. The operator is 2absolutely summing by Theorem III.F .29, so by the Pietsch theorem III.F .8 there exists a probability measure JL on the set N of natural numbers and an operator u such that the above diagram commutes. Let = In · g  I , we us put hn = fn( l u i JJL({n } )g) 1 . Since u(en) = get for an arbitrary sequence of scalars (an);;:='= 1 00 00 hn = an � � t2 [0, 1] � � l u l � u(en) t2 [0, 1] 00 1� � I � virr;;}(n i L 2 (N ,JL) = (2: l an1 2 )
u
L N,
u
u
u(en)
•
In order to complete the proof we appeal to the following.
If functions (hn);;:='= 1 L [0, 1] are such that ( 2:�=1 l an l 2 ? /2 for every finite2 sequence of scalars l �exists 2::,then I(an), 1 anhnthere an orthonormal system Wn in L 2 [0, 2 ] such that wn i [O, 1] = hn for n = 1, 2, . . . . 18 Lemma.
C
This lemma follows ( see below ) from the following general result about operators on Hilbert spaces:
I Ti l
19 Theorem. (Dilation Theorem) . Let H be a Hilbert space and let T: H + H be a linear operator with � 1. Then there exist a Hilbert space 1t :::> H and an isometry U: 1t + 1t such that PU I H where P is the orthogonal projection from 1t onto H.
T=
273
III.H. Factorization Theorems §20.
Let us define = ( 2:: ;:'= 1 H) 2 and embed H into onto the first coordinate. Let U(h1, h , . . . ) = (Th1, Sh1, h , ha, . . . ) where S = (id  T*T) 1 12 . Clearly PU I H 2= T. 2In order 2to check22 that U is an isometry it is enough to check that 1 Th l + I S h l = l h l for all h E H. But 1 Th l 2 + I Sh l 2 = (Th, Th} + ( (id  T* T) 1 /2 h, (id  T* T) 1 12 h} = (T* Th, h) + ((id  T* T)h, h) = (h, h) = l h l 2 . Proof of Lemma 18. Let (
Proof of Theorem 19. 1t
II
1t II
n
C
in
9
21. 0rno's theorem 17 can be very useful in transferring properties of general orthonormal series to series unconditionally convergent in mea sure. There are several instances known when this is possible. One such case is the extension of the MenchoffRademacher theorem on almost everywhere convergence to general series unconditionally convergent in measure (see Corollary 25) . The proof of the MenchoffRademacher theorem which we will present now uses the theory of pabsolutely sum ming operators. Let us state the following proposition whose proof will be given later.
(an)�= 1 be such that 2::;:'= 1 l an l 2 log2 (n + 1) < u:£2 + ioo defined by u(ej ) = (2:j= 1 ei ai):'= 1 is 2absolutely summing and 11'2 ( u) c JI:� 1 l an l 2 log2 (n + 1).
Proposition. Let Then the operator
22.
result:
oo .
:5
Using the above proposition we will show the following classical
274
III.H. Factorization Theorems §23.
Let Cfn)':'= l be an orthonor L2 (n, J.L) and let E:=l l an l 2 log2 ( + 1) < Then the E:=l anfn converges J.Lalmost everywhere. Proof: Let us define functions � 8 : n + £2 , = 1, 2, . . . by the formula �8 (w) = (fi (w ), h (w), . . . , f8 (w), 0, 0, . . . ) . Theorem. (MenchoffRademacher) .
n
mal system in series
oo.
s
From Proposition 21 and III.F.33 ( b ) we get
J l a�8 (w) l !, dJ.L(w ) 1) } � c(� l an l 2 log2 ( sup j l e ( � 8 (w)) l 2 dJ.L(w). Using the definitions of a and � 8 and (11) we get n +
(E£2 , 11(11 9
(11)
Since s was arbitrary from (12) we infer that
For a fixed sequence
(an), ( 13) obviously yields
This easily implies that
E:=l anfn(w ) converges J.Lalmost everywhere.a
In the proof of Proposition 21 we will use the following two lemmas.
Tn: Xn + Yn sequence of operators between T: (:EXnh + (:EYnh by T((xn)) = 11'2 (T) ( E 11'2 (Tn ) 2 ) 1 12 .
23 Lemma. Let be Banach spaces and let us define Then �
(Tn(xn)).
a
275
III.H. Factorization Theorems §24.
The proof of this lemma easily follows from Theorem III.F.8 and is left as an exercise.
u:if + f.�1 2be defined u(e3) = (E;= 1 a3e3 ) := 1 • Then 1r2 (u) :5 C (Ej': 1 l a3 1 2 ) 1 log( N 1). Proof: One can easily compute that u is given by a matrix [a n, m l:, m = 1 where an,m = { am0 ifif >:5 n.n, Let us consider the matrix valued function A(9) = ( E �= O e i k9 ) [ei(n  m) e amJ:.m= 1 . We easily see that for every 9, A(9) represents a onedimensional operator. When we treat A( 9) as an operator from if into f.� we get 1 N N 1r2 (A(8)) = I A (9) 1 = I � eikS I · ( &, l am l 2 ) /2 . 24 Lemma.
Let
as
+
m
m
This implies
1r2( 2� 12� A(9)de) :5 ( t1 l am l 2 y 12 2� 12� � t, eike l de
Integrating coordinatewise we check that (14) gives the lemma.
(14)
(27r)  1 f02� A(9)d9 = u so
Let us fix a sequence
a
(an)�= 1 . Let
IN = {n: 2N < n :5 2N+ 1 } and let l2 (IN) ( resp. l00(1N )) denote the subspace of £2 ( resp. 1.00 ) consisting of vectors supported on lN. Let VN = (an)�= 1I IN . Let us write £2 = X 1 Xo where X1 = span{ vN } �= o· Clearly Xo = ( E�=0 f.g (JN ) ) 2 where lg (IN) = {x E l2 (IN): x..lvN} . We will show the desired estimates for 1r2 (ui Xo) and 1r2 (ui X I) . We have l u (vNI I vN I  1 ) 1 :5 ( _L l a3 1 2 ) 2 :5 1/N ( _L l a3 1 2 Iog2 (j + 1) ) 2 • Proof of Proposition 21.
Ef)
1
1
J E [N
J E [N
276 Thus
Ill.H. Factorization Theorems §26.
a i X1 admits a natural factorization (3
where
(vNI I vNI I 1 ) = N1 eN, (3(eN ) = N · l vN I  1 I a (vN1) I · eN, v (eN) = a(vN ) · l u (vN ) I ;, . a

Since a is a diagonal operator between Hilbert spaces with the diagonal (1/N)/l= 1 it is HilbertSchmidt, so by III.G.12 we have � C. Since 1 12 2 2 for desired log 1 + the estimate (n an ) 1 1 !3 11 � o ::: := 1 l l follows. In order to estimate let us note that C From Lemma 24 we get
) 1r2 (ai Xo)
1r2 (a) 1r2 (a i X1) a(lg(IN))
l� (IN ). 7r2 (a l lg (�N )) � log 2N ( jLElN l aj l 2 )" 2 � c ( jL laj l 2 log2 (j 1) ) ElN 000 (1N ) ) 2 has 2This implies (use Lemma 23) that ai Xo: Xo o::: £ N 1 2 summing norm at most c o ::: := 1 l an l 2 log2 (n 1 ) ) 1 . A fortiori, the same holds for a: X0 £00 • 1
+
1
2.
+
+
+
a
Let 2:: := 1 fn be a series unconditionally convergent in measure. Then 2:: := 1 Un/ log(n + 1)) converges almost everywhere. 25 Corollary.
Proof:
Use Theorem 17 and Theorem 22.
a
We will prove a result already announced in the previous chapter, namely we want to show that Theorem III.G . 14 is best possible in the following very strong sense. 26.
Let 0 < a � 1 and let (gn)�= 1 be any complete or thonormal system. Then there exists a function f E Lipo: (Y) such that L I ( f, 9n W = oo for p = ( 2 o:� 1 ) . Theorem.
277
III.H. Factorization Theorems §26.
G: Lipa .. lp defined as G(f) = ((/,gn))n?.I ·
Proof:
case 0 < o:
£2
If this is not the case then we have a continuous map Let us consider the < 1 . In this case we have the commutative diagram
id
i
F
loo
Lp
a
id
2 lE £2 L
Fi l l c id lp £2 where F(en) = E ::'=  I (n + 2 )  cr  I / 2 en/n and Un)':'=  I denotes the orthonormal Franklin system. Theorem III.C.27 shows that F is a con tinuous map. The operators FI and � form a factorization given by Corollary 16 applied to the operator GF. Thus � is a diagonal operator with � = (8n) E lr where 1/2 + 1/r = 1/p. The operator E is defined as E(en) = Een9n· The operator id o F o id is a diagonal operator on £2 given by the sequence (n + 2 )  cr  I / 2 , thus id o F o id ¢ O'p (see III.G. 10). On the other hand FI o id E 0'2 (see III.G. 12) and id o � E O'r · Proposition III.G.4 gives that id o F o id = E o (id o �) o (FI o id) E up . This contradiction shows the claim. In order to show the Theorem for = 1 we consider the diagram LipI id L2 £2 D L� (T) 1 VI l c id l E £2 £2 £2 / 3 where G and E have the same meaning as before; L� (1l') is the subspace of L00 (1l') consisting of all functions whose integral is zero; (f)( ) = J; f(t)dt and VI and � are the factorization given by Corollary 16 of the operator G o v . The operator D is given by the formula D(en) = of positive numbers such that Lni'O dnene2,.int where (dn) isI a2sequence 3 Ln2 i'nOtd;_i'= 1 and Lni'0(dnn ) 1 = Using the orthonormal system (e ,.i )n I one checks that id o o D u2;3 but analogously as in the previous case we have E o (i d o �) o ( o D) E u2 3 . This contradiction 1 completes the proof of the theorem. o:
v
v
¢
s
oo.
v
vi
a
278
III.H. Factorization Theorems §27.
As one more example of an application of our results we would 27. like to present a description of certain multipliers. For s > 0 we define
Xs
= {f(z): f is analytic for l z l < 1 and Ill / Ill s = sup l f (z) l (1  l z W < oo } . JzJ< 1
We have encountered these spaces in the remark after the proof of The orem III.A. l l and we know that each of them is isomorphic to €00 • This fact will be important in our considerations. We are interested in coeffi cient multipliers from Xs into the Bergman space Bv (D) , 0 < p :S 2, i.e. we look for sequences of complex numbers such that E X8 • is in Bp [) for every
(.Xn)�=O 2:::;:'=0 anz n Proposition. Suppose = (.Xn)�= O is a multiplier from Xs into Bp (D) , 0 < < oo , 0 < p < 2. Then we can write An J.Ln · Vn, n = 0, 1, 2, . . . ' in such a way that (J.L n )�= O is a multiplier from Xs into e2 and ( vn)�= O is a multiplier from €2 into Bv (D) . Proof: Since X8 rv €00 and Bp ( [) ) Lp ( D) we can apply Corollary 16, so there exists a function g1(z) 0, l z l < 1 such that I A�;u)) I L2 ( 1D) C lll f lll s · Since the operator commutes with rotations of the disc and the norms involved are rotation invariant, one can choose g ( z) = g( l z l ) (simply average over rotations) . Since the sequence ( 9(Gn ) :'=o is 2:::;:'=0 Ananzn
( )
A
=
s
2:
C
:S
A
1
orthogonal in L 2 ([)) we obtain
(15)
f 2:::;:'=0 anzn E Xs and f3n = I 9(GJ) I L2 (1D) ' This means (.Xnf3n)�=O (J.Ln )�=O is a multiplier from Xs into e2 . Vn = n = 0, 1, 2, . . . , determines a multiplier
for every = that the sequence One easily checks that from £2 into Bv (D) .
= /3;; 1 ,
a
This proposition splits the original problem into two. We will ad dress those two in the next two propositions. 28 Proposition. The sequence into e2 if and only if
00
A =
(.Xn)�=O is a multiplier from Xs (16)
III.H. Factorization Theorems §28.
279
l z l  l z l )8
Let X2 = {f(z) : lim l z l + 1 f ( ) (1 = 0}. This space X2 is clearly a closed subspace of X8 containing all polynomials. The first observation we need is that A is a multiplier from X2 into £2 if and only if it is a multiplier from X8 into £2 . Since convolution with the Fejer kernel has norm 1 in supnorm (see I.B. 16) we see that for f E X � and f we have E X2 . Thus we see that if A: X2 + £ then A also maps X8 into £2 . The converse is obvious. Since X "' £00 we infer from Theorem III.F.29 that A: X + £ is 2absolutely summing, so A: X2 + £ is also 2absolutely summing. Let us consider the isometric embedding i: X2 + defined as i(f) = ( z ) · (1 The Pietsch theorem III.F.8 and III.F. (13) give the factorization Proof:
:Fn Ill ! * :Fn lll s III I I II s 2 8 f lzl )8• 8
xo
·J
C(D)
8
* :Fn
2
A
C(D)
£
8 2
2
iA
(17)
L2 (D,JL) where JL is a rotation invariant measure on D. Since i(X2) vanishes on aD, we can additionally assume JL( { z: l z l = 1}) = 0. So there exists a probability measure JLI on [0,1) such that for E C(D) we have id
cp
(i(zn ))�=O is an orthogonal sequence in L2 (D, JL) so
Now we see that is bounded if and only if
A
(18) The above reasoning is clearly reversible so we see that A is a multiplier from X8 into £2 if and only if (18) holds for some probability measure
n n sup2k:5n < 2k+l n8 I .Xn l nf: I .Xn8k l, 2 n� 1 Ank l 8ak , 8ak n
Now assume that {16) holds with K = 1 . Let us fix k , 2 k � k < l + k 2 such that for k = 0, 1, 2, . . . . Let = is the ak = ( 1 ;; 1 ) and let us define v = L:;: o where Dirac measure concentrateq at a k . This is clearly a probability measure on [0, 1).
280 have
Ill.H. Factorization Theorems §29.
n
Take arbitrary and let ko be such that
2ko n < 2ko + l . Then we �
1 1 r2n ( l  r) 2s dv(r) kL=O a%n (1  ak ) 2s n%s i .Xnk l 2 L {1  n;; 1 ) 2 n i .Xn k l 2 k =O 1 2n 2 > {1  nko ) i .Xnk0 1 > C I .Xn 1 2 · so {18) holds. Conversely, assuming {18) we have 00
0
=
00
=
Now we want to show that the integrand is a bounded function. For 1 2 N r 1 2 N  1 we have for some q independent of N, 0 < q < 1 k N 22sk {1  2 N  1 ) 2· 2k k 2 2· 2 s 2sk {1  2 N  1 ) 2· 2k 2 2 r L L L k=O kN=O k=N+ 1 k  +l) L 22s k L 22s k q2 (N k =O k = N + 1 C22sN C22sN kL= 1 22sk q2k C22sN C{1  r)  28 so the integrand is really bounded and so {16) holds. 29 Proposition. The sequence = {.Xn)�= O is a multiplier from i2 into Bp(D), 0 < p 2 if and only if 1 1 1 r 2 p {20) �
oo
0
�
�
+
�
+
�
+
�
oo
oo
00
�
�
a
A
 + 
=
 .
281
Ill.H. Factorization Theorems §29.
nk
(nk )�0
be any sequence such that 2 k :5 < 2k+ 1 . Let Passing to polar coordinates and using Khintchine's inequality I.B.8 (see also Exercise III.A.9) we get that there exists a constant such that for any such sequence and any sequence of scalars Proof:
C (ak )�0
(nk )�0
(21) From (21) we see that for a multiplier A: £2  Bp (D) we have
(an)�0
(nk )�0
Since (22) holds with the same constant for all scalars and all sequences as above we get (20) . In order to prove the other implication we use the following inequality
I t. .z· l . , ( � 1 ··�· .z· ID
*
n, n L:::'=o l an l 2
·
(23)
For p :5 1 this is just the pconvexity of the space Lp (D) (see I.B.2) . For p = 2 it follows directly from orthogonality of z = 0, 1, 2, . . . , in L 2 (D). The remaining cases follow by standard interpolation. = 1 and such that For a sequence satisfying (20) we obtain from (23) and the HOlder inequality
(an)�=O
(.Xn)�=O
282
III.H. Factorization Theorems §30.
a
If A satisfies (20) this shows that A is a multiplier.
Now we are able to prove the description of multipliers from X8 into Bp(ID) . From Propositions 27 and 28 and 29 and Holder's inequality we get 30 Theorem.
The sequence A =
(.An)�=O is a multiplier from X8 into
Bp(ID) , 0 < s, 0 < p � 2 if and only if
n< 2• + 1 ns � I .An i )P < k=O 2•:s:;sup 00
L(
oo .
Notes and Remarks.
Various sublinear operators are of paramount importance in modern harmonic analysis; they include a variety of maximal operators, square functions or area functions etc. Texts such as de Guzman [1981] , Garcia CuervaRubio de Francia [1985] , FollandStein [1982] , Torchinsky [1986] etc. make their importance absolutely clear. In our presentation we give only the most general results which fall naturally into the scope of Banach space theory. Kolmogorov [1925] has shown the weak type (1,1) of the trigonometric conjugation operator ( see 8 Example 1). This was probably the first paper where finiteness almost everywhere was shown to imply weak type (1,1). The principle was generalized in Stein [1961] where a version of our Corollary 7 was proved. E.M. Nikishin was led to consider his general theorems by problems connected with the structure of systems of convergence in measure for £2 . His main results in this area are published in Nikishin [1970] . This paper basically contains Theorem 6. Later Maurey [1974] gave a more abstract presentation. It is his approach that we follow in this book. A sequence of functions C Lo [O, 1] is called a system of convergence in measure for £2 if every series with E converges in measure. Clearly there is a correspondence between £2
(¢n)�= l
L::=l an¢n
(an)
283
III.H. Factorization Theorems §Notes.
systems of convergence in measure for £2 and continuous linear operators T: £2 + L 0 • Some other similar notions have been investigated (see Exercises 13, 14 and 16). 8 Example 2 is an old theorem of Calderon (see Zygmund [1968] XIII. 1.22) . The fundamental theorem due to Carleson [1966] (for p = 2) and extended by Hunt [1968] to 1 < p < oo asserts that for every f E Lp(T) its Fourier series converges almost everywhere. An example of an L 1 function whose Fourier series diverges a.e. was given by Kolmogorov [1923] . The example was improved in Kolmogorov [1926] to yield an L 1 function with everywhere divergent Fourier series. Very recently the Armenian mathematician Kheladze gave a remarkbly simple construction of an L 1 function for which condition (c) of 8.Example 2 fails. Inspired by Nikishin [1970] and Rosenthal [1973] B. Maurey under took his study of operators from X into Lp(O, which factors strongly through Lq (O, His results are presented in Maurey [1974] . Our presentation of 1012 and 15 follows that monograph, with the excep tion that the proof of (b)=>(a) in Proposition 1 0 is taken from Pisier [1986a] . We recommend the reader to consult this paper. It contains many additional results, also in the setting of C* algebras. Its main interest is to present necessary and sufficient conditions for the operator T: X + Lp (n, to factorize strongly through Lq,oo (n, that is the topic between Nikishin's and Maurey's theorems. Corollary 13 is one of the main results of Rosenthal [1973] . This paper played a very important role in the development of the theory. We would like to mention that the notion of type used in Maurey [1974] is different from the one used in this book. We use the type and cotype which is sometimes called in the literature 'Rademacher type' and 'Rademacher cotype', while Maurey uses the socalled stable type. Let 1 � p � 2 and let ei be a sequence of independent, identically distributed standard pstable random variables. A Banach space X is called of stable type p if there exists a constant C such that for any finite sequence (xi ) in X we have
J.L) .
