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(Pm(x}= ~(am,nlxni)P < +oo),) n
equipped with the topology determined by the sequence of £pseudonorms {llxllm} in M(am, n), {Pm(x)} in LP(am,n) for 0
1. Example 1.3.IO Let X be a linear space. Let (x,y) be a twoargumentscaiarvaluedfunction satisfying the following conditions: (il) (x,x) ~ 0, (i2) (x,x) = 0 if and only if x = 0, (i3) (x1 +x2 ,y) = (xl>y)+(x2 ,y), (i4) (ax,y) = a(x,y) for all scalars a, (i5) (x,y) = (y,x), where adenotes the conjugate number to the number a.
Chapter 1
18
The function (x,y) is called an inner product. The space X with the inner product is called a preHilbert space. It is easy to verify that llxll = (x, x) is an Fnorm.
y
1.4.
COMPLETE METRIC LINEAR SPACES
Let X be a metric space with a metricp(x,y). A sequence {xn} of elements of X is said to be a Cauchy sequence or to satisfy the Cauchy condition, or to be fundamental if lim
p(xn, Xm)
= 0,
n,m+co
The space (X,p) is called complete if each Cauchy sequence {Xn} is convergent to an element x 0 EX, i.e., lim p(xn, x 0)
= 0.
It is easy to verify that a subset A of a complete metric space (X,p) is complete if and only if it is closed. A set A contained in a metric space X is called nowhere dense if the closure A of the set A does not contain any open set. A set A is said to be of the first category if it can be represented as a union of a countable family of nowhere dense sets ; otherwise, it is said to be of the second category. From the definition of the set of the category it trivially follows that a closed set of the second category contains an open set. Theorem 1.4.1 (Baire), A complete metric space (X,p) is of the second category. Proof. Suppose that X is of the first category. Then, by definition X= 00
=
U Fn, where the sets Fn are nowhere dense and Fn C Fn+I· Since the
n=l
set F 1 is nowhere dense, there exists an open set K 1 such that K 1 and the diameter of K1 d(K1) =sup p(x, y)
<
fl
F1 =
0
1.
z;uek1
K2
The set F 2 is nowhere dense, hence there exists an open set K2 such that C K~> d(K2) < 1/2 and K 2 fl F 2 = 0.
Basic Facts on Metric Linear Spaces
19
Continuing this process, we can find by induction a sequence of open sets Kn such that Kn C Kn+l, d(Kn) < 1/n, and
Kn
fl
(1.4.1)
Fn = 0.
Let Xn E Kn. Since d(Kn):;..0, {Xn} is a Cauchy sequence. The space X is complete; thus th:=e is an element x 0 E X such that {xn} tends to x 0 • Since Kn C Kn+l, x 0 E Kn, n = I, 2, ... Hence, by (1.4.1) x 0 ¢ Fn, n = 1, 2, ... GO
This implies that x 0 ¢
U Fn =
X and we obtain a contradiction.
D
a=l
1.4.2. Let (X,p) be a complete metric space. Let E C X be a set of the first category. Then the set CE = X"" E is of the second category. Proof. X= E u CE. Suppose that the set CEis of the first category. Then X as a union of two sets of the first category is also of the first category.
CoROLLARY
This contradicts Theorem 1.4.1.
D
A metric linear space X is said to be complete if it is complete as a metric space. A closed linear subspace of a complete metric linear space is also complete. Let (X,p) be a complete metric linear space. Let p'(x,y) be another metric on X equivalent to the metric p(x,y). Then (X,p') is not necessarily complete, as follows from Example 1.4.2 Let X be a real line and letp(x,y)
= JxyJ. Obviously (X,p) is complete. Letp'(x,y) = JarctanxarctanyJ. It is easy to verifythatthemetricsp and p' are equivalent. The space (X,p') is not complete, since the sequence {xn} = {n} is a Cauchy sequence with respect to the metricp'(x,y) but, evidently it is not convergent to any element x 0 E X. The situation is different if we assume in addition that the metric p'(x,y) is invariant. Namely, the following theorem holds 1.4.4 (Klee, 1952). Let (X,p) be a complete metric linear space. Let p'(x,y) be an invariant metric equivalent to the metric p(x,y). Then the space (X,p') is complete. THEOREM
Chapter 1
20
The proof is based on the following lemmas : LEMMA 1.4.5 (Sierpinski, 1928). Let (E,p) be a complete metric space. Let E be embedded in a metric space (E', p'). Suppose that on the set En E' the metrics p(x,y) andp'(x,y) are equivalent. Then Eisa Go set (i.e. it is an intersection of a countable family of open sets) in E'. Proof Since p and p' are equivalent, for any x E E there exists a positive number r11(x) < 1/n such that y E E, p'(x,y) < rn(x) implies p(x,y) < < 1/n. Let
Un(x) = {y E E': p'(x, y) Gn
=
<
rn(x)},
U Un(x),
xEE
The sets Un(x) are open. Thus the sets Gn are also open. Therefore G0 is a G0 set. Evidently E C G0 • It remains to show that E :::> G0 • Let x 0 E G0 • Then x 0 E Gn, n = I, 2, ... By the definition of the sets Gn, there exist elements Xn E E such thatp'(xn,x0) < r(xn). Since rn(x) < 1/n, the sequence {xn} tends to x 0 in the metric p'(x,y). Let e be an arbitrary positive number. Let n be a positive integer such that 2/n < s, and let k 0 be a positive integer such that
fo < rn(Xn)p'(xn, x Then, for k
>
0 ).
k 0,
p'(xk, Xn)
< fo +p'(x
0,
Xn)
<
rn(Xn).
Hence, by the definition of the number rn(xn),p(xk,Xn) < 1/n. This implies that the sequence {xn} is a Cauchy sequence with respect to the metric p(x,y). Since (E,p) is complete, there is an element x 0 E E such that lim p(xn,xO) = 0. The metrics p(x,y) and p'(x,y) are equivalent on the set E. Therefore x 0 = xO and x 0 E E.
0
Basic Facts on Metric Linear Spaces
21
LEMMA 1.4.6 (Mazur and Sternbach, 1933). Let(X, p) be a complete metric linear space. Let the metric p(x,y) be invariant. Let X 0 be a dense linear subset of X. If X 0 is a Go set, then X 0 = X. ct)
Proof If X 0 is a G0 set, then by definition X 0 =
U G,, where Gn are open n=l
sets, G, C Gn+ 1 . Since X 0 is dense in X, Gn are also dense in X. This impliesthatthesetsX""Gn arenowheredense. ThusX""X0 is the set of the first category. Then the set X 0 is of the second category. Suppose that the set X"'X0 is not empty. Let x EX"' X 0 • Since X 0 is a linear subset,
This leads to a contradiction, since the invariance of the metric p(x,y) implies that x+Xo is of the second category and X"'X0 is of thefirstcate
D
~~
LEMMA 1.4.7 Let(X, p)be a metric linear space. Let p(x, y) be an invariant metric. Then there is a complete metric linear space (Y,p') such that X is a dense subset in Y and the metrics p(x,y) and p'(x,y) are equivalent on X. The space Y is called the completion of the space X. Proof We complete X in the classical way, defining Y as the space of all sequences {xn} satysfying the Cauchy condition. We identify two sequences, {xn} and {Yn}, iflim p(xn, Yn) = 0. A metric in Y is defined by the formula p'({xn}, {Yn}) =lim p(Xn,Yn). 11r+ct;J
It is well known that Y is a complete metric space. X can be embedded in Y as the set of stationary sequences x = {x, x, ... } ; X is dense in Y and on X the two metricsp(x,y) andp'(x,y) coincide. The fact that the metric p(x,y) is invariant implies that Yis a linear space and that the operations of addition and of multiplication by scalars are continuous in the metric p'(x,y). D
Proof of Theorem 1.4.4. Let (Y,p') be the completion of(X,p). By Lemma 1.4.5. Xis a dense G0 set in Y. Hence, by Lemma 1.4.6, X= Y. D
Chapter 1
22
1.4.8. Let {xn} be a Cauchy sequence of elements of a metric space (X,p). If it contains a subsequence {xnk} convergent to x 0 eX, then {Xn} also converges to x 0 • Proof. Let e be an arbitrary positive number. Since Xnk+x0 , there is an index k 0 such that fork > k 0 , p(xn.,x0) < ef2. On the other hand, thesequence {xn} is a Cauchy sequence. Therefore, there exists a number N such that, for n, m > N, PROPOSITION
p(xn, Xm)
e
< 2·
Putting m = nk, k
> k 0 , we obtain
p(xn, x 0) ~p(xn,Xn.)+p(xn., X0)
e
e
< 2+ T =e.
The arbitrarines of e implies the proposition. PROPOSITION
1.4.9. Let (X,
II II) be an
D
F*space. Let {ei} be an arbitrary 00
2
fixed sequence ofpositive numbers such that the series
ei is convergent.
i=l
If for each sequence {Xi} of elements of X such that
llxill < C:i the series
00
2
Xi is convergent, then the space (X,
II II) is complete.
i=l
Proof. Let {Yn} be an arbitrary Cauchy sequence of elements of the space X. We can choose a subsequence {Ynk} such that llxkll < ek, where Xk 00
=
Ynk+lYnk· The assumption implies that the series
2
Xk is convergent
k=l
to an element xeX. We shall show that {Yn} tends to x+y 111 • In fact Sk k
=
2 X;= YnkYn
1
tends to x. Thus Yn. tends to x+yn 1 • By Proposi
i=l
tion 1.4.8 {Yn} tends to x+yn1 • D A complete F*space is called an Fspace. A complete preHilbert space (Example 1.3.10) is called a Hilbert space. 1.4.10. Let (X, II II) be an Fspace. Let Y be a subspace of X. Then the quotient space X/Y (see Section 1.1) is an Fspace (i.e. it is complete). THEOREM
Basic Facts on Metric Linear Spaces
23
Proof Let {Zn} be an arbitrary sequence of elements of X/Y such that IIZ11II < 1/211• By the definition ofthe norm in the quotient space there are X11 e Z11 such that llx11ll < 1/211 •The space X is complete, hence the series 00
L X11 is convergent to an element x eX. Let Z denote the coset containing n=l
x. Then, by the definition ofFnorm in the quotient space k
k
112z,zll~li2x,xll < i=m i=m
221·
00
Therefore, the series
2Z
11
is convergent to Z, and by Proposition 1.4.9
n=l
the space X/ Y is complete.
D
1.4.11. The product (X, II II) of n Fspaces is an Fspace. Proof Let xm = (xi) be a Cauchy sequence. Then the sequences {xi}, i = 1, ... , n also is a Cauchy sequence. Since (Xt, II liD are complete, there are Xi eXt such that !!xixt 11+0, i = 1, ... , n. Let x = (xt). Then PROPOSITION
n
l!xmx!!
=
};!!xixt lli+0.
D
i=l
1.5. COMPLETE METRIC LINEAR SPACES. EXAMPLES 1.5.1. The spaces N(L(Q, L:, p,)) are complete. Proof Let {xn} be a Cauchy sequence in N(L(Q, L:, p)). It is easy to verify that the sequence { 11 } is a Cauchy sequence with respect to the measure (this means that, for each a > 0, PROPOSITION
x
lim p({t: !xn(t)xm(t)! >a})= 0). n,m~co
Therefore, by the Riesz theorem, the sequence {xn} contains a subsequence {Xnk} convergent almost everywhere to a measurable function x(t). Let e be an arbitrary positive number. Since the sequence {x11 } is a Cauchy sequence, there is a positive integer N such that for n, m > N PN(XnXm)
=
JN(!xn(t)xm(t)!)dp ~B.
a
24
Chapter I
Put m
=
nk and
let k+oo. By the Fatou lemma, we obtain
PN(XnX)
~e.
This implies that XnX E N(L(Q, r, p). Since N(L(Q, r, /1)) is linear, X E N(L(Q, r, /1)). The arbitrariness of e implies that Xn+X. 0 1.5.2. The space M(Q, £, f.l) is complete. Proof Let {Xn} be a Cauchy sequence in M(Q, £, p). Then the sequence {Xn(t)} is convergent for almost all t. Let x(t) denote the limit of the sequence {xn(t)}.lt is easy to verify that x(t) E M(Q, £, p). Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n,m > N PROPOSITION
llxnXmll
= esssup lxn(t)xm(t)l <e. tED
Hence, when m tends to infinity, we obtain llxnxll = esssup lxn(t)x(t)l
~e.
tED
The arbitrariness of e implies that Xn+X.
0
1.5.3. The space C(Q) is complete. Proof Let {xn} be a Cauchy sequence in C(Q). This implies that at each t the sequence of scalars {Xn(t)} is also a Cauchy sequence. Its limit x(t) is continuous as a limit of a uniformly convergent sequence of continuous functions. Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n, m > N PROPOSITION
llxnXmll = sup ixn(t)xm(t)i <e. tED
Hence, when m tends to infinity, we obtain llxnxll = sup lxn(t)x(t)i
~e.
tED
The arbitrariness of e implies the proposition. 1.5.4. The space C(QjQ0) is complete. Proof C(QjQ0) is a closed subspace ofthe space C(Q).
0
PROPOSITION
0
Basic !<'acts on Metric Linear Spaces
25
e
1.5.5. The space 0(!J) is complete. Proof Let {Xn} be a Cauchy sequence in eo(D). Then {Xn} is also a Cauchy sequence in each pseudonorm II lit, i.e., PROPOSITION
i = 1, 2, ...
llxnXmlli = 0,
lim m,~oo
The sequence {Xn(t)} is convergent for each t.lts limitx(t) is a continuous a)
function on each Qi, hence it is continuous on the set Q =
U Q« and i=l
lim llxnxllt = 0,
i = 1, 2, ...
n>a>
0
1.5.6. The space C"'(Q) is complete. Proof Let {xn} be a Cauchy sequence in the space C"'(Q). Then {xn} is a Cauchy sequence in each pseudonorm II Ilk, i.e., PROPOSITION
llxnXmilk = 0.
lim m,n+co
Let k = (0, ... , 0). This implies that the sequence {xn(t)} tends uniformly to a continuous function x(t). In a similar way we can prove that the sequence of derivatives tends uniformly to the corresponding derivative of x(t). Thus C 00 (!J) is complete. 0 PROPOSITION 1.?.7. The space c5(EP) is complete. Proof Let {Xn} be a Cauchy sequence in c5 (EP). Then
lim
llxnXn' llm,k = 0
for all m and k.
(1.5.1)
n,n~oo
Putting m = (0, ... , 0), we infer by Proposition 1.5.6 that the sequence {xn(t)} is uniformly convergent to an infinite differentiable function together with all its derivatives. Let s be an arbitrary positive number. Formula (1.5.1) implies that there is a positive integer N such that, for n, n' > N,
26
Chapter 1
Hence, when n' tends to infinity, we obtain Jlxnxll = sup tEEP
It
k,
a~·,
1 •••
kp
tp
The arbitrariness of e implies that
(xn(t)x(t))
n ittim'
/
p
~e.
t=l
D
Xn+X.
1.5.8. The spaces M(am,n) and V'(am,n) are complete. Proof Let {xk} = {{x!}} be a Cauchy sequence. Since supam,n > 0, the
PROPOSITION
m
sequence {x!} converges for any fixed n to a limit x~. Write x0 = x~. The sequence {xk} is a Cauchy sequence, hence it is a Cauchy sequence in each pseudonorm llxllm, i.e. for any e > 0, there is a positive integer K such that for k,k' > K Jlxkxk'llm = supam,n lx!.x!'l n
< e,
(for0
Jlxkxk'llm =};(am, n lx!x!'I)P
and for 1
<
+=
<
8
n
llx!x~\lm = (};Cam, nl x!x!' I)P) 11P <e). n
Hence, when k' +oo we obtain [[xkx0 ll
~e.
The arbitrariness of e implies the proposition.
1.6.
[]
SEPARABLE SPACES
A metric space (X, p) is called separable if it contains a countable dense set. PROPOSITION 1.6.1. A metric space (X, p) is separable if and only if for an arbitrary positive e there is a countable set E. such that for an arbitrary x eX, there is an element y e E. such that p(x, y) <e.
27
Basic Facts on Metric Linear Spaces
Proof. Necessity. Let (X,p) be a separable space and let E be a dense countable set. Let E8 = E for all e. Sufficiency. Let E8 be a family of sets satisfying the property described 00
above. Then the set E =
U E 11n is a dense countable set in X.
D
n~I
CoROLLARY 1.6.2. Let (X,p) be a metric space. If there are a noncountable set A and a number {J > 0 such that
p(x, y)
>
{J
for all x, yEA, x # y,
(1.6.1)
then the space (X,p) is nonseparable. Proof. Suppose that (X,p) is separable. Then by Proposition 1.6.1 there is a countable set E812 , such that for each x E A, there is an x E E.12 , such thatp(x,x) < e/2. Thus by (1.6.1) there is a onetoone correspondence between x and This leads to a contradiction since A is noncountable and E,12 is countable. D
x.
PROPOSITION 1.6.3. A subset X 0 of a separable metric space (X,p) is a separable metric space (X0 ,p'), where p' is the restriction of the metric p to X 0 • Proof. Let e be an arbitrary positive number. By Proposition 1.6.1 there is a countable set E,12 such that for any x E X there is a zx E E,12 such that p(x,zx) < ef2. We associate with each element z E E,12 an element y(z) E X 0 , such thatp(z,y(z)) < e/2, provided that such an y(z) exists. Then
p(x, y(zx))
< p(x,
zx)+p(zx, y(zx))
e
<2
+ 2e
Thus, by Proposition 1.6.1 X 0 is separable.
=e.
D
Let (.Q, L:, p) be a measure space. The measure J.1 is called separable if there is a countable family of sets {An}, An E £, such that, for any set BEL: of finite measure and for an arbitrary positive e, we can find a set An0 such that p(B""'An 0)+p(An.~B)
<e.
PRoPOSITION 1.6.4. A space N(L(Q, L:, p)) is separable measure p. is separable.
if and only if the
28
Chapter 1
Proof. Sufficiency. Let {An} be a countable family of sets with the property described above. Let m: be the set of all simple functions K
x(t) =
L alcXAnk
k=l
where, as usual, XB denotes the characteristic function of the set B, a1c are rationals in the real case and complex rational (i.e., of the form an = bn +en i, where bn and en are rational) in the complex case. It is easy to verify that the set m: is countable and that it is dense in N(L(Q, E, p)). Necessity. If the measure J1 is noncountable, then there are a noncountable family of sets {A"} of finite measure and a positive constant b such that
Let x" = XAa· ThenpN(x",xp) > N(l)b for cx:j=fl. Since the set {x"} is noncountable, by Corollary 1.6.2 the space N(L(Q, E, p)) is not separable. 0 1.6.5. A space M(Q, E, p) is separable if and only if the measure J1 is concentrated on a finite number of atoms PI> ... , PTe· Proof. Sufficiency. Suppose that the measure J1 is concentrated on a finite number of atoms PI, .•. , PTe· Then the space M(Q, E, p) is finite dimensional, and thus separable. Necessity. If the measure J1 is not concentrated of a finite number of atoms, then there is a countable family of disjoint sets {An} (n = I, 2, ... ) of positive measure. Let ex= {ni> n2 , • •• } be a subset of the set of positive integers. Let A = Ani u An 2 u . . . The family {A"} is noncountable and for ex =I= {3, p(A"""Ap)+p(Ap""A")> 0. Let x.,. = XAa· Then llxaxpll = I for ex =F fl. Therefore by Corollary 1.6.2 the space M(Q, E, p) is not separable. 0 PROPOSITION
1.6.6. A space C(Q) is separable if and only if the topology in the compact set Q is metrizable (i.e., it can be determined by a metric d(t, t')). PROPOSITION
Basic Facts on Metric Linear Spaces
29
Proof. Let (Q,d) be a metric compact space. Then for each n = 1, 2, ... , there is a finite system of sets {An,k}, k = 1, ... , Kn, such that An,k (') An,k'
=
fork ::j::. k',
121
Kn
UAn,k= Q, n=l
sup{d(t, t'): t, t' E An,k}
<
1/n.