J.L)
J.L)
J.L) ,
Actually the use of the exponent 1/2 in the left hand side integral is irrelevant. It can be replaced by any number q < p. With this, one checks that Lp, 1 � p < 2, is not of type p but is of type s for any s < p. The following fact analogous to Theorem 6 holds.
284
III.H. Factorization Theorems §Notes.
If X is a Banach space of stable type p, 1 :S p :S 2, then every linear operator from X into Lr(O., JL ) , r < p factors strongly through Lp (n, JL ) .
Theorem A.
We have practically proved this theorem during the proof of The orem 12. The usefulness of the notion of stable type can be seen from
this proof. There is an obvious analogy between Definitions 4 and 9 and be tween Proposition 5 and Proposition 1 0. Let us note that conditions (b) of those propositions can be interpreted as vector valued inequalities (see the remark after Proposition 5) . This point of view is explained in detail in GarciaCuervaRubio de Francia [1985] , as is the equivalence between factorization and weighted norm inequalities. We do not discuss this important subject here. Our informal discussion in 14 is more or less folklore. It can be found in full detail in Maurey [1974] . Proposition 14 is a special case of the following result due to Maurey [1974] . Theorem B.
The following conditions on the Banach space X are
equivalent:
(a) co is not finitely representable in X; (b) there exists a q < oo such that ll q (C(K), X )
= L(C(K), X ) .
This result and its consequences for Banach space theory are dis cussed in great detail in Rosenthal [1976] . The MenchoffRademacher Theorem 22 was proved in Menchoff [1923] and Rademacher [1922] improving many earlier results. This is the best result. Menchoff [1923] constructed an orthonormal system on [0,1] such that for every sequence with 1 :S :S . . 2 and (log divergent al n there exists a series o most everywhere and such that < oo . The connec tion between the theory of psumming (or radonifying) operators and the MenchoffRademacher theorem was noted in Schwartz [1970] and KwapienPelczynski [1970] . This last paper also contains some general izations of the MenchoffRademacher theorem in the spirit of Corollary 25 which was proved in Maurey [1974] . Later Bennett [1976] gave an other, more elementary, but in fact closely related, treatment of such generalizations. Theorem 1 7, proved in 0rno [1976] , shows that se ries unconditionally convergent in measure (in particular unconditionally convergent in Lp ) are closely related to orthogonal series. Our Lemma
Wn =
)
(wn)�=l (wn)�=l = w 1 w2 . L::=l an Wn 2 L::=l l an l wn
Ill.H. Factorization Theorems §Notes.
285
18 is a classical result of I. Schur, published first in Rademacher [1922] (see also KashinSaakian [1984] ) . The dilation Theorem 19 is classical by now and is a basis of a large part of the theory of operators on Hilbert spaces (see NagyFoias [1967] ) . Our proof of the MenchoffRademacher theorem is a mixture of various published proofs like K wapietiPelczytiski [1970] , Bennett [1976] , Nahoum [1973] , Schwartz [1970] . Our Corollary 20 was proved in 0rno [1976] . It improves an earlier result of Kashin [1974] . Theorem 26 for a = 1/2 was proved by Mitiagin [1964] and the general case was shown by Bockariov [1978] . Our proof follows Woj taszczyk [1988] . Actually it is possible to obtain analogous results for systems more general than orthonormal and for more general moduli of smoothness. We refer the interested reader to Wojtaszczyk [1988] for formulations, proofs and the history of the subject. Theorem 30 and its proof are taken from Wojtaszczyk [P] . We would like to mention also the paper Bichteler [1981] where factorization theorems are applied to the theory of stochastic integration. The factorization theorems are basically a type of Tauberian the orem; they assert that the operator is actually better that it seems to be. This is useful both ways; we get stronger information once we prove something weaker or conversely we show the 'very' bad behaviour once we show a 'moderately' bad one.
286
1.
2.
3.
4. 5.
6.
7.
8.
Ill.H. Factorization Theorems §Exercises
G f) x)
Exercises
Let ( ( = [M( I / 1 2 )] 1 1 2 where M is the HardyLittlewood max imal operator. Show that is a sublinear operator on £ 2 [0, 1] which is of weak type (2,2) but not continuous on £ 2 [0, 1] .
G
T: X* Lp (O, J.L) factors strongly Lq (O, J.L) X* p < 1) then IIp (X, lq ) = IIq (X,lq ),p Let (cpj ) �1 be a sequence of independent, pstable random vari ables in LI(J.L ) and let X = span(cpj ) �1 � lp . Show that i*: L00(J.L)�X*, where i is the identity embedding of X into L 1 (J.L), is not p'absolutely summing, � + � = 1 . Show, without using Proposition 15, that every operator from L00 into Lq , 1 � q � 2, factors strongly through L 2 . Let T: Lp Lo, 1 � p < be a continuous sublinear operator. Assume moreover that T is monotone (i.e. if 1 / 1 � I Y I then also I T/ I � I Tg l ) . Then T factors strongly through Lp, oo · Let T: Lp Lo(O, J.L) be a positive (i.e. if f � g then Tf � Tg) linear operator and assume J.L(f2) = 1 and p � 1. Then T factors strongly through Lp (O, J.L ) . Show that there exists a positive linear operator (see Exercise 6) T: lp + Lq (O, J.L) , with 1 > p > q > 0, and (0, J.L) a probability measure, which does not factor strongly through Lp (O, J.L ) . Let X be a Banach space and 0 < < p < Show that the fol lowing properties of the bounded linear operator T: X L r ( O, J.L) are equivalent: (a) there exists a constant C such that for all finite sequences (xi ) X we have I s�p I T(xi ) ll l p � c ( � l xi l p) (b) there exists a constant C' such that there exists a function I E L1(f2, J.L), f � 0, J0 fdJ.L = 1 such that for all x E X and all measurable subsets E f2 have I T(x) · XEI I r � C'l l x ll ( l fdJ.L) �  � ; Show that if every operator + through (and has b.a.p. if < q, q � 1 .
+
oo ,
+
r
oo .
+
C
1
•
c
we
p;
287
III.H. Factorization Theorems §Exercises
(c
9.
10. 11. 12.
13.
)
the operator T admits a factorization of the form T
M
X +Lp,cxo (O, fdJ.L) +Lr (J.L) with f bounded, f E L 1 (0, J.L) , J � 0, In fdJ.L = 1 where M is an operator of multiplication by f�. is Every operator from Lv [O , 1] into iq , 1 < q < 2 < p :::; compact. Let (fn)':'= 1 be an orthonormal system in L 2 [0 , 1]. Show that N 1 2:::= 1 fn + 0 almost everywhere. For a given number x, 0 < x < 1 let (cj (x))j�1 be its dyadic expansion (cj = 0 or C"j = 1). Show that for almost all x E [ 0 , 1] we have N 1 2:: := 1 €j (x) + 1/ 2 , i.e. almost every number has asymptotically equal number of O's and 1 's in its dyadic expansion. Let Un)':'= 1 be a complete orthonormal system in L 2 [0 , 1]. Show that 2:: := 1 I fn i = on a set of positive measure. Let Aa , 0 < a < 1, be the space of all functions in the disc algebra such that l f (e i 9 )  f(e i ( 9 h ) ) i :::; Cj h j <> . Show that there exists a f E Aa such that f' (j. N ( N denotes the Nevanlinna class) . A system of functions Un)n>1 L0[0 , 1] is called a system of con vergence in measure for £2 if every series E n>1 anfn with (an) E £2 converges in measure. Show that Un)n�1 iS a system of conver gence in measure for £2 if and only if for every c > 0 there exist a set EE [0, 1] with l Ee I > 1  c, a constant CE and an or thonormal system (cpn)n�1 on [0 ,1] such that fni EE = CEcpn i EE for n = 1, 2 , . . . Show that the following conditions on a system of functions Un)n>1 L0[0 , 1] are equivalent. a.e. ( a) For every (an) E £ 1, we have E i anfn l < ( b) For every c > 0 there exist EE [ 0 , 1] with l Ee I > 1  c and a constant CE such that supn IE, I fni :::; CE. Show that there exists a function f E L 1 (Y2 ) such that n+m j(n,m)ein9 eim

oo
C
C
14.
.
C
oo
C
15.
""" L....J
does not exist on
a
set of positive measure .
288
III.H. Factorization Theorems §Exercises
r.p r.p(t)dt = 0. Show that the following condi (a) fn(t) = r.p(nt), n = 1, 2, 3, . . . , is a system of convergence in measure for i2 ; (b) I I E n>1 anfn l 2 � C E n>1 l an l 2 for some constant C and all sequences of scalars. Use this to show that fn(x) = sgn sin (nx) , n = 1, 2, . . . is not
16. Let E L 2 ('1r) with I tions are equivalent:
a system of convergence in measure of i2 . This means that the smoothness of trigonometric functions, and not only the distribu tion of signs, plays a role in the almost everywhere convergence of trigonometric series of £ 2 functions.
(r.pn )n�1 is a uniformly bounded orthonormal system on [0, 1] and (dn)n�1 is a sequence 1of positive numbers such that En>1 d� = oo where q = 2p( 2  p) and 0 < p < 2, then there exists � function f E C[O , 1] such that En�1 l dn (!, I{Jn } I P = oo. Show that the map f = E �oo J(n)einB �+ ( .Xn J(n))';= oo maps C(1r) into lp, 10 < p < 2, if and only if E�oo 1 Xn l q < oo where q = 2p( 2  p)  .
17. If
18.
·
19. Show the von Neumann inequality using Theorem 1 9. Let us recall (see Exercise III.B.8) that the von Neumann inequality says that for a contraction on a Hilbert space (i.e. II T II � 1 and any polynomial = we have II � IIPIIA · 20.
T ) n p(z) En=O anzn anT E n=O l (a) Suppose Y :::> X and assume that both X and YIX have some type p > 1. Show that Y has some type q > 1 . (b) Let X, Y, Z be Banach spaces with Y :::> X and let T: X + Z. Show that there exist a space V, an isometric embedding j: Z + V and an operator T1:Y + V such that jT = T1 I X and the spaces YI X and VIZ are isometric. (c) Suppose that X C(S) is a subspace such that C(S)IX is reflexive. Show that every operator from X into ip, 1 � p � 2, is 2absolutely summing. (d) Suppose A is a Apset for some p > 1 (see I.B. 14) and sup pose that (r.pn );:'= 1 is a complete orthonormal system in L2 (Y) . Show that there exists f E C(Y) such that j (n) = 0 for n E A and 2::= 1 I ( !, I{Jn ) I s = for all < 2. c
c
7l
oo
s
289
III.H. Factorization Theorems §Exercises
21. ( a) Describe the coefficient multipliers from and 0 < p :::; 2.
Xs
into f.p, for s > 0,
(b) Describe the coefficient multipliers from Bp (D) into Bq (D) for 0 < q :::; 2 :::; p < 00 .
111.1. Absolutely Summing Operators On
The Disc Algebra
We start this chapter with the construction of a noncompact, !absolutely summing operator from any proper uniform algebra into £2 . This shows in particular that such an algebra is never comple mented in C(K) . Then we study pabsolutely summing operators on the disc algebra A. We construct an 'analytic projection' which maps some weighted Lp('l', �d.X) spaces onto the closure of analytic polyno mials and has properties analogous to the properties of the classical Riesz projection n. Then we show that every pabsolutely summing operator from the disc algebra is pintegral, p > 1 . We also show that A* has cotype 2 and derive some corollaries of these results. Next we study reflexive subspaces Y C Ld H1 and show that any linear operator T: Y + H00 extends to an operator T: L I /H1 + H00 • This is applied to some interpolation problems on D x D. In this chapter we present the detailed study of psumming and re lated operators defined on the disc algebra A. Such a study is motivated both by the intrinsic beauty of the problems and by important appli cations. Actually we have already seen one application. In Proposition III.F.6(a) we have exhibited an absolutely summing operator P: A + £2 which was later used to prove the Grothendieck theorem III.F.7. This example suggests that absolutely summing operators on A may have some rather unexpected properties. Actually this phenomenon is not restricted to the disc algebra, but is shared (to a certain degree) by all proper uniform algebras. Let B C C(S) be a proper, pointseparating subalgebra of C(S) , with 1 E B. 1.
2 Proposition.
operator T: B
There exists a noncompact, 1absolutely summing
+ £2 .
Proof: Take J.L
E M(S) such that J.L E B l_ and J.L ¢. B l_ where B =
E
C(S) : f E B}. It follows from the StoneWeierstrass theorem that B =I B so such a J.L exists. Let v denote the HahnBanach extension of J.L I B to C(S) . We assume that I v ii = 1 . Thus we can take a sequence ( /n)':=l C B with 11/n ll oo $ 1 and such that J fn dv + 1 for n + oo . Considering {!
292
III.I. Absolutely Summing Operators On The Disc Algebra §2.
this sequence in L 2 (ivi + i JL i ) we can pass to a subsequence and convex combinations to get a new sequence (still denoted by Un)�= 1 ) such that In E B for n 1, 2, . . . and ll lloo � 1 and limn _, oo I 1 and In (ivi + I JL I )almost everywhere for some F (see III.A.29 but in fact easier) . Clearly we get I Fdv = 1 = ll v ll so = 1, l vla.e. Now let us consider V Bl_ n L1 ( l v l + i JL i) and define T1 : V  by T1 In order to show that T1 is continuous it is (J enough to show that
ln
F, =
fndv = IFI i
= 2 k (a) = F da)k:: 1 •
2
Rn: C(Y)  C(S) R�:
for every n with the constant independent of n. Let be defined by = H(cp) o I where H(cp) is the harmonic extension of C B we get that Bl_  Al_ = HP cp. Clearly � 1. Since so by the Paley theorem (see I.B.24)
Rn(cp) I Rn l
n Rn(A)
2 = � i (R� (a) , z:2k W z2 J;. r 1 l k k a Rn( a )d J d = J � �I � CI I R� ( a) l i 2 � C l i a ii 2 . Now we look for the operator i : B  V given by i( f ) = I · p for some p E V, such that T = T1 i will be noncompact. Clearly) such T is 1absolutely summing (see III.F.4) . We take p = JL (v  p F2 • Since for i E B J ldp = J ldJL J F2 ldv  J F2 ldP, = 0 J F2 ld(v  JL) = 0 we see that p E V. Moreover lim sup i ( TI, e k ) i = lim sup I JF2 k ldp l k k 11/11 :::::: 1 11!11 9 ;::: lizn s�p I J F2k 1�k  1 dp I ;::: lizn J I F2k p2k  l dp I  I f{ I FI =l } Fdp l = I J{ IFI = l } FdJL J{ I F I = l } Fd;;  J{ I FI =l } Fdp,l F dJL F dJL  [ = 11 ( [ ) I· J{ I F I =l } J{ I FI = l l } o
+
+
+
+
+
·
293
III.I. Absolutely Summing Operators On The Disc Algebra §3.
Since the bracket above is purely imaginary we see that sup lim k
11 ! 11 9
I (Tf,
ek
) l ?: 1
a
so T is not compact. 3.
From this proposition we get
Corollary. Let B be a proper, closed subalgebra of C(S) separating points and with 1 . Then B is not isomorphic (as a Banach space) to any quotient of any C(K)space. In particular B is not complemented in C(S) . Proof:
onto
T: To
Suppose that there exists a map q: C(K) + B . Let B + f.2 be a 1absolutely summing, noncompact operator. Then q: C(K) + f.2 is a 1absolutely summing, noncompact operator. From III.F.8 we infer that we have the factorization
Since L1 (K, 1L) has the DunfordPettis property ( see III.D.33,34 ) and both S and id are obviously weakly compact we get that q is a compact. This contradiction finishes the proof.
To
Remark. The above corollary shows that some algebraic properties of a uniform algebra are determined by its Banach space structure. In particular there does not exist a multiplication on the disc algebra A which makes it into a commutative C* algebra.
There are some very natural limitations to what can be proved about the Banach space structure of a general uniform algebra. We have the following.
4.
Proposition. For every complex separable Banach space X there exists a uniform algebra Ux such that X is isomorphic to a 1complemented subspace of Ux .
294
III.I. Absolutely Summing Operators On The Disc Algebra §5.
Proof: Let K be the unit ball in X* equipped with the a ( X* , X topology. We define Ux to be the smallest closed subalgebra of K containing all functions for E X and constants. For a = function E Ux we define = 2 rr Since the i elements E�= l ( · , Xr ) r for Xr E X and ir E N are dense in Ux we easily see that for every E Ux , Pf ( k ) = for some E X. Since clearly a 1 we get the claim ( see II.A. 10 ) . �
f
I PI
'Px (k) (k,x) x1 i i Pf ( k) ( ) J e 8 f(e 8 k)d0. f (k,x) x T: A +
A.
) C( )
5. Now we return to the disc algebra Suppose we have a pabsolutely summing operator X, 1 � p < oo. The Pietsch theorem III.F.8 gives that there exists a measure on the circle such that � which is the In other words extends to the space closure of the polynomials in Thus it is natural that we start the detailed investigation of pabsolutely summing operators on with some observations about spaces
C(f l f !PdJ.L) ; .
J.L
J.L T Lp(1I', J.L) . Hp(J.L) .
1I'
Hp(J.L)
l i T/I I
A
J.L J.Ls fd
be a probability measure on 11', with the Let 6 Proposition. Lebesgue decomposition = >. where >. is the normalised + Lebesgue measure. Then
91 Lp (J.Ls ) 92 Hp( fd ) h A (1I') l 9t 9 h i Lp (tt. ) l 92 h i Hp ( fd>.) 1I'� 1 92 A. J.Ls l_). � (JT\t�. l 9t! PdJ.Ls ) to ( J1f\L:I. I 92 I PdJLs ) to · () cp A cp l � 9tl � (JT l cp i P fd>.) � J'lf\ L:I. I cp ! PdJ.Ls '¢ A 1 '¢ 1 '¢ 1 ( J I '¢92 I P fd>.) � h cp '¢)92 · h.
What we really have to show is the following: given E such that and E >. we have to find E � c. We can obviously assume that is � c and continuous and E Since there exists a closed set � C with and < < >. � = 0 such that Using Proposition III.E.2 we find a E such that = and � e / 10. Let us use Proposition � 1E0 and III.E.2 once more to get a function E such that = 1, � = 1 and < e / 10. We put = + ( 1 One checks that a we have imposed enough conditions to make it the right Proof:
T: A
Note that Proposition 6 in particular implies that a psumming operator + X whose Pietsch measure is singular with respect to X. extends to a psumming operator T: measure the Lebesgue
C(1I') +
7. Now we want to give a heuristic indication of what will be done in the subsequent sections 8, 9 and 10. We would like to reduce the inves tigation of psumming operators on A to the study of certain operators
295
III.I. Absolutely Summing Operators On The Disc Algebra §8.
6
on C(1r) or Lp (1r) . From Proposition we see that this requires inves tigating a projection from Lp (fd>.. ) onto Hp(fd>.. ) , where f is a positive function and we impose the normalization J fd>.. = 1 . In general we cannot assume any properties of f. The only freedom which we retain is that we can replace f by any It � f (because if f d>.. was a Pietsch measure for an operator T, then ft d>.. also is) . However in order to keep things under control we have to control J ft d>.. . The classical case when f 1 , i.e. Hp(fd>.. ) = Hp (1I') , correspond, to operators which are rotation invariant (see Exercise 2) , and in this case there is the Riesz projection whose properties are well known (see I.B.20) . The main result of this chapter, Theorem 10, will give a very useful analogue of the Riesz projection in our general case. Technically, our projection will be built by patching up pieces of the Riesz projection More precisely (but not really precisely) we will construct a sequence of Hcxofunctions (depending on f) such that 1 and we put = Formally is a projection onto an alytic functions. In Theorem 9 we will construct such a sequence of functions having very elaborate properties, and the projection will be constructed in Theorem 10. =
'R
'R.
P
('Pi )� Pg 2::� 1 'Pi1'R(cpi g).
1
2::� 1 'PJ
P
=
P
G £1
There exist constants C and M such that for every f E E (1r) = 1 we have a function and a sequence of Hcxofunctions such that
8 Lemma.
('Pi )�0
L (1r) such that f � 0 and J f d>.. CX)
""" rn  1' � rJ 
i =O
I � I 'PiiP I = <
f
'.5,
CX)
(1)
·
G,
oo
for every p , 0
<
p
< oo ,
(3)
(4) (5) (6)
i =D I 'Pii G CMi , I G I I '.5, c. '.5,
'Pi
(2)
Comment. We are trying to get analytic functions to imitate the sit = X { M1 9 � MH' } · In order to be sure that (1) holds we uation when define first auxiliary functions and then the functions The com plicated formula (8) defining is needed to make sure that (2) holds. Note also that formulas like (9)( 1 1) to define a sequence satisfying (2)
'1/J '1/Ji i
'Pi .
296
III.I. Absolutely Summing Operators On The Disc Algebra §8.
have already been used in the proof of Theorem III.D.31 formulas and
{27).
{17)
M, £, 8 in such a way that 0 < e < 2 0 8 M 2, M8 16, M83 < 1. Assume additionally that 1 / l oo J = max{j: 1 / l oo > Mi } . We define sets Ai = { t E f ( t ) > Mi } , j = 0, 1, . . . , J. We define outer functions (see I.B.23) ( '1/Ji ) f=o by the conditions {7) 1 '1/JJ I = 1  {1  e) XAJ , 1 '1/Ji l = {1  {1  e) XA; ) [ sup { 1, 8 1 1 1  '1/Ji +l l , 8 2 1 1  'I/Ji +2 1 , . . . , {ji  J I 1  '1/JJ I} r 1 for j = 0, 1, 2, . . . , J  1. {8) The desired functions (cpj ) f;!J are defined as {9) 'PJ+l = 1  '1/JJ , {10) 'Pk = '1/JJ 'l/Jk (1  '1/Jk  d k = 1, 2, . , J, {11) cpo = '1/JJ '1/JJ  1 '1/Jo . Proof: Let us fix numbers �' < < �' > > < oo and put 'If:
0
·
0
0 0
°
0
· . . . ·
0
We define G to be
G=
{1  e)  3M j=O L Mi l 1  'I/Ji l 3 + 1.
With the above definitions we have to check conditions mostly routine. Ad Follows directly from Ad {2) . Observe that implies �
{1 ).
{12) {1) (6). This is
{9){11). {8) 1 '1/Ji l {8  8 1 1  '1/Ji +s l ) 1 so I 'I/Ji l l 1  '1/Ji +s l � 88 for j = 0, 1, . . , J  1 and s = 1, 2, . . . , J  j. {13) For a given t E '][' let n be the smallest integer such that 1 '1/Jn ( t )l > £. We see from {10) and {11) that I 'Pi ( t )l � 2£n  i for j = 0, 1, . . . , n, so n P { 14) L j =O I 'Pi (t)I � C(e) . .
III.I. Absolutely Summing Operators On The Disc Algebra §8.
297
(I ) I I  ,Pn+s (t) l � 6; so (9) and (IO) yield J+l P j=Ln+l l
> n
we get from 3 that
k. Y ( M83 � CM k + M k L � CM s2:: 3 Ad (6). This condition is the most difficult to check. Let us recall (see I.B.23) that ,Pj = I ,Pj l · exp ilog I ,Pj l · Integrating the obvious numerical inequality I I  a ei.B I � I I  al + I .B I we get I ,Pj I � I
I  I ,Pj I �
I ,Pj I I I L 2
Since we get and using the fact that the j log trigonometric conjugation operator has norm in (1r) we get
Using (8) we estimate it further as
I I I  ,Pj l 2 � 8 (/ l lo (I  (I  c) XA;w g
(I 5 )
298
III.I. Absolutely Summing Operators On The Disc Algebra §9.
J log sup{ 1, 8 1 1 1  ,Pj + ll , . . . ,8i  J I 1  ,PJI } 2 ) J :::; 8(log c) 2 1 Ai l + 8 L j 82 (j  k ) l 1  ,Pkl 2 . k=j +l Summing ( 15) over j we get (16) L Mi J 1 1  ,Pi l 2 j=O J J J L Mi k=jL+l 82(j  k) j 1 1  ,Pkl 2 L Mi i Ai l + 8 j=O :::; 8(logc)2 j=O :::; C + 8 f: Mk � (M82 ) i  k J 1 1  ,Pk l 2 . k= l j=O 00 Since 8 I: (M8 2 )  i < 1, ( 16) gives j=l ( 17) L Mi J 1 1  ,Pjl 2 :::; c. j=O But J I G I :::; M(1  c) 3 2:L.:f= 0 Mi J l 1  ,Pjl 2 + 1 so (6) follows from (17). Now we have to remove the assumption 1 / l oo < This can be done as follows. Given f with 1 / l oo = we apply the previous construction to the functions /J = min(!, M J ). Since no constant in the previous construction depends on J we see that for each j the functions ,Pj and 'Pi (dependent on J) will tend to a limit (almost uniformly in D or a(L00 , LI /HI ) ) as J + Obviously the limits will also satisfy (1)(6). +
oo.
oo
•
oo.
9.
Now we will improve this decomposition.
Given a function � E £1 ('][') with � integer r 2:: there exist a sequence of scalars of H00functions such that and
Theorem.
2
2::
0, J � = 1 and an
(Cj )�0 and two sequences (Oj )�0 (rj )�0 (18) II Oj I l eo :::; 1 for j = 0, 1, 2, . . . , (19) I � l rjl l oo :::; C, L:oi rj = 1, (20) j=O
III.I. Absolutely Summing Operators On The Disc Algebra §9.
299
F = L�o Cj !ri I we have D. � F, (21) (22) !rii F � Cci , (23) I F i h � c. for some constant C independent of Ideally we would like to have hi = l �i l � . In this way, Comment. however, we lose control over I Ti l · F and I F i h · given by Lemma 8 and write Proof: Let us take ( �j )�0
and for
D. .
as
j
j
a(j1 , . . . , j3seer _1 );that¢i ) we 1 ¢i l � C(j. rWe � also have
where the above equalities actually define functions and Since and . . , ]3 r d l conditions ( 1 ) , (2) , hold with replaced by ( using
� C(r) l a (j1 ,(5),. �j. �j (4)) t, Mi l �jl � � C � Mi ( � � �i l r · M �l �i l ) 3 M f (L � c j=O ll? i?_ j i l �il 3 M M � C j=O LL ll? i?_j i l �il 3 ) ( � M ll? ) M = c(f t=O J '5,_t � C · G.
(4')
(X)
Once more we write 1
= ( Jf=O �1 r = j=O f: �1 j , , . . .L:: b(j1 , . . . , Jr d �j , . , �ir 1 ,jr l '5_j = j=O L:: �j 1Yi = L:: �'J j=O .
(X)
(X)
·
·
.
300
III.l. Absolutely Summing Operators On The Disc Algebra § 1 0.
where the above equalities actually define functions b(j 1 , . . . , ir  d , ¢j and cp'j. Since l b(j b . . . , ir  t ) l :5 C(r ) one gets l¢j l :5 C(r ) so conditions (1), (2) , hold for (cp'j) �o · We also have (using
(4')
5)
(r 1 ) r
( � I 'P� i ) :::; c ( � I 'P� i ) G :::; c � Mi :::; cMi . ·
l cpj' I � G :5 l cpj l � l¢j i � G :5 C l cpj l �
(5')
G
Now we use the innerouter factorization to write j = 0, 1, 2, . . . with (}j inner (so (18) holds) and rj outer. Since rj is outer Tj = (rj ) � makes sense. Since (cp'j) satisfies (1) and (2) we get (19) and (20) . Since 1 :5 I:�o I Oi l lri l r = I:�o h l r :5 C we get also 1 :5 I:�o h l 2 :5 C so by and we have
(5') (4')
00
00
j=O
j=O
a :::; I: h i 2 G :::; I: cMi h i :::; ca. This shows that for Cj = CMi we have G :5 F :5 CG so (21) follows • from (3) , (22) follows from and (23) follows from
(6).
(5')
10.
Now we are ready to state the main result of this chapter.
J
J
Theorem. There exists a C such that given � E £1 (1r) such that � � 0 and � :5 1 there exists � � � with � :5 C and a projection P such that
(a) P is a continuous projection in Lp (ll', �d>.), 1 < p < image of P is Hp ( �d>. ) and li P I :5 C max(p, (p�l ) ) ,
oo
and the
(b) P is of weak type (11) with respect to the measure �>., (c) the dual projection P* is also of weak type (11) with respect to the measure �>..
First we reformulate what it means for an operator T to be of weak type ( 11) . We have
III.l. Absolutely Summing Operators On The Disc Algebra § 1 1 .
301
T : £ 1 (JL)  Lo (JL), JL a probability measure, a E £ 1 (JL) and every (3 E
11 Lemma. An operator is of weak type (11) if and only if for every we have
L00(JL)
(25)
l f3 l oo � l f3 1 1 ·J ) JL A T( { I T(a) l I 17 }a ! !Ai f3 i d { I T(a ) l 17 } 17 JL(A) � · f I T (a) l2l f3 i dJL I T (a ) l2 Cl l a l l A  2 d).. C l a l r � 2 Cl l a l r l f3 1 � 1 f3 1 r .
Assume that T is of weak type (11) and take (3 with =1 and put = This is enough since (25) is homogenous in (3. Under is maximal if (3 = X A where is a set these restrictions such that > for some and :::> :::> � = So 1 1 [ 1 [e 1 =} � lo A 1 1 1 1 1 = � Conversely, let us define (3 = X { I T( a: ) l >>. } and we obtain from (25) Proof:
JL{ I T(a ) i >.} � f I T (� D I ! f3dJL � c>. ! l a l f l f3 1 i l f3 1 f = c>.  2 i l a l l JL{ I T ( a ) l >. } 2 >
1
!
>
1
so T is of weak type (1 1 ) .
J�
•
Obviously it is enough to assume = 1 so let us apply Theorem 9 to � with r = 8 and put � = F. We define the desired projection by Proof of Theorem 10.
'R
00 P(f) = j=O L Ojrj'R (rj · f)
where is the classical Riesz projection (see I.B.20.) . Clearly algebraically a projection onto analytic functions. For 1 < p < oo we have
j I Pf iP�d).. = j I t, oi rj'R(rJ! f �d).. � C Jt I Oi ll ri l 4 i 'R (rjJ ) I P �d).. j=O C j=O t j i 'R(rJ f ) I P i rj I �d).. :S
P is
by ( 19)
( )
by 1 8
302
III.I. Absolutely Summing Operators On The Disc Algebra § 1 1 .
:::; c if=O cj f i 'R(rjJ ) I Pd>.. f Cj j l rf JI P d>.. :::; CI I 'R I � j=O :::; CI I 'R I � j l f i P · � ci hi 4Pd>.. :::; CI I 'R I � j I J IP � Cj l ri i d>.. :::; CI I 'R I � f I J I P t:::.d>.. .
by (22) by (I.B.20)
3
3
P f3 L00(Y, >..)
by (19)
Lp (!:::.d>.. ) I 'R I I J3 I oo f3
This shows that is continuous in and the estimate for the norm follows from the well known estimates for ( see I.B.20) . = 1. Then for Let us take E with ;:::: 0 and we have ('][', C
f L1 t:::.d>.. )
00
j I P/I ! f3!:::.d>.. = j I � Oi rj'R(rf f) 1 2 f3!:::.d>.. by (18) :::; / if=O hi 2 I 'R (rj! ) I ! J3t:::.d>.. :::; by (22) :::; c 'f,=O ci j !riii 'R (rj! ) I ! J3d>.. i ! by (25) � C t, c; (! l rfil d>. ) (! l r; l /3 d)y ! � C ( t, c; j l rf f l d>. ) ( t, c; j l r; I /3d>. ) ! ! � C (! 1 1 1 ( t, c; l r; I ' ) d>. ) (! /3 ( t, c; l r; 1 ) d>. ) :::; c ( j l f l t:::.d>..) ( j f3t:::.d>..) Since this holds for arbitrary f3 we infer from Lemma 1 1 that P is of weak type (11) with respect to the measure t:::. d>.. . 1
l
1
2
1
2•
III.I. Absolutely Summing Operators On The Disc Algebra §12.
Given
303
f E Lp(l:!.d>..) and h E Lp'(!:!.d>..) we have j hP(f)!:!.d>.. = Jf=O j hOi rjR(rj f)!:!.d>.. = f J rj JR (hOjrj f:!. ) d>.. j=O 4 = j f L 7i R ( Bjrj · !:!. · h) i:!. d>.. j=O f:!. (X)
so
(X)
4 P*(h) = j=O L 7 R(Bjrjl:!. h) . f:!. 1
P* is of weak type (11) let us take f3 E L=(1f, .X) with f3 1 /3 1 = = 1. Then we have J 1 P*h l ! f31:!.d.X :S C J Jf=O 1 ri i 2 I R(Bjrji:!.h) l ! f3�d>.. by (22) :S C f .jCj J l ri 1 3/ 2 f31 R (Bjrj f!.h) l ! d>.. j=O ::; c f; vs ( / I Bjrjl:!. h l d>..) 2 ( / hl 3 1 2 f3d.X) 2 by (25) ::; c f; ( / l rf h l i:!. d>.. ) 2 ( / l rjl 3 1 2 cjf3d.X) 2 ! ( / ( t, l r; l 'i' c; ) f3dA) ( h ( ) C ) d I / ! i t, A t. ' lr; ,; ! ! by (19) . ::; c ( J l h l !:!.d>.. ) ( J f31:!.d.X) From Lemma 1 1 we infer that P* is of weak type (11) with respect to the measure f:!. d.X. To check that � 0 and
1
1
00
!
00
1
a
12. Now let us see how we can apply our considerations to estimate integral norms of absolutely summing operators on the disc algebra.
Theorem. Let T: A p < oo. Then
>
X be a pabsolutely summing operator, 1
<
304
III.I. Absolutely Summing Operators On The Disc Algebra § 1 2.
(a) T is pintegral and (b) for every q with
ip(T) � C max(p, p� l ) · 7rp(T) ,
oo >
q
>
p, and every (), 0 < () < ¢ where ¢ =
we have
1� (26)
!:id>..
Let J.L = + J.L8 , be a Pietsch measure for T (see Theorem III.F.8) . Applying Theorem 10 to the weight we get a measure jl = + J.Ls which is also a dominating measure for T with � C. Proposition 6 and Theorem 10 show that there is a projection from and also of weak type Lp (jl) onto of norm � C max(p, (11) . Let us consider the commutative diagram Proof:
!:id>..
!:i
(p � l ) )
Hp (jl)
id i
l il lP
where and with various subscripts denote the formal identity acting between spaces indicated, and is the natural extension of T which exists because T is pabsolutely summing. This diagram clearly shows that T is pintegral with
T
1 ip (T) � I P I P · I T I � C max(p, p)7rp (T) 1 so we have (a) . In order to show (b) let us note that
iq(T) � l i T o iq,p o P o ioo ,q o id l � l i T iq,p o P I = l i P* i�,p T* I · Given x* E X * with l x * l = 1 we define = P*i � , p T*(x*) E Lq' (jl) , so 0
0
0
(27)
Ill.l. Absolutely Summing Operators On The Disc Algebra § 1 3.
305
Applying the Holder inequality we get
where � + � = 1, q'()u = a and q'(1  O)v = p'. Note that a < 1. Since clearly (see the diagram) cp = P* T* (x*) we get (29) Every a * E A* can be extended to a measure v on '][' and if v = gjl + v8 we define an operator Q: A* + La (fl) by Q(a* ) = P* (g )jl. Since any other extension of a * to a measure on '][' is of the form Y1 fl + V8 with P* (g ) = P* (g1 ) we get that Q is a continuous, linear operator from A* into La (fl) with II Q II � � (use weak type (11)). One also sees ( 1  a ) "' that as an element of L a (ii.) the function cp equals QT* ( x *). This gives
ll cp ll a �
c
c
II T * (x * ) ll � II T II (1  a) .l.a (1  a) .1.a ·
(30)
Putting together (27) , (28) , (29) and (30) , we get
i q (T) � Cp (1  a)  ! II T II 8 11'p (T) 1  8 . Since a = p(p  1 + � )  1 (this requires a small calculation) a routine a
but a bit tedious calculation gives (26) .
Y
13 Corollary. Let p � 2 and let be a space such that Then also L(A, = ITp (A,
IIp (C(Y),
Y).
Y
Y)
Y).
L(C(Y),
Y) = Y
Let T: A + be a finite rank operator and let "Yoo (T) denote the infimum of norms of extensions of T to an operator f: C('][') + (see III.B.3) . Clearly for every q > p, we have "Yoo (T) � i q (T). Using (26) we get "Yoo (T) � (c(p)/(¢  O)) II T II 87l'p (T) 1 8 . Note that for fixed p and q + oo we have ¢ + 1 so passing to the limit we have
Proof:
(31) Since L(C(T) , Y) = IIp (C(T), Y) there is a constant C such that � Cy00 ( T ) for every finite rank operator T: A + Y. Thus ( 3 1 )
11'p (T )
306
III.I. Absolutely Summing Operators On The Disc Algebra § 14.
yields 7rp (T) ::; C II T II . Since A has the bounded approximation prop erty (see II.E.5(b) ) the same inequality holds for an arbitrary operator •
T: A + Y.
14 Corollary.
The space Ld H1 (and so also A*) has cotype 2 and
Corollary 13 and Theorem III.F.29 give that II2 (A, i l ) L(A, £ 1 ) so the standard localization (using the fact that A has b.a.p. (see II.E.5(b) ) and Proposition III.F.28 give II 1 (A* , £2 ) = L(A* , £2 ). This fact implies that A* has the Orlicz property. Take any uncondi tionally convergent series L: :'= l Xn in A* and let 'Pn E A** be such that I I 'Pn ll = 1 and 'Pn(xn) = ll xn ll for n = 1 , 2, . . . . Take any a = ( an )�= l E £2 and define Ta : A* + £2 by Ta (x*) = (an'Pn(x* ))�= l · Clearly II Ta l l $ ll a l l 2 so 7rl (Ta) $ C ll a ll 2 · Thus L: :'= l lan l ll xn l l $ L::'= l II Ta (xn) ll $ C so L::'= l ll xn ll 2 < oo. Remark III.E.13 and Propo sition III.A.24 show that A* has cotype 2. •
Proof:
f: C(T)
Every rank n operator with l iT II ::; C II T I I · log n.