The family {An, ~c}, n = I, 2, ... , k = 1, ... , Kn, is of course countable. Let X be the space offunctions x(t) of the form M
x(t) =
L amXAnm'
(1.6.2)
km'
m=l
where am are scalars and, as usual, XY denotes the characteristic function of a set Y. Let X be the completion of X with respect to the norm flx!l = sup [x(t)[. tEO
Let m: be the set of all functions of the form (1.6.2) with coefficients am either rational in the real case or complex rational in the complex case. The set m: is countable and it is dense in X. Thus Xis separable. Kn
Let x(t) E C(Q). Let Xn(t)
=
L an,kXAn,k k=l
in such a way that inf [x(t)an,k[
<
E
m:, where an,k are chosen
I/n. Since the function x(t) is con
tEAn,•
tinuous, the sequence {Xn(t)} tends uniformly to x(t). Thus it is fundamental in X. Therefore C(Q) can be considered as a subspace of the space X Thus, by Proposition 1.6.3, C(Q) is separable. Necessity. Suppose that the space C(Q) is separable. Let {xn} be a sequence dense in the unit ball K = {x: llxll < 1}. For t,t'
E
Q, let d(t,t')
1 . = ~ ~ 2 n [xn(t)xn(t')[. Smce [xn(t)[ < 1, n=l
d(t, t') is always finite. It is easy to verify that d(t, t') is a metric on Q. We
shall show that the topology determined by this metric is equivalent to the original topology on Q.
Chapter 1
30
Let t 0 e Q and let e be an arbitrary positive number. Let m be a positive integer such that 112m < e/4. Since the functions Xn(t) are continuous, there is a neighbourhood V of the point t0 such that, for t e V, lxn(t)xn(to)
e
I<2
(n
=
1,2, ... ,m)
Then, for t e V, 00
d(t, t 0 ) =
.2,; 2~ lxn(t)xn(to)l n=l m
~ .2,; 2~
oo
lxn(t)xn(to)l +
.2,; 2~ lxn(t)xn(t )1. 0
n=m+l
n=l
Conversely, let V be an arbitrary neighbourhood of t 0 e !J. Since Q is compact, there is a continuous function x(t) such that lx(t)l < 1 x(t0 ) = 0, x(t) = I
forte V, fort¢ V.
The set {xn} is dense in the unit ball of the space C(!J), therefore there is Xn such that llxnxll = suplxn(t)x(t)l tEV
This implies that, if lxn(t)l lxn(t)xn(t0)1
1
< 4 .
< 3/4, then t e Vand lxn(t0)1 < 1/4. Therefore 1
<2
(1.6.3)
implies that t e. V. Let d(t, t0 )
< 1/2n+I. Then (1.6.3) holds and t E
V.
0
PROPOSITION 1.6.7. A space C(!Ji!J0) is separable if and only if the set Q~Q0 is metrizable. Proof The proof follows the same line as the proof of Proposition 1.6.6. 0
e
PROPOSITION 1.6.8. A space 0 (Q) is separable if and only metrizable. Proof Sufficiency. According to the definition of the space
if the set
Q is
e (Q), the set 0
Basic Facts on Metric Linear Spaces
31
Q is the union of an increasing sequence of compact sets Dm. For each
m and n we can find a finite system of sets {A!::k} such that
A:.~c n
A:.1c' =
UAmL k n,,.. =
fork# k',
0
Qm,
sup{d(t, t'): t, t' E
A:,,J <
1/n.
Let X be the space of functions of type (1.6.2). Let X be the completion of the space X with respect to the metric induced by the Fnorm.
where llxllt = sup lx(t)l. Let m: be defined in the same way as in Proposition 1.6.6. The set m: is countable and dense in X. Therefore X is separable. Then by Proposition 1.6.3. e0(Q) is separable. Necessity. Let {xn} be a dense sequence in the space e 0(D). Let for t,t' E Q A
A
00
d(t t')
'
=
~ _1 lxn(t)xn(t')! L.J 211 l+lxn(t)xn(t')l'
n=l
It is easy to verify that d(t, t') is a metric. In the same way as in the proof of Proposition 1.6.7 we can show that the metric d(t, t') induces a topology
equivalent to the original one.
D
Let (Xt, II lit) be a sequence of F*spaces. Let X= (Xt)(s) be the space of all sequences x = {Xt Xt EXt}. The topology in X is given by a sequence of pseudonorms llxll; = llxtllt. Of course the space X is an F*space. 1.6.9. It the spaces (Xt, II lit) are separable, then the space X= (Xt)(s) is separable. Proof Let denote a dense countable set in Xt. Let m: be the set of
PROPOSITION
m:i
• all sequences of the form
(a, a?, ..• , a., 0. 0 .... }.
Chapter 1
32
mis countable and that it is dense in the
It is easy to verify that the set
o
~~x
Let X be a linear metric space with topology determined by a sequence of Fpseudonorms llxlli· Let XJ = {x: llxlli = 0}. Let Xi be the quotient space X/XJ. The pseudonorm llxlli induces in the space Xi the Fnorm
11x11;. PROPOSITION 1.6.10. If all the spaces Xi are separable, then the space X is separable. Proof Let l: = (Xt)(s)· By Proposition 1.6.9 the space l: is separable. The space X can be identified with the subspace X 0 C l: of all sequences
{[x]i}, where [y]i denotes the coset in Xi containing y. Obviously llxlli = ll[x]ill;, i.e. the topology of X 0 inherited from l: and the original topology of X coincide under this identification. Therefore X, as a subspace of a separable space, is, by Proposition 1.6.3, also separable. 0 1.6.11. The spaces C (.Q) are separable. Proof Let x E C"'(.Q). Then PROPOSITION
00
= II a~,'
xllk
.~~k~a~;:
x(t)llo.
Hence, the space Xk can be considered as a linear subset of the space C(.Q). Since .Q is a bounded domain in the ndimensional Euclidean
space, .Q is compact. Thus by Proposition 1.6.6 the space C(.Q) is sepaD rable. Hence by Proposition 1.6.10 the space c=(.Q) is separable. PROPOSITION
Proof Let x
1.6.12. The space c5 (En) is separable. c5 (En). Then
E
llxllm, k
=II t'/.'' ··· t'::" ·
a~,' ... a~;: a1k1
II
x(t) o,
0
and lim t~' ...
t+oo
t:"
a1k1
ak,
t1 • • •
akn tn
x(t) = 0.
Therefore Xm,k can be considered as a linear subset of the space C(.QI.Q0), where Q is the one point compactification of En and .Q0 is the added
Basic Facts on Metric Linear Spaces
33
.Q"'
point. The space .Q0 is metrizable. Hence by Proposition 1.6.6. C(.Qj.Q0) is separable. Therefore, by Proposition 1.6.3 c5 (E11) is also separable. D 1.6.13. The spaces LP(am,n) are separable. Proof Elements of the type {x1,x2 , ••• , Xn,O,O, ... },where Xt are rational
PROPOSITION
in the real case and complex rational in the complex case, constitute a dense countable set in LP(am,n). D PROPOSITION 1.6.14. A space M(am,n) is separable provided that, for any m, there is an m' such that
.
am,n
I1 m   = noo Om',n
0.
Proof We construct a dense countable set in the same way as in the proof of Proposition 1.6.13. D
1.7.
TOPOLOGICAL LINEAR SPACES
This book deals principally with metric linear spaces but, in many cases the notion of topological linear spaces can be a very useful tool. A linear space X is caJied a topological linear space if it is a Hausdorff space and if the operation of addition of elements and the operation of multiplication by scalars are continuous. Since the addition is continuous, the set of neighbourhoods of the form x+ U, where U runs over the set of neighbourhoods of 0, determines a topology in X equivalent to the original topology. Of course, each metric linear space is a topological linear space with a countable basis of neighbourhoods of 0. As follows from the K.akutani construction (see Theorem 1.1.1 ), if there is a countable basis of neighbourhoods of 0, then there is a metric determining a topology equivalent to the original one. Moreover, this metric is invariant. In this case the topological linear space is called metrizable. A point a is called a cluster point of a set A if, for any neighbourhood U of the point a the intersection U n A is not empty. We say that a is a cluster point ofa family ofsets {A .. } if it is a cluster point of each member of the family.
Chapter 1
34
Let X be a linear topological space. A family ~ of nonvoid subsets of X is called fundamental, if for every two sets M, N e ~. there exists an E e ~ such that E c M n N, and for each neighbourhood of zero U there is a set Me ~ such that M M C U. Example 1.7.1 Let (X,p) be a metric space. Let {Xn} be a Cauchy sequence in X. The family of sets {An}, where An= {xn,Xn+I• . .. },is fundamental.
1. 7.2. A fundamental family~ has at most one cluster point. Proof Suppose that x and y are two cluster points of the fundamental family ~ Let U be an arbitrary balanced neighbourhood of 0. Since ~ is fundamental, there is an Me~ such that MM CU. Since x,y are cluster points of ~. they are also cluster points of the set M. This implies that there are x1 ,y1 EM such that xx1 , yy1 eM. Hence PROPOSITION
xy = (xxi)(yyi)+(xiy1 )e U+U+U.
The arbitrariness of U implies that x = y.
D
A subset A of a topological linear space (in a particular case, the space itself) is said to be a complete set if every fundamental family of subsets of A has a cluster point a eA. 1.7.3. A subset E of a complete topological linear space X is complete if and only if it is closed. Proof Sufficiency. Let ~ be an arbitrary fundamental family of subsets of the set E. Since X is complete the family ~ has a cluster point a e X. The set E is closed, and thus a e E. Necessity. Let a be a point in the closure of E. The family {(a+ U) n E}, where U runs over all neighbourhoods of 0, is a fundamental family of subsets of the set E. Since E is complete, it has a cluster point a' e E. By Proposition 1.7.2 a fundamental family has at most one cluster point. Thus a' = a. Therefore the set E is closed. D THEOREM
1. 7.4. For any topological linear space X, there is a complete topologicallinear space X such that X is a dense linear subset of the space X, THEOREM
•.
A
A
35
Basic Facts on Metric Linear Spaces A
A
and the topology in X and the topology in X coincide on X. The space X is called the completion of the space X. Proof We define the points of X as fundamental families of subsets of the space X. We shall identify two fundamental families mand m if 0 is a cluster point of mm. Further steps are similar to the proof in the metric case (cf. Lemma 1.4.7). 0
Chapter 2
Linear Operators
2.1.
BASIC PROPERTIES OF LINEAR OPERATORS
Let (X, II llx) and (Y, II IIY) be two Fspaces. We shall denote the norms II llx and II IIY by the same symbol II II whenever no confusion result. A mapping A transforming a linear subset D.4. e X into Y is called an additive operator if A(x+y)
=
for all x, y
A(x)+A(y)
E
DA.
The set D A is said to be the domain of the operator A. An additive operator A is called a linear operator if A(tx) = tA(x)
for all x
E
DA and all scalars t.
A linear (additive) operator is called q continuous_ linear (additive) operator if it is continuous. If the spaces X and Yare linear spaces over reals, then each continuous additive operator A is a linear operator. Indeed, the additivity of the operator A implies that A(nx) = nA(x)
for every integer n. Since
)+ ... +A (: ),

A(x) =A (:
nfold
we obtain that A (:) =
~
A (x). Consequently, for arbitrary rational r,
A(rx) = rA(x). 36
Linear Operators
37
Let 8 be an arbitrary positive number. Lett be an arbitrary real number. The continuity of the operator A and the continuity of multiplication by scalars imply that there is a rational number r such that //A((tr)x)//
8
<2
and
//(tr) A(x)ll
8
<2 .
Hence //t A(x)A(tx)// ~ //(tr)A(x)//+//rA(x)A(rx)//+//A(rxtx)/1
<
8.
Therefore, the arbitrariness of 8 implies the linearity of the operator A. Let X and Y be complete metric linear spaces. Let A be a continuous linear operator defined on a domain DA C X with the image in the space Y. Let x 0 belong to the closureDA of the domain D A· Then, by the definition of closure, there is a sequence {xn}, Xn EDA• tending to x 0 • The sequence {xn} is obviously a Cauchy sequence. Since the operator A is continuous, the sequence {A (xn)} is also Cauchy. The space Y is complete, therefore, there is a limit y E Y of the sequence {A (xn) }. Write A(x0 ) = y. It is obvious that A(x0 ) is uniquely determined in this way. We have thus extended the operator A from the domain DA onto its closureDA· This is the reason why in the theory of continuous linear operators we shall restrict ourselves to the operators defined on the whole space X. Let X be a metric linear space. A set B C X is said to be bounded if, for any sequence of scalars { tn} tending to 0 and for any sequence {Xn} of elements of B, the sequence {tnxn} tends to 0. In other words, a set Biscalled bounded if, for any neighbourhood of zero U, there is a number b such that B C bU. A sequence {xn} is called bounded if the set formed by its elements is bounded. A linear operator A mapping an F*space X into an F*space Y is called bounded if it maps bounded sets upon bounded sets. THEOREM 2.1.1. Let X and Y be two F*spaces. A linear operator A mapping X into Y is bounded if and only if it is continuous. Proof Sufficiency. Since no confusion will result, we denote the Fnorms in X and in Y by the same symbol II 11. Suppose that the operator A is
Chapter 2
38
not bounded. Then there exist a bounded set E and a positive number e such that sup xeE
llaA(x)ll >
s
for all scalars a different from 0. The set E is bounded, therefore, for an arbitrary positive(> there is a number b such that sup X
llbxll
EE
This implies that, for an arbitrary positive b, there is an element that IIA(xo)ll > s. llxoll < b aPd
x" such
Therefore, the operator A is not continuous. Necessity. Suppose that the operator A is not continuous. Then there are a positive number(> and a sequence {xn} of elements of X tending to 0 such that
IIA (xn)ll >
(>
> 0.
Let ' X n =
By the subadditivity ofFnorm we immediately obtain
llx~ll :::;;;; llxnll
v'llxnll
+sup
lltxnll.
o,;;;t,;;;l
x:
Hence Xn +0 implies +0. Let tn = llxnll. Evidently the sequence {tn} tends to 0. On the other hand, lltnA(x~)ll = IIA(tnx~)ll = IIA(xn)ll > b > 0. Therefore, the bounded set {x~} is transformed onto an unbounded set {A (x~) }. This implies that the operator A is not bounded. 0 Let B 0 (X+ Y) denote the set of all continuous linear operators mapping X into Y. The set B 0(X +X) we shall denote briefly by B 0 (X). It is easy to verify that B 0(X+ Y) is a linear set.
Linear Operators
39
Let u be a family of bounded sets in X. By Ba(X? Y) we denote the set B 0(X? Y) with the topology determined by neighbourhoods of the following form : U(A 0 ,B,s) = {AEB0(X?Y): supiiA(x)A 0(x)ll <s,BEu, 8
> 0}.
XEB
The space Ba(X? Y) with this topology is a topological linear space, i.e. the operation of addition and multiplication by scalars are continuous. If the family u is the family of all bounded sets, then the topology generated by u is called the topology of bounded convergence. In this case the space B,y(X? Y) will be denoted briefly by B(X? Y). Linear operators mapping X into the field of scalars will be called linear functionals. Continuous linear operators mapping X into the field of scalars will be called continuous linear functionals. Sometimes, when there is no danger of misunderstanding, continuous linear functionals will be called briefly linear functionals or even simply functionals. The space B(X?K), where K is the field of scalars (i.e. the field of reals in the real case and the field of complex numbers in the complex case), is called the conjugate space to the space X. We shall denote it by X*.It may happen that X* = {0}. In this case we say that X has a trivial dual.
2.2. BANACHST_EINHAUS THEOREM FOR FSPACES Let (X, II llx) and (Y, II lly) be two F*spaces. Let X be complete (i.e. be an Fspace). Since there is no danger of misunderstanding, we shall denote the both norms by II 11. A family ~ of operators belonging to B0(X? Y) is said to be equicontinuous iffor each positive e there is a positive (J such that sup {IIA(x)ll: A E ~' llxll ~ b} <e. The following theorem is an extension of the BanachSteinhaus theorem (Banach and Steinhaus, 1927) on Fspaces. THEOREM 2.2.1 (Mazur and Orlicz, 1933). Let~ be a family of linear operators belonging to the space B0(X? Y). For each x E X, let the set {A (x): A E ~} be bounded. Then the family ~ is equicontinuous.
Chapter 2
40
The proof is based on the following lemma: LEMMA 2.2.2. Let (X, II ID be an Fspace. Suppose that a closed set V is absorbing, i.e. for each x EX there is a positive number a such that, for b, 0 < b
U nV.
X=
ft1
By the Baire theorem (Theorem 1.4.1) the space X is of the second category. Therefore, there is a positive integer n0 such that n 0 V is of the second category. Since n0 Vis closed, it contains an open set U. Thus V conI
D
tains the open set  U. no Proof of Theorem 2.2.1. Let e be an arbitrary positive number. Let
ul =
n {x EX: IIA(x)ll < e}.
AE'll
Since the operator A is continuous, the set U1 is closed. We shall show that U1 is an absorbing set. Indeed, let x be an arbitrary element of X. By assumption, the set {A (x): A E ~} is bounded. Hence there is a positive number a such that, forb, 0 < b < a, llbA (x)ll < s for all A e ~. Thus bxe u1. By Lemma 2.2.2 the set ul contains an open set u2. Let Xo E u2. The set {A (x0): A E ~} is bounded. Hence there is a positive number b, 0 < b < 1 such that llbA(xo)ll <e. Thus bxo Eul. Let u = b(U2Xo) Of course U is a neighbourhood of 0. Let x E U. Then x = bybx0 • where y E u2. Therefore IIA(x)ll
< llbA(x0)il+llbA(y)ll =
s+sup {llbzll: lbl
< 1, llzll<s}.
Hence the continuity of multiplication by scalars implies the theorem. D CoROLLARY 2.2.3 (Mazur and Orlicz, 1933). Let X, Y be two Fspaces. Let sequence of continuous linear operators {An} be convergent at each point x to an operator A,
lim A 71 (x) = A(x). n+~
Linear Operators
41
Then the operator A is a continuous linear operator. Proof The linearity of the operator A follows trivially from the arithme
tical rules of limits. For each x the sequence {An(x)} is convergent, and thus bounded. By Theorem 2.2.1 the family of operators {An} is equicontinuous, i.e., for an arbitrary positive e, there is a {J > 0 such that llxll < {J implies IIAn(x)JI
e
< 2·
Let JlxJI
JJA(x)An (x)JJ < 2" 0
Hence e
JIA(x)JI ~ IIA(x)An.(x) li+IIAn.(x)ll < 2
e
+ 2 =e.
Therefore, the operator A is continuous at 0. The linearity of the operator A implies that it is continuous everywhere. D 2.2.4. If (X, is complete.
CoROLLARY
B(X~Y)
II ID
and (Y,
I ID
are Fspaces, then the space
Proof Let \!£ _be a fundamental family of subsets of the space B(X~ Y). This implies that for any x eX the family \!l(x) of the sets { {A (x): A e M} Me \!£} is a fundamental family of subsets of the space Y. The space Y is complete, and thus the family W(x) has a cluster point A 0(x). Corollary 2.2.3 implies that A 0(x) is a continuous linear operator. Let B be a bounded set in X and Jet U be a neighbourhood of 0 in
Y. Let E be an arbitrary member of \!£. Let M C E be a member of \!£ such that, for A, A1 e M, x e B, A(x)A 1(x) e U.
Then A(x)A 0(x) e U.
The arbitrariness of Band U implies that A 0 is a cluster point of M. Thus it is also a cluster point of E, and since E was an arbitrary member of \!£, A 0 is a cluster point of the family\!£. D
Chapter 2
42
In the particular case where X is an Fspace the conjugate space X* is complete.
2.3. CONTINUITY OF THE INVERSE OPERA TOR IN FSPACES , THEOREM 2.3.1 (see Banach, 1932). Let (X, II II) and (Y, II II) be two Fspaces. If a continuous linear operator A maps the space X onto the space Y, then the image of any open set G is open. Proof. Let Ube an arbitrary neighbourhood ofO. We shall first show that the closure A(U) of the set A(U) contains a neighbourhood of 0. Since ab is a continuous function of its arguments, there is a neighbourhood of zero M in the space X such that M M C U. Since M is a neighbourhood of zero, X=UnM. n=l
Thus Y
= A(X) =
U nA(M). n=l
By the Baire category theorem (Theorem 1.4.1) there is an n0 such that the set n0 A (M) contains a nonvoid open set V. Hence A(U) ) A(M)A(M) ) A(M)A(M) ) _I (V V),
no
1
and the set ( V V) is of course a neighbourhood of 0. no For any positive s we shall write Xe = {xE X: llxll < s}
and
Y. = {y
E
Y: IIYII < s}.
Let s0 be an arbitrary positive number. Let {Bi} be a sequence of positive numbers such that }; Bi < s0 • As we have already shown, there is C1
a sequence {?Ji} of positive numbers such that i=0,1,2, ...