15 Corollary.
+ Y
T: A
+
Y
has an extension
It follows easily from the Auerbach lemma II.E. ll (or use III.G Exercise 14(a) ) that 1r2 (T) ::; n(T) ::; II T I I · n. From (31) we get Proof:
•
Taking () = 1  lo! n we get the claim.
Note that this is an optimal result. It follows from Theorem III.B.22 that every extension to C(Y) of the projections defined in III.E.15 has norm greater that C log n. Before we proceed further we will present some modest applications of the results obtained so far. Every linear operator T: A factorizes strongly through L 2 (f!, J..L ) . 16 Proposition.
Proof:
III.H. 15.
+
Lp (fl, J..L ) , 0 < p ::; 2,
This is a direct consequence of Corollary 14 and Proposition a
III.!. Absolutely Summing Operators On The Disc Algebra § 1 7.
(.Xn )�=O is a coefflcient multiplier (.Xn )�=l E l!...k.... .
The sequence A = A into l!p, 0 < p � 2 if and only if
17 Corollary.
from
307
(2p)
Proof: From Proposition 16 we obtain that A is a coefficient multiplier from A into l!p, 0 < p � 2, if and only if A is a multiplier from 1!2 into • l!p . The rest is a clear application of the Holder inequality.
(IPn )�=l
Let Then there exists f
18 Corollary.
L 2 (Y) .
all p < 2.
E:'= 1 i (�Pn . !W
be a complete orthonormal system in such that = oo for
E A
Proof: Repeat the proof of Theorem III.A.25 using the fact that every operator from A into l!p, p < 2, is compact. This follows from Proposition 16.
19. Let us now consider the space A@A. ( The definition of X@Y is given in III.B.25 ) . There is a natural 'identity' map A@A + C(T)@C(Y) . We want to show that this is an isomorphic embedding. Let us start with id: A@A + A@C(T) . By Corollary III.B.27 this map is an isomorphic embedding if and only if every operator from A into A* extends to an operator from C (Y) into A* . To show this it is enough to show L(A, A*) = II 2 (A, A*) ( cf. III.F. 9 ( c )) . Corollary 13 gives that it is enough to check that L(C(Y) , A*) = II 2 (C(Y) , A*) so by the duality Theorem III.F.27 and Corollary III.F.25 it is enough to check that every operator of the form
A * S oN00 with 1r2 (S) � 1 is nuclear with n 1 (ST) � C II T II (C independent of N and T) . But S admits a factorization through a Hilbert space ( see III.F.8 ) so from Corollary 14 we get 1r 1 (S) � C. Thus by the remark oN
{; 00 t
T
t{;
after III.F.22 and III.F.24 we get
So A@A is closed in A@C(T) . Repeating basically the same argument we get A@C(Y) closed in C(Y)@C(T) so putting things together we get A@A closed in C(T)@C(Y) . We do not however, have, equality of the norms. Since for any tensor t E A@A we have lltii A ®A � lltli A®c('Jl') � iitll c('Jl')@c('Jl') it is enough to check that id: A@A + A@C(Y) is not an isometry. By III.B.27 we have to exihibit an operator T: A + A* such that every extension T: C(Y) + A* will have IITII � ciiTII for some
308
III.I. A bsolutely Summing Operators On The Disc Algebra §20.
A*
c > 1. Since £i is !complemented in it suffices to consider the operator T: A . £i given by = (j( O ) , � ]( 1 )) . We have
T (f)
1 II T IIA �R� = sup { l f ( O) I + 2 1/ (1) 1 : 11/ll oo S 1 } '
'
{ 2� j f(O) (a + �/Je  i0)de : lal 1 , 1/JI 1 , 11 /lloo = 1 } = I l + � ei O I 1 / 1 I G e  i O + 1 + � ei O ) t 1 = 1 . l LH = sup
S
S
S
O n the other hand if f : C (1I') . £i is an extension of T then, by a standard averaging argument we get IITII � II T1 II where T1 = ]( 1 (J ( o), � ) ) for E C(1I') . But
(f)
f
20. We would now like to discuss the reflexive subspaces of LI/ H1 . We start with the following proposition which is analogous to Corollary III.C. l8.
X
c LI/ H1 be a closed subspace which does not Proposition. Let have type p for any p > 1. Then contains £ 1 .
Proof:
X
This follows directly from Theorem III.C.16 and III.D.31.
XC
a
LI / H1 be a reflexive subspace and let q denote the natural quotient map from L 1 onto LI/H1 • There exists a reflexive subspace L 1 such that q i is an isomorphism from onto 21 Theorem. Let
XC
X
X
X
X.
It follows immediately from Proposition 20 that has type for some p > 1. We see from Proposition III.H.l4 and Corollary 13 that L(A, = IIq(A, for some q < oo. In particular the operator A . defined by = for is qabsolutely summing, so by Theorem 12 qintegral. Thus extends to an operator . M (1I'). For E A and Let us take C (1I') L 1 (1I') and so = = = we have is an isomorphic embedding and we This shows that qo = 1:1 can put Proof:
p
X* ) X*) r.p: X * r.p(f)(x) x (f) x E X r.p f
III.I. Absolutely Summing Operators On The Disc Algebra §22.
309
22. Our next major theorem will be Theorem 25 which says that every operator defined on a reflexive subspace X C Ld H1 going into H00 extends to an operator from Ld H1 into H00 • We will prove this using III.B.27 so our analysis will deal with functions of two variables. Suppose (!1 1 . J.Ll ) and (!1 2 , J.L2 ) are two probability measure spaces. Assume that Ri : Lp (n i , J.Li ) � Lp (n i , J.Li ), i = 1 , 2, are continuous linear operators. By R 1 ® R2 we mean the operator from Lp (nl X n 2 , J.L l X J.L2 ) into itself defined as 'an operator R 1 acting in first variable and R2 acting in the second'. More precisely on a function of the form J(w 1 . w2 ) = 2:i g] (wl ) · gy (w2 ) with the series convergent in L 2 (S1 1 x S12 , J.L 1 x J.L2 ) we write R 1 ® R2 (f)(w 1 . w2 ) = E i R 1 (g])(wl ) · R 2 (gy) (w2 ) · One easily checks that R 1 ® R2 is bounded so, since functions of the above form are dense in Lp (nl X n 2 , J.Ll X J.L2 ) , it extends to the whole space. This also shows that given J(w 1 . w2 ) we can compute R 1 ® R2 (f)(w 1 . w2 ) applying first R2 to each function f ( w 1 , ·) , treat the result as a function of two variables, F(wb w2 ) , and apply R 1 to each function F (·, w2 ). The result will be the same if we reverse the order in this procedure, i.e. first apply R 1 and next R2 • We have the following
J
Proposition. Let .6 i E L 1 (11') be such that .6 i ;::: 0 and .6 i = 1 for i = 1 , 2. Let .6 i and Pi be given by Theorem 1 0 and let us define Qi = I  Pi and Q = Q 1 ® Q 2 . If 1 < p < q < oo and ! = () + (l ; ll)
then
(32) II Q(cp) ll p � C II Q(cp) ll: · inf{ ll cp  /lh : supp J (n , m) c (Z x Z)\(Z _ x Z_ ) } 1  11 where all the norms are with respect to the measure .6. 1 (x)dx x .6. 2 ( y)dy on T2 •
Let us recall that Z denotes the set of integers and Z_ the set of negative integers. The proof of this proposition follows from the explicit form of pro jections Pi and the analogous fact for Riesz projections. So we start our analysis of the Riesz projection with 23 Lemma.
1
IE R denotes the Riesz projection and R_ then < oo and � = () +
{l ;ll)
=
I

R and
(33)
310
III.I. A bsolutely Summing Operators On The Disc Algebra §23.
Proof: Fix a number .\ > 0 to be specified later and consider the outer function r with lrl = min{I, 1 ; 1 ). Since
=
n _ ( t.p) n _ ( rt.p) + n _ (( f r ) t.p ) = n _ (rt.p) + 'R  (( I  r ) 'R  ( t.p )) {34)
so
II'R  (
.\ } I � C .\P  1 IIt.p ll l,oo llr �.p ll � � f J{lcpl 9}
because
f
J{lcpl 9}
1
A
I
I I1 1
{35)
{36)
A A xPdmcp (x) � xPmcp (x) 0 + P xP  1 mcp (x)dx 0 0 A � _\P  1 II �.p l l l,oo + Pllx mcp (x) ll oo xP  2 dx.
l �.p i P =
l
l
From the Holder inequality we get
II { I  r ) 'R  { t.p ) ll � �
( I I I  rl � r� ( I I'R (t.p) l q ) �.
{37)
Using the explicit form of r (see I.B.23) and the boundedness of the trigonometric conjugation (see I.B.22) we have
r I I I  rl d!!!p) � c l I I  ri P � c ( l { lt.p l > .\} I + J{lcpl I I  ri P ) 9} � C l { l �.p l > .\ } I + c f
J{lcpl 9}
l log l r l l p
I I log lr i i P) {38) ( ) log 1 � 1 f � Cp .\  1 l l�.pll l, oo � cp ( l { lt.pl > .\ } I + r ( j{lcpi> A } � Cp l { l
because
=
31 1
III.I. Absolutely Summing Operators On The Disc Algebra §24.
Putting together (35), (36), (37) and (38) we get
Choosing A = ll
•
Define K. = n�> ® n�> where n�> is the operator n _ acting in the ith variable, i = 1, Then for 1 < p < q < oo and � = (} + ( 1 � 9) we have
2.
24 Lemma.
I IK. (
f
Since for arbitrary
f)
Proof:
(40)
If
in the second variable we infer from Lemma 23 that for each x 1 we have
where all the norms are computed with respect to the second variable. Integrating (41) with respect to x1 and applying the Holder inequality (note that ;IJ > 1 ) we get 11 n � > (
�
J ll
,
9
) :::; C j l
III.I. A bsolutely Summing Operators On The Disc Algebra §24.
312
Applying (42) for
R�)(
But the weak type (11) inequality for
so integrating
so Since K
R gives that for every x2
(43)
(44) with respect to x2 we get
(45) is continuous in L q (1I' x 1I') we see that (43) and (45) give (40) .•
Now we are ready for the Proof of Proposition 22. Using the explicit definition of the pro jection P (see the proof of Theorem 10) we get
Q1 Q9 Q2
s
Thus applying (18) and (22) and then Lemma 24 we get
Jf i Q 1 Q9 Q2 ('P)(x , y)I Pfl l (x)fl2 (Y)dxdy � C L L c�l )cf) J I R � ) Q9 R�)(r;, 1 (x) rt 2 ( y )
l=
s
•
s
•
Using the Holder inequality and the definitions of get
fl 1 and fl2 we further
III.I. Absolutely Summing Operators On The Disc Algebra §25.
313
•
25 Theorem.
If Y
c
LI /H1 is a reflexive subspace then every H00 extends to an operator T : LI/ H1 +
bounded linear operator T: Y +
Hoo .
The reflexivity assumption is needed as is indicated in Exercise 6. Proof: Since (LI/ H1 )* = H00 we infer from Corollary III.B.27 that it is enough to show that i: Y@(LI/ H1 ) + (LI/ Hl )@(LI/ H1 ) is an isomorphism, i.e. for every tensor L:: 7= l Yi ® Xj we have
But clearly if xi = q( /j ) , j = 1 , . . . , n ( note that q is the natural quotient map from £ 1 onto LI/ Hl ) then
Let Y c £ 1 (1l') be the subspace given by Theorem 21. We see ( see III.B.28) that
I � Yi ® L®Lt('Jr) e l � h
s;
s;
C
ih
®
h
I Y®L t('Jrl
=
Jj lili (x) · /j (y) l dxdy.
(48 )
So comparing ( 46 ) , ( 47 ) and (48 ) we see that to prove the theorem it is enough to show that J� inf
{ JJ lfh (x)fj (Y) I dxdy: q (fj )
=
Xj
}
s;
c � II Yi ll · ll xi ll · 1
( 49 )
314
III.I. A bsolutely Summing Operators On The Disc Algebra §25.
From Rosenthal's theorem (Corollary III.H.13) we infer that there exists Ll 1 (x) > 0 with I Ll 1 = 1 and r > 1 , such that
r .£l1 ) ( I ( 1!)
for all y E Y. Let us fix the sequence ( /j )'J= 1 1 , 2 , . . . , n and
C
�
( 50 )
� C II 1J I I 1
L 1 (Y) such that q ( fi )
=
Xj ,
II IYi (x) /j (y) l dxdy � 21.
j
=
( 51 )
·
We define Ll 2 (y) = J  1 I I L".: Yi (x)fi ( Y ) I dx. j
Applying Theorem 10 to the weights Ll 1 (x) and Ll 2 ( y) we obtain weights Ll 1 (x) and Ll 2 ( y ) and projections P1 and P2 . We put Qi I  Pi , i 1 , 2. We also define outer functions ¢i , i = 1 , 2 such that on '][' we have l¢i l = Ll i . We put =
=
j = 1 , 2, . . . , n
(52)
and
j = 1 , 2 , . . . , n. ( 53 ) ¢2 Q 2 ( ¢2 1 Yi ) , Since /i  fj = ¢1 ( ¢1 1 /j  Q 1 ( ¢1 1 /j )) = ¢1 P1 ( ¢ 1 1 /j ) we get q (fj) = q( fi ) , j = 1 , 2, . . . , n, and analogously we see that q (yj ) = q(yj ) Yi • j 1 , 2 , . . . , n. In particular for every sequence of scalars ( aj ) j 1 we get
yj
=
=
=
I � a l 1 � e l � ai Yi i Ll /Hl c l L ajyj ) I ( Ll / H1 � c � � ai yj t j yj
=
q
j
J
( 54)
Thus we have (see (49) and (54) )
I � j j I � fh (x)fj (y) i dxdy � C jj I � yj (x)fj (y) l dxdy . J
J
(55)
315 If we write F(x, y ) = 'E f.h (x)fi ( y) and Q = Q 1 ® Q 2 then from (5 2 ), (53) and (55) we get j III.!. Absolutely Brimming Operators On The Disc Algebra §25.
1
Now let us fix q, < q < r and () such that � = () + ( 1 � 8 ) . Proposition and the continuity of Q in Lr ( I:J.1dx ® I:J. 2 dy) give
22
8
I� ·
c ( II I
inf
{ II I
 f(x, y) ji:J.1 (x) I:J. 2 ( y)dxdy: supp j C A
where A denotes the set (Z x Z) \(Z_ x Z_ ) . Using ( and the definition of I:J. 2 the first factor in by
50)
}
18
(56) is majorized
) ) (I (I I � l � c ( l (I I L iJ ( y ) j]j (x) l dxr I:J.;  r ( y)dy ) c ( l r"K;(y)I:J.;  r (y)dy) ;: � CI8• 3
r /j (y ) f_ij (x) I:J. �  r (x)dx I:J.;  r ( y )dy
�
�
8
(57)
=
56
Since
{ II � inf
I F(x, y)  f(x, y ) j dxdy: supp j C A
{ I I I � (t]j (x)  h} (x)) (IJ (y) 3
I ) 18 · .
 hJ (y)) dxdy: h} , h] are analytic � L II Yj ii II IJ II
(
3
}
18
1
} 8
(58)
316
III.I. A bsolutely Summing Operators On The Disc Algebra §26. a
Putting together (56) , (57) and (58) we get (49) .
26. One of the reasons the above theorem may be relevant to the more classical analysis is the following.
Given ¢ E H00 (D x D) we define T4>: Ll /H1 + H00 by T4> (u) (z) = (u, ¢(z, ·) ) . The correspondence ¢ ++ T4> establishes an isometry between L(Ll /HI . Hoo ) and H00 (D x D) . Proposition.
Clearly T4> is a well defined linear operator into H00 • Since (Ll / H1 ) * = H00 we have
Proof:
IIT
z EID
z E D llull9
so the map ¢ �+ T4> is an isometry. If we are given T: Ll/ H1 + H00 then we define ¢(z1 , z2 ) = T(Oz1 ) (z2) where for z E D the symbol Oz denotes the functional on A 'value at z ' . The Poisson formula shows that Oz E Ll/H1 . Obviously ll ¢ll oo � IITII · Since Oz1 E Ll/H1 for Z1 E D we get that ¢(z1 , z2 ) is analytic in z2 . But T(6z1 ) (z2 ) = T* (Oz2 ) (zl ) where T* : H� + H00 so ¢ is also analytic in z 1 . This shows that ¢ E Hoo (D
x
•
D) .
27. Let us recall that a set of integers A is called a Ap set, 1 < p < oo , if there exists a constant C such that for all sequences of scalars ( aj ) j E A we have
Corollary. If A is a A2 subset of positive integers and sequence of H00 functions such that
sup L lcpk (z) l 2 <
z E D kE A
(cp k (z)) kE A is a (59)
oo
then there exists a function ¢(z, w ) E H00 (D
x
D) such that
One easily checks that Y = span{ e  i k iJ h E A c Ll / H1 is iso morphic to £2 • The operator T: Y + H00 given by T( I:: kE A a k e  ik1J ) =
Proof:
III.I. Absolutely Summing Operators On The Disc Algebra §Notes 317
L kE A ak'Pk (z) is well defined by (59) and by Theorem 25 admits an ex tension T : Ld H1 + H00 • Using Proposition 26 this extension yields the a desired function ¢. Notes and Remarks.
Clearly this chapter contains the most recent and the most advanced results of this book. It shows the very intricate interplay between the general theory and concrete analytical problems. The main results pre sented in this chapter are due to J. Bourgain. It is my feeling that they are not fully understood yet and their power remains to be explored. Proposition 2 and Corollary 3 are due to Kislyakov [P] . This is the solution of the problem posed by Glicksberg [1964] , whether a closed, proper, pointseparated subalgebra of can be complemented. The earlier work on the Glicksberg problem is presented in PelczyD.ski [1977] , Ch. 5. Building on ideas of Kislyakov [P] , Garling [P] has generalized Corollary 3 and has shown that a proper uniform algebra is not a quo tient of any algebra. Proposition 4 is an observation of Milne [1972] . Theorem 1 0 and its proof (i.e. Lemma 8 and Theorem 9 and Lemma 11) are due to Bourgain [1986] but build on his earlier work, Bourgain [1984] . This paper contains also our Theorem 12(b), Corollaries 13, 14, 15, and Theorem 21 . Theorem 25, its proof and consequences are taken from Bourgain [1986] . Actually later J. Bourgain developed still another approach to these results. He proved the following theorem (see DechampGondim [1985] or BourgainDavis [1986] ) .
C(S)
C*
Theorem A . Let 1 < p < oo and 0 < a < 1 and let (!1, E, JL ) b e a probability measure space and let R denote the Riesz projection. The operator Id ® R acts as a continuous operator from Lp('ll' , L 1 (S1)) into
Lp('ll', L0 (f2) ) .
This Theorem implies Corollary 14 (see DechampGondim [1985] or BourgainDavis [1986] ) and Theorem 25 (see Kislyakov [1987] ) . It is perfectly possible and even shorter to present the main results of this chapter avoiding Theorem 10 and Proposition 22 and using Theorem A instead. We have chosen to present longer arguments for the following reasons. (a) We feel that the approach to psumming operators on the disc al gebra A described in 7 is very natural. (b) We feel that Theorem 1 0 is very interesting in its own right (but this is clearly also true about Theorem A) .
318 III.I. Absolutely Summing Operators On The Disc Algebra §Notes (c) There are some interesting applications of Theorem 9. To be more precise, Bourgain [1984a] uses Theorems 9 and 10 to prove the fol lowing Theorem B. Assume n is a positive integer and biorthogonal sequence in A x A* such that
11
for
((/Jk , x k )�=l
form a
k = 1 , 2, . . . , n, n
II 'L:>k x k ll � M · ( L l ak l 2 ) ! for all sequences of scalars.
k =l
Then for some z E
k =l
lzl
(1) (2)
< 1 we have
This is the disc algebra version (and generalization) of the results of Olevskii (see Olevskii [1975] ) . The papers Bourgain [1984] and [1986] contain answers to prob lems asked in Pelczynski [1977] and by N. Varopoulos. Varopoulos, motivated by his important theory of tensor algebras (see e.g. Graham McGehee [1979] Chapter 11) asked the question whether A�A is closed in C(1I')�C(1I'). We saw in 19 that the positive answer routinely follows from Corollary 13. The observation that id: A�A + C(1I')�C(1I') is not an isometry is due to Kaijser [1978] . Naturally the work of Bourgain subsumes many earlier particular results. We will not discuss these here in any detail, but let us point out that the whole effort in Theorem 10 is directed at obtaining the weak type estimate. Projections bounded in Lp, 1 < p < oo, are much easier to construct but they lead at best to the conclusion that Lt f H1 has cotype q for any q > 2 (see Kislyakov [1981b] ) . The concrete applications of our main results which we present are mostly routine. Some more are to be found in the Exercises. Corollary 27 is a twodimensional analogue of III.E.9. Let us also recall that Theo rem 1 0 has applications to operators on Hilbert space which we discussed in Notes and remarks to Chapter III.F. It was also instrumental in con structing counterexamples to some old conjectures of Grothendieck (see Pisier [1983] ) . The very important technical aspect of our work in this chapter is the extrapolation argument based on interpolation inequalities. The first time this appeared in this book was in the proof of Theorem III.F.27.
III.I. Absolutely Summing Operators On The Disc Algebra §Notes 319
In this chapter we have two interpolation inequalities, the abstract one namely Theorem 12(b) and a very concrete one namely Lemma 23. It was recognized by Kislyakov [1987a] that inequalities like in Lemma 23 hold for other natural operators and can sometimes serve as a substitute for the weak type (11) .inequality. This has some nice consequences. It is also interesting to note that our work is essentially restricted to spaces of analytic functions of one variable. Bourgain [1985] and [1986] constructed for every p > 2 an operator on A(IBa), d > 1, and on A([)n ) , n > 1 , which is pabsolutely summing but not qintegral for any q. This shows that most of the results presented in this chapter about the disc algebra are false both for A(IBa) and for A([)n ) with d > 1 and n > 1 . It seems to be a very interesting problem to figure out what goes on in several variables. Added in proof (May 31, 1 990} After this book was submitted I received a very interesting work of Kislyakov [P1 ] . It contains a new, simple proof of Theorem 12 b) . The main truncation lemma (refered to in the title) asserts the following: If b. 2:: 8 > 0 is such that b 1 1 2 ::; cb 1 1 2 for some c then every
H1 (b.d>.. ) can be represented (i) /I E Hoo and II /I lloo :=; 24c2
fE
as
f
= /I + h
:=; s < oo if f E Hs (b.d>.. ) ;s 1 . (f l h I s bo d>.. ) /s ::; 1 06c4 �� � �1 I ll s b.d>..
(ii) for every s , 1
(
r
where
then /2 E
Hs (b.d>.. )
and
Some further extensions are contained in [P2 ] . Exercises
1. Show (without appeal to Theorem 1 0 ) that every multiplier T: £2 + A (i.e. any operator of the form T(�n) = 2:: ;:'= 0 �ntnz n ) has !absolutely summing adjoint. Let T: A + H2 be a multiplier (i.e. T(2:: ;:'= 0 anz n ) = 2:: ;:'= 0 antnz n ) . Show that the following conditions are equivalent: 2.
(a) T is pabsolutely summing for some p < 1 . ; (b) T is 1nuclear;
(c) T factors through an L 1 (/L) space;
320 III.I. Absolutely Summing Operators On The Disc Algebra §Notes Note that it follows from this exercise and Paley's inequality I.B.24 ( or Hardy's inequality I.B.25 ) that the map i d : A (1r) + H1 (Y) does not factor through any £1space. A second such example is in Ex ercise III.G. lO ( d ) . 3.
Let X be a finite dimensional Banach space and let us fix numbers K and p, 1 < p < oo. Show that the following conditions are equivalent: ( a)
ip(T) ::; K1rp (T ) for every operator T: X + Z, where Z is an arbitrary Banach space;
( b ) for every subspace
Y C Lp' (JL) , � + � = 1 , and for every linear operator T: Y + X there exists an extension f: Lp' (JL) + X with II T II ::; K II T II ·
4.
Let X C C(S) be a closed subspace. Show that the following con ditions are equivalent: ( a)
X@C(S) is closed in C(S)@C(S); ( b ) every operator T: X + £1 (JL) extends to an operator T : C(S) + Ll (JL); ( c ) II 2 (X, £1 ) = L(X , i l ) and X* has cotype 2. 5. Show that every reflexive subspace of C 1 (1r2 )* embeds into some Lp space for some p > 1. 6. ( a) Define an operator T: H1 (Y) + H00 (1r) by T(L:�...., 0 anz n ) = L::'= o � zn . Show that T is continuous and does not have an extension to f: L1 (Y) + Hoo (1r) .
( b ) Show that, if
X
c
L 1 (T) is a nonreflexive subspace, then there exists an operator T: X + H00 (1r) which does not have an extension to f: £1 (1r) + Hco (1r) .
( c ) Show that, if
X c Ld H1 is a nonreflexive subspace, then there exists an operator T: X + H00 (1r) which does not have an extension to f: Ld H1 + Hoo (1r) .
7.
8.
Show that Theorem 25 holds with the space Ld H1 replaced by the space £1 (0, JL, Ld HI ) of Bochner integrable (Ld H1 ) valued functions ( see III.B.28 for definitions ) . Let R be a reflexive subspace of £1 (0, JL) and let T: R + Hoo (T) . Show that T extends to an operator T: L1 ( f2 , JL) + H00 (1r) .
III.I. Absolutely Summing Operators On The Disc Algebra §Notes 321
9.
(a) Suppose that ( /j ) f= 0 H00 (T2 ) into £1 by
c
Lt (T) and define an operator T from 00
T(g) = ( (gj , fi} ) 'f= 0 where g(z, w ) = L Yi (z) wi . j =O Show that T is bounded if and only if
1 2 ) 2 : hi E Hr for j = o, 1 , 2, ... } oo . { h ( f; Show that the matrix (an, m ) n, m ;::: 0 is a coefficient multplier from A(D2 ) into £1 (N x N) if and only if:L": n, m ;::: o l an, m l 2 oo . Suppose that M (mk 1 , k N )k 1 , ,kN ;::: o is a coefficient mul N tiplier from A(D ) into £1 (N N ) . Show that there exists a inf
(b) (c)
00
l.
I Ii + hi
<
<
=
•••
•••
constant C such that for every K E N
( kl >···•LkN :S, K
lak 1 , ,k N 1 2 ..•
)
1
2 :::::;
C II M II (N log Nk) ! .
10. Let A C 7l be { 2 k }� 1 . Let X C C(T) be the span{einB }nEAUN · Define an operator R on X by R(f) = (j(  2 k ))'k= t · (a) Show that R maps X onto f2 (A) . (b) Show that R is pabsolutely summing for p < 1. (c) Show that there is no projection from X onto A. (d) Note that this gives an alternative proof of Corollary III.F.35.
11. If R is a reflexive subspace of L1 (T) , then R is isomorphic to a subspace of Ll /Ht and to a subspace of Ht (T) .
12. An n dimensional projection P: A(T) + A(T) is called interpolating if there exist distinct points (tj)'J= 1 C T such that P l (tj ) = l (tj ) for j = 1, 2, . . . , n and every I E A(T) . Show that there does not exist any sequence of interpolating projections Pn : A(T) + A(T) such that Pn (f) + I for every I E A(T) . 13. Show that A(D) is not isomorphic to 14. Show that the ball algebra A(ffid) for d > 1 is not isomorphic to the disc algebra A(ffit ) ·
A(D2 ).
Hints For The Exercises
1. If there is a metric, the balls have to be unbounded in norm. This would give a weakly null sequence which is not normbounded. 2. Each basic neighbourhood U(O; c, xi , . . . , x�) restricts only countably many coordinates from r. This implies that if U1 ::) U2 ::) · · · are weakly open sets in B£2 ( r ) and 0 E U; for j = 1, 2, . . . then n;: 1 U; has contin uum cardinality. 3. Use the Riesz representation theorem I.B.ll and the dominated convergence theorem. 4. Take X = eo and ( x �)�= 1 the unit vectors in £ 1 = c0 . X = £ 1 and x� E f00 , x� = L:;:n e; also works. 5. Both topologies are metrizable so it is enough to check the convergence of sequences. Work with Taylor coefficients. 6. Use Exercise 5. Note also that T is 11 so r  1 (BA (D) ) does not contain a line, so it is not u(H00 , Ld Hl )open. 7. Show that the products are positive and have integral 1 . Use the Fourier coefficients to show that the cluster point is unique. 8. This is basically the same as Exercise 7. 9. First show that for 'Y E r we have T7 = a"'f'Y · Take as f..L the w*limit of T(gn) where 9n = II XA,. II 1 1 XA,. where An are neighbourhoods of the neutral element in and I An l + 0. 10. Put fr (e i6 ) = f(r ei6 ) and note that functions fr are uniformly bounded in L 1 (T). Take the u{M(T) , 0{11'))limit. II.A.
G
II.B. 1 . See Proposition 3. 2. (a) Think what it means in terms of the unit balls. The proof is in Pelczynski [1960] . {b) Note that if ll x + Y ll2 = ll x ll2 + I IY II 2 then x = >.y for >. � 0. (c) Show that I l l · I l l of (b) is strictly convex. (d) Consider the subspace £0 = { x E foo (r): card{"': x ('Y ) =f. 0 } � No } . For x, y E fD such that ll x ll oo = IIYIIoo = 1 write x < y if y ('Y ) = x('Y) for all 'Y E r such that x('Y) =f. 0. Put Fx = {y E £0 : x < y}. Let I l l · I l l be any equivalent norm on £0 . Put mx = inf{ I I IYI I I : x < y } and Mx = sup{ I I IYI I I : x < y}. Use transfinite induction to get z E £0 , ll z ll oo = 1 such that for all y > z we have my = mz and My = Mz . Then mz = Mz . This is only a glimpse of renorming theory. For detailed exposition the reader can consult Diestel [1975] . 3. Look at the partial sum projections. Consider fn = :L:;= 1 e; in CQ . 4. Examine how in the proof of Corollary 18 we used the assumption that (xn) is weakly null. 5. If the sum X + Y is not closed the map ( x + y) �+ x is unbounded so there are x E X and y E Y with ll x ll = I I Y I I = 1 and ll x  Y ll arbitrarily small. 6. (a) Approximate the FaberSchauder system. (b) Approximate the Haar functions, the nth function in Lp,. where Pn /' oo. Examine
324
Hints For The Exercises
what happens if 8 � 1 in Proposition 15. 7. (a) The Haar system is basic in L00 [0, 1] . (b) Interpret Haar functions as functions on 6. . 8. Find the coefficient functionals explicitly. This shows l an l < en "' for [O, 1] . Conversely write f = E :'= l an'Pn = 2::, 1 !k where /k = f E�Lipa 1 E n = 2 k + l an'Pn and for t, s such that I t  s l 2  N estimate separately E �=l !k and E'; fk . Compare with 111.0.27. 9. (a) Permute the trigonometric system and use the Riesz projection. This was shown by Boas [1955] . (b) The derivative maps Lip1 [0, 1] onto L00 [0, 1] and is almost an isomorphism. (c) , (d) Look at (b) . 10. (a) Show that contains a closed subset homeomorphic to 6.. This is done in Kuratowski [1968] 111§36.V and Lacey [1974] . (b) Compare the dual spaces. 11. Use the decompositoin method. For (a) represent (EC[O, 1])o as a subspace of C[O, 1] . 12. Note that for 1 � p < oo, p =f. 2 two functions J, g such 1 that II / + ag ii P = ( II / II � + IIYII � ) :P for all scalars a with lal = 1 have to be disjointly supported. 13. Use Theorem 4 and Exercise II.A.3. 14. Start first with finite E1 . For existence in the general ca.Se use the RadonNikodym theorem. Such projections are called conditional expectations, and are of fundamental importance in probability theory. They are studied in almost every introductory book on probability. 15. (a) Take any countable dense set in Bx and map the unit vectors onto this set. (b) f1 (r) is a subspace of C[0, 1] * . Use (a) . (c) On f00 there exists a sequence of functionals (x�);;"= 1 such that if x� (x) = 0 for n = 1 , 2, . . . then x = 0. 16. Try to repeat the proof of Proposition 6. "'
S
1. Define P(x*** ) = x*** l i(X) E X* . 2. Use the domi nated convergence theorem. Construct Rademacherlike functions. 3. Find a sequence (xn);;"= l C X with ll xn ll = 1 for n = 1, 2, . . . such that (Txn);;"= 1 is a basic sequence in Y and there is an y* E Y* such that y* (T�n) � 8 > 0 for n = 1 , 2, . . . . This was proved in LindenstraussPelczynski [1968] . 4. (c)::::} (b) follows from the Fejer the orem I.B.16 and for (a)::::} (c) consider fe = (2e:)  1 X (  e , e ) and show that w*lime o Tp. (fe) E L1 (Y) . 5. Note that TK : L2 [0, 1] + Loo [O, 1] . To see that TKo is not weakly compact look at the images of the Haar functions. 6. Suppose lp, (ni ) l > E: for ni + oo. Let f..Loo be the w* cluster point of { e in;ll f..L }�1 and let v be the w* cluster point of {e  in; 9 Vn; * f..L }�1 . The F.M. Riesz theorem yields v absolutely con tinuous. Also [1,00 (n) = D(n) for n � 0, so f..L oo is also absolutely con tinuous. On the other hand writing f..L = fdt + f..Ls one checks that f..L oo is singular. This is a result of Helson [1954] . Compare with Exercise 7. 7. If not, take n(p) and m(p) in N tending to oo with p so that p, (n(p)  j ) = p, (m(p)  j) for j = 1 , 2, . . . , p but p, (n(p)) =f. p, (m(p) ) . II.C.
.....
325
Hints For The Exercises
Write fL = fdt + /L s and look at { (e  i n (p}l1  e  im ( p ) li )/Ls } �1 C L1 ( l l"s l) . The weak cluster point /Loo exists, belongs to £1 ( 1 /Ls i ) and is not zero. On the other hand F.M. Riesz Theorem gives that /Leo is absolutely continuous. This is from Helson [1955] . Compare with Exercise 6. 8. (a) The very definition of a shrinking basis gives that (x�)�= 1 is a basis in X* . It is boundedly complete by the Alaoglu theorem. (b) Boundedly completeness give the *weak compactness of the unit ball. (c) Put together (a) and (b) . (d) By (a) (x�)�= 1 is a basis in X* so every x** E X** can be identified with a sequence of scalars. This is old and well known to specialists (see LindenstraussTzafriri [1977] ) . Some parts are already in Karlin [1948] . 9. (a) Cauchy sequences in I · I I J are coordinatewise Cauchy. (b) Consider vectors (1, . . . , 1, 0, 0, . . . ) . (c) Show that if n1 < m1 < n2 < m2 < · · · and Xk = '£';�n k O.jej then ll '£ := 1 xk ii J :s; ( '£ := 1 llxk ll ) ) ! . (d) Use Exercise 8 (d) . (e) Look at the isomorphism between Co and c given in II.B.2(a) . (f) The number dim ( J** f J) is an isomorphic invariant. All this except (f) can be found in James [1950] . (f) is due to BessagaPelczynski [1960a] . 10. This and much more can be found in Davis, Figiel, Johnson, PelczyD.ski [1974] . 1 . Use the form of the partial sum projection as given in the 2 k+l proof of II.B. 10. 2. Look at L 00 k =O '£ 2 k + l an'Pn · 3. Note that the series is we;tkly unconditionally convergent and use Proposition 5. 4. Use � £1 is not compact then Theorem 13 and Theorem 6. 5. If T: '£:'= 1 T* (en) is a weakly unconditionally convergent but not uncondi Proposition 5 and Theorem 6 lead tionally convergent series in to a contradiction. 6. Consider the Orlicz property. 7. (a) It is enough to consider Hilbert space. It is possible to prove it by induction on n. (b) Show that if x E X is a limit of some subsequence of partial sums of the series '£:'= 1 Xn then x E U(xn) · Both (a) and (b) are due to Steinitz [1913] . (c) Take functions {± 2� l hn i } �=O · Show that they all can be ordered into a series whose sum is 0 and into another series whose sum is 1 . Since each function takes only values 0 and 1 every sum will take integer values. (d) The example is in KadecWozniakowski [P] . It is a bit too complicated to repeat it here. 8. (a) This is almost obvious. (b) Show that for every N there is a measure preserving transformation of [0,1] which transforms { gn,k} , n 1, . . . , N, k = 1, . . . , 2 n onto the first Haar functions. (c) Like in the proof of Theorem 10 produce a blockbasic sequence as in (b) . (d) Find blocks of (cpn)�= 1 behaving like those considered (gn,k) in (b) . This basically reduces the problem to the Haar system. This exercise shows the fundamental role played II.D.
C(K)
M(K).
=
326
Hints For The Exercises
by the Haar system in the study of Lp [O, 1]spaces and in theory of or thonormal series. (d) is a result of Olevskii (see Olevskii [1975] p. 75). A Banach space theoretical version of these phenomena is presented in LindenstraussPelczyD.ski [ 1 97 1 ] . 9. (a) Look at the formulas and com pute carefully. (b) Apply to the dk 's the procedure applied to the Haar functions in order to get the 9n,k 's of Exercise 8(b) . Note that the con clusions of (a) hold. Represent it as blocks of the Haar system. (c) This follows directly from Exercise 8(c) . All this is due to Burkholder [1982] and [1984] . II.E. 1 . (a) This is just reformulation of the definition. (b) Take x** E X** , llx** ll = 1 and a net (x'Yh E r C Bx tending to x** in the a ( X** , X*)topology. Use (a) to show that this net converges in
norm. This is a classical result of D.P. Milman. This proof is due to Ringrose [1959] . (c) One has to show the Clarkson [1936] in equalities I I (u�v) 11: + II (u;v) 11: :::; ! ( llull � + llv ll � ) for 2 :::; p < oo ' l and II (u� v) 11:' + II (u;v) 11:' :::; ( ! llull � + ! llv ll � y  for 1 < P :::; 2. The first one is the integration of the corresponding numerical inequal ity while the second follows from the appropriate numerical inequal ity and the inequality ll lul + l v l ll q 2:: llull q + llvll q valid for q :::; 1 . We apply it t o q = p  1. (d) Observe that if liz + vii :::; 1 and liz  v ii :::; 1 then ll z ll :::; 1  cp(l l v ll ) . Apply this observation induc tively to the finite sums. This is due to Kadec [1956] . (e) This is quite obvious. Use (b) to show the existence of the best approxima tion. 2. (a) Replace max by the average in the definition and estimate 2 from below J0 '��" 1 1 + be i9 ld0. (b) Use the ideas of the proof of Exer cise l.d. The notion of complex uniform convexity was first studied by Globevnik [1975] . 3. Take (Pn );:"= 1 and ( qn );:"= 1 two disjoint se quences, dense in [1, 1.5] such that P1 = 1. Take X = ( :L: :'= 1 .e�J 2 and Y = ( :L::'= 1 .e�J 2 • Show that Y does not contain .e� isometrically. 4. (a) Adapt the proof of Theorem 9. This is a correct estimate (see III.B.22) . (b) Consider everything on [ 1r, 1r] . Look at the translation of the square of the Dirichlet kernel. If Vr (t) = sin(r + 2)t/2 sin !t then Vr (O) = r + ! and Vr (08) = 0 for 08 = ( 2;s_;l ) , 8 = ± 1, . . . , ±r. Put /r,s (t) = [Vr (t  08 ) ] 2 , 8 = 0, ± 1 , . . . , ±r. Then :L: :=  r /r,s = (r + ! ) 2 and this helps to show that { (r + � )  2 fr, s } �=  r is isometrically equiv alent to the unit vector basis in .e� + 1 . This is a classical interpolation problem (see Natanson [1949] ). 5. Apply Theorem 9 twice. 6. Every open ball contains infinitely many disjoint balls of equal radii. 7. This is a compactness argument. For each o: E r and x* E X* define a function
327
Hints For The Exercises
on Y by the formula IP
{
a (x * ) (y ) = x * (Sa0 (Y )) ifif Yy ¢E Ya, Ya.