(2.3.1)
Linear Operators
43
Let y be an arbitrary element belonging to Y110 We shall show that there is an element X E X2,, such that A(x) = y.
(2.32.)
Formula (2.3. I) implies that there is an element x 0 E X,. such that 1/yA(xo)ll < 1J1, i.e. yA(x0 ) E Y11,. Putting i = 1 in formula (2.3.1), we can find that there is an element x 1 E X, 1 such that
llyA(x0)A(xl)ll <
1J2·
Repeating this reasoning, we may define a sequence {Xn} of elements of X such that Xn e X, and n
llyA
"
(2.: Xt) II < 1Jn+1·
(2.3.3)
i~O
Since the space X is complete, the series}; Xn is convergent to an elen=l
ment x and, moreover, llxll < 2s0 • Formula (2.3.3) implies (2.3.2). Hence the image of a neighbourhood of 0 contains a neighbourhood of 0. Let G be an open subset of the space X. Let x E G. Let N be a neighbourhood ofO such that x+N C G. Let Mbe a neighbourhood ofO in Y such that A (N) :) M. Then A(G) :) A(x+N) = A(x)+A(N):) A(x)+M.
Therefore, A(G) contains a neighbourhood of each of its points, i.e. A(G) is open. D THEOREM 2.3.2 (Banach, 1932). Let a continuous linear operator A map an Fspace X onto an Fspace Y in a onetoone manner. Then the inverse operator A 1 is continuous. Proof The operator A = (A1)1 maps open sets on open sets. Hence the D operator AI is continuous. COROLLARY 2.3.3. Suppose we are given in an F*space X two different Fnorms II I and II 11 1 • Suppose that the space X is complete with respect to the both Fnorms. Suppose that the Fnorm II III is stronger than the Fnorm II 11. Then the two Fnorms are equivalent. Proof The operator of identity mapping (X, II 11 1) into (X, II I[) is con
Chapter 2
44
tinuous. Then by Theorem 2.3.2 the inverse operator is also continuous. 0 This implies that the Fnorms II II and II 11 1 are equivalent. Let X be a linear space. A family of linear functionals F defined on X is called total if the fact thatf(x) = 0 for allfe F implies that x = 0. 2.3.4. Suppose we are given in an F*space X two different Fnorms II II and I lit· Suppose that X is complete with respect to both Fnorms. Suppose that there is a total family F of linear functionals which are simultaneously continuous with respect to both Fnorms. Then the Fnorms II II and II lit are equivalent. Proof We shall introduce a new norm in the space X by the formula
CoROLLARY
llxll2 = llxll+llxllt and we shall show that the space X is complete with respect to the norm
II 11 2. Let {Xn} be a fundamental sequence in the Fnorm II 11 2 • Then it is also a fundamental sequence in the Fnorms II II and II III> because the Fnorm II 11 2 is stronger than the Fnorm II II and the Fnorm II lit· Since X is complete in II II and I 11 1 , the sequence {xn} tends to xO in the norm II II and to x1 in the norm I lit· Suppose thatx0 =/= x 1 • Since the family Fis total, there is a functional f E F such that f(x 0 ) =/= f(xt). The continuity off in the both norms implies that
f(x 0 ) = lim f(xn) = f(x 1 ), and this leads to a contradiction. Therefore, x 0 is a limit of the sequence {xn} in the norms II II and II 111 • Thus it is a limit of the sequence {xn} in the norm II 11 2. Hence the space (X, II 11 2) is complete. By Corollary 2.3.3 the norm II 11 2 is equivalent to the norm II II and to the norm II 11 1 . Hence the norms II II and II 11 1 are equivalent. D
2.4.
LINEAR DIMENSION AND THE EXISTENCE OF A
UNIVERSAL
SPACE
Let (X, II llx) and (Y, II IIY) be two F*spaces. We say that the space X and Yare isomorphic if there is a linear operator A mapping X onto Y in
Linear Operators
45
a onetoone manner such that the operator A and the inverse operator A1 are continuous. We say that an Fspace (X, 1/ llx) has a linear dimension not greater than an Fspace (Y, II IIY), and we write dimzX:::::;; dimz Y
if the space X is isomorphic to a subspace of the space Y. If dimzX:::::;; dimzY and simultaneously dimzY:::::;; dimzX, we say that the spaces X and Y have the same linear dimension (Banach and Mazur, 1933) and we write it as dimzX = dimzY. If two F*spaces X and Y are isomorphic, then obviously they have the same linear dimension. On the other hand, we shall see later that there are nonisomorphic spaces with the same linear dimension. We say that the linear dimension of an F*space X is less than the linear dimension of an F*space Y and we write dimzX
<
dimzY
if dimzX:::::;; dimzYand dimzY =I= dimzX. Let X be a family of F*spaces. An F*space X 0 is called a universal space with respect to isomorphism, or briefly a universal space, for the family X if for every X EX there is a subspace of X 0 isomorphic to X (i.e. dimzX:::::;; dimzX0). A universal space is called sometimes universal with respect to linear dimension. For each family X of F*spaces there is a trivial universal space defined in the following way. We enumerate all the spaces belonging to X. Let A be the set of indices. Let X = {(Xa, II lla), a E A}. Let Xu be the space of generalized sequences x = {xa}, a E A, Xa E Xa, such that
llxll =
sup llxa
a E.AJ
lla <
+oo.
We determine the addition and the multiplication by scalars as follows {xa}+{Ya} = {xa+Ya}, t{ xa} = {txa}. It is easy to verify that (Xu, II II) is an Fspace. Space Xb is isomorphic to the following subspace of Xu X~= (xEXu:
Xa=Ofora=:j=b}.
Chapter 2
46
Unfortunately the space Xu obtained in this way does not necessarily belong to the family ~. The problem of universality is a problem of finding a universal space belonging to the family~. Kalton (1977) has shown that there is a universal separable Fspace for the family of all separable F*spaces. We shall present his proof here. The proof is based on several notions, lemmas and propositions. To begin with, following Kalton, we shall present what is called the packing technique. This technique was introduced by Pelczynski (1969) for another purpose. Let (E, ~) be a partially ordered set. A subset A C E is called a chain if it is totally ordered. A subset A C E is called a section if b ~a for an element a E A implies that bE A. Let E[c] ={a E E: a~ c}. If, for each c E E, the set E[c] is a finite chain, we say that Eis treelike. PROPOSITION 2.4.1 (Kalton, 1977). If E is countable and treelike, then there is a onetoone nondecreasing mapping of the set N of positive integers onto E. Proof We form a sequence {an} such that each e1ementof E occurs in the sequence infinitely many times. We shall now define by induction an increasing sequence {mn} of positive integers as follows. As m 1 we take the smallest n such that E[an] = {an}. Suppose that elements am 1 , ... , amk 1 are chosen. As mk we shall take the smallest integer n such that n > mkl {am 1, ••• , amk_ 1, an} is a section of E, an¢ {ami> ... , amk_ 1 }. Since E[c] are finite for all c E E, it can be proved that each c E E is in the range of a, where a(k) = amk· D Let ':l(E) be the set of all real valued functions defined onE and vanishing everywhere except a finite number of points (the functions of this type will be called finitely supported). By :l(A) we denote the set of those elements of :l(E) which have a support contained in A. By PA we shall denote the natural projection PA: :l(E)~:l(A), PAx = xxA. By a consistent family ofFnorms we mean a collection {II lie, c E E} ofFnorms such that
II lie is an
Fnorm on :l(E[c]),
if a~ c, then for x
E
:l(E[a]), [[x[[a = [[x[[e,
[[tx[[e is nondecreasing for
t
> 0.
(2.4.1.i) (2.4.1.ii) (2.4.1.iii)
Linear Operators
47
For a given consistent family II lie, we define the limit Fnorm II II on 7(E) as follows n
llxll
=
inf
{L
n
llutllc1 :
i=l
L
Ut
=
x,
UtE
:=l(E[ct]), n
eN}.
i=l
Observe that II II satisfies the triangle inequality llx+yll ~ llxii+IIYIIIndeed, n1
n1
llx+yll = inf
{L
i=l
n,
n8
+ inf
x+y}
llwtllc,: .}; Wt =
i=l
{L
llvtllct:
i=l
L
Vt =
Y} = llxii+IIYII·
i=l
Since (2.4.l.iii) holds, lltxll is nondecreasing for t > 0. Observe that for x E :=l(E[c]), llxll ~ llxllc. Thus llxll is an Fpseudonorm (i.e. satisfies conditions (n2)(n6). PROPOSITION2.4.2(Kalton, 1977). Let Ebe treelike and let {II lie, c E E} be a consistent family ofFnorms. Let II II be the limit norm of this family. Then If A = suppx (the support of x), then llxll = inf
{,l llualla: 1
Ua E
If x
E
(2.4.2.i)
:=l(E[a]),.}; Ua = x}.
aEA
aEA
:=l(E[c]), then (2.4.2.ii)
llxll = llxllc·
llxll is an Fnormon ::l(E),i.e.llxll = 0 ifand only ifx = 0. (2.4.2.iii) Proof (2.4.2.i) Lets be an arbitrary positive number. Let (1 ~ i ~ n) be a minimal collection such that u1
+ ... +u, =
x,
Ui E
:f(E[c,])
Chapter 2
48
2:" llutllc, ~
llxll+s.
t=l
Since the family {II lie, c E E} is consistent, we may assume without loss of generality that c, = max(supput). To prove {2.4.2.i) it is sufficient to show that for each i there is an atE A = supp x such that c, ~at. Suppose that the above does not hold. Then there exists a maximal c1 = c {j being some index not greater than n) such that c =I= a for any a E A. n
This implies that c f} supp x. Thus
2 Ut(c) =
0. Since UJ(c) =I= 0 and cis
i=l
maximal, there is an index k =I= j such that Ck = c = c1 • Therefore
L ut+(uJ+uk) = x
i*J, k
and
}; llutllc1+llui +ukllc ~ llxll+s i*j, k
contradicting the minimality of the collection (ul> ... , un). (2.4.2.ii) This follows immediately from {2.4.2.i); if ua 2 Ua = x, a .s;;; c, then
aEA
E
C): (E[a])
and
L llualla =}; lluallc;? 11J; Ua lie= llxllc· aEA aEA
aeA
(2.4.2.iii). If llxll = 0, then for each positive integer n there exist u~ E r:f(E[a]), a E A = supp x such that ~ n LJ Ua = X
aEA and
2..: ll~lla < ! ·
aEA
Then, for each fixed a, lim lim
u~(a)
llu~lla
= 0. Since Sl(E[a]) is finitedimensional,
= 0. Since the support A = suppx is finite, x(a) = 0 for all
a E A. Thus x = 0.
An Fnorm II II on r:f(E) is called monotone if section A C E.
D
IIPA(x)ll ~ llxll for any
Linear Operators
49
2.4.3 (Kalton, 1977). Suppose that E is tree/ike and that {II lie. c E E} is a consistent family ofFnorms such that IIPE(aJ (x)llc ~ llxllc for a~ c. Then the limit Fnorm II II is monotone. Proof Suppose that A is a section of E. Let x E :!(£). For any B > 0 we can find Ut,Ci, i = I, ... , n such that Ui E :l(E[ct]), x = u1 + ... +u11 and llutllcl+ ·· · +llunllc11 ~ llxll+e. Since A is a section, A n E[ci] = E[at] for some at ~ Ct. Hence PROPOSITION
n
11
IIPA(x)ll
~}.; IIPA(Ut)llc1 ~}.; llutllc1 ~ i=l
llxll+e.
i=l
The arbitrariness of s completes the proof.
0
Let N denote as before the set of all positive integers. By :l(N) we de· note the set of all finitely supported sequences. For each n eN we define en={O,O, ... ,O,I,O, ... }. nth place
The space :l(E[n]) will be denoted more traditionally by R 11 • Of course R 11 =lin{ e~> ... , e11 }. Let 4J be the family of all Fnorms defined on :l(N) and let 4Jn be the family of all Fnorms defined on R 11 • Now we shall determine the distance between two Fnorms defined on an Fspace X as follows. Letp(x), q(x) be two Fnorms on X. Let
d*(Ji, q) =sup arctan llog p((x))l· XEX q X x,.
It is easy to verify that d*(p,q) is well determined and that it is a metric. Indeed, (I) d*(p,q) = 0 if and only if p(x) = q(x),
(2) d*(p,q) = d*(q,p) since /log p(x) I= 1log p(x) =!tog p(x) \. q(x) q(x) q(x) \ 1'
1
(3) d*(p,q) =sup arc tan /log p(x) xex q(x)
I
11"0
=sup arc tan Ilog p(x) +log r(x) XEX
x,.
1
r(X)
q(X)
I
Chapter 2
50
~sup arc tan Ilog p(x) XEX
Y(X)
X#O
~
I
\+sup arc tan log r(x) xEX
I
q(X)
X#O
d*(p,r)+d*(r,q).
Thus we can treat ((/),d) and ((/)n,d)as metric spaces. Let ln denote the restriction mapping ln: (/J+(/)n· It is easy to see that ln is not expanding, i.e., d*(Jnp, lnq)~d*(p,q). In further considerations we shall not use the metric d*(p,q) but a function
I
p(x)) . d(p, q) =sup log( xEX
I
q X
X#O
The function d(p, q) is not a metric. It can be equal to +oo. Observe thatforanys >Othereis ab >0 such that d(p,q)
<s. We shall now describe the general construction of universal spaces. We shall suppose that 9 is a set ofFnorms defined on CJ(N) satisfying the following conditions: 111(9) is separable for each n e N
for each p e 9, ln(q)
=
qe 1
11
(2.4.3.i)
(9) there is q e 9 such that
q and d(p, q) =
d(Jn(p), q).
(2.4.3.ii)
Take a dense countable subset EI of JI(9). Then, by induction, we pick a countable subset En of ln(9) as follows: if Entis chosen, choose for each p e Ent a dense countable subset A(p) of ln(9) n ln(Jn_:t (p)). Then En=
U
A(p).
pEEnt
Let E
=
U En. We define a partial order ~on E as follows, p ~ q n~l
if p is a restriction of q, i.e., p e Em, q E En, m ~ n and p(x) = q(x) for xeRm. Next we define a consistent family {II liP, p e E} ofFnorms. If p e Em, E[p] = {PI, ... , Pm}, where PI = p IR'' ... , Pm = PIRm = p. We define m
llxlljl =
P( ,2; x(p,)e,). i~l
51
Linear Operators
Let llxll be the limit Fnorm of the family
{II llv,p e
E}.
PROPOSITION 2.4.4 (Kalton, 1977). For each p e 9 and e > 0 there exist a maximal chain C in E and an order preserving onetoone mapping rp of C onto N such that the linear operator T: ~(N)~~(C) defined by
T(x) Ia = x(tp(a)) satisfies the following inequality :
(1e) p(x) :::( IIT(x)ll :::( (l+e)p(x).
(2.4.4)
Proof Let {f)n} be a sequence of positive numbers such that
n 00
(l+rJn)
< 1+e.
(2.4.5)
n=l
Let p 0
=
p. We define by induction a sequence Pn e 9 such that
(1 +rJn)1PnI(x) :::( Pn(X) :::( (1 +rJn)PnI(x) for n lnlPn)
=
for n ;;?: 2,
lnI(Pn1)
for n? 1.
> 1, (2.4.6.i) (2.4.6.ii) (2.4. 6.iii)
Suppose that {Pk: k < n} are chosen. Then, by (2.4.6.iii) ln 1(pn 1 ) E e En1 (this condition is vacuous if n = 1). Find q E A (PnI I Rn 1) such that (1 +rJn)1PnI(x) :::( q(x) :::( (1 +rJn)Pn 1(x) By (2.4.3.ii) there is a Pn(x)
E
for X E Rn.
9 such that
(1 +rJn)1PnI(x) :::( plt(x) :::( (1 +?Jn)PnI(x) and ln(pn) = q.
Since q E A(PnIIRn1), by (2.4.6.ii), {ln(Pn)} is a chain in E and it is clearly maximal. Let C = {Jn(Pn): n eN} and we define (Jn(Pn)) = n
for n
> 1.
For xe Rn
II T(x)ll = Pn(x) and by (2.4.5) (1e)p(x) :::( pn(x) :::( (1 +e)p(x).
D
Chapter 2
52
For the construction of a proper consistent family ofFnorms we need the following notions and lemmas. We say that an F*space (X, II ID is a {JF*space if X does not contain arbitrarily short lines, i.e. inf sup lltxll > 0. xeX teR
x;CO
The {JF*spaces are also called spaces without arbitrarily short lines (see Bessaga, Pelczynski and Rolewicz, 1957b). LEMMA 2.4.5 (Kalton, 1977). Let X be a separable F*space. The product space Xx / 2 contains then a dense {JF*space. Proof(Pelczynski, see Kalton, 1977). Let {Yn} be a dense sequence in X. Let en = {0, ... , 0,1,0, ... }. Let {xn, m} be the sequence {en} enumerated n·th place
as a double sequence. Let Zm,n
=
(yn,+Xn, m)· Let X 0
=
lin{zn,m,
n = 1, 2, ... , m = 1,2, ... }. Elements Zn,m are linearly independent and each z E X 0 is a finite sum N Z =
.}; tn,m n,m=!
Zn,m.
Therefore, if z o:j::. 0, then the natural projection of z on l 2 ,Pz, is different from zero. This implies, by the definition of the norm in the product of spaces, that inf
suplltzll
~
z;eo teR zeXxN
inf
suplltPzll
=
+oo.
z;e 0 teR zeXxN
Thus X 0 is a {JF*space. Observe that (ym, 0)
=
lim Zm, n·
Hence (ym,O) E X 0 • Therefore (O,xm,n) is dense in Xx / 2 •
=
n(zm,n(ym,O)) E X 0 • Thus X 0 D
II ID be a finitedimensional Fspace. Let the norm II I be nondecreasing, i.e., let lltxll be nondecreasing for t > 0 and each x o:j::. 0. Let
LEMMA 2.4.6 (Kalton, 1974). Let (X,
c = inf suplltxll· xex teR
x;eo
53
Linear Operators
Then,Jor all a < c, the set K = {x: llxll ~ a} is compact. Proof Since X is finitedimensional it is enough to show that the set K is bounded. Suppose that K is not bounded. This means that there is a sequence Xn+oo, Xn E K. By the BolzanoWeierstrass theorem we can Xnk choose a subsequence {xn } such that1  1 +x0 , where IYI denotes the k Xnk Euclidean norm in X. We shall show that
(2.4.7)
suplltx0 ll
Since Xn k E K and the norm II II is nondecreasing, txn k E K for It I < 1. Let e = (ca)/2. Let () > 0 be such that lxl < r5 implies l!xll <e. Then, for ally such that lytyXn k I < r5 for a certain ly, ltul < 1, we have ltyl ~
(c+a)/2. By the homogeneity of the Euclidean norm
Itx t 0
;:: 1 1
I I, we obtain
I < r5
if and only if ()
It I < ,.....
l x0~l !xnkli
Since the righthand side of the inequality tends to infinity as k'?oo, we obtain (2.4.7) and this leads to a contradiction with the definition of c. 0 THEOREM 2.4.7 (Kalton, 1977). There is a universal separable Fspacefor all separable F*spaces. Proof We shall first prove the theorem for separable (3F*spaces. Without loss of generality we may assume that we consider (3F*spaces with norms p(x) such that p(tx) are nondecreasing and sup p(tx) = 1
for all
X
EX,
X=/=
tER
Indeed, if p does not satisfy (2.4.8), we write p*(x) =min ( 1,
p~) ),
0.
(2.4.8)
Chapter 2
54
where
o
inf sup p(tx). zeX teR z;o
The norm p*(x) is equivalent to the norm p(x) and satisfies (2.4.8). Let '}J be the class of all Fnorms on :f(N) satisfying (2.4.8) and such that for each n eN there exists a neighbourhood Un of Rn such that if x e Un and it/ < 1 thenp(tx) = /t/p(x).
(2.4.9)
We shall show that the family '}J satisfies conditions (2.4.3.i) and (2.4.3.ii). Let n eN and let Vm = {x eRn: i~
/xi/~ ~}·Let Q(m,n)
be the set of all Fnorms p(x) defined on Rn such that
supp(tx) = 1 for all x =F 0, x eRn, if xe Vm, /t/ Observe that Jn('JJ) C
< 1, thenp(tx) = /t/p(x).
U Q(m,n).