Taking a pointwise cluster point we find cp: X* + Y* which is bounded and linear and T*cp is a projection onto T* (Y* ). 8. (a) Note that ( E :'= 1 �) 00 = ( E :'= 1 �) ; and use Exercise 7. (b) If d(X, Y) is small one can represent X and Y as norms on Rn (or ccn ) such that the unit balls are close, so the norms are close as functions on Rn (or ccn ) . Now we see that the limit exists, so we have completeness. For total boundedness use the Auerbach Lemma. (c) Use (b) and Exercise 7. This and Exercise 7 can be found in Johnson [1972] . Exercise 7 is an improvement of an earlier result of C. Stegall. 9. (a) On each finite dimensional subspace of X we have a uniformly convex norm, uniformly close to the original. Use a compactness argument. (b) Similar to (a) . 10. Show that if E C Lp [O, 1] is finite dimensional, 1 � p � oo, then there exists F C Lp [O, 1] , F � t;: such that E is close to a subspace of F, where both n and 'closeness' are controlled. The case p = oo is relatively easy (use Lemma 1 1 or Proposition 10) . For the case p < oo show that f(t) = sup { J x (t) J : x E E, ll xJJ � 1} is in Lp [O, 1] and consider g(t) max(f(t) , 1). Take the isometry J: Lp [O, 1] + Lp( [O, 1] , gPdt) defined by Ih = h · g  1 . Note that IBE C BLoo ([o , 1j , gP dt) · Now we can follow the case p = oo. This is taken from PelczynskiRosenthal [1975] . 11. Take the quotient map from £1 onto £� (see Exercise II.B.15 (a) and dualize. Or use II.B.4. Show that a finite dimensional subspace of co is also a subspace of £� (for some so has a finite number of extreme points. =
N),
J
1. Note that if f � g and f =1 g in Lp (J.L) then I f  g J P dJ.L > 0. Since everything is below g we can reach the max in a countable number of steps. 2. Estimate J x l q using the Holder inequality. 3. Assume Lp (J.L) = Lp [O, 1] and find in Y a block basis (Yn )�= 1 of the Haar system equivalent to the unit vector basis in fw Take (y�)�= 1 , the sequence of biorthogonal functionals such that y�( Yn ) � epJJ y� J I IIYn ll and such that y� are in the same block of the Haar system as Yn for n = 1, 2, . . . . The projection P(f) = E :'= 1 y� (f)yn works. 4. Consider sets Me = {! E Lp [O, 1] : I f l i P � cl l fl l 2 } for c > 0. If X C Me for some c > 0 then X rv £2 and is complemented . If f E Me then there exists a set A c [0, 1] with I AI < c and II ! · XA l i P � (1  cP) � II f l i P · From this, if X is not in any Me we can find a sequence in X close to the sequence of disjointly supported III.A.
J
328
Hints For The Exercises
Lv (J.t)
functions. This is a result of KadecPelczyiiski [1962] . 5. Each norm1 projection in is a conditional expectation projection {for definition see Exercise II.B.14) . To see this is a rather tedious process. We check that if E ImP and supp g C supp then supp Pg C supp We also check that P ( h sgn = I P ( h sgn !) I sgn for E ImP and h � 0. The details and references are in Lacey [1974] . 6. Use the finite dimensional version of Proposition 7 and Theorem 6. 7. Use Exercise 6. 8. If P1 , . . . , Pn E P, Q1 , . . . , Qn E Q with PkPj = 0 and QkQj = 0 for k # then the norm of Ej ,k PkQj can be estimated by twice applying the Khintchine inequality. Note that Ej= 1 ± Pj is uniformly bounded. This is due to McCarthy [1967] . 9. From the Khintchine inequality 1 we get II E := 1 f; . To estimate it ( f0 ( E:= 1 from above we use the � convexity of the norm {for p $ 2) or the Holder inequality {for p � 2). To estimate from below we replace 1 by = { t: 1 $ t $ 1  2n } . 10. X where Follow the proof of Theorem 8 (P = Po ) . For (a) note that P is a selfadjoint (and so orthogonal) projection. This is classical, due to Bergman. A similar exposition on 1Bd can be found in Rudin [1980] . 11. The operator T9 : Bp (D) + Bp (D) is compact (see Exercise 16) . Consider the spectrum of T9 . This is due to Axler [1985] . 12. Start with n = 2 and write P explicitly, then use the multiplier theorem I.B.32. 13. Apply Proposition 9. 14. Apply Proposition 9c with y(x) = X01 for right a. 15. Use Proposition II.B. 17 (or see Exercise II.B.4) or its modifications for p < 1 to show that the existence of a noncompact operator T: lp + lq implies that lp + lq is bounded. 16. All except (e) are variants of Theorem 25 and can be found in Wojtaszczyk [1988] . For (a) , {b) , {d) repeat the proof. For (c) apply (a) inductively. For (e) take a system such that E := 1 J < oo (e.g. a subsequence of the Haar system. 17. Apply definitions. 18. Use Remark 20. 19. (a) is an example of Schreier [1930] . The original construction requires some familiarity with ordinal numbers. The other way is to invent any Banach space with the sequence violating (a) and use Theorem II.B.4. One such example is to define ll (xj ) �1 11 = sup{ E ;= 1 I x : n = 1 < < For {b) use (a) and apply Theorem II.C.5 to the h < operator T: £ 1 + C[O, 1] given by = n = 1, ... .
f
f
f)
f.
f f
j
l an l 2r2k+ 1 )! rdr) Lv ; 2 2n
anfn l v "'
r2n+l r2n+l · En
En
id:
l fPn l
· · · jn} · 2
T(en) fn,
2,
ik l
j
1. Show that L00 [0, 1] embeds into £00 (see Exercise II.B. 15(a) , use Theorem and the decomposition method. This is due to Pelczynski [1958] . 2. span { z k w  k : k = 0, 1, . . . , n } is such a subspace. 3. Identify t k with cos k (} E T:O , k = 0, 1, . . . , n. This is a classi cal device; see Natanson [1949] . 4. Compute the relative projecIII.B.
n
Hints For The Exercises
329
tion constants of subspaces of polynomials of degree at most n. 5. (a) Use estimates for d(.e; , .e�) and d(.e;, � ) . (b) Dualize. 6. Let �n = {  1 , 1 } n . Note that li is isometric to the span{ri }J= l C C(� n ) , where rj (ct . . . . , en) = cj · Observe that this can b e identified with the span of the first n Rademacher functions in C(�) . Apply Theo rem 13. Projection constants of £; spaces can be found in Theorem VII. 1.9 of TomczakJaegermann [1989] . 7. Identify Y with eo and put (x�)�= l the normpreserving extension of coordinate functionals. Find z� E X* n y.L such that x�  z� + 0 in a(X* , X)topology. Put P(x) = (x� (x)  z� (x))�= l · This fact is due to Sobczyk [1941] and the proof indicated here to Veech [1971] . 8. (a) Use the following set theoretical result due to W. Sierpinski. If N is a countable set then there exists a family { A'Y } 'YE [O , l ] of infinite subsets of N such that A'Y, n A'Y2 is finite for all 'Yl # ')'2 • For the proof of this identify N with the set of rationals in [0,1] and put A'Y any sequence of rationals tending to 'Y · (b) Use Exercise II.B.15c. This is a classical result of Phillips [1940] . The argument indicated here is taken from Whitley [1966] . (c) Suppose P is a projection onto X and i: eo � X is an isomorphism. Extend i to j: £00 + £00 and show that i 1 Pj is a projection onto CQ . 9. Start the induction in the proof of Theorem 21 with the polynomial p. 10. Show Proposition 19 for cp being a lower semicontinuous function on Bd and with the inequality in (a) holding on Bd. This can be found in Rudin [1986] . 11. Take f E H1 ( Bd ) , f = E':= o fn where fn is a homogeneous polynomial of degree n. First note that Rf = E':=o nfn and next show that 1:. ';:'= 1 n  d ll fn lloo :::; ll flh · For this use the Hardy inequality on onedimensional complex subspaces of ccd and estimate the ratio between ll fn lh and ll fn l l oo like in the proof of Proposition 18. This is taken from AhernBruna [1988] . 12. Dualise and use the weak type (11) of the Cauchy projection. For details see Wojtaszczyk [1982] . 13. (a) This is a direct calculation. (b) Use functions from (a) . This can be verified by the direct calculation or by appeal to Corollary III.H.16. 14. Use the ideas from Exercise 13 and the polynomials constructed in Proposition 18. Better results can be found in Ullrich [1988a] . 1. Reduce to the case T: eo + X, I Te n I :::; n 2 • This is not a semiembedding because 1:. ';:'= 1 Ten E X. This can be found in BourgainRosenthal [1983] . 2. For each t consider ( 2c)  1 X [t  e , t+e ] and let x; E X* be a a( X*, X)cluster point. Show that there are uncount ably many t's so that xi are far apart. This is a classical result of Gelfand. Modern generalizations can be found in DiestelUhl [1977] . 3. Take two sequences convergent to different limits. The desired sequence Ill. C.
330
Hints For The Exercises
consists of long stretches of one sequence separated by long stretches of the other. 4. If it is not so, build in H the unit vector basis in £ 1 . 5. Ob serve that spanT{Ll {fl, JL)) = spanT{L2 {fl, JL)) . 6. Simply a uniformly integrable sequence convergent in measure converges in norm. 7. One 2n .!. possible candidate is { ! E LI [O, 1] : L n ( J n 1 l l (t) i l + ;;1 dt) ( n + 1 ) $ 1 } . (n + 1 ) 8. Factor In = BnFn and show that a subsequence of Bn and Fn converge weakly. Consider ffn in H2 {Y) . This is due to Newman [1963] . 9. Modify the proof of Lemma 15. This is a result of James [1964] . 10. If £1 were finitely representable in X* then the � 's would be uniform quotients of X. But X has type p, p > 1 so by Exercise III.A. 17 � would also have type p. 11. Instead of characteristic functions use their smooth approximations. 12. Find closed, disjoint sets Fn C 1] and functions hn E H such that infn JFn l hn (t) ldt > 0. Find In E C O , 1] such that JFn l hn (t) ldt rv JFn hn (t) ln(t)dt, ll ln l l oo $ 1 and In I Fk = 0 for k < n. Put 'Pn = IJ�;:: (1  I Ik l ) ln · This is due to Pelczyll.ski More general results are in III.D. 13. Use Lemma 10 and Proposition III.A.5.
[0,
[ [1962] .
III.D. 1. Write T E L(lp ) as T(x) = E : 1 l'{ (x)e i . For F E L(lp ) * define G E L(lp )* by G{T) = E: 1 F( Ti ) where Ti (x) = f'{ (x)e i . Show II G II = II F II and on compact operators G agrees with F. Show that II F II = II G II + II F  Gil · This is due to Hennefeld [1973] . For p = 2 see AlfsenEffros [1972] . 2. For h* E H * define Eh* = JL I S where JL is any measure on T which extends h* to C(T). Use the def initions to check that it makes sense. 3. For K = [0, 1] the Faber Schauder system shows this. For the general case use the same ideas. 4. Show that dist(f, Irp(C[O, 1] )) = ! sup{ l l(s')  l{s") l : cp(s') = cp(s")} so C(�)/Irp(C[O, 1] ) � eo , with unit vectors corresponding to points s' # s": cp(s') = cp(s"). If there is a projection we can lift these unit vectors to In E C { � ) . On the other hand since those points are dense in � we can find a subsequence such that I I E;= l In; I I 2: en . This was proved by M.l. Kadec. The proof is in Pelczynski §9. 5. One example is: K1 is a disjoint union of the interval [0,1] and the inter val [0,1] with circle attached at each end. K2 is the disjoint union of two intervals each with one circle attached at one of the ends. Check that it works. The details are in Cohen [1975] . 6. For (a) and (b) reduce to the case of selfadjoint operators and write (Ax, x ) explicitly. (c) follows from {b) . More details can be found in KwapienPelczynski [1970] and Bennett [1977] . 7. If nk is very lacunary then you can analyse sgn ei n k B and conclude that (e i nk 9)�= l is in supnorm equiv alent to the unit vector basis. As a model think about Rademacher
[1968]
331
Hints For The Exercises
functions. A more efficient way is to use Riesz products (see Exercise II.A.7) . 8. Assume II Tn ll = 1 and take p(x) such that p(xo) = IIPI I · Then T� (8x0 ) (p) + 1 so T� (8x0 ) > 8x0 in w*topology. Note also that the mass of T� ( 8x0 ) has to concentrate around xo . From this get the con vergence of Tn (f) for smooth f's. This is an improvement of the original Korovkin theorem (see Korovkin [1959] or Wulbert [1968] ) . 9. (a) Note that I;�= O (� ) (1  x) n  k xk = 1 . (b) Use the Korovkin theorem (Exerx x2 cise 8) and compute that Bn (1) = 1 , Bn (x) = x, Bn (x 2 ) = x 2 + ( � ) . This is a modern version of S.N. Bernstein's proof of the Weierstrass approximation theorem. 10. Use the remark after III.A.12 to show that X8 rv £00 and that (X2)** = X8 • The fact that X2 rv Co is more involved. Analogously as in the proof of Theorem III.A. l l show that X2 is isomorphic to a complemented subspace of co. 11. If not then R: A(JBd ) + B 1 (JDd ). Like in III.A. l l we show that B1 (1Bd ) is iso morphic to a subspace of £1 . Thus (use DP and Pelczynski property) R is compact. But for the polynomials Pn (z) constructed in III.B.18 we have II RPn ll 2: c > 0 for all n. This contradicts the compact ness. 12. (a) The desired embedding of T::O into £� + 1 is given by p t> (p(exp(27rk8/(4n+ 1))k� o (see II.E.9) . This is due to Marcinkiewicz [1937a] (see also Zygmund [1968] chapter X §7) . (b) Take small 8 and a maximal Jn separated subset of the unit sphere §d C <ed considered with the quasimetric p((, ry) = 1  l \ ( , ry ) l . The embedding into £00 is given via the point evaluations (see Wojtaszczyk [1986] ) . (c) Identify f� with L00 ( { 1 , . . . , N}, J.t ) where J.t ( {k}) = 11 for k = 1, 2, . . . , N. Take a maximal set (x1 ) j= 1 C E such that
k
II L i xj l ll oo :::; 1 + 8. j=l
Then span(xj) j= 1 = G. To estimate k show that if I { i: I:;= l l xj (i) l > 8} I :::; �n then III.B.9 implies that (xj ) j= 1 is not maximal. This is Corollary 6.2 of FigielJohnson [1980] . 13. Instead of Aj 's consider the basic splines bj , i.e. functions such that b1 is continuous and b1 l (s k , Sk + l ) is a quadratic polynomial for all k and b1 ( s1 ) = 1 and b1 is nonzero only in three of the intervals (s k , S k+ d · (The numbers sk are those defined in 20. ) Show that such b1 's exist and check their properties. Follow the proof of Proposition 21. 14. (a) Note that B c £1 [0, 1] . (b) Show that ll fn ii B :::; C(n + 1) ! . To do this expand fn into the Haar series. The antiderivative of 2:::7= 1 Un, hi )hi is a piecewise linear function. Write it in terms of A1 's as defined in 20 and differentiate back. Conversely it is
332
Hints For The Exercises
enough to estimate the Franklin coefficients of a special atom. Estimate separately small coefficients ( n �  log 2 I IJ) and big ones. For details see Wojtaszczyk [1986a] . 15. The general strategy is similar to the proof of Theorem 27. The details and generalizations can be found in Ciesielski [1975] . 16. As a simple model show directly that the Haar system is not an unconditional basic sequence in L00 [0, 1] . Next use the same idea to show that derivatives of the Franklin system are not an unconditional basic sequence in L00 [0, 1] . 17. Use Proposition 21b) to show that N
f �+ sup I L ( !, fn ) fn l N
n= O
is a weak type 11 map. This can be found in Ciesielski [1966] . 18. (a) We can follow the proof for the Franklin function or perform a direct and rather explicit calculation. (b) The orthonormality follows easily from the definitions. The completeness follows from the fact that continuous functions are dense in L 2 (R) . This and much more can be found in Stromberg [1983] . 19. Suppose Xn�O and Yn are bounded and such that ll xn ll = Xn ( Yn) · Find a weakly Cauchy subsequence Ynk and consider T: X t c defined as x �+ x(ynk ) . 20. Use the Dunford Pettis property. 21. One example is ( L:: := l .ey) 1 ; use Exercise II.E.8. 22. Modify implications (e)=?(d) and (d)=?(b) of Theorem 31. This requires the use of nets. The argument is in Bourgain [1984b] . 23. Use the Pelczyiiski property of £00 and Exercise III.B.7 to show that the existence of an operator that is not weakly compact would imply that £00 has a complemented subspace Y isomorphic to eo . This is impossible; see Exercise III.B.8(c) . 24. Use the Ascoli theorem. 25. Identify C 1 (11'2 ) with a subspace X of C(T2 , £�) by f �+ (j, {h f, lhf) . Consider the annihilator X .L of X, X .L c M(T2 , £�) , where M(T2 , £�) is the space of measures with values in £� . Consider the space G C M(T2 , £�) of all measures J.L such that limn +oo J fn dJ.L = 0 for all sequences fn E X such that if fn = (gn , 01 9n, fhgn) then 01 Yn and fhgn tend to zero pointwise on T2 • Show that G is complemented in M(T2 , £�) (use that it is a C(T2 ) module) . Show that G/X.L separable. Also show that the kernel of a projection onto G is isomorphic to M(Y) . The details can be found in Pelczynski [1989] . 26. The closedness follows from Lemma 6 like in the proof of Corollary 7. To show that it is an algebra, use III.B.20 to show that for a Lipschitz function cp E C(S) and h E Hoo (S) the function cp · h E H00 + C. This is from Rudin [1975] . 27. The proof of closedness is similar to the case n = 1 (see Corollary 7) . To show that H00 (11'n ) + C(r) is not an algebra for n > 1 take f E H00 (11')\A(11')
Hints For The Exercises
333
and show that zn f ( z 1 ) E L00 {T"" ) but is not in H00 {T"" ) + C{T"" ) . This is from Rudin [1975] . that one can assume they are both in £ 1 (11') . This implies that there is cp E H00 {11') such that both cpp, 1 � 0 and 'Pf.£ 2 ;:::: 0. Since for every 'ljJ E H00 {T), J 'ljJcpdp, 1 = J 'ljJcpdp,2 we infer f.£ 1 = f.£2 · 3. If you have a set of extensions that is not relatively weakly compact {it is enough to assume these extensions are in £1 (11')) then use Theorem III.C.12 and Lemma III.C.20 (or Exercise III.C. l l ) to produce a eosequence showing that the original set of functionals was not relatively weakly compact. Use the methods of Theorem III.D.31 or Exercise III.C.l2. 4. (a) Use the Hardy inequality. {b) Easily follows from the fact that diagonal multiplication by 7.: in £2 is not 2absolutely summing (see III.G. 12) . To find an elementary example look at lacunary series. 5. The isometry in both cases is given as [/ ] tt { G.ij )i,j � O with G.ij = j(  (i +j)) . For f E L00 (T) consider the operator HJ (g) = P(f ·g) where g E H2 (T) and P is an orthogonal projection from £2 (11') onto H2 (11') . To evaluate II HJ II use the canonical factorization I.B.23. These are classical results of Nehari and Hartman (see Nikolskii[1980] ) . 6. (a) Use condition {b) of Theorem 4. These examples are due to Hayman and Newman and can be found in Hoffmann [1962] . For {b) take { A n );:'= 1 such that {An } c D is such that {An } n 11' has positive measure but is not T. 7. (a) For F( z ) = � (z + � ) we have F{1Pr) = Er is an ellipse, so A( E) � A(D) . The function F induces an isometric embedding of A( E) into A(1Pr) and the image is !complemented. {b) Consider the map z tt e z from an appropriate strip onto 1Pr. Consider A{1Pr) as a space of functions on this strip. Use the ideas of the proof of Theorem 12. This still requires some effort. For details of (a) and {b) see Wolniewicz [1980] . (c) The exact computation of this norm is in Voskanjan [1973] . Consider a very thin annulus. 8. (a) Use the canonical factorization, Theorem I.B.23. {b) Show that J::_'lr (limr +1 J::_'lr log l w'l'( r ei t) l dt)dcp = 0 and use (a) . (c) Use {b) to approximate inner functions by Blaschke products. Use Exercise III.B.9 to approximate an arbitrary function by the inner functions. {d) From (c) follows that functions f ( z ) = B ( r z ) with B a finite Blaschke product and r < 1 are dense in BA . Represent explicitly Ba ( r z ) = s�;;J) as a convex combination of Blaschke products. (e) Show that for a Mobius trans format ion p (z) = (z  A) (l  Xz)  1 we have ll p (T) II :::; 1. Use ( d ) to show that it extends to any f E A. (b) is a classical result of Frostman (see Koosis [ 1980] IV.9 or Garnett [1981] ). (c) is even older, it goes back to Nevanlinna. ( d ) is a result of Fisher [ 1968] . ( e ) is a classical and important result of von Neumann [1951] . The proof we indicate here is from Drury [ 1983] and is close to the original. For more about this inequality see III.F . 15. A different
334
Hints For The Exercises
{t
{!
proof is indicated in Exercise III.H. 19. 9. (a) Put Max{!) = E T: 1/1 = 11/11 } . Show that E ImP: I Max{f) l = 0} is dense in E. Find a sequence (et , . . . en) C E, ( e i , . . . , e�) C E* , n = dim E such that Max(ej ) = O, j = 1, . . . , n, ll ei ll = ll ej ll = ll ej (ei ) ll = 1 and the matrix (ei {ej)) f.i = 1 is nonsingular. Then ft = ei P, i = 1, . . . , n, o
A
span ImP* . (b) If the disc algebra is a 1r1space then (a) and the F. M. Riesz theorem show that locally looks like l� but this is not the case. This is taken from Wojtaszczyk [1979a] (see also Exercise III.I. 12.). 10. (a) Use the u(L00 , L1 )compactness of the closed ball in H00 • (b) Regularize using the Poisson kernel. (c) Use duality to find F E HP ('I') such that II F II = 1 and (21r)  1 Jy F / = di s {f , H00 ) . This gives that for any best approximation g E Hoo to f we have f  g = 1�1 • (d) Take f = L:: :'= 1 Un'Pn where II 'Pn lloo = 1 and supp 'Pn C ( ( n� 1 ) ' �) C (11", 11"] = T and an + 0 slowly enough. All this is quite old. A nice presentation, references and much more can be found in Garnett [1981] .
t
1 . Work with the definitions and the Holder inequality. 2. Think of l1 as a span of Rademacher functions in L00 [0, 1] and l2 as a span of Rademacher functions in Lp [O, 1] , 1 ::; p < oo. 3. Consider the Haar system. This gives (a) and can be used to get the lower es timate in (b) . To get the upper estimate in (b) you can follow the proof of III.H.24. 4. Use the fact that ( v';+ 1 )�= 1 ¢ l2 . 5. Pietsch's theorem shows that T must map some L1 ( [0, 1] , JL) into C[0, 1] . What can be said about JL? 6. Use the Pietsch theorem and the fact that Lp [O, 1] is not equal to any L q [O, 1] . 7. (a) Use the factorization. (b) First note that every T E Ip (X, Y) is compact (use the Dunford Pettis property) . Next use the Pietsch theorem and arguments like in Lemma III.A.12 to show that T is a sum of absolutely convergent se ries in Np (X, Y). This is due to Persson [1969] . 8. Use Corollary 9. 9. (a) Since TIL (L1 (m) ) C L1 (m) , we see that TIL is 1integral. Since fl('y) + 0, TIL is compact (look at L2 (G)). If TIL is 1nuclear, then the definition yields (look also at T; ) that TIL (f) (x) = fa K(x, y)f(y)dm(y) for some K E L1 ( G x G, m x m) , but this is impossible for singular JL · (b) Use Proposition 12. Show also that translation invariant, nuclear operator on C('l') is a limit in the nuclear norm of operators of convolu tion with a polynomial. Compare with III.G. 18. 10. Take (xj)j= 1 c lf such that L::;= 1 1 x* (xi ) l ::; C ll x* ll and apply Theorem 14 to the matrix k (xj ( i )) i,i = 1 • 1 1. Estimate ( L:: := 1 e  i 2 9 ) f(O)d0 from above using the Holder inequality. The Khintchine inequality will yield the esti mate from below for 11/ll oo· The details are in Bourgain [1987] . 12. III.F.
J1r
335
Hints For The Exercises
(a) Note that if x = I:�= l anXn then I:�= 1 Jan J JJTxn J I � CJJx J J . (b) id: C[O, 1] + £1 [0 , 1] does not factor through £1 . (c) Let X be ei ther L: or CF . Fix x E X, � E X* and (cn)n E F , En = ±1. Define A: Lf + X by A ( I: n E F aneinO) = I: n E F ani:(n)en einO and B: X + Lf by B(x) = 'L: n E F �(ein°)i:(n)ei nO . Show that 1r1 (A* ) � CJJxJJ and 1r1 (B) � CJJ�JJ and next JtrBAJ � 1r1 (A* )7r1 (B) . This is a weak form of a result of Pisier [1978] . 13. Show that if idF is pintegral then the orthogonal projection from Lp(lr) onto L: is bounded. Then use II.D.9.
III.G. 1 . Use Proposition 4 in one direction. For the other use the Schmidt decomposition. 2. For selfadjoint A1 , A2 , . . . , An E a 1 (f2 ) such that I:7= 1 a1 (Aj ) 2 = 1 take selfadjoint Bb · · · , Bn E L (£2 ) such that I:7= 1 JJBi J I 2 = 1 = 2::7= 1 tr(Ai Bi ) · For k = 2k 1 +2 k2 + · . · +2 k; with o � k1 < k2 < · · · < kj � n we define (t) = I: k Bk 1 Bk; rk1 (t) . . . rk; (t) . 1 Estimate JJ ( t ) JJ and evaluate J0 tr ( (t) I:7= l rj (t)Aj )dt. This is taken from TomczakJaegermann [1974] . 3. (a) is obvious. For (b) compare the apnorm of an n x n unitary matrix all of whose entries have absolute value Jn with the apnorm of the matrix all of whose entries are Jn · (c) follows from Exercise III.A.8 and (b) . 4. Use 3(a) and gliding hump arguments. Note that if Uj E a00 (£2 ) are such that PA; UjPA; = Uj =/= 0 for some disjoint sets Ai C N, j = 1, 2, . . . then span(uj ) �1 "' c0 and is complemented. On the other hand if PB uj PA; = Uj =/= 0 for some finite set B C N and disjoint Aj C N then span( Uj ) �1 "' £2 and is complemented. This is due to Holub [1973] . More results of this type are in ArazyLindenstrauss [1975] . 5. Use the projections PA of Exercise 3 and the decomposition method II.B.23. 6. Consider the diagonal operators. If A 1 (£00 , £ I ) had an equivalent norm, then every diagonal operator would be in A 1 (£00 , f I ) . To see that this is not so, show that for idn: � + if we have ak (idn) 2: (n  k + 1)  1 for k = 1 , 2, . . . , n. 7. Write An = O:n · f3n with (o:n) E £2 and (f3n) E co . Define T(x, y) = (o:(y) , f3(x)) where o: (y) = ( o:; yn );;::'= 1 and f3(x) = (f3;xn);;:"= I · This example is due to KaiserRetherford [1984] . 8. If K(x, y) is square integrable this follows from Proposition 13 or Theo rem 19. The general operator differs from this case by one dimension. 9. Show that TK admits a factorization Lq ' �Loo .!Lq ' where f3 is an oper ator of multiplication by a function. Use Theorem 19. 10. (a) For the lower estimate take ej ®e i and see that I: i , j Jx* (ej ® e i ) J = I: i ,j Jx* (i, j) J where x* is really an operator on f2 with n1 ( x* ) � 1. Use Exer cise 12. For the upper estimate note that a2 (u) � cn ; J (u(ut), vt) Jdt where Ut = n ! (r i (t) , . . . , rn (t)) and Vt = n ! (rn + I (t) , . . . , r2n(t)) where •
•
•
J
336
Hints For The Exercises
(rj (t)) J!! 1 are Rademacher functions. (b) Let U be an n x n unitary matrix with lu (i, j) l = Jn and let It have diagonal r1 (t) , r2 (t) , . . . , rn (t) , zeros otherwise and let Jt have diagonal rn + l ( t) , . . . , r2n ( t) , zeros oth erwise. Show that a2 (u) :5 3yn J; ltr(ultUJt ) l dt. c) Passing to the adjoint note that /'l (Jn) = n00 (id: a2 (£�) + a00 (£�)). Use the trace duality III.F.24. (d) Glue together the finite dimensional operators from (c) . (e) This is almost the same as Exercise III.F.12a. (f) Use (e) and (c) . All this is taken from GordonLewis [1974] . 11. Use III.G (13) . 12. Consider onedimensional operators. 13. (a) Fac torize T as C(K) � L2 (J.L) � Ll (J.L) �.e2 and write a id: L2 (J.L) + £2 as L:: :'= l an with an of finite rank and a2 (an) :5 2 n . Show that n 1 (an id) = i1 (an id) :5 2 n . (b) Consider the factorization C(S)�C 1 ('1'2 ) �Wf (T2 )...i.. L 2 ('1'2 ) where i is the identity and j is also the identity (see I.B.31) . Show that it is not nuclear and use (a) or show that it fails (c) . (c) Given V9 C L 1 (J.L) and T: L1 (J.L) + L2 (v) consider M9 : L00 (J.L) + Ll (J.L) , M9 (!) = g · f and use the fact that TM9 is nuclear by (a) . 14. (a) First do it for .e� . For general E factor id: E + E as E�£� L E with 1r2 (a) = 1 and 1 1.811 = 1r2 (id) . But then ,Ba = idt� . This is a result of GarlingGordan [1971] but the proof indicated is due to Kwapien. It can be found in Pisier [1986] . (b) Use (a) , (c) and (d) . Use also III.F.8. 15. Assume that one norm is given by the usual scalar product on Rn (or CC") and show that the other can be chosen to be (x, y ) 2 = L::j= l ajxj'jjj with aj > 0. Take X1 spanned by an appropriate block basis. 16. (a) Take x to be an extreme point in BE . (b) Identify £� with L 2 (N, J.L) where N = { 1 , 2, . . . , N} and J.L is a probability counting measure. On E we have two Hilbertian norms, from L 2 (N, J.L) and the one given by the distance. Use (a) and Exercise 15. 17. (a) , (b) Use Exercise 16b. 1. Use the well known fact (see Katznelson [1968] , Zygmund [1968] etc.) that M f is of weak type 11 but not continuous on Ll [O, 1] . 2. It is enough to work with finite dimensional spaces. Apply III.F .33 and dualize. This works fine for q > 1 . The case q = 1 requires more care. This shows that Proposition 15 is actually an equivalence. Like Proposition 15 this is from Maurey [1974] . 3. If i* is p' summing use Pietsch theorem and dualize. This shows that i factors through Lp . Use Proposition 10 to show that this is impossible. This was observed by Kwapien [1970] . 4. Use III.F.29 and Corollary 1 1 . 5. and 6. Use Propo sition 5. These can be found in Maurey [1974] . 7. For p < 1 there exist positive stable random variables (where stable is understood in a more III.H.
Hints For The Exercises
337
general sense than in III.A. 14) for which III.A.16 holds. The construc tion can be found in Feller [1971) or Lukacs [1970] or in other books on probability theory. 8. Modify the proof of Proposition 10. The details are in Pisier [1986a) . 9. Use Proposition 16 and Exercise III.A.15. 10. Show that if (ai )� 1 are numbers such that Cn = E �= l ai converges then E�= l (*) ai  0 as n  oo. Apply this and the MenchoffRademacher theorem to an = n  1 fn (w) . This is due to Banach [1919] . The argu ment indicated here was shown to me by Mr Wojciechowski. 11. If E;:"= 1 I fn i < oo a.e. then the map (�n)�= l 1+ E;:"= 1 �nfn is a contin uous operator from £00 into Lo [O, 1] . Use Corollary 16. 12. If every f' E N then the map f �+ f' (ei8) would be well defined into L0 (T) , so would admit a factorization through Hp (T) , p < 1 (use Corollary 7) . This is impossible. 13. Note that a system Un)n �l is a system of convergence in measure if and only if there exists T: £2  £0 [0, 1] , a continuous, linear operator such that fn = T ( en ) · Use Proposition 5, Corollary 1 1 and the dilation theorem 19. This is due to Nikishin [1970] . 14. Use Proposition 5. 15. If so then the double Riesz projec oo n n n tion " L.. n , m � O anm ei B eimcp would be of weak L.. n+, m=  oo anm ei B ei cp 1+ " type (11) (use Corollary 7) . The Marcinkiewicz theorem (see I.B.7) gives a bound on the norm in Lp (ll'2 ) , p > 1 , which is false. 16. Use Corollary 1 1 and show using the structure of the ,system (more precisely the ergodicity of measure preserving maps t �+ nt) that the multipli cation operator equals the identity. Compute the Fourier coefficients of sgn sin x. This yields the coefficients of sgn sin nx. Use it to estimate the L2 norm of E := l sgn sin nx. 17. Use Corollary 16. 18. Use Corollary 16. 19. Check the von Neumann inequality for unitary maps. 20. (a) Use Theorem III.C.16 and Lemma III.C.15. (b) Put V = (Y E9 Z) t /H where H = { (x, Tx) : x E X } . (c) If T: X t £2 use (b) to extend  V and use (a) to show that V* has some type > 1 . T to T1 : Use Proposition 1 4 and Lemma III.F.37. (d) Use similar arguments to III.A.25. An analogue of Lemma III.A.26 follows from (c) and the Pietsch factorization theorem. 21. This is like Theorem 30. For (b) ob serve that Bp (D)* = Br (D) where � + � = 1 , 2 :5 p < oo (this follows from the boundedness of the Bergman projection in Lp (D) , 1 < p < oo ; see Exercise III.A. 10) . We use Proposition 29 twice.
C(S)
1. By III.F.36 it is enough to show 1r2 (T* ) < oo , so by dual 111.1. ity (see III.F.27 and III.F.25) we have to show that aT* is nuclear for summing multiplier. Factorize a and use a: £2 t Ltf Ht . a 2absolutely 1 2 the fact that ( E ::'= o !tn l ) 2 :5 !I T ! ! . This is due to KwapienPelczyiiski [1978] . 2. (d) => (c) and (b) => (c) are obvious while (c)=> (a) follows from
338
Hints For The Exercises
A
III.F.35. For (a)=> (d) take a �+ fa. E with fa. (t) = E;= O eiit eii a. , apply III.F.33 and use the Kolmogorov theorem I.B.20. For (d)=>(b) note that T extends to L1 (T) , so use Exercise III.G. 13(a) . This is due to KwapienPelczynski [1978] . 3. The proof is based on duality the ory and clever diagram chasing. This is due to Maurey [1972] (see also Pelczynski [1977] ) . 4. (c)=> (b) follows from elementary properties of 2absolutely summing maps. (b)<=?(a) by III.B.27 and (b)=>(c) follow from the Grothendieck theorem III.F.29. 5. Use Exercise III.A.12 and Theorem III.D.31 to show that such a subspace embeds into some Lp for p < 1. 6. (a) Show by averaging that if there is an extension, then the invariant extension is bounded. This is false. (b) By III.C.l8 and III.C.l6 the space X contains l1 almost isometrically, so comple mented, even in L1 (T) . Use the finite dimensional version of (a) . (c) Is almost the same as (b) but one has to use different theorems. This is noted in Bourgain [1986] to show that the reflexivity assumption in 111.1.25 and III.I.Ex.8 is needed. 7. Use Remark III.E.13 and observe that all arguments in the proof of Theorem 25 are local. 8. Follow the proof of Theorem 25. One does not need Proposition 22. This is due to Bourgain [1986] . 9. (a) the 'if' part is easy. For the 'only if' part use Exercise 7 and follow the ideas of sections 26 and 27 with L1 / H1 replaced by L1 (T, LI/H1 ) · For (b) use (a) . The details are in Bourgain [1986] . (c) Like in Exercise III.D.l2 show that span{z � 1 z�_,.N } k 1 :5K is uniformly a subspace of f� with s = {lOOK N) N . Next show that if X c f� and T: X + £1 then 1r2 (T) ::; CJlOgS II T II · This uses III.F.37 and the estimate Cp ::; cy'P for the constant cp appearing in III.F.37. This estimate follows from the estimate given without proof in Remark III.A.20. This is due to Kislyakov [1981] . 10. (a) Dualize and use the F.M. Riesz theorem and Paley's projection. (b) Take Xp , the closure of X in Lp (T) , p < 1 , and show that Xp = span{ e in 9 } n E A E9 Hp (T) . (c) Average and note that X/ "' £2 but span{ e in 9 } n E A C C(T) is isomor phic to £1 . This can be found in Kislyakov [1981a] . 11. Use III.H.13 and Exercise III.A.2 to show that R is isomorphic to a subspace Y of Hp (T) , p > 1 , such that for y E Y we have II Y II v ::; c ll y l h · 12. If such a sequence exists then sup d(Jm Pn , f�) < oo. This would imply that !absolutely summing operators on behave like !absolutely sum operators on C(T). (see also Exercise III.E.9b) . 13. Show that ming There are many other Theorem 12(a) fails for id: + H1 ways. The result was first proved in Henkin [1967a] . 14. The idea is to observe that Proposition 6 implies that if T: + £2 is a !absolutely summing operator onto then T* ( £ ) is 'essentially contained' in LI/H1 . On the other hand in we have a continuum of complex lines La. such • • •
A
A(D2 )
([2
2
A(D)
(D2 ). A(D)
Hints For The Exercises
339
that � n La are disjoint, so we have a continuum of !absolutely sum ming operators A(JB2 ) + £2 defined by Ta(f) = P( ! I S2 n La) where P is a Paley operator. Also 82 n L a are disjoint peak sets so T� (£2 ) tend to behave like an £ 1 sum. This is too much to fit into Ld H1 . The details (quite complicated) are in MitiaginPelczyiiski [1975] .