(2.4.8') (2.4.9')
We shall now show that Q(m,n) is
m=l
separable with respect to the metric d defined previously. Choose in Q(m,n) a countable set Q which is dense with respect to the topology of uniform convergence on compact subsets of Rn (see Proposition 1.6.8). Letp e Q(m,n). LetO > 0. Let K=
{x:
1
x eRn, p(x)
<
e2 1}
By (2.4.8') and Lemma 2.4.6 the set K is compact. We assume that 0 is chosen so small that Vm C K. Let
r=
inf p(x).
(2.4.10)
zelnt Vm
Then choose q e Q such that for x e K (2.4.11) Now we consider three cases:
Linear Operators
55 1
1
(a) x¢ K. Then there is at< 1 such that p(tx) = e211 . Since y and ex is a convex function _!_11
q(tx)~e
ye
2
_!..11( _!:_11) 2 1e 2
_!_11+ ~6
....__
:;:::: e
e
2
2
~e
_!_11 2
e
11 (
< e2 6 
16 )
1e 2
e6....__ :;:::: e6 .
Therefore q(x) ~ e 8 and p(x) ~ o q(x) """'e '
since p(x)
~
I. On the other hand
:~? ~ e~/2 = e}o <eo. (b) x E K~ Int Vm· Then, by (2.4.1 0), (2.4.11) and the convexity of the function ex,
1
, < e2 8 < e 8 • 1e
(c)
!..o 2
+ eo
Int vm· Then there is s
X E
logp(sx) q(sx) and by (2.4.9') l
> 1 such that SX E avm· Hence
I~ fJ
/log~~:~ I~ fJ. Thus in all three cases d(p,q) ~ fJ and Q is dense in Q(m,n). This implies that Jn(1>) is separable and (2.4.3.i) holds. We shall now show (2.4.3.ii). Let E 7> and let E Jn(1>) be such that
p
d(Jn(p),
q)
= fJ
> 0.
q
(2.4.12)
Chapter 2
56
Let p*(x)
Of course p*
E
=
min(eBp(x), 1).
(2.4.13)
']J. Let
q(x) = inf{q(y)+p*(xy): y ERn}. q(x) is an Fnorm. Indeed, if x = 0, taking y = 0 we obtain that q(x) = 0. On the other hand, if q(x) = 0 then there is a sequence {Yk} such that q(yk)+0 and p*(XYk)+0. Since q is a norm in Rn, Yk+0. Then, by the continuity of p*(x), p* (x) = 0 and x = 0. It is easy to observe that q(x) satisifes conditions (n2)(n6) of the Fnorm. Indeed, q(x)can be identified with the quotient norm of the space Xx Rn with the norm ll(x,y)ll = q(y)+p*(xy), with subspace (0, Rn). Observe thatp*(xy);;:::: q(xy) for x ERn. This follows immediately from (2.4.12) and (2.4.13). Hence for x ERn, q(x) = inf {q(y)+p*(xy): y ERn} ~ inf {q(y)+q(xy):y ERn}= q(x).
The converse inequality is obvious. Thus q(x) Jn(q) = q. By (2.4.13) q(x) ~p*(x) ~ e8p(x). On the other hand, for x E c:;: (N), y E Rn, q(y)+p*(xy)
teR
q(x) for x eRn, i.e.
> e0p(y)+p(xy) ;;:::: e 0(p(y)+p(xy));;:::: e8p(x).
so that q(x) ~ e 8p(x). Thus d(p,q) = 0. We have to show that q E ']J. Clearly, for x sup q(tx)
=
E
c:;: (N),
~sup p*(tx) ~I. teR
Suppose that supq(tx)
~b
<
I. Then for any t
E
R there are Ut and Vt
teR
such that tx = Ut+Vt, UtE Rn, q(ut) ~ b and p(vt) ~b. The set V ={zeRn: q(z)~b} is compact by Lemma 2.4.6. Let W={z E Iin(Rn,x: p*(z) ~ b}. This set is also compact for the same reason. However, lin{ x} C W + V and we obtain a contradiction. Thus sup q(tx) = I tER
for all x =I= 0, and (2.4.8) holds.
Linear Operators
Let m
~
n. By condition (2.4.9') there exist em
57
> 0 such that
if y ERn, ij(y) ~Em, [t[
~
1, then q(tx) = [t[q(x),
Rm,p(y) ~Em, [t[
~
1, then p*(tx)
if y
E
=
(2.4.14.i)
[tfp*(x). (2.4.14.ii)
Now suppose that y E Rm and q(y) ~Em. Suppose that [t[ = u+v, u ERn, q(u)+p*(v) ~Em, then
<
1. If y
q(tu)+p*(tv) = [t[ (ij(u)+p*(v))
and q(tv)
~
[t[q(y).
We shall show that the strong inequality does not hold. Indeed, suppose that ty = u+v and q(u)+p*(v) < t q(y). Then q(u) < em[t[ and p*(v) < em[t[. Lets denote the largest real number such that q(su) ~Em. Then q(u) = s1Em, and so s ~ [t[ 1 • Thus q(t 1u) ~Em and p(t1 v) ~Em. Hence q(t 1u)+p*(t 1 v) < q(y) and this contradicts with the definition of q. Thus, by Proposition 2.4.4, there exists a countable dimensional 1 F*space (Z, II [[) such that, for all p E ']>, (Y(N),p) is isomorphic to a subspace of Z. Now letp be any Fnorm on Y(N) such that( Y(N),p) is a {JF*space. We shall show that p is equivalent to some q E 'J>. Without loss of generality we may assume that p(tx) ~ [t[p(x)
for
ltl
~
(2.4.15)
1.
Indeed, if p(x) does not hold (2.4.15), we can replace p by a new equivalent norm pl(x) = sup p(tx) . t~l t
Let p*(x) = (p(x)) 112 • Of course p*(tx)
~
[t[ 112p(x)
for [t[
~
1.
(2.4.16)
Now select any strictly increasing sequence On, 1 ~On, On+2. For any x E (N), let 1 We recall that a linear space X is called countable dimensional if X= lin { a 1 , ... } for a sequence {a,.} oflinearly independent elements.
Chapter 2
58 n
p(x) =
inf{~Okp*(uk): Uk e Rk, u1 + ... +un = x, n eN}. k=l
By the triangle inequality we have
p(x)
~p(x) ~
2p(x).
If x e R 11 , it is easy to show that n
p(x) = inf {~ Okp(uk): u1 + ... +un =
~}
k=l
and
p(x)
~ Onp(x).
Let { Mt} be an increasing sequence of positive numbers. We define co
q(x) = inf{p(y)+ ~Mcix(i) y(i)l: y e c:! (N)}. i=l
It is easy to show that q(x) is an Fpseudonorm. If
p *(M~l en )
~
1 2n,
(2.4.17)
where en= {0, ... , 0, 1,0, ... }, then q(x) ~p(x) ~ 2p*(x). Suppose that nth place
for a sequence {Xn}, q(xn)J>0. Then, from the definition of q, Xn can be =
represented as a sum Xn = Un +vn such that p(un).1>0 and ,1; Mtlvn(i)iJ>0. i=l
00
If ,1; MtiVn(i)i
< 1 then
lvn(i)l
< M? for all
i, and hence, for any r, we
i=l
have under condition (2.4.17) r
p*(vn)
~p*(l; Vn(k)ek)+ k=l
;r.
Hence lim sup p*(vn) ~ 1/2r. Thus p*(vn)J>0. Since p(un)J>0, p(xn)J>0. This shows that under condition (2.4.17) the Fnorms p and q are equivalent.
Linear Operators
59
We shall now refine the selection of M n by the following induction procedure. We define qk(x) on Rk as follows: k
qk(X)
= inf {iJ(y)+}; Mt/x(i)y(i)/: y
E
Rk}.
1=1
Let o be chosen in such a way that for each m the set Rm n {x: p*(x) < o0 } is compact. The existence of such a 00 follows from the fact that (c:f(N),p) is anp .F*space. Now we shall choose an increasing sequence of positive numbers {Mn} and a decreasing sequence of positive numbers {On}, so that *( 1 ) p Mn en 1
1 < 2n'
if u e Rn1 and p*(u) ~
= 1, 2, ... ,
00 , n = 1, 2, ... ,
On<4 for n
n
<
!
(2.4.18.a) (2.4.18.b)
00 , then qn(u) = qn1(u),
2,
if u eRn, p*(u)
(2.4.18.c) (2.4.18d)
To start the induction pick an M 1 such that (2.4.18.a) holds and 01 such that (2.4.18.b) holds. Condition (2.4.18.c) ought to be satisfied for n ~ 2. Condition (2.4.18.d) follows from (2.4.18.a), and the definition of q and (2.4.16). Now suppose that M 1 < M 2 < ... < Mn 1 and 01 > 02 > ... > On1 have been chosen. The set {xe Rn: On 1
llxJJ! =
};lx(i)/, i=1
we may choose Mn so that (2.4.18.a) holds, and moreover if On1
<
!
00 • 1
We shall now show that (2.4.18.c) holds. Let x e Rn1 andp*(x)< 4 00 •
Chapter 2
60
Suppose that qn(x)
<
qn_ 1(x). From the definition of q, we have
qn(x) = inf{qn1(u)+Mnllvii~+Onp*(w): u+v+w = x, ueRnt,v,weRn}.
And by our hypothesis there are u, v, w such that x = u+v+w and (2.4.19) Hence Mnllvll~ +Onp*(w)
<
qn1(x)qn1(u) ::::( qn 1(v+w).
(2.4.20)
On the other hand, qnl(x) ::::( On1P*(x) ::::( t bo.
Hence llvll~ :::(
M;; 1 bo
and, by the choice of Mn,p*(v)::::;;; { b0 • Therefore p*(v+w) :::(p*(v)+p*(w) :::(
First suppose that p*(v+w) and hence
>
fbo+t b0 = b0 •
bn~> then p*(w) ~ 0;; 1 (;ln_ 1p*(v+w),
Mnllvll~ +Onp*(w) ~ On1P*(v+w) ~ qn1(v+w),
which contradicts (2.4.20). Next suppose p*(v+w) < bn_ 1 and choose t ~ I so that p*(t(v+w)) = bn_ 1. Then either lltvll~ > M;; 1lJ 0 or p*(tw) ~ On 10;; 1lJn_ 1. In both cases
and hence, as v+w E Rnt, Mnllvll~ +Onp*(w) ~ ltl 1On1 bn1 ~
ltl 1qnl(tv+tw) = qn1(v+w)
(by (2.4.18.d)) and this contradicts (2.4.20). Thus (2.4.18.c) holds. It remains to choose bn to satisfy (2.4.18.d). Let n
L =max {~Mtlx(i)J: x ERn, p*(x) = b0 }. i~1
Linear Operators
Choose bn < min(bnI> .Lb~). If p*(x) ~ 2bn and if y E Rn and
61
< bn, then
qn(x) ~p(x) ~ 2p*(x)
n
p(y)+ .rMtix(i)y(i)l
~ 2<5n, p*(y)~2<5n.
i=l
Hence, for some t ~ I, p*(ty) =<5 0 • Then by (2.4.I6) <5 0 ~
ltl 1/ 2 p(y) and
n
.rMtiY(i)i
~ Liti1 ~ L<50 2(p*(y)) 2 ~ 2<5nH0 2p*(y) ~p*(y).
i=l
Hence
.r n
.r n
Mtix(i)i
~
i=l
.r n
Mtiy(i)i+
Mtix(i)y(i)[
i~l
t=l
n
~p*(y)+ l~ Mtix(i)y(i)l i=l n
~p(y)+
_r Mtix(i)y(i)l. i=l
n
Then, by the definition of qn(x), we obtain qn(x)
=
2Mtix(i)l, provided i=l
p*(x)
~
bn, and so qn(tx)
=
it Iqn(x) for
(2.4.I8.d) is also satisfied. To conclude, we observe that if x
E
It I ~
I and p*(x) ~ bn. Thus
Rm, then q(x)
=
limqm+n(x). n+oo
Hence ifp*(x) ~ bm and It I~ I, q(tx) = ltiq(x) by conditions (2.4.18.c) and (2.4.I8.d). Thus q satisifies condition (2.4.I4.ii). To ensure (2.4.14.i), observe that, as q is equivalent top, inf sup q(tx) = d
>0
x;CO teR
and hence q(x) = min(d1 q(x), 1) belongs to 9 and it is equivalent to q. The space (Z, II II) constructed above is countabledimensional and it is universal for all countabledimensional {JF*spaces. Now we shall show that the completion of the space Z is universal for all separable F*spaces. Let X be an arbitrary separable F*space. By Lor.ma 2.4.5 X X / 2 contains a dense countable dimensional subspace X 0 , being {JF*space. Thus there is a subspace X0 , 1 of the space Z isomorphic to X 0 • Since X 0 is
Z
62
Chapter 2
dense in Xx 12, the completion X0 , 1 of the space X 0 , 1 contains a subspace isomorphic to the space Xx / 2 • Therefore the completion Z of the space Z contains a subspace isomorphic to the space Xx / 2 and it trivially implies that Z contains a subspace isomorphic to X. D 2.5.
LINEAR CODIMENSION AND THE EXISTENCE OF A COUNIVERSAL SPACE
In this section a notion in a certain sense dual to the notion of linear dimension will be considered. We say that an Fspace X has the linear codimension not greater than an Fspace Y, codimzX ~ codimz Y,
if Y contains a subspace Y0 such that the space X is isomorphic to the quotient space Y/Y0 • PROPOSITION 2.5.1. Let X andY be two Fspaces. codimzX~ codimzY if and only if there is a linear continuous operator T mapping Y onto X. Proof Necessity. By the definition of linear codimension, there is a subspace Y0 of the space Y such that the space X is isomorphic to the space Y/Y0 • Let T 0 denote this isomorphism. Write T(y) = T0 ([y]), where [y] denotes the coset containing y. Since T0 1 is continuous, Tis also continous. It is easy to observe that T maps Y onto X. Sufficiency. Let Y0 = T1 (0). The transformation T induces the continuous linear operator T 0 mapping Y/Y0 onto X defined as follows
T 0([y])
= T(y).
T 0 is onetoone, and hence, by the Banach theorem (Theorem 2.3.2) its inverse is continuous. Therefore X and Y/Y0 are isomorphic. D
If codimzX ~ codimz Y and simultaneously codimz Y ~ codimzX, we say that the spaces X and Y have the same linear codimension and we write itcodimzX = codimzY. If two spaces X and Y are isomorphic, then obviously. codimzX = = codimzY. We say that the linear codimension of an Fspace X is less than the linear
Linear Operators
63
codimension of an Fspace Y, codimzX < codimz Y, if codimzX ~codimz Y, but codimzX =F codimz Y. Problems connected with linear codimension are in general more difficult than problems connected with linear dimension. For example the quotient space LP([O, 1] x [0, 1])/H, 0 < p < 1, where H denotes the subspace of the space LP([O, 1] x [0, 1]) ofthe functions depending only on the first variab e is not isomorphic to the space LP([O, 1]) (see Kalton, 1981b). Problem 2.5.1. Let (Y, II II) be an Fspace. Does there exist a subspace Y0 C Y such that the quotient space is an infinite dimensional separable space ?1
Let X be a family ofFspaces. We say that an Fspace Xu is universal for a family X with respect to the linear codimension (or briefly couniversal for family X) if, for each X eX codimzX ~ codimzXu. In the same way as in the case of linear dimension we can construct a trivial couniversal space for any family X ofFspaces. Namely we enumerate all members offamily X. We construct the space Xu as in the previous section. Let ..h = {{xa}: Xb = 0}. The space Xu/Xb is isomorphic to the space Xb. The problem of couniversality for a family X consists in finding a space Xu eX couniversal for the family X. THEOREM 2.5.2 (Kalton, 1977). There is a separable Fspace which is couniversal/or all separable Fspaces. The proof is based on several notions and propositions. To begin with, we shall extend the notion of N(l) spaces to the case where N(u) depends also on n. Namely, suppose we are given a sequence {N1t(u)} of continuous 'nondecreasing functions (defined for u ~ 0). Suppose that Nn(O) = 0 if and only if u = 0, n = 1,2, ... Suppose that the sequence Nn(u) satisfies a uniform (~ 2) condition, i.e. there is a K > 0 (which does not depend on n) such that Nn(2u) ~ KNn(u). (2.5.1) 1
Added in proof· The answer is negative (Popov, 1984).
Chapter 2
64
It is easy to verify that co
PN(X)
= }; Nn(Xn) n=l
is a metrizing modular on the space of all sequences. Let Nn(/) be the space of all such xthat PN(x) < +oo. By Theorem 1.2.4 the space Nn(/)is linear. Moreover, the modular PN(x) induces a metric on this space such that it is a metric linear space. In examples which wiii be interesting for us the functions N, will satisfy a stronger condition than (2.5.1), namely (2.5.2)
Nn(u+w):::;;;;; Nn(u)+Nn(w).
In this case the modular PN(x) is an Fnorm. It is easy to verify that Nn(/) is complete. Indeed, let {xk} = {x7} be a fundamental sequence. Then, for each i, the sequence {x~} is also fundamental. This implies that there is a limit i =I, 2, ...
Observe that for an arbitrary s k,k'>K
>0
there is an index K such that, for
co
PN(xkxk')
= }; Nn(_x!x~):::;;;;; s. n=l
Then passing, with k' to infinity, we get PN(xkx):::;;;;; s,
where x = {xi}. The arbitrariness of s implies the completeness of the space Nn(l). PROPOSITION 2.5.3 (Turpin, 1976). For any separable Fspace (X, II II), there are a space Nn(l) and a continuous linear operator T such that T maps Nn(l) onto X. Proof Let {Xn} be a dense sequence in X. Let Nn(u) = lluxnll. LetT({ at}) n
=.I; atXt. By the definition of N
11
(1) it follows that the operator Tis well
i=l
determined and continuous. The proof will be complete when we show
Linear Operators
65
that T maps Nn(l) onto X. This immediately follows from the fact that each x EX can be represented in the form ct)
X=
2:
Xnt,
i=l
1
where llxnill
< lf•
D
Modifying the construction of Turpin we can obtain a stronger result. 2.5.4 (Bessaga, 1982, oral communication). For any separable Fspace (X, II ID there is a space Nn(l) such that for any subspace Y C X there is a continuous linear operator T mapping N n(/) onto Y.
PROPOSITION
The proof is based on the following LEMMA 2.5.5. Let (Y, II II) be a separable Fspace. Letisup {IIYII: y E Y} > b. Then there is a sequence {Yn} dense in Y such that sup lltYnll > b for tER
n = 1,2, ... Proof Let y
E
Y be an element such that suplltyll
>
2<5. Let {xn} be an
teB
arbitrary dense sequence in Y. We put if suplltxnll
Xn
Yn =
!
xn+
tER
~
y
> b,
otherwise.
It is easy to verify that Yn has the requested property.
D
Proof of Proposition 2.5.5. Without loss of generality we may assume that the norm II II is nondecreasing, i.e. lltxll ::::;;; llxll for It I ::::;;; 1. Basing on Lemma 2.5.5 we can find a number b > 0 and a sequence {Yn} C Y such that IIYnll > b for n = 1,2, ... and 1
UU n
teR
{tyn} = Y.
(2.5.3)
Chapter 2
66
Let the space Nn(l) be constructed as in Proposition 2.5.3. For each Ym we can find Xn 111 such that 1
IIYmXnm II~ 2m
(2.5.4)
•
Let T: Nn(l)+ Y be defined as follows 00
T({tn})
=}; tnmYm· m1
Since the norm II II is nondecreasing and IIYnll < lJ, inequalities (2.5.4) imply that the operator Tis well defined and continuous. Since (2.5.3), for every y E Ythere is a sequence ofreals {am} such that 00
y=}; amYm
(2.5.5)
m=l
and 1
(2.5.6)
llamYmll < 2 m · By (2.5.6) the sequence t = {tn},
am tn = {O
for n = nm, otherwise
belongs to the space Nn(l) and T(t) = y.
D
Proof of Theorem 2.5.2 (Bessaga, 1982, oral communication). By Theorem 2.4.7 there is a universal separable Fspace (X, II II). Using for this space D Proposition 2.5.4 we obtain a couniversal space.