List of symbols General symbols
8x 8n, m
IAI
] (n) [x]
XA
u
v
rn ( t)
_
the Dirac measure at the point x the Kronecker symbol; 1 if n = m, 0 otherwise the absolute value of a number; the cardinality of a finite set; otherwise the Lebesgue measure the nth Fourier coefficient of a function f integer part of a real number x indicator function of a set A the normalized Lebesgue measure on Sd when it denotes a measure on JBd it is the normalized Lebesgue measure Rademacher functions, I.B.8. Sets
open unit (euclidean) ball in (Cd complex plane unit disc in
D.
Functional analytical symbols
X EB Y x®Y X rv Y
direct sum of Banach spaces, II.B.19 projective tensor product of Banach spaces, III.B.25 isomorphic Banach spaces, II.B.l
342
X�Y ( , .) xl. span conv
Bx
u(T)
List of symbols
isometric Banach spaces, II.B. l inner product in a Hilbert space and sometimes the duality between linear spaces annihilator of a subspace X c Y, i.e. { y* E Y * : y* I X = 0 } the linear closed span of a set of vectors, I.A. I the convex hull of a set of vectors, I.A.21 the closed unit ball in the space X I.A.2 the spectrum of an operator T Spaces
A(·) Ao()
c co Hp (·) Lp (·) Lo ( ) Loo () lp t; ioo i� Lipa ()
the disc or ball or polydisc algebra, I.B.26 and I.B.28 subspace of A(·) of functions vanishing at 0, I.B.26 or I.B.28 continuous functions, I.B.9 continuous functions vanishing at oo , I.B.9 functions on 1l'8 all of whose derivatives up to order k are continuous, I.B.30 convergent sequences, I.B.9 null sequences, I.B.9 Hardy space, I.B.19 and I.B.21 pintegrable functions, I.B.2 all measurable, a.e. finite functions, I.B.2 essentially bounded functions, I.B.2 psummable, infinite sequences, I.B.5 Rn or ccn with the lpnorm, I.B.5 all bounded sequences, I.B.5 Rn or ccn with the supnorm, I.B.9 functions satisfying the Holder condition of order a, I.B.29 regular, Borel measures on K, I.B. l l . Sobolev space o f functions on 1l'8 all of whose derivatives up to order k are in Lp, I.B.30 Norms
I · II
II ll x ·
norm in general norm in the space X
List of symbols
I · li P ll · ll oo ll · ll p,oo ap(T) ip (T) np (T) 7rp(T) Up(T)
343
norm in some Lp the supremum norm norm in Lp,oo , LB. 7 papproximable norm of an operator T, III.G.3 pintegral norm of an operator T, III.F.21 pnuclear norm of an operator T, III.F . l9 pabsolutely summing norm of an operator T, III.F.2 norm in the Schattenvon Neuman class up, III.G.6
Sets of operators
Ap (X, Y) Ip (X, Y) K(X, Y) L(X, Y) Np(X, Y) ITp(X, Y)
papproximable operators from X into Y, III.G.3 pintegral operators from X into Y, III.F.21 compact, linear operators from X into Y continuous, linear operators from X into Y pnuclear operators from X into Y, III.F . l9 pabsolutely summing operators from X into Y , III.F.2 the Schattenvon Neuman class, III.G.6
Operators the the the the the
Cauchy projection, I.B.21 Dirichlet kernel or corresponding projection, I.B . l5 Fejer kernel or corresponding operator, I.B. 16 Poisson kernel or corresponding operator, I.B.l8 Riesz projection, I.B.20
Constants
be(·) ubc(·) d(X,Y)
l'oo (·) .\(X) e(X) Tp(X) Cp(X)
basis constant, II.B.6 unconditional basis constant, Exercise II.D.8 BanachMazur distance, II.E.6 constant of factorization through L00 , III.B.3 projection constant, III.B.3 extension constant , III.B.3 type p constant , III.A. l7 cotype p constant, III.A.17
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Index This index contains concepts and results which appear in the main text of this book. A name appears here only if it is commonly associated with a concept or a result. The Exercises are indexed only when a new concept (or definition) is introduced there. Alaoglu Theorem II.B.9 algebra uniform I.B. l O . , III.I. l4. almost everywhwere convergence III.A.29, III.H.22ff approximation numbers III. G . l , III. G.7 asymptotic distribution of a sequence III. C. 7. Auerbach Lemma II.E. l l , II.D.29 ball Algebra III.D.30 Banach Algebra I.B. lO. BanachMazur distance II. E.6.ff BanachMazur theorem II.B.4. BanachSaks theorem III.A.27. BanachSteinhaus Theorem I. A. 7 basis (Schauder)
II. B . 5 . , II.E.4., III.B.24. , III.D. 19. ,25,26, III.E. 1 7.
basis boundedly complete II. C.Ex.8 basis constant II.B.6. basis shrinking II. C.Ex.8 basis unconditional II.D.Sff, II.D. 13, III.F . lO basic sequence II.B. 13ff Bergman projection III.A.Ex . l O Bergman space I . B.28 . , III.A.S.ff, III.H.27.ff big subspace of £
! III.D.Ex . 1 2
Biorthogonal Functionals II.B. 7. Blaschke product I.B.23 . , III.E.4,5 Bloch function III.B.Ex. 1 3 , 1 4 block basic sequence II.B. 16.ff block basis II.D. l l Bochner integrable functions III. B . 28. boolean algebra of projections III.A.Ex.8 bounded approximation property II.E. 2 . , II.E. l 2 . , III.D. 1 6. , III.H. 15. bounded extension property III.D. l l . Bourgain's analytic projection III.I. lO canonical factorisation I . B . 23 . , III.E.4. capacity III.E.6. Carleman theorem III.A.25. character I . B . 14. closed graph theorem I.A.6.
fp III.A.6
complemented subspace II.B.5., I I I . F . l O complemented subspace o f convex set I . A . l .
Index
378 convolution I . B . 1 3 . convolution operator II.A.Ex.9, II.C.Ex.4, III. F . 1 2 , 30,32, Ex.9, III. G . 1 8 cotype III. A . 1 7.ff, III.F.36,37 cotype constant III. A . 1 7 ootype o f
L I / H1
111. 1 . 1 4
Demko lemma III.D.23 decomposition method II.B.23 . , III.E. l l . derivative radial III.B.Ex. l l , III.D.Ex. l l dilation theorem III.H. 19. direct sums of Banach spaces II. B . 19.ff Dirichlet kernel I . B . 1 5 . Dirichlet space III.E.6. disc algebra I.B.26., II.E.5 . , III.D .25. dual space I.A. I .
DunfordPettis property III.D.33 . , 111.1.3. DunfordPettis Theorem III. C . 1 2 . EberleinSmulian theorem II.C.3. eigenvalue of an operator I.A . 1 8 . , III . G . 6 ,8 , 1 0 , 1 4, 1 5 , 1 7, 1 9 , 2 1 eigenvector I . A . 1 8 . embedding canonical II.A. l O . , II. C . 2 .
embedding isomorphic I I . B . I., II.D . l 0 , 1 2 , II.E . 1 3 . energy III.E.6.
extension constant III.B. l . extreme point I.A.2 1 . FaberSchauder system II.B. 1 2 . factorisation canonical I.B.23 . , III.E.4. Fejer kernel i . B . 1 6 . finite representability II.E. 1 5 . , III.C. 16. finite representability of
£ 1 III. C. 1 6 , 1 8
Fourier series I I . D . 9 , III.C.8, I I I . E . 9 , III. F . 1 3 , III. G. 14,22, III.H.8, III . F . 1 7 Franklin system III.D . 2 1 .ff, III.E. 1 7 . , III . G . 14. function inner I . B . 23. function of weak type I.B. l 7. function outer I.B.23. Goldstine theorem II.A. 13. Grothendieck's inequality III.F. l4. Grothendieck's theorem III.F. 7,29. III. G . 1 2 . GrothendieckMaurey theorem III.F.35 group, locally compact I . B . 1 3 . Haar measure I . B . 1 3 . Haar system II.B.9 .ff, II.D. 1 3 . , III . G . 13. HahnBanach theorem I.A.8.ff Hankel matrix III.E.Ex.5 Harmonic conjugate function I.B.22 . , III.E. 1 6 . , III.H.8. HausdorfYoung inequality III.G.22. Ravin lemma III . C . 20. HilbertSchmidt operators III. G . l 2 , 1 3 HilleTh.markin operator III.G.22 Holder's inequality I.B.3. ideal III.D. l . ideal o f operators III. F . l . II I . G . 3 infinite divisibility o f A III.E. l 2
Index
379
infinite divisibility of infinite divisibility of
HCX! III.E. 1 3 . , III.E.18. LI / H1 III . E . 1 3
injective space III. B . l . inner function I.B.23. inner function on
IBd III. B . 2 1
interpolating sequence III.E.4. interpolation by polynomials III.B.23 interpolation of Fourier coefficients III.E.9. invariant projection III. B. 13. isomorphism II. B. l . Kahane's inequality III.A . 1 8 . Khintchine's inequality I.B.8. KreinMilman theorem I.A.22. lacunary sequence III.E.9. linear extension operator III. D . l O. III.E.3,5. linear extension theorem III . D . 1 6 . linear operator, positive III.H.Ex.6,7 linear topological space I.A. I.
locally convex linear topological space I.A. I. local reflexivity principle II.E . 1 4 . , III . G . 5 .
LozinskiKharshiladze theorem III. B . 2 2 . , III.E. 15. Marcinkiewicz interpolation theorem I. B. 7. matrix banded III.D.23. Mazur theorem II.A.4. measure space separable I.B. l . Menchoff theorem III.C.6. MenchoffRademacher theorem III.H.22. Mideal III.D.2 .ff, III.D. 8 . , III.E.2. Milutin theorem III. D . 1 9 . multiplicity o f eigenvalue I . A . 1 8 . multipliers III.H.27.ff, III.I. 17. Nikishin theorem III.H.6. norm, equivalent II.B.Ex.2 open mapping theorem I.A.5. operator adjoint I.A . 1 2 . operator compact I.A. 1 5 . , III.A. 1 2 . , III. B . 20. , III. G.5,6. operator dual i . A . 1 2 .
L2 III.H. 1 1 , 16 . , III. I . 1 6 . Lp III.H.9.ff, III.H. 1 5 operator factors strongly through Lp, CX! III.H.4.ff operator factors strongly through
operator factors strongly through
operator HilbertSchmidt III.G. 1 2 , 1 3 operator HilleTamarkin III.G.22. operator invertible I.A. 17.
operator integral II.C.Ex.5 , III.A.8,9,Ex . 1 0 , 13,14, III.F.Ex. 3,5, III. G . 1 3 ,22,Ex.8,9 operator of weak type (pp) I.B.7. operator of weak type ( 1  1 ) III.I. 1 1 . operator pabsolutely summing III . F . 2.ff, III.G. 1 2 , 1 6 , 19. , III.H. 1 4 , 1 5,23, 24. , operator pintegral III.F. 2 1 . , III.I. 12. operator pnuclear III.F . 1 9 . , III. G . 1 7 , 18. operator polynomially bounded III.F. 1518. operator power compact I.A. 1 6 . , III.G.6, 1 5 .
operator power bounded III.F. 1 5 1 8 operator sublinear III.H. l . ff
380
Index operator weakly compact II. C.4.ff, 111.0.34 . , III.F.9. operators related III. G . 1 5 . operators Bernstein III.O.Ex.9 Orlicz property 11.0.7. , III.A . 24. Orlicz theorem 11.0.6. Orlicz, Paley, Sidon theorem III.F.30 0rno theorem III.H. 1 7 orthogonal series (general) III .A.25, III . G . 1 3 , III.H. 18,22,26, III.F. 18 outer function I.B. 1 2 . Paley's inequality and operator I.B.24. , III.F.6. Pelczyiiski's property III.0.33. Pietsch factorisation theorem III.F.8. , III.F.22. , III.H. 1 7. Poisson kernel I.B. 18. principle of local reflexivity II.E. 14, III. G . 5 projection I.A. 20. projection constant III.B.3.ff , III. B . 1 5.ff, III.B.22. projection constant of
£2 III.B. lO.
projection constant of ndimensional space III.B. lO. projections in A III.E. lO, l l projections onto ncodimensional subspaces III.B. 1 1 . projections onto ndimensional subspaces o f projective tensor product III.B.25.ff
Lp III. B . l O.
quasinorm I.A.2. Rademacher functions I.B.8. reflexive space II.A. 14. , II.C.5., III. C . 1 , 18., III.H. 13. reflexive subspaces of reflexive subspaces of
L 1 (J.L) III.C. 18. , III.H. 13. LI /H1 111.1.20,21 ,25
related operators III.G. 15.
relative projection constant III.B.3.
rich subspace of C ( K ) III.0.29.ff
relatively weakly compact set II. C . l .ff, III.C.9.ff, III.0. 3 1 ,34. Riesz products II.A.Ex. 7
Riesz projection I.B.20. , III.I. 23,24. Riesz theorem I.B. 1 1. Riesz F.M. theorem I.B.26. , III.E. l . RieszThorin theorem I.B.6. RudinCarleson theorem III.E.2. Schattenvon Naumann classes III.G.6, 1 1 . , III.H.26 Schmidt decomposition III. G.6, 12. Schur lemma III.A.9. Schur theorem III. C.9. semiembedding III.C.2.ff sequence weakly Cauchy II.A. l . ff sequence weakly convergent II.A. l.ff, II.B.18. series unconditionally convergent II.O. l . ff, 11.0.6. , III.A. 2 1 . series unconditionally convergent i n measure III.H. 17,20,25. series weakly unconditionally convergent 11.0. 3 . , 111.0. 3 1 , 35 . , III . F . 2 . set
Ap 111.1.27.
Sidon set III. F . 3 1 . singular numbers 111.68, 10.
Sobolev spaces I.B.30. , III.A.3. , 111.0.30. , III. F . l l. space C(K) III.f. 8 , 1 2 . , III. G . 1 2 . , III.H. 1 4 . , III.I.3.
Index space space
381
C[O, 1] or C(ll') II. B . 4 . , II.D. l 2 . , II.E. 5 . , III . B . 24. , III.D. 2 5 . , III.E. 14. C k (11'8 ) III.D.30.
space Dirichlet III.E.6. space F I.A.2, III.H.3 space Hardy I.B.l9. space space
H00 III . C . 1 9 . , III.D.6 . , III.E.4, 1 3 , 1 8 . , III.I.25,26. H00 + C III.D.6.ff
space James' II. C.Ex.9 space
Lipo. I.B.29., III.D.27. , III. G . 14. , III.H.26 L 00 I.B. 2 . , III.B.2 ,5. , III.C. 19., III.H. l . ff space Lp I.B.2 . , II.D. 1 3 . , II. E . 5 . , III. B . l O . , III.D.26 . , III.H.9.ff
space
space space space
t; II.E.8.
Lp,oo III.H.4.ff £1 II,D . l O . , III.F . 7 , 1 0 . , III. G . 1 2 . ,
space of homogenous polynomials on
III.H. 13.
C[d III.B. 14.ff
space of trigonometric polynomials II.E.9 . , III.B.22.ff, III.E. 1 5 . space 1l"A II.E. 3 . , III.D. 1 5 . space quasinormed I.A.2.
space reflexive II.A. 14., II.C . 5 . , III.C. 1 , 18 . , III.H. 13. space uniformly convex II.E.Ex. 1 , 2 space
W;' (T8 ) III.A.3 . , III. F . l l .
space weakly sequentially complete III. C . 14., III.D. 35. spectrum I.A.18. splines III.D.Ex . 1 3 stable law III.A . 1 3 . , III.H. 1 2 . Steinhaus theorem III. C. l4. StoneWeierstrass theorem I . B . 1 2 subspace complemented I.A.20. system of convergence in measure for
£2 III.H.Ex. 1 3 , 1 6
tensor product, projective III.B.25 .ff, III.I. 19,25 theorem BanachMazur II.B.4. theorem BanachSaks III.A.27 theorem BanachSteinhaus LA. 7. theorem closed graph I.A.6. theorem Carleman III.A.25. theorem dilation III.H. 19. theorem DunfordPettis III. C . l 2 . theorem EberleinSmulian II.C.3. theorem Goldstine II.A. 13.
theorem Grothendieck III.F.7. , III.F.29. , III. G . 1 2 . theorem GrothendieckMaurey III.F.35. theorem HahnBanach I.A.S.ff theorem Korovkin III.D.Ex.8 theorem KreinMilman I.A.22. theorem local reflexivity II.E. 1 4 . , III.G.5. theorem LozinskiKharshiladze III . B . 2 2 . , III.E. 15. theorem Marcinkiewicz on interpolation I.B.7. theorem Mazur II.A.4. theorem Milutin III.D . 1 9 . theorem Menchoff III.C.6. theorem MenchoffRademacher III.H.22. theorem Nikishin III.H.6.
382
Index theorem open mapping I.A.5. theorem Orlicz 11.0.6. theorem Orlicz, Paley, Sidon III.F. 30. theorem OrliczPettis II.O.Ex.3 theorem 0rno III.H. 17. theorem Pietsch factorisation III.F.8,22 . , III.H. 17. theorem Riesz I.B. l l . theorem Riesz F.M. I.B.26. , III.E. l . theorem RieszThorin I.B.6. theorem RudinCarleson III.E.2. theorem Schur III.C.9. theorem Steinitz II.O.Ex.7 theorem StoneWeierstrass I.B. 1 2 . theorem weak basis II.B.Ex. 1 6 Toeplitz matrix III.O.Ex.6 topology weak II.A. l . ff * II.A.6.ff
topology weak
trace duality III.F.26 trigonometric conjugate function I.B.22. type III. A . 1 7.ff, III.C. 16.ff, III.H.6, 1 1 , 1 2 , 14. , 111.1.20. type constant III.A . 1 7. unconditional basis constant II.O.Ex.8 uniform algebra I.B. lO. , 111.1. 14. uniform approximation property II.E.Ex. lO uniform integrability III.C. l l . Vallee Poussin d e Ia, kernel i.B. 1 7. weakly compact subset of weakly compact set
L 1 111 . 0 . 1 2 .
weak sequential completeness III.C. 1 2 . , 111.0.35. Weyl set III . C . 7. Weyl's inequality III.G.8, 1 1 . Zygmund class III.O.Ex. 15