For a given separable Fspace (X, II II) the construction of Turpin can give two nonisomorphic spaces. Indeed, if {Xn} and {Yn} are two dense sequences such that inf sup lltxnll > 0 and inf sup lltYnll = 0. then obn
teR
n
teR
viously the spaces induced by the sequences {Xn} and {Yn} are not isomorphic. On the other hand if {Xn} and {Yn} are dense sequences such that inf sup lltxnll > 0 and inf sup iitYnll > 0, then the spaces induced by sen
teR
"
teR
Linear Operators
67
quences {Xn} and {Yn} are isomorphic. Indeed, there are b > 0 and sequences of scalars {rn} and {s11 } such that llrnXnll > b, llsnYnll > (), n = I, 2, ... (2.5.7) Now we define by induction a permutation of positive integers k (n) in the following way: k(n) is the smallest positive integer different than k(l), ... ... , k(nI) such that I llrnXnSk(n)Yk(n) II< 2n · (2.5.8) Formulae (2.5.7) and (2.5.8) imply that the spaces induced by the sequences {Xn} and {Yn} are isomorphic.
2.6.
BASES IN FSPACES
Let (X, II ID be an Fspace. A sequence {en} of elements of the space X is called a Schauder basis (Schauder, I927) (or simply a basis) of the space X if every element x E X can be uniquely represented as the sum of a series (2.6.1)
c
A sequence {g11 }
X is called a basic sequence if it is a basis in the
space Y generated by itself, i.e. it is a basis in Y = lin {gn}. Evidently, if an Fspace has a basis, then it is separable. For any x of the form (2.6.1), let n
Pn(x) =
.2; ftet. i=l
2.6.1. The operators Pn are equicontinuous. Proof Let X1 be the space of all scalar sequences y = {ni} such that the
THEOREM co
series }; 'l']t et is convergent. The arithmetical rules of the limit trivially i=l
imply that Let
xl is a linear space. n
IIYII* =sup II n
X'l']tetll·
i=l
Chapter 2
68
The space X1 with the norm II II is an F*space. We shall show that it is complete, i.e. that it is an Fspace. Suppose that a sequence {yk} E X 1 is a Cauchy sequence. Let yk = {nn. Since {yk} is a Cauchy sequence, for an arbitrary e > 0 there is a positive integer m0 such that, for m,k > m0 , n
llymykll* = sup n
/I}; (n'[' nf)et
/J
<e.
(2.6.2)
i=l
Hence nI
ll(n~n!)enll ~ 1/l\n'l'n~)ei 1/ i=l
n
+ Jj}; Cn'l'nf) ei [/ < 2e i=l
Consequently lim (n';'n~) m,k+oo
=
o,
n =I, 2, ...
Thus the sequences of scalars {'11~} are convergent for all n. Write 'l']n
= lim
'11~·
~00
Passing with k to infinity in inequality (2.6.2), we obtain that for m>m0 n
sup\\}; n
(n'!'nt) et II ~ 2e.
(26.3)
i=l
Let and Taking into account inequality (2.6.3) we obtain
llsrsnll ~ lls';'s~ll+2e for all m > m0 and all n and r. Fix m1 > m0 • The sequence {s;:''} is a Cauchy sequence. Therefore there is a number n0 such that, for n,r > n0 ,
lls::''s"/''11 <e.
69
Linear Operators
Hence, for n, r > n0 ,
00
Thus the series 2 rJt et is convergent andy = {nt} e X1 • i1
From inequality (2.6.3) follows n
sup n
II? (rJi nt)e,IJ ~ 2e \=1
for m > m0 • Hence the space X1 is complete. Let A be an operator mapping X1 into X defined as follows A(y)
=}; rJtet. i=l
By the definition of the space ~1 the operator A is welldefined on the whole space X 1 • The arithmetical rules of the limit imply that the operator A is linear. Since {en} is a basis, the operator A is onetoone and maps X1 onto X. The operator A is continuous, because · n
oo
IIA(y)ll = [[}; rJtet II~ sup i=l
n
II}; niet I = IIYII* · i=1
By the Banach theorem (Theorem 2.3.2) the inverse operator A1 is also continuous. Hence n
IIPn(x)ll
=If LrJtetl[ ~ IIYII* =IIA {y)ll 1
i=1
and the operators Pn are equicontinuous.
D
Let x eX be reperesented in the form (2.6.1). Let fn(x)
= tn.
It is easy to see thatfn are linear functionals. They are called basis junetionals.
Observe that fn(x)en =· Pn(x)Pn' 1(x):
Thus from Theorem 2.6.1 immediately follows
Chapter 2
70
2.6.2. The basisfunctionals are continuous. Suppose we are given two Fspaces X and Y. Let {en} be a basis in X and let {fn} be a basis in Y. We say that the bases {en} and {in} are equivalent COROLLARY
00
00
i=I
i=l
if the series 2 ttet is convergent if and 'only if the series 2 tdi is convergent. Two basic sequences are called equivalent if they are equivalent as bases in the spaces generated by themselves. 2.6.3. If the bases e{n} and {fn} are equivalent, then the spaces X and Yare isomorphic. Proof Let Tn: X+Ybe defined as follows
THEOREM
oo
Tn(x)
n
= r(~ ttet) = ~ ttfi. i=l
i=l
By Corollary 2.6.2 the operators Tn are linear and continuous. The limit T(x) = lim Tn(x) exists for all x. Thus, by Theorem 2.2.3, T(x) is continuous. Since the bases are equivalent, the operator, Tis onetoone and maps X onto Y. Thus, by the Banach Theorem (Theorem 2.3.2), the 0 inverse operator r1 is continuous. The following theorem is, in a certain sense, converse to Theorem 2.6.1. THEOREM 2.6.4. Let (X, II /D be an Fspace. Let {en} be a sequence of linearly independent elements in X. Let XI be the set of all elements of X which can be represented in the form
Let n
Pn(x) =
.J; ttet. i=l
be a sequence of linear operator defined on X 1 • If the operators Pn are equicontinuous, then the space XI is complete and the sequence {en} constitutes a basis in this space.
Linear Operators
71
Proof. Since the operators P, are equicontinuous, each element x of X 1
can be expanded in a unique manner in the series
Let X2 be the space of all sequences {ti} such that the series
is convergent. Let n
ll{tt}ll*
=sup n
Jj}; t,e,Jj. i=l
In the same way as in the proof of Theorem 2.5.1, we can prove that II II* is an Fnorm and that (X2 , II II*) is an Fspace. Observe that
llxll ~ ll{tt}ll*.
(2.6.4)
On the other hand, the equicontinuity of the operators P, implies that if X+0 then II{ ti}ll*+0. Hence the space x2 is isomorphic to the space X]. Therefore X 1 is. an Fspace. By (2.6.4) the sequence {e,} constitutes a basis in X1 • D 2.6.5. Let X be an Fspace. A sequence of linearly independent elements {e,} is a basis in X if and only if
CoROLLARY
(1) linear combination of elements {e,} are dense in X, (2) the operators n
Pn(x)
= }; ttet i=l
are equicontinuous on the set lin{ e,} of all linear combinations of the set
{e,}.
Chapter 2
72
2.6.6. Let X be an Fspace with a basis {en}. Let t 1 , t 2 , ••• be an arbitrary sequence of scalars. Let p 1 ,p2, ••• be an arbitrary increasing sequence of positive integers. Let
COROLLARY
n = 1, 2, Let Pn+t
e~=}; t,e,. i=pn+l
Let X1 = {line~} be the space spanned by the elements {e~}. Then the space X 1 is complete and the sequence {e~} constitues a basis in X1 • n
oo
Proof Let x EX, x =
2
ttet. Let Pn(x) =
i=l
2
ttet.
i1
For ye Xb let n
P~(y)
=}; a,e·,. i=l
Then P~ (y) = PP
n+l
(y) for ally E X1• Therefore, by Theorem 2.6.1, the
operators P~ are equicontinuous and, by Theorem 2.6.4, X1 is complete and {e~} is a basis in X1 • D A basis {e~} of the type described above is called a block basis with respect to the basis {en}· PROPOSITION
2.6.7. Let (X,
II
jl) be an Fspacewith a basis {en}. Let {xn} be
a sequence of elements of X of the form 00
Xk = }; fk,tet
where lim tk,t = 0, i = 1, 2, ...
i=l
k>ro
If {en} is an arbitrary sequence of positive numbers, then there exist an increasing sequence of indices {Pn} and a subsequence {xkn} of the sequence { Xk} such that Pn+t
llxkn :L>kn,tetjj <en. i=p.. +l
Linear Operators
Proof (by induction). Let p 1 = 0,
Xk1
=
73
x1. We denote by p 2 an index
satisfying the inequality
Suppose that the element Xknl and the index Pn are already chosen. The hypothesis lim fk, t = 0 implies the existence of an element Xkn such k>oo
that
Let Pn+l be an index satisfying the inequality Pn+t
llxkn}; tkn,tetll
Then obviously Pn+t
!ixkn:};tkn,tetJI
<
D
8n.
'=Pn+l
Observe that in Proposition 2.6.7 we can additionally require that for an arbitrary given sequence {qn} there should be {Pn} satisfying the conclusion of Proposition 2.6. 7 such that qn < Pn· Let X be a complex Fspace formed of complex valued functions (in particular, complex sequences). We say that X is a C*space if x = x(t) E X implies that the conjugate function = x(t) belongs to X and the norm in the space X depends only on lx(t)l. Let X,. be the space of realvalued functions belonging to X. Since X is a C*space, each x E X may be uniquely represented in the form
x
where
x\ x2 E X,..
PROPOSITION 2.6.8. Let X be a C*space. Let {en} be. a basis in X,.. Then {en} is a basis in X.
74
Chapter 2
Proof Since {en} in a basis in X,, there are unique expansions of elements x 1 and ,x2 with respect to this basis:
Hence x can be uniquely extended with respect to the basis {e11 }
:
D We shall now give examples of bases. Example 2.6.9 The sequence

en= {0, ... , 0, 1, 0, ... },
n
=
1, 2, ...
n th place
is a basis in the spaces c0 , lP(O < p < +oo), (s). This basis is called a standard basis. Example 2.6.10 (Schauder, 1927) There is a basis in the space C(O, 1]. It is constructed in the following manner. We define a function Uk,,(t) (0 ~ i < 2", k = 0, 1, ...) in the following way:
Uk,t(t) =
0
for 0 ~ t
2k+l ( t ;k)
for
2k+l
(t i~1)
i
i+1
< 2k and zk ~ t ~ 1,
i
2i+1 2k+ 1
2k ~ t <
2i+1 for 2k+ 1 ~ t
,
i+1
< v·
Each continuous function x(t) defined on the interval [0, 1] can be uniquely written in the form
• x(t)
=
00
a0 t+a1 (1t)+
2~1
.2; }; ak,IUk,,(t), k=O
io
Linear Operators
75
where a0 = x(l), a 1 = x(O) and the coefficients ak,t are defined as follows. We draw the chord l(t)ofthe arcy = x(t)through the points ofthe i i+I abscissae 2k , ~ and we define ak, t
=
x
(2i+l) ( 2i+l) 2k+l 1 2k+l .
Therefore, the sequence {t, It, u0 , 0(t), u1 , 0(t), u1 , 1 (t),
U 2 , 0(t),
... }
constitutes a basis in the space C[O, 1]. Example 2.6.11 Suppose we are given the following system of functions on interval [0, 1]:
r(2n/fp
h.,,(t)
~ 1(2")~' to
n
j1 2jI for~< t < 2n+l 2jI for 2n+l
,
j 2n '
elsewhere,
= 0, 1,2, ... ,j = 1,2, ... , 2n. The system of functions
is called the Haar system. It is a basis in the spaces LP[O, 1] (1 ~ p < +oo) If there is a basis in an Fspace X, then obviously the space X is separable and by Corollary 2.6.2 there are continuous linear functionals different from 0 in X. In sequel linear functionals different from 0 will be called nontrivial. Let X be an F*space. Let {xn} be a sequence of elements of X. We say that the sequence {Xn} is linearly independent if for each finite system of scalars a1 , ••• , an the equality (2.6.5) implies a 1 = a2 = ... = an = 0.
(2.6.6)
Chapter 2
76
We say that sequence {xn} is topologically linearly independent if for each sequence of scalars {a 1 , a 2, ••• } the equality (2.6.7) implies al
= ... =
an
= ... =
0.
(2.6.8)
We say that the sequence {xn} is linearly mindependent (see Labuda and Lipecki, 1982) if for each bounded sequence of scalars {an} equality (2.6.7) implies (2.6.8). We say that a sequence {Xn} of elements of an F*space X is a Hamel basis (quasibasis, mquasibasis) if (1) the linear combinations of {xn} are dense in X, (2) the sequence {Xn} is linearly independent(resp. topologically linearly independent, resp. linearly mindependent). Peck (1968) showed that in each separable Fspace there is a quasibasis. (For locally convex spaces it was proved by Klee (1958)). Drewnowski, Labuda and Lipecki (1982) extended this result on topological linear spaces. In the same paper they also showed, that if a sequence {xn} is a Hamel basis in X, then there is a sequence {b~> b2, ••• } such that the sequence {b1 x 1 , b1 x 1 +b 2 x 2 , ... , b1 x 1 + ... +bnxn, ... } is a quasibasis. Labuda and Lipecki (1982) showed that in each separable F*space ti:iere is an mquasibasis which is not a quasibasis. A sequence {en} in an Fspace X is called an Mbasis sequence, if there are continuous linear functionals e! defined onE= lin{ en} such that fori= j, fori =j::.j,
and if x E E and e: (x) = 0, n = 1, 2, ... , then x = 0. We say that a sequence {en} is equicontinuous if e!(x)en+0 for all x E E. Ka1ton (see Kalton, 1979) showed that every Mbasis sequence contains an equicontinuous sequence. Let X be an Fspace. We say that a sequence {xn} tends weakly to x if, for each continuous linear functional J, f(xn)+f(x). Since, in general
Linear Operators
77
there is no possibility of extension of continuous linear functionals from a subspace Y of the space X it may happen that a sequence {xn} C Y tends weakly to x in X, but does not tend weakly to x in Y. Drewnowski (1979) constructed an Fspace X which, for each basis {e11 }, each of its bounded block bases tends weakly to 0 in X, but no basis in X has the property that its bounded block basis tends weakly to 0 in Y which is a subspace spanned on that block basis.
2.7. SOLID METRIC LINEAR SPACES AND GENERAL INTEGRAL OPERATORS
Let (.Q, E, p,) be a afinite measure space. Let A be a subset of the space L 0(.Q, E, p). We say that the set A is solid if u.E A and lvl ~lui implies v EA. Let (X, II I[) be an F*space contained settheoretically in L0 (.Q, E, p). We say that the space X is solid if there is a basis of neighbourhoods of zero consisting of solid sets. 2.7.1 (Szeptycki, 1980). If (X, II llo) is a solid £_space, then there is an Fnorm II II equivalent to II llo such that
PROPOSITION
llxll =II lxl II, (ii) llull ~ 11~11 if lui (i)
~
Ivi.
Proof. Observe that if A, Bare two solid sets in L0 (.Q, E, p), then the set. A+B is also solid. Thus the construction of the norm II II follows from
the construction of an invariant metric (Theorem 1.1.1).
0
We say that X is continuously imbedded in L0(.Q, E, p) and we write X Cc L 0(.Q, E, p) if the identity mapping from X into L0 (.Q, E, p) is continous. PRoPOSITION 2.7.2 (Luxemburg)./f(X, II I[) is a solid F*space contained in L 0(.Q, E, p), then it is continuously imbedded, X Cc L 0(.Q, E, p). Proof. (Szeptycki, 1968). Suppose that the proposition does not hold. Then there are a neighbourhoo9 of zero U in L 0(.Q, E, p) of the form
U = U(E, a)= {u
E
L 0(.Q, E, p): p{t E E: lu(t)[ >a}< a},
Chapter 2
78
where E is a set of finite measure and a is a positive number, and a sequence {un} C X such that Un ¢ U and 1
llun// ~ 2n. Let
En= {teE: /un(t)/ >a}. Since Un ¢ U, by the solidity of the space X we have JJ.(En) Let •
> 0.
m=n
The measure of E is finite. Thus we can find an m11 such that ?](E~"'.E~)
<
2n 1 a,
where mn
E==
U E~. m=n
E'=
n E~,
Let
n E~. ct)
ct)
E"
n=l
=
n=l
Then ct)
JJ.(E'"'.E") =
00
ct)
JJ.(n E~"'. n E~) ~ JJ.(U E~"'.E:) n=l n=l n=l 00
~ Lfl(E~"'.E:) < ~.
(2.7.1)
n=l
Since E~ is a decreasing sequence of sets and fl(E~) ? fl(En)
we have JJ.(E')
> a,
> a. Thus by (2.7.1)
JJ.(E");?: ~.
(2.7.2)
Let x denote the characteristic function of the set E'. Then
ax~
2: /um(t)/ ta=A
(2.7.3)
Linear Operators
79
almost everywhere. The space X is solid. Thus, without loss of generality, we may assume that the norm II II satisfies the properties (i) and (ii) of Proposition 2.7.1. Therefore by (2.7.3) mn
llaxll
mn
~IlL lum(t)l II~ m=n
L llumll ~ 2n+l.
m=n
Hence llaxll = 0, which contradicts (2.7.2).
D
Let A be a linear space of measurable functions. A set E is called an wifriendlyset(Luxemburg and Zaanen, 1971) for A iff[E = 0 for all/eA. PROPOSITION 2. 7.3 (Aronszajn and Szeptycki, 1966). Suppose we are given a afinite measure space (.Q, E, p). Let A be a solid F*space in L 0(!J, E, p). Then there exists a maximal unfriendly set for A, unique up to sets of measure 0. The proof of the proposition is based on the following 2.7.4. Let (.Q, E, p) be a afinite measure space. Let u, v eL E, p). Then, except an at most countable number of reals ~. the sets E = {t: lu(t)l+lv(t)l > 0} and E~; = {t: u(t)+~v(t) =/= 0} differ by a set of measu~e 0. Proof The sets E n (!J"'.,E~;) are disjoint for differenU. Since Jl is afinite, only a countable number of these sets has positive measure. D LEMMA
0 (.Q,
Proof of Proposition 2.7.3. Let qJ e L 0 (.Q, E, p) be a positive function such that qJdJl = 1. The existence of such functions follows from the fact that u Jl is afinite. Write p(E) = qJdqJ. Let
J
J
E
a= sup{,U({t: u(t) =/= 0}), u e A}.
Let {Un} be a sequence of elements of A such that an= ,U({t: un(t) =/= 0})+a.
By Lemma 2.7.4 there are reals ~2 ,
... ,
~n
such that
En= {t: lu1(t)l+ ... +lun(t)l > 0} = {t: Vn(t)=/=0},
Chapter 2
80
where Vn(t)
=
u1(t)+;2u2(t)+ ... +;nun(t).
(2.7.4)
Observe that p(En) increases to a. 00
Let EA =
D"'.U
En. The setEA is the required unfriendly set. Indeed,
n~l
suppose that EA is not an unfriendly set. Then there is a function u(t)E A such that 7t({t E EA: u(t) # 0});;:::, b
> 0.
By the definition of {un} we can find an Un such that aan< b.
Using Lemma 2.7.4, we can construct a function of the form u+;un such that
f,t({t: u(t)+;un(t) =F 0}) = b+an >a, which contradicts to the definition of a. Now we shall show that EA is a maximal unfriendly set up to the sets of measure 0. Indeed, P, (EA) = 1a. Hence E A is unique up to the sets of pmeasure 0. Thus it is unique up to the sets of .ameasure 0. D PROPOSITION 2.7.5 (Aronszajn and Szeptycki, 1966). Let a measure space (Q, .E, .u) be afinite. Let A C L 0(Q, .E, .u) be a solid Fspace. Let EA be a maximal unfriendly set for A. Then there is a v e A such that EA = {t: v(t) = 0}.
Proof Let {un} be chosen as in the proof of Proposition 2.7.3. Let positive numbers Mn and An be chosen so that
,U({t: lun(t)l > Mn} < 2n,
(2.7.5)
where p(E) is defined as in the proof of Proposition 2. 7.3, and
ll;unll < for;,
1~1
2n
< An.
Having { Mn} and {An} we shall choose by induction two sequences {;n} and {1Jn} in the following way. We define~~ = 1, 0 < 171 ~ 1. Sup
Linear Operators
81
pose that ~to ••• , ~nand 'f/~o ... , 'f}n are chosen. Using Lemma 2.7.4, we can choose c;n+1 such that 0 < ~n+l < min(A.n+l• (2Mn+I)~n) (2.7.6) and
{t: Vn(t)+~n+IUn+I(t) ::1= 0}
= {t:
lvn(t)l+lun+l(t)l
> 0},
where Vn(t) is given by formula (2.7.4). We take
0
<
< i 'fjn ·
'f/n+l
(2.7.7)
By (2.7.5), (2.7.6) and (2.7.7) the sequence {vn} is convergent on D. Let co
v(t) = lim Vn(t) = ,._,. 00
L ~nun(t).
n=l
Observe that
~
lv(t)l
00
L
lvm(t)l~nlun(t)l n=m+l
~ 'fjm ~
00
}; 'fjn ~ }'fjm n=m+l
>0
on the set
Em= {t; lvm(t)l
> 'fJm} n {t: luj(t)l < Mj,
j = 1, 2, ... }.
It is easy to verify that
p(EA"\.Em)
< aam+ 2m+ 2m.
Hence v(x) ::1=.0 almost everywhere on EA.
D
Let (D, E, /1) be a measure space. We write En '\,.0, if En 1s a decreasing sequence of sets, En E E, such that J1(E n En)+ 0
for every set E of finite measure. Let (X, II ID be a solid F*space contained in L 0 (D, E, J1). By Xa we denote the set ofthose u EX for which (2.7.8)
[luxE.[I+0
for every sequence En '\,.0. If Dis of finite measure, (2.7.8) is equivalent to lim
UXE =
p(E}+0
0.
82
Chapter 2
It is easy to verify that Xa is a closed solid subspace of X and, if {En} is co an increasing sequence of sets such that [J = En, then xx + x for
U
n=l
En
all x E Xa If X= N(L(Q, .E, Jl), then Xa =X. If f1 is nonatomic and X = L co (Q, .E, Jl), then Xa = {0}. '
PROPOSITION 2.7.6 (Luxemburg and Zaanen, 1963). Suppose that f1 is afinite and that (X, II II) is a solid Fspace contained in L 0 (Q, 1:, Jl). Then a set C C Xa is compact in X if and only if Cis compact in L 0 (Q, 1:, Jl). For any sequence En~" for u E C.
(2.7.9.i)
lluxE,.II tends to 0 uniformly (2.7.9.ii)
Proof Necessity. The necessity of(2.7.9.i)immediately follows from Proposition 2.7.2. Suppose that (2.7.9.ii) does not hold. Then there ares > 0, a sequence En ~o, and a sequence Un E C such that
llunXE.. II >
(2.7.10) Since Un E C C Xa, we can construct by induction subsequences { Un.i}, { EnJ} such that s.
8
llun,XE,.1 II < 2
fori <j.
(2.7.11)
By (2.7.10) we get fori <j.
Hence the set C is not compact. Sufficiency. Let {vk} be any sequence of elements of C. By the compactness of C in L 0(Q, 1:, Jl) and the Riesz theorem we can find a subsequence {un} = {Vkn} of C tending to u E C almost everywhere. Now we shall show that {un} tends to uin X. Without loss of generality we may assume that X does not have an unfriedly set. Then by Proposition 2.7.5 there is a rp EX positive almost everywhere. Let En,k
=
{r
E
[J:
suplum(t)u(t)l;:;:::, m~n
+rp(t)}.
Linear Operators
83
Observe that, En, k ~0 for each fixed k andfor any e > 0 we can find by (2.7.9.ii) an index such that llvxn,kll
e
<3
for n
>
nk, v e C,
(2.7.12)
where, for brevity, we write Xn,k = XEn,.· Then llunull ~ llxn,k(unu)ll+ll(lxn,k)(unu)ll ~ llxn,kUnll+llxn,kull(1xn,k)(unu)ll
~ ; + ; + I ~ ~II <e. provided that k is chosen so that
I ~ ~11 < ; .
0
2.7.7 (monotone convergence theorem, Luxemburg and Zaanen, 1963). Let XC £ 0(!2, 1:, p) be a solid F*space. Then u e Xa if and only iffor each sequence {vn} such that PROPOSITION
(2.7.13) tending to zero almost everywhere, llvnll+0. Proof Let u eX, En ~0, Vn = UXEn· By our hypothesis llvnll+0. Thus ueXa. Conversely, let u e X a and let {Vn} be a sequence satisfying (2. 7.13) and tending to 0 almost everywhere. Let Em,n
=
{t:
Vn(t)
~ ~
u(t) }·
Let Xm,n denote the characteristic function of Em,n· For a fixed m, !2"'._Em,n ~0 and II(Ixm,n)ull+0 since u e Xa. We have llvnll
~II~ Xm,nUII +11(1xm,n)ull ~II;
I
+11(1xm,n)ull.
(2.7.14)
Chapter 2
84
For each s
> 0 there is an m such that
II; II<~·
(2.7.15)
Formulae (2.7.14) and (2.7.15) imply llvnll < s. This completes the proof.
D
PROPOSITION 2.7.8 (dominated convergence theorem, Luxemburg and Zaanen, 1963). Let (X, II II) be a solid Fspace contained in L 0(Q, E, p). Then u E Xa if and only if for each sequence {un} C X such that lun(t)l ~ lu(t)l almost everywhere and un(t)+u(t) almost everywhere, we have llunull+0. Proof Sufficiency. If En '\.e, then un = (1XEJ u has the property indicated in the statement. Thus u e X a. Necessity. Suppose that u e Xa and a sequence {un} has the indicated properties. Let Vn(t) = sup{lum(t)u(t)l, m
>
n}.
The sequence {vn}(t)} is not increasing. It tends to 0 almost everywhere and vn(t) ~ 2lu(t)l. Then by Proposition 2.7.7 llvnll+0. By the solidity of the space and Proposition 2.7.1 this implies that llunull+0. D Let (Q, E, p) and (QI> E 1 , p 1) be two afinite measure spaces. Let u e L0 (!21 , E 1 , p 1 ). Let k(t, s) be a function defined on the product space Q X !21 measurable with respect to the product measure fl. Xp1 • Let K(u) =
J k(t, s) u(s)dp (s) 1
D,
be a linear integral operator acting from L 0{Q1 , El> p 1) into L 0 (Q, E, p) By a proper domain DK of the operator Kwe shall mean the set DK = {ue L0 (!21 ,E1 ,p1):
Jlk(t, s)l lu(s)ldp (s) < + oo 1
D,
palmost everywherej. 00
We recall that (Q, E, p) and (.Q1 , El> p 1) are afinite, i.e . .Q =
U .Qn, n=l
0,
Linear Operators
85
00
!/1 =
U Dn, n=l
I>
where f1.(!/,n 0)
< + oo and fl.i!Jn, 1) < + oo. Thus the to
pologies in L 0 (Q, .E, fl.) and in L 0(Q1 , .E1 , f1.1) are described by sequences of pseudonorms, II lin in L 0(Q, E, fl.) and II II~ in L 0 (QI> .El> f1.1) (see Example 1.3.5). We shall introduce Fpseudonorms on the space Dx in the following way
llullm,n =
lluii~+IIIKIIulllm,
where
IKIIul =
J lk(t, s)llu(s)l dfl.l·
(2.7.16)
n,
It is easy to see that the space D xis a solid F*space with the topology defined by the double sequence of pseudonorms II llm,n· THEOREM 2.7.9 (Aronszajn and Szeptycki, 1966). The space Dx is complete. Proof Let {uk} be a Cauchy sequence in Dx. This means that, for each fixed m, n, lim llukUk•ll = 0. (2.7.17) k,k'+co
m,n
The sequence {uk} is obviously also a Cauchy sequence in the space L 0 (Q1 , .E1 , Jl.J. Thus there is a u e L 0 (Ql> El> J1.1) such that {un} tends to u in the space L 0(Q1 , E 1 , Jl.J. By (2.7.17), {uk} contains a subsequence {u;} = {uk,} su«h that 00
}; llu~+lu~llm,n <
+oo.
i=l
Thus the series 00
}; lu~H(t)ut(t)l i1
and
00
}; IKI lu~+lu,l i=l
cqnverge almost everywhere on sets Q,., 1 and !Jn, 0 , respectiveiy. We have CIO
lui~ I~ I+}; lu~+l~u~l. i=l
86
Chapter 2
Thus by the Lebesgue theorem 00
JKJ JuJ ~ JKJ Ju~J+}; JKJ Ju~+~u~J i=l
almost everywhere on Dn,o. Similarly we can estimate 00
JKIJuz4J ~}; JKJJu~+lu~J~O i==r
almost everywhere on Dn,o• as r~=. And, by Proposition 2. 7.8,
II JKJJu;uJ llm~O. In a similar way we can show that llu;ulln~O.
Thus
llu;ullm, n~O. Therefore u; tends to u in Dx· In this way we have shown that each Cauchy sequence contains a convergent sequence. This implies the com0 pleteness of Dx (cf. Proposition 1.4.8). THEOREM 2.7.10 (Aronszajn and Szeptycki, 1966). The operator K is a continuous linear operator from Dx into L 0 (Q, E, p). Proof IIKuJJm ~ II JKJJuiJJm ~ llullm,n for all n. 0 THEOREM 2.7.11 (Aronszajn and Szeptycki, 1966; see also Banach, 1931). Suppose that X C c L0 (D1 , E 1 , p 1) is a solid Fspace and suppose that XC Dx. Then the integral operator K maps X into L 0(Q, E, p) is a continuous way. Proof By Theorem 2.7.10 it is enough to show that the identity is a continuous mapping from (X, II ID into (Dx, II llnx), where II llnx is the Fnorm induced in the space Dx bythe sequence ofpseudonorms II llm,n (see Section 1.3). Accordingly, we shall introduce a new norm on X, namely
Linear Operators
87
and we shall show that the space X is complete with respect to this new norm. Indeed, let {un} be a Cauchy sequence in (X, II lit). Then it is a Cauchy sequence in (X, 1/ //)and in (Dx, II 1/vK). Since both spaces are complete, the sequence {Un} has limits in both of them. Let u0 be the limit in X and let u1 be the limit in Dx. Since XC c L0 (!:J1 ,I1 ,fit) and Dx C c L 0(Qb 1:1 , fit), the sequence {un} is convergent in the space L 0(Q 1 , It> jt1) simultaneously to u0 and to u1 • Thus u0 = u1• Hence l/unu0 1/ 1 = l/unu0 1/+l/unU0 1/vx+0.
Therefore (X, 1/ 1/J is complete, and by Corollary 2.3.3 the two norms are 0 equivalent. This implies the continuity of K. THEOREM 2.7.12 (Aronszajn and Szeptycki, 1966). Let (X, 1/ 1/x) CcL0(f:J 1 ,l:t>Jit) and (Y,/1 I/Y)CcL0(Q,l:,Ji) be given Fspaces. If XC Dx and KX C Y, then the operator K: X+Yis continuous. Proof. We introduce a new norm on X, llxll
= llxl/x+IIK(x)l/y,
and, in the same way as in the proof of Theorem 2.7.11, we show that 1/ //)is complete. We take a fundamental sequence {un} in (X, II //).It is alsofundament~lin(X, II llx). Thus there is a ue X such that llu11 ul/x+0. By Theorem 2.7.11 llunullvx+0 and by Theorem 2.7.10 Kun+Ku in L 0 (Q, 1:, Jl). Suppose that 1/KunVIIy+0. Since Y C c L 0 (Q, 1:, Jl), Ku11 +v in L 0 (Q, 1:, Jl). This implies that Ku = v. Hence (X, II //)is complete. Thus, by Corollary 2.3.3, the norms llxll and llxl/x are equivalent. 0 This implies the continuity of Ku. (X,
PROPOSITION 2.7.13. (Aronszajn and Szeptycki, 1966). Let K be an integral transformation. Then (Dx)a = Dx. Proof. Letfe Dx. Let Et~0. Thus Jl(Qn,t n E,)+0,
n = 1,2, ...
Therefore
n = 1,2, ...
88
Chapter 2
By Proposition 2.7.7, IKIIIXE,I tends to 0 almost everywhere and, by the definition of the Fpseudonorm llxllm,n,
II/XEI!Jm,n ~ 0 •
D
Example 2.1.14 Let D = [0,27t] with the Lebesgue measure. Let D 1 = Z be the set of all integers with the discrete measure
Pl{y}) Let k(t, s)
=
1,
y
=
= e'"'. Then Dx =
Example 2. 7.15 Let D = Z and let D 1
0, +I, ±2, ... L(Z)
= fl.
= [0,27t]. Let k(t,s) =
eil8 •
Then Dx
= £1[0,27t].
Example 2.1.16 Let D = D 1 = R with the standard Lebesgue measure. Let k(t,s) = eilB. Then Dx = £1( oo, +oo).
Chapter 3
Locally Pseudoconvex and Locally Bounded Spaces
3.1.
LOCALLY PSEUDOCONVEX SPACES
Let X be a metric linear space. A set A C X is said to be a starlike (starshaped) set if tA C A for all t, 0 < t ~ I. The modulus of concavity (Rolewicz, I957) of a starlike set A is defined by c(A)
= inf {s > 0:
A+A C sA},
with the convention that the infimum of empty set is equal to +oo. A starlike set A with a finite modulus of concavity, c(A) <+oo, is called pseudoconvex. PROPOSITION
3.1.1. Let A be an open pseudoconvex set. Then (3.1.1)
A+A C c(A)A.
Proof. Suppose that (3.1.1) does not hold. Then there is an x ¢ c(A)A such that x e A:+A. Since the set A+ A is open, there is an r > I such that rx ¢ A+A. Then A+A ¢ c(A) A r '
because x ¢ c(A)A. Since r >I, we obtain a contradiction of the defiD nition of the modulus of concavity. Observe that we always have c(A) ;::;::: 2. A set A is said to be convex if x, yEA, a, b;::;::: 0, a+b = I imply ax+bye A. Of course, for each convex set A the modulus of concavity c(A) of the set A is equal to 2. The condition c(A) = 2 need not imply convexity, but the following proposition holds : 89
90
Chapter 3
3.1.2. Let A be an open starlike set. If c(A) = 2, then the set A is convex. Proof Let x,y EA. Since A is an open starlike set, there is a t > 1 such that tx,ty EA. Since c(A) = 2, A+A E 2tA. Thus tx+ty E 2tA. PROPOSITION
This implies that (x;y) EA. Therefore, for every dyadic number r,
rx+(1r)y EA.
(3.1.2)
The set A is open. Then the intersection of A with the line
L = {tx+(1t)y: t real} in open in L. Therefore there is a positive numbers such that
x 0 = t0 x+(1t0 )y E A
for
ltol < E
and Applying formula (3.1.2) for x = x 1 andy= x 0 , we find that, for every a such that Ia  rl < s, ax+(la)y EA. (3.1.3) Since r could be an arbitrary dyadic number, (3.1.3) holds for an arbitrary real a, 0 ~ a ~ I. Thus the set A is convex. D PROPOSITION 3.1.3. Let A be a starlike closed set. If c(A) = 2, then the set A is convex. Proof Since the set A is closed, 2A = sA. Thus c(A) = 2 implies
n
•>2
x+y that A+A C 2A. Therefore, if x,y E A, then  2 EA. This implies (3.1.2) for every dyadic number r. Since A is closed, (3.1.3) holds. D A metric linear space X is called locally pseudoconvex if there is a basis of neighbourhoods of zero { Un} which are pseudoconvex. If moreover c(Un) ~ 2 11P, we say that the space X is locally pconvex (see Turpin, 1966; Simmons, 1964; Zelazko, 1965). THEOREM 3.1.4. Let X be a locally pseudoconvex space. Then there is a sequence of Pnhomogeneous Fpseudonorms {II lin}, i.e. such that
lltxlln = ltiPnllxlln,
(3.1.4)
Locally Pseudoconvex and Locally Bounded Spaces
91
determining a topology equivalent to the original one. If the space X is locally pconvex, we can assume Pn = p (n = 1,2, ... ). Proof Let {Un}be a basis ofpseudoconvex neighbourhoods ofO. Without loss of generality we may assume that the sets Un are balanced (cf. Section 1.1). From the definition of pseudoconvexity it follows that there are positive numbers Sn such that Un+Un C SnUn.
Let (q = 0, ±1, ±2, ... ). t
For every dyadic number r > 0,, = }; at 2i, where at is equal either to i=B
0 or to 1, we put Un(r) = asUn(28)+ ... +atUn(21). In the same way as in the proof of Theorem 1.1.1, we show that Un(r 1 +r2)
Un(r 1 )+ Un(r 2)
)
and U11 (r) are balanced. Moreover, the special form of U11 (r) implies U11 (2qr) = s~ U11 (r). Let //xi/~= inf {r > 0: x E Un(r) }. The properties of the sets U11 (r) imply the following properties of 1/xl/~: . (1) 1/x+y//~ ~ 1/x/I~+IIYII~ (the triangle inequality), (2) 1/ax/1~ = 1/x/1~ for all a, /a/ = 1, (3) 1/s~x/1~ = 2ql/xl/~.
Let log 2 Pn =log sn" Let 1/xl/n = sup 1>0
1/tx/1~ . fPn
By (3) 1/xl/n is well determined and finite since 1/xl/n = sup 1>0
1/tx/1~ fPn
=
sup 1,;;;1,;;;sn
1/tx/1~ • fPn
92
Chapter 3
We shall show that llxlln are Fpseudonorms. Indeed, if lal = 1, then
llaxlln =sup lltaxll~ =
sup
t>O
t>O
fP•
lltxll~
= llxlln
fP•
and (n2) holds. Let x,y e X. Then II x + y II 11 = sup llt(x+y)ll~ P• = sup t
t>O
lltxll~
=S;; sup
tPn
I.,;;t.,;;sn
llt(x+y)lln
I.,;;t.,;;s,.
+ sup I.,;;;t.,;;sn
lltyll~ tP•
t
P•
=S;; llxlln+IIYIIn
and (n3) holds. Observe that lltxlln = sup llrtxll~ = sup llrltlxll~ r>O
jtjP•
r>O
lf/P• sup llsxll~
=
sP•
•>O
(r It i)P•
=
lt/P•
/t/P•
llxlln.
Hence llxll is Pnhomogeneous. This implies (n4)(n6). Now we shall prove that the system of pseudonorms {llxlln} yields a topology equivalent to the original one. Indeed, from the definition of llxlln, llx[[~ =S;; llxlln, in other words, {x: llxlln
<
(3.1.5)
r} C Un(r).
On the other hand, the sets Un(r) are starlike. Therefore, the pseudonorms llxl/~ are nondecreasing, which means that 1/txl/~ are nondecreasing functions of the positive argument t for all n and x e X. Then lltx[i~ [[tx[[~ , , llxlln =sup= sup   =S;; llsnXIIn = 211xlln, t>O
tP•
I.,;;t.,;;sn
tP•
in other words Un(r)C {x: llxlln
< 2r}.
(3.1.6)
Formulae (3.1.5) and (3.1.6) imply the first part of the theorem. If the space X is locally pconvex, then by the definition there is a basis of neighbourhoods of 0 {Un} such that c(Un) =S;; 2 11P. Since Un are open it implies that Un+UnC2 1 iPUn.
Locally Pseudoconvex and Locally Bounded Spaces
93
Putting s11 = 21 /P and repeating the construction above, we obtain a system of phomogeneous pseudonorms determining a topology equivalent to the original one. D 3.1.5. Let X be a locally pseudoconvex space with a topology given by a sequence of Pn·homogeneous Fpseudonorms {II 11 11 }. A set A C X is bounded if and only if PROPOSITION
On=
sup {JJxJJn:
XE
A}
<+oo.
(3.1.7)
Proof If a set A is bounded, then for each n = 1,2, ... there is a tn such that tnA C {x: JJxJJ,
>0
< 1}.
Thus I tn
On~.
Conversely, let U be an arbitrary neighbourhood of 0. Then there are e > 0 and n such that {x: JJxJJn
< e}
C U.
By (3.1.7) I
e4c u. On
0
A topological linear space X is called locally convex if there is a basis of convex neighbourhoods of 0. The construction given in Theorem 3.1.4 'leads to the fact that if X is a locally convex metric linear space, then the topology could be determined by a sequence of homogeneous (i.e. !homogeneous) Fpseudonorms. Homogeneous Fpseudonorms will be called briefly pseudonorms whenever no confusion results. Locally convex metric linear spaces are called B~spaces. As we shall see later, for locally convex spaces there is also another much simpler construction of a sequence of homogeneous pseudonorms determining the topology. A complete B~space is called a B0 space. Let X be an F*space. If the topology in X can be determined by a sequence of Pn·homogeneous pseudonorms, then the space X is locally
Chapter 3
94
pseudoconvex. In the particular case where the pseudonorms are homo
geneous, the space X is locally convex. Indeed, the sets  1 Kn, where m Kn
= {x: //x//n < 1},
constitute a basis of neighbourhoods of 0. The modulus of concavity ofthe set Kn is not greater than 2Pn. Indeed, if x,y E Kn, then llx+y//n ~ 2. Hence
i.e. x+ y E 2P• Kn. Thus Kn +Kn C 2P• Kn. In the particular case where the pseudonorms are homogeneous the sets Kn are convex. It is easy to verify that the following spaces are B0spaces: LP(Q,E,J1) for 1 ~p ~+oo, C(Q), C(QjQ0 ), the Hilbert space, 0 (D), C 00 (Q), c:S(En). We say that a set A in a metric linear space X is absolutely pconvex, 0
e
ax+bye A. Z1,
z
Fig. 3.1.1
If a set A is absolutely pconvex, then it is pseudoconvex and its modulus of concavity is estimated by the following formula c(A) ~ 21/P. Indeed, from the definition of the absolute pconvexity it follows that
if x,y E A, then
x+y 211P
E
A and
Locally Pseudoconvex and Locally Bounded Spaces
95
A+A C2 11P A. On the other hand, there are pseudoconvex sets which are not absolutely pconvex for any p. For example, the following open set on the plane
{(x,y): 0<x<2,0
Kr = {x:
llxll < r}
are absolutely pconvex (absolutely pconvex with p dependent on r).
3.2. LOCALLY BOUNDED SPACES Let X be an F*space. The space X is said to be locally bounded if it contains a bounded neighbourhood of 0. THEOREM 3.2.1 (Aoki, 1942; Rolewicz, 1957). Let X be a locally bounded 1, there is a phomogeneous Fnorm
space. Then, for a certain p, 0 < p II II equivalent to the original one.
<
Proof Let U be a balanced bounded neighbourhood of 0. The set U is pseudoconvex. Indeed, the set U U is bounded and, on the other hand, the set U is open. Therefore there is a number a such that
+
96
Chapter 3
U+U C aU. Hence U is pseudoconvex. Since U is a bounded set, the system of sets 
.
1
m
U constitutes a basis
of neighbourhoods of 0. Basing ourselves on Theorem 3.1.4 for each n we can construct a Pnhomogeneous pseudonorm II lin· Observe that the Fpseudonorms II lin differ from one another by a constant coefficient. D Therefore II 11 1 is an Fnorm with the required properties. Let us remark that Pn is equal to
log c(U) log 2 . Let c(X) = inf {c(U):
U runs
over the open bounded balanced sets}. The number c(X) is log c(X) called the modulus of concavity of the space X. Let Po = log 2 · Looking at the proof of Theorem 3.2.1 we can see that this theorem may be formulated more precisely. Namely THEOREM 3.2.1' (Rolewicz, 1957). Let X be a locally bounded space. For each p, 0 < p < p 0 , there is a phomogeneous Fnorm II II equivalent to the original one. The strict inequality p < p 0 in Theorem 3.2.1' cannot be replaced by the conditional inequality p ~ p 0 , as follows from Proposition 3.4.8 given later. For locally convex spaces the construction of the norm implies THEOREM 3.2.2 (Kolmogorov, 1934). Let X be an F*space. If there is in X a bounded convex neighbourhood of 0, then there is a homogeneous norm II II equivalent to the original one. An F*space in which there is a bounded convex neighbourhood of 0 is said to be a normed space. A complete normed space is called a Banach space. Let (X, II ID be an F*space. Let II II be phomogeneous. Then the space X is locally bounded. Indeed, the unit ball K = {x:
llxll ~ 1}
Locally Pseudoconvex and Locally Bounded Spaces
97
is a bounded set, because if Xn e K and {tn} is a sequence of scalars tending to 0, then lltnXnll = ltniP llxnll+0. In the particular case where the norm II II is homogeneous we find the set K is convex (cf. Section 3.1). Those facts are inverse to Theorems 3.2.1 and 3.2.2. As follows from the definition, the spaces M(D,.E,J1), C(.Q), C(DID0), l:P(.f.J,.E,J1) for 1 < p < +oo and the Hilbert space are normed spaces. Since they are complete (see Section 1.5), they are Banach spaces. Examples of locally bounded spaces which are not Banach spaces will be given in the next section. 3.2.3. Let X, Y be two locally bounded spaces. If the space X is isomorphic to the space Y, then they have equal moduli of concavity, c(X) = c(Y). Proof Let T be a linear operator mapping X into Y. If A+ A C sA, then T(A)+ T(A) C sT(A). Therefore c(T(A)) ~ c(A). If T is an isomorphism, then T and T1 map open bounded sets onto open bounded PROPOSITION
sets. Thus c(Y)
= c(T(X)) ~ c(X) = c(T1(Y)) ~ c(Y).
D
Let (X, II ll).be a locally bounded space. Let II II be a phomogeneous norm. As a trivial consequence of Proposition 3.1.5 we find that a set A C X is bounded if and only if sup llxll <+oo. :tEA
3.2.4. Let (X, II II) be a locally bounded space. Let Y be a subspacf! of the space X. Then c(Y) ~ c(X). Proof Without loss of generality we may assume that the norm II II is phomogeneous. Let A be an open set in Y. Let PROPOSITION
B=
U {x eX:
IlEA
llxyll ~ t inf {llyzll: z e Y"'A}}
The set B is open and bounded. Moreover, A= Y ()B.
Chapter 3
98
Thus A+A = (Bn Y)+(Bn Y) C (B+B)n YC c(B) Bn Y
Hence c(A)
~
c(B) and c(Y)
~
c(X).
= c(B) A. 0
II II) and (Y, II IIY) be two complete locally bounded spaces. LetT be a continuous linear operator mapping X onto Y. Then c(Y) ~ c(X).
PRorosmoN 3.2.5. Let (X,
Proof Without loss of generality we can assume that II II is phomogeneous. Let A be an arbitrary open bounded starlike set in Y. Since T induces a onetoone operator T mapping X/ker T onto Y, by Theorems 2.3.2 and 2.1.1 RA =sup inf {llxll:
X E
T1(y)} <+oo.
lfEA
Let B = T1 (A)n{x:
llxll <
2RA}·
It is easy to see that the set B is an open bounded starlike set and T(B) =A. Then c(A) = c(TB)) ~ c(B). Therefore
c(Y) ~ c(X).
0
CoROLLARY 3.2.6. Let (X, II ID be a locally bounded space. Let Y be a subspace of the space X. Then
c(X/Y) ~ c(X). Proof Basing ourselves on Lemma 1.4.7 we can assume without loss of generality that X is complete. Thus we use Proposition 3.2.5. 0
As an immediate consequence of Propositions 3.2.3, 3.2.4 and Corollary 3.2.6 we obtain THEOREM
3.2.7. Let X andY be two locally bounded spaces.
If dimz X~ dimz Y, then c(X) ~ c(Y). IfdimzX = dimzY, then c(X) = c(Y).
99
Locally Pseudoconvex and Locally Bounded Spaces
THEOREM 3.2.8. Let X and Y be two complete locally bounded spaces. If codimzX::::;;; codimzY, then c(X)::::;;; c(Y). If codimzX = codimzY, then c(X)
= c(Y).
THEOREM 3.2.9 (Kalton, 1977). There is a separable locally bounded complete space (X, II ID with an ahomogeneous norm II II which is universal for all separable locally bounded spaces with ahomogeneous norms. Proof: Let ']>be the set of all ahomogeneous norms defined on c:J(N)
(for the notation see Section 2.4). The set ']J satisfies conditions (2.4.3.i) and (2.4.3.ii). Condition (2.4.3.i) is obvious. For the verification of (2.4.3.ii) let us observe that putting 3 = d(J11(p),q) we have eBp(x)::::;;; q(x)::::;;; eBp(x)
for x ERn. Let q(x)
= inf {q(y)+eBp(xy):
y ERn}.
For x ERn we have q(x) = q(x) and for all x
E
c:J(N) we have
eBp(x)::::;;; q(x)::::;;; eBp(x).
Hence (2.4.3.ii) holds. Let {Xn} be a sequence of linearly independent elements such that lin{ Xn} is dense in X. Then lin{ Xn} is isomorphic to (c:J(N),p) for a certain p. Now we coU:struct the space (c:J(E), II ID as in Section 2.4. For each a E E we denote by ea an element of c:J(E) such that ea(x)
= {~
for x =a, for x =1= a.
Choose a: N~E to satisfy Proposition 2.4.1 and write Wn = ea(n> Then, by Proposition 2.4.4, for any e > 0 there is an increasing sequence of indices (nk) such that for any m m
m
m
i=l
i=l
i=l
(1e)p{~ tcxc) ::::;;;!l.r tcwn,//::::;;; (1+e)p ~ tcx,). Observe that the theorem.
II II
(3.2.1)
is ahomogeneous. Since lin {x11 } = X (3.2.1) implies D
Chapter 3
100
Using a different method Banach and Mazur (see Banach, 1932) showed that the space C[O, 1] is universal for all separable Banach spaces, hence also for !homogeneous Fspaces. From Theorem 3.2.9 follows the existence of universal spaces for separable locally pconvex (and locally pseudoconvex) spaces. To show this fact we introduce the following notion. Let (Xt, II lit) be a sequence of F*spaces. By (Xt)c,> we denote the space of all sequences x = {xt}, Xt eXt. The topology in (Xt)c,> is defined by Fpseudonorms n
llxll~ = };llxtllt. i=l
In the case where all Xt are identical, Xt = X we denote briefly (Xt)c,> by (X)(s)• It is easy to verify the following facts. (Xt)cs> is an F*space. If all spaces (Xt, II lit) are complete, then the space (Xt)<•> is also complete. If all spaces (Xt, II lit) are locally pconvex (locally pseudoconvex), then (Xt)<•> is also locally pconvex (locally pseudoconvex). 3.2.10. There is a separable locally pconvex space, which is universal for all separable locally pconvex spaces. Proof Let (XP, II ID be a separable locally bounded space with a phomogeneous norm universal for all separable locally bounded spaces with phomogeneous norms. Let Xt = XP, i = 1,2, ... The space (XP)<•> is a separable locally pconvex space. We shall show that it is universal for all separable locally pconvex spaces. Indeed, let X be an arbitrary separable locally pconvex space. By Theorem 3.1.4 there is a sequence of phomogeneous pseudonorms {II lin} determining a topology equivalent to the original one. Let THEOREM
Xn,o = {x: llxlln = 0} and let Xn = X/Xn, 0 be the quotient space. The pseudonorm llxlln induces a phomogeneous norm in Xn. Since no misunderstanding can arise, we shall denote this induced norm also by II lin· The space (Xn, II lin) is a locally bouded space with a phomogeneous norm II lin· Let
Locally Pseudoconvex and Locally Bounded Spaces
101
The space X is locally pconvex. The space X is isomorphic to a subspace of the space X consisiting of elements {[x]n}, where [x]n is the coset in Xn induced by x. On the other hand by the universality of XP, X is isoD morphic to a subspace of the space (XP)<•>· Basing themselves on the conclusion of Banach and Mazur 0933), that C[O, I] is universal for all separable Banach spaces, Mazur and Orlicz (1948) showed that C(oo,+oo) is universal for all separable B0spaces. THEOREM 3.2.11. There is a separable locally pseudoconvex space which is universal for all separable locally pseudoconvex spaces. Proof Let (XP, I ID denote a separable locally bounded space with aphomogeneous norm, being universal for all separable locally bounded spaces with phomogeneous norms. Let {Pt} be a sequence tending to 0, for example Pi = 1/i. The space (XP')<•> is the required universal space. It is easily seen to be a separable locally pseudoconvex space. Let X be an arbitrary pseudoconvex space. Without loss of generality we may assume that the topology in X is determined by a sequence of Pthomogeneous pseudonorms {II lit}. In the same way as in the proof of Theorem 3.2.10 we define Xt and X. The space X is locally pseudoconvex. The rest of the . proof is the same. D A
A
THEOREM 3.3.12 (Shapiro, 1969; Stiles, 1970; for Banach spaces see Banach, 1932). Every separable locally bounded space X with a phomogeneous norm II II is an image of lP by a continuous linear operator A. Proof The proof proceeds in the same way as the proof of Proposition 2.5.3. It is only necessary to observe that in this case the space Nn(/) is just the space lP. D
THEOREM 3.2.13. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be Pxhomogeneous and let II IIY be pyhomogeneous. A linear operator A from X into Y is continuous if and only if
IIAII =
sup
IIXIX.;;l
IIA(x)IIY
<+oo.
(3.2.2)
Chapter 3
102
Proof The theorem is a trivial consequence of the fact that a set E is bounded if and only if sup {llxiiY: x E E} < +oo. D
If px = py then the definition of IIAII implies IIA(x)IIY:::::;; IIAII llxllx. In this case the number IIAII is called the norm of the operator A. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be pxhomogeneous and let II IIY be pyhomogeneous. Then we can always introduce phomogeneous norms, p:::::;; min(px, py), in X and Y equivalent to the original norms. Indeed, the norms
11x11:r = (llxllx)PIPx,
IIYII~ = (IIYIIY)P!Py
have the required properties. We shall now show that the norm IIAII is a pyhomogeneous Fnorm in the space B(X+Y). It is obvious that IIAII = 0 if and only if A= 0. Let A,BE B(X+Y). Then IIA+BII = sup IIA(x)+B(x)IIY ::( sup IIA(x)IIY+ sup IIB(x)liY 11x11X:;;;1
11x11X:;;;1
11x11X:;;;1
= IIAII+IIBII· The original topology in B(X+ Y) is equivalent to the topology induced by the norm IIAII. Indeed, for arbitrary e > 0, {A: IIAII
< s} J
U(O,B,s) = {A: IIA(x)IIY
< e for
x
E
B},
where the set B = {x: llxllx:::::;; 1} is bounded. On the other hand for an arbitrary positive and an arbitrary bounded set B C X, U(O,B,s) J U(O,Kr,e) ={A: IIAII
<_ _!_I_\. rPy Px (
where r = sup{llxllx: x E B} and Kr = {x: llxllx ::( r}. If the spaces X and Yare complete, then by Corollary 2.2.4 the space B(X+ Y) is complete, i.e. it is an Fspace. In this particular case the conjugate space X is always a Banach space. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be pxhomogeneous and let II IIY be pyhomogeneous. From the defini
Locally Pseudoconvex and Locally Bounded Spaces
tion or'the norm IIAII it follows that a family equicontinuous if and only if sup{IIAII: A
Em:}
103
m: of linear operators is
<+oo.
Then Theorem 2.6.1 implies THEOREM 3.2.14. Let Z be a complete locally bounded space with a phomogeneous norm II II and with a basis {en}. Then K = sup n
liPnil
< +oo,
where, as before, ""
Pn
n
(,2 ttet) =}; ttet. i=l
i=l
The number K is called the norm of the basis {en}. As a consequence of Corollary 2.6.5. we obtain THEOREM 3.2.15. Let (X, II ID be a locally bounded space. Let II II be phomogeneous. A sequence {en} of linearly independent elements is a basis in X if and only if the following two conditions are satisfied: (1) the linear combinations of {en} are dense in X, (2) there exists a constant K > 0 such that n
n+m
[[}; ttetj[
~K[[2ttet[[
i=l
i=l
for all positive integers m and n. THEOREM 3.2.16 (Krein, Milman and Rutman, 1940). Let (X, II II) be a complete locally bounded space. Let II II be phomogeneous. Let {e11 } be a basis in X of the norm K. Let llenll = 1, n = 1,2, ... If, for a sequence
{gn} of elements of X, <X>
C
=
L llgnenll <
n=l
2K,
(3.2.3)
Chapter 3
104
then {gn} is a basis in the space X 0 generated by {gn}, X 0 = alent to the basis {en}. Proof. The triangle inequality implies n
n
i=l
i=l
lin{gn}, equiv·
n
112 tcecJJ}; Jt,JPJJgcecJI ~ IJ}; tcgcl[ i=l n
n
~II}; t1etll +}; JtcJPIIgtecJI. i=l
i=l
From the definition of the norm of the basis it follows that : j
ltiiP =
n
j1
llt1e1ll~ l!l' ttecll + 1\.L ttet// ~2K ll};ttetJI. i=l
i=l
i=l
Therefore n
n
_l
1
Jt,JPIIgtetll
~ m~x Jt,JP}; llgtetll I.,;~.;;n
i=l
i=l n
~ 2KC 112: ttetll i=l
Hence (3.2.3) implies n
o!5)
n
n
112: t,e,ll ~ 112: t,gi II~
where b = 2KC < 1. Thus, the elements basis {en}.
i=l
{gn} consitute a
i=l
basic sequence equivalent to the D
PRQPOSITION 3.2.17. Let (X, I II) be a complete locally bounded space. Let I II be phomogeneous. Let {en} be a basis in X. Suppose that X is infinitedimensional and that X 0 is an infinitedimensional subspace of X. Then there is in X 0 a sequence of elements {xn}, llxnll = I, Xn 00
=
~ ~
t n,1..e.t such that
i=l
lim tn,i = 0
(i
=
1,2, ... ).
Proof. Suppose that the conclusion does not hold. Then there are a
Locally Pseudoconvex and Locally Bounded Spaces
positive integer k and a positive e such that, for each x J
co
x =
2
E
105
X0,
llxl/ = 1,
t~,e,
i=l
llxll =max lttl >e. lo;;;io;;;k
Hence there exists a linear homeomorphism of X 0 and kdimensiona space. This contradicts the fact that X 0 is infinitedimensional. D By Proposition 2.6.7, Theorem 3.2.16, Proposition 3.2.17 we obtain PROPOSITION 3.2.18. Let (X, II II) be an infinite dimensional complete locally bounded space with a basis {en}. Let X 0 be an infinitedimensional subspace of the space X. Then X 0 contains an infinitedimensional subspace Y with a basis {in}, which is equivalent to a block basis in X. PROPOSITION 3.2.19 (Stiles, 1971). There is a locally bounded space (X, II II) such that there is no phomogeneous norm in X determining a topology equivalent to the original one, but every infinitedimensional subspace Y contains an infinitedimensional subspace Y 0 in which that topology can be determined by a phomogeneous norm. Proof Let qn
= ~ ( 1  log :ogn ) '
n
= 30, 31, ...
Let Nn(u) = uqn. Let X 0 be an infinitedimensional subspace Qf the space Nn(l). Then by Proposition 3.2.18 there is an infinitedimensional subspace Y of X 0 such that there is in Y a basis {/11 } equivalent to a block basis 1/e~l/
= 1.
Without loss of generality we may assume that p 11 > ee 2n (see the remark after Proposition 2.6.7). Then, for It I < 1
ltiP ~ 1/te~l/ ~ ltl 11 (1 
2: ).
Chapter 3
106
Since sup ltP(l 2';.)tPI:::;;; 21n,
o.;;;t.;;;I
by Theorem 3.2.16 {e~} and the standard basis {en} in/Pare equivalent. On the other hand the space Nn(/) is not isomorphic to /P. What is more, there is no phomogeneous norm II II in Nn(l). Indeed, suppose that such a norm II II exists. It is easy to verify that that the sequence en= {0, ... , 0, 1,0, ... }

n th place
is bounded. Hence supllenll
<
n
Let Xn
= (nl: 2n)I/p. Then g
+oo.
.i; lxniP
n=30
00
Therefore the series }; Xnen is convergent in II II· This leads to a contran=so diction since
n~lxniPn
=go 00
~
(n lolg2 1
n) ~ P ( 1 log n
2 loglogn =L.Jee n=30
and
log
~og n)
21og logn
__ +oo
n
{xn} ¢ Nn(l).
D
Pelczynski (see Rolewicz, 1957) has shown that in the space Nn(l) given above for each q with p < q :::;;; 1 there is a qhomogeneous norm determining the original topology. This has been the first example showing that the inequality p > p 0 in Theorem 3.2.1' is strict. Let X be a complete locally pseudoconvex space. As follows from Theorem 3.1.4, there is a sequence of Pnhomogeneous pseudonorms {II lin} determining the original topology. Without loss of generality we can assume that [[xllt :::;;; [[xll2 :::;;; llxlls :::;;; ... Suppose that {en} is a basis in X. Let
[[x[[~ =
r
sup II~ r
m=l
tmemllt,
Locally Pseudoconvex and Locally Bounded Spaces
107
where
By Theorem 2.6.1 the system of pseudonorms original topology. Let Xt,o = {x: 1/x/1; = 0}. Let
{II //;} determines the
Xt = X/x,,,
be the quotient space. The pseudonorm norm 1/ /It on Xt in the following way
1/ //~
induces a Pihomogeneous
1/[x]i/1~' = 1/x/1;,
where {x}t denotes the coset containing x. By the definition of 1/ 11;' we obtain l/t1[e1]i+ ... +tn[en]i //;' = l/t1e1+ ... +tnen/1; ~ l/t1e1+ ... +tn+men+m/1; = l/t1[e1]t+ ... +tn+m[en+m]/1~'
for all positive integers nand m. Therefore, by Theorem 3.2.15, we obtain: 3.2.20. The cosets [en]i consitute a basis of norm one in the space Xt, wfzich is the completion of the space Xi, provided we omit all those [en]i which are equal to 0.
PROPOSITION
For locally convex spaces, Proposition 3.2.20 was proved by Bessaga and Pelczynski (1957).
3.3.
BOUNDED SETS IN SPACES
N(L(fJ, 1:, tt))
Let a space N(L(fJ,l:,J1)) be given. Let
n(t) = inf {a> 0: N(at) Of course n(t)
~ N~t) }·
> 0 for all t, 0 < t <+oo.
(3.3.1)
Chapter 3
108
THEOREM 3.3.1 (Rolewicz, 1959). inf
r=
If
n(t)>O,
O
then the space N(L(D,E,J1)) is locally bounded. Proof From (3.3.1) it follows that PN(rx) ~ PN(X) for each 0 < r
t
<
r.
Then by induction PN(rnx)
~ ;n PN(x),
n
= 1,2, ...
Therefore, the open set Ka = {x: PN(X)
< d} is bounded.
0
We can weaken the hypothesis on N(t) by assuming additional properties on the measure fl. Namely THEOREM 3.3.2 (Rolewicz, 1959). Let 11 be a finite measure. If liminfn(t) > 0, t.oo
then the space N(L(D,E,fl)) is locally bounded. Proof Let for t > 1, · {N(t) N'(t) = tN(1) for 0 ~ t ~ 1.
Since the measure 11 is finite, PN(Xn)*0 if and only if PN'(xn)70. Let n'(t)=inf{{a>O:
N'(at)~
Nit)}·
It is easy to verify that infn'(t) > 0. Therefore, Theorem 3.3.1 implies
Theorem 3.3.2.
0
THEOREM 3.3.3 (Rolewicz, 1959). Let the measure 11 be purely atomic and let (3.3.2) r = inf fl(Pl)> 0, where p, runs over all atoms. If
liminfn(t) > 0, t+0
then the space N(L(D,E,/1)) is locally bounded.
Locally Pseudoconvex and Locally Bounded Spaces
·109
Proof. Let
for 0:::;:;; t:::;:;; 1,
N(t) { N'(t) = tN(1)
for t
> 1.
By (3.3.2), PN(Xn)+0 if and only if PN• (xn)+0. The rest of the proof is the same as the proof of Theorem 3.3.2. 0 The following fact is, in a sense, inverse to Theorem 3.3.2. THEOREM
3.3.4.
If the
measure Jl is not purely atomic
and
lim inf n(t) = 0,
(3.3.3)
L•
then the space N(L(Q, fl)) is not locally bounded. Proof By (3.3.3) there is a sequence of numbers {tn}+oo such that an= n(tn)+0. Let e be an arbitrary positive number. The measure fl is not purely atomic, hence for sufficiently small e there are sets e A 71 , n = 1,2, ... such that JL(An) = 2N(tn). Let Xn = tnXA.• where, as
usual, XA. denotes the characteristic function of the set An. It is easy to e verify that PN(Xn) = 2. On the other hand, · PN(anxn) =
J
N(n(tn)tn)dJl =
An
1
2
J
N(tn)dJl =
e
4·
An
Hence {anxn} does not tend to 0. Therefore no set K = {x: PN(x)
< e}
is bounded. Since e is an arbitrary positive sufficiently small number, the space N(L(D,E,JL)) is not locally bounded. D 3.3.5. If the measure is not purely atomic and the function N(u) is bounded, then the space N(L(D,E,JL)) is not locally bounded. Proof Since N(u) is bounded, (3.3.3) holds. 0
CoROLLARY
3.3.6. If there is a positive number k such that, for a sufficiently small r, there is a set Ar such that
THEOREM
110
Chapter 3
<
k
rJl(Ar)
< 2,
and if
lim inf n(t) = 0
(3.3.5)
t+oo
then the space N(L(Q,l:,Jl)) is not locally bounded. Proof By (3.3.4), there is a sequence tn+0 such that an= n(tn)+0. Let
be an arbitrary positive number. The assumption on the measure implies that for sufficiently large n there are An such that ks < N(tn)Jl(An)
The following theorem gives us the connection between local pseudoconvexity and local boundness in the case of spaces N(L(Q,l:,Jl)). 3.3.7. Let the measure Jl be finite and not purely atomic. Then the space N(L(Q,l:,Jl)) is locally bounded provided it is locally pseudoconvex. Proof Suppose that the space N(L(Q,l:,Jl)) is not locally bounded. Then, by Theorem 3.3.2. liminf n(t) = 0, i.e. thereis a sequence {tn}+oo THEOREM
t+co
such that an = n(tn)+0. Let A be an arbitrary open set such that r
= SUPPN(X) <+oo. XEA
We shall show that the set A has infinite modulus of concavity c(A)
= +oo. Let p = inf PN(x). Since the set A is open, p
> 0. Let k be an integer
XEA
k
4r
< p . The measure
Jl is not purely atomic, hence for a sufficiently,
large n, we can find sets An,i, i = 1,2, ... , k such that
p
Jl(An, t)
= 2N(tn)
and the sets A 11 ,t and An,J are disjoint for i =f=j, provided lim N(t) = +oo. t+oo
111
Locally Pseudoconvex and Locally Bounded Spaces
Let Xn, ( = lnX.An,l" Then PN(Xn,t) = On the other hand,pN(Xn, 1 + Therefore
p
2 < p,
... + Xn,k) =
whence Xn,(
kp/2
E
A.
> 2r.
+ ...
PN(n(tn) (xn, 1 +xn,k))> r and n(tn (xn, 1 + ... +xn,k) ¢A.
Since n(t?i)+0, we do not have A+A+ ... +ACKA
for any K> 0.
This implies c(A) = +oo. Now we shall consider the case where R = lim N(t) < +=. t+co
Let !21 be a subset of Q on which the measure ll is nonatomic. Let A be an arbitrary open set in N(L(Q, E, It)) such that r = suppN(x) ze.A
<
Let p = inf PN(x). Since the measure ll is nonatomic
N(1)~t(!2 1).
a:'A
on the set Ql> then there are disjoint subsets At> ... , Ak of !21 such that ~t(At) < p/R (i = 1,2, ... , k) and A 1 u ... u Ak = !21 • Let n = 1,2, ... , i = 1,2, ... ,k. Xn,i = nx.A., Then PN(xn) < p and xn,i eA. On the other hand, Xn,I+ ... +xn,k = nxa,
and PN(
~
(xn,l+ ... +xn,k))
>
r.
The rest of the proof is the same as in the case where lim N(t) = +oo.
D
t+co
3.4.
CALCULATIONS OF THE MODULUS OF CONCAVITY OF SPACES
N(L(Q,
~.
p,))
We recall that a function N(u) is called convex if, for arbitrary nonnegative numbers a, b such that a+b = 1 and arbitrary arguments ul> u2 , we have
Chapter 3
112
3.4.1. Let N(u) = N 0(uP), where N 0 is a convex function and 0
THEOREM
PN(ax+by)
JN(!ax(t)+by(t)!)dp ::s;; JN(!a! !x(t)! +!hi [y(t)!)dp u = JN ((!a!!x(t)!+!h!!y(t)!)P)dp n < JN (!a!P!x(t) !P+ [bjP[y(t)fP)dp u ::s;; la!P JN (jx(t)jP)dp+ lb!P JN ([y(t)jP)dp u u =
u
0
0
0
= jajP
J
0
J
N([x(t)!) dp+ jbjP N(!y(t)!) dp u u = ja[PpN(x)+jbjPpN(Y) <e.
The functionN0(u) is convex, whence if N 0(au) ~ }N0(u), then a~ 1/2. Therefore if N(au) ~ }N(u), then a~ (}) 1/P. Thus n(u) ~ (}) 1/P and the sets K. are bounded (cf. the proof of Theorem 3.3.1). Hence the sets K. are bounded and absolutely pconvex. This implies (see Theorem 3.2.1' and Theorem 3.1.4) that there is a phomogeneous norm equivalent to the original one. 0 Suppose we are given two continuous positive nondecreasing functions M(u) and N(u) defined on the interval (O,+oo). The functions M(u) and N(u) are said to be equivalent if there are two positive constants A, B such that A
::s;;
M(u) N(u)
::s;; B
(3.4.1)
for all u. We say that M(u), N(u) are equivalent at infinity if there are a,A,B > 0 such that (3.4.1) holds for u >a. We say that M(u), N(u) are equivalent at 0 if there are b,A,B > 0 such that (3.4.1) holds for 0 < u
Locally Pseudoconvex and Locally Bounded Spaces
113
By simple calculation we obtain PROPOSITION 3.4.2. Suppose we are given two continuous positive nondecreasing/unctions M(u) and N(u) defined on the interval (O,+oo). If one of the following three conditions holds : the functions M(u) and N(u) are equivalent, (3.4.2.i) the functions M(u) and N(u) are equivalent at (3.4.2.ii) infinity and the measure f.l is finite, the functions M(u) and N(u) are equivalent at 0 and the measure f.l is purely atomic and such that (3.4.2.iii) inf{f.l(A): A being atoms} > 0, then the spaces M(L(Q,E,f.l)) and N(L(Q,E,f.l)) are identical as sets of functions and, moreover, PM(xn)*0 if and only if PN(Xn)*0, i.e. the topologies in this spaces are equivalent. Proposition 3.4.2 and Theorem 3.4.1 imply THEOREM 3.4.3 (Rolewicz, 1959). Let M(u)
=
0
M 0(uP),
1,
where M 0 is a convex function. Let N(u) be a function such that one of conditions (3.4.2.i), (3.4.2.ii), (3.4.2.iii) holds. Then there is a phomogeneous norm in N(L(Q,E,f.l,)) equivalent to the original one.
The followitrg two theorems are, in a sense, inverse to Theorem 3.4.3. THEOREM 3.4.4. Suppose that the measure f.l is not purely atomic. lim inf N(u) uP
=
0,
If (3.4.3)
u...... oo
then the space N(L(Q,E,f.l,)) is not locally pconvex. Proof By Corollary 3.3.5 we can assume that lim N(u)
=
+oo. By
U>00
Un
(3.4.3) there is a sequence {un}*oo such that N(un)IJp *OO. Let kn be the greatest integer such that kn ~ N(un), kn = [N(un)]. Since the measure f.l is not purely atomic for sufficiently small s > 0 there are disjoint s sets An,i, i = 1, 2, ... , kn, such that J.l(An,t) = 1 +kn ·
Chapter 3
114
Let Xn,t =
UnXAn,t•
Then
<
PN(Xn,i)
8.
On the other hand, 1 N ( kn1fp
4~ Xn,i ) =
N
( Un ) k,,_!IP
8 kn 1+kn +oo.
t=1
The arbitrariness of 8 implies the theorem.
0
THEOREM 3.4.5. Let there be a positive number rand ani nfinite family o disjoint sets Aa such that
1
< Jl(Aa) <
(3.4.4)
r
If lim sup N(u)
+=
=
(3.4.5)
uP
U>oo
then the space N(L(Q,E,Jl)) is not locally pconvex. Proof Let 8 be an arbitrary positive number. By (3.4.4) there is a b > 0, such that for each u, 0 < u < b, there is an infinite family of disjoin sets {An,u} such that 8
rN(u)
<
(An,u)
<
8
N(u)'
Let Xn,u
=
UXAn,u•
Then PN(Xn,u)
<
8.
On the other hand, PN (
x1,u+ ... +xk,u) piP
> ~
(_!!_) _ ~ rN(u)
"" rN(u) N k 11P 
N(k 11Pu) (k 1 /Pu)P
and by (3.4.5) . sup sup I 1m
8k ) N ( k u ) N( 11 u P . cuP N(k 1/P u) = hm sup sup N( ) (k_ 11 ) ++oo.
k~oo
k>oo
O
O
u
PuP
The arbitrariness of 8 implies the theorem.
0
Locally Pseudoconvex and Locally Bounded Spaces
115
3.4.6. If the measure fl satisifes either the assumption of Theorem 3.4.4. or the assumption of Theorem 3.4.5, then in the space LP(Q,£,fl), 0 < p:::;;; 1, there is a phomogeneous norm determining a topology equivalent to the original one, and there is no qhomogeneous (q > p) norm with this property. In other words, c(LP(Q,£,p)) = 211P and there is a starlike bounded open set A C LP(Q,£,p) such that c(A) = 2 1/P. CoROLLARY
From Corollary 3.4.6 follows THEOREM 3.4.7. There is no locally bounded space universal (couniversal) for all separable locally bounded spaces. Proof Suppose that such a universal (couniversal) space X exists. Then, by definition,
dimz/P:::;;; dimzX, (codimz/P:::;;; codimzX). Hence, by Theorem 3.2.7 (Theorem 3.2.8) 21 /P = c(lP) :::;;; c(X) Thus c(X) bounded.
=
+=
for all p, 0 < p:::;;; 1.
and this contradicts to the fact that X is locally
C
Another consequence of the properties of spaces N(L(Q,I,p)) is PROPOSITION 3.4.8. There is a locally bounded space X such that the topology in X can be determined by a phomogeneous norm for p, 0
21h•. Proof Let h(u) be a positive decreasing continuous convex function defined on the interval [O,+oo) such that h(O)
Let
N(u) = uPoh(u). Let J1 be a finite atomless measure. By Theorem 3.4.3 for any p, 0
Chapter 3
116
determining a topology equivalent to the original one. Theorem 3.4.4 implies that there is no p 0 homogeneous norm with this property. D For locally convex spaces Theorems 3.4.4 and 3.4.5 can be formulated in a stronger way. THEOREM 3.4.9 (Mazur and Orlicz, 1958). Let N(L(Q,E,fl)) be a locally convex space. If the measure f1 is not purely atomic, then the function N(u) is equivalent to a convex function at infinity. Proof. Since the space N(L(Q,E,fl)) is locally convex, there is a positive number e such that PN(Xk) < e, k = 1, ... ,n, implies PN (
x 1 + ... +xn) < I· n
Let q, p, n be positive integers. Let p :::( n. Let t be a positive real number. Since the measure f1 is not purely atomic, then, for sufficiently large q, there are disjoint sets El> ... , En such that fl(Et) = 1/nq, i = 1, ... , n. The smallest such q will be denoted by q0 • Let for s E EJ, wherej = k, k+l, ... , kl+p(mod n), otherwise. Then 1 PN(Xk) = pN(t) nq
and PN
(
X1
+ ... + Xn) _1 N (p t ). n
Hence I p   N(t) :::( e q n
implies
q
n
Locally Pseudoconvex and Locally Bounded Spaces
117
The continuity of the function N(u) and the arbitrariness of p and n imply that co  N(t) ::'( e (0
q
Let N(t) :;::: eq0 and let q be the greatest integer less than coN(t)fe. From the definition of q follows sq < coN(t) ::'( e(q+ 1). Then 1 1 q+lcoN(t)"(s, whenceq+IN(cot)::'(l and N(cot)"((q+l)
2
q+l
= ·eq ::'( N(t). Putting C = 2/s, we obtain eq e N(cot) ::'( CcoN(t)
Recall that, by Theorem 3.4.4, lim infN(t) t
> 0.
1+<Xl
Choose{) > 0 and T > 0 such that N(t)/t ;;?.: {) for t > T and Tb Then for cot > T, 0 < co < 1, N(t) coN(t). = cot1 ;;?.: Tb
>
>
eq0 •
eq 0 •
Since cot ;;?; Timplies t;;?; T, N(cot) ::'( CcoN(t)forcot Let (Ncot) supfor t > T, O<wo;;;I CO P(t) = wt>T
>
T, 0 "(co::'( 1.
1
for 0 ::'( t ::'( T.
N(t)
Let 0
< a ::'( 2, at > P(at) =
T. Then
sup
O<wo;;;l
N(coat) N(coat) =a sup co O<wo;;;l coa
wat~T
wat~T
N(st) ::'(a sup  0<~1
st>T
s
= a P(t).
(3.4.6)
Chapter 3
118
Formula (3.4.6) implies that the function P(t)ft is nondecreasing for t > T. Indeed, let T < t' < t"
t' ) t' ( p " t" I I PUU) t ::::::::: _t_:_ t' """ t'
P(t') t'
Let
l
P(t)
Q(t)
=
for t
>
for 0
< t~T
;(T)
tT2
P(t")
=ru··
T,
and u
M(u)
=
JQ(t)dt. 0
Since Q(t) is a nondecreasing function, the function M(u) is convex. We shall show that the functions M(u) and N(u) are equivalent at infinity. To begin with, we shall prove that P(t) and N(t) are equivalent at infinity. Indeed, P(t);;?: N(t) and for sufficiently large t, N(wt)/w ~ CN(t), hence P(t) ~ CN(t). Since P(t) is equivalent to N(t) at infinity and N(t) satisfies condition (A 2), P(t) also satisfies that condition, i.e. there is a positive constant K such that for sufficiently large t, P(2t) ~ KP(t). The function Q(t) is nondecreasing, whence~ for sufficiently large t, M(t)
< tQ(t) =
P(t)
and M(t);;?o
J'Q(s)ds;;?o ~ Q(~)=P(~). t/2
Hence, for sufficiently large t, P(t)
~ KP ( ~) ~ KM(t)
and the
functionsM(t) andP(t) are equivalent at infiriity. Therefore the functions M(u) and N(u) are equivalent at infinity. 0 THEOREM 3.4.10 (Mazur and Orlicz, 1958). Suppose that there is a constant K > 0 and an infinite family of disjoint sets {Et} such that 1 ~JJ.(E,)
~
K.
Locally Pseudoconvex and Locally Bounded Spaces
119'
If the space N(L(Q,E,p.)) is locally convex, then the function N(u) is equivalent to a convex function at 0. Proof. Since the space N(L(Q,E,p.)) is locally convex, there is a positive e such that PN(Xk) < e, k = 1,2, ... , n, implies
PN ( x 1 + ·~· +xn)
~
1.
Let p be an arbitrary positive integer and let
{~
xk(s) =
We get PN(Xk) and
~
for sEEk+tn (i=0,1, ... ,p1), otherwise. (k=1,2, ... ,n)
KpN(t)
__.. pn N (  t ) ~PN n
(x + ...n +Xn) · 1
Hence N(t)~ efKp implies N(t)/n ~ 1/pn. Let 0 < N(t) < efK, and let p be chosen so that efK(p+1) ~ N(t) < efKp. Then N(tfn) ~ 1/pn ~ 2Kfe N(t)fn. Given p > 0. Choose a q ~ 1 such that N(tfq) < e for it I < p. Condition (LlJ implies that there is a constant D~ such that N(qt) ~DqN(t)for Jti
N(!_).
1 )~ 2 KDq_!_N(t). N(qt )~DqN(qn qn e n
n
Let 0 < w ~ 1 and let n be such that 1/(n+ 1) 1/n < 2w, putting C = 4KDqfe we obtain N(wt)
~
CwN(t)
Let P (t)
N(wt)
= sup   , O
Q (t) =
P~t)' t
M(t)
W
= JQ(s)ds.
·•·
for 0
<w~
< w ~ 1/n. Since
1, It!
Chapter 3
120
In the same way as in Theorem 3.4.9 one may prove that M(u) is a convex function equivalent to the function N(u) at 0. D 3.5. INTEGRATIONS OF FUNCTIONS WITH VALUES IN FSPACES
Let X be an Fspace. Let x(t) be a function defined on an interval [a,b] with values in the space X. As in the classical definition of the Riemann integral, we define a normal sequence of subdivisions of the interval [a,b] as such a sequence of subdivisions a = t~ < t~
that
. 11m
~oo
< ... <
sup
o.,;;;;i.,;;;;k(n)1
tk(n)
= b,
n n 0. 1ti+ 1 ti 1 =
We say that a function x(t) is Riemann integrable if, for each normal sequence of subdivisions of the interval [a,b] and arbitrary sk, tJ:_ 1 < < s~ < t~, there exists a limit of the sum k(n)
~ x(s;)(t;tk1). k=1
As in the calculus, one may show that this limit does not depend on the sequences of subdivisions. It will be called the Riemann integral of the function x(t), and we shall write it as b
Jx(t)dt. a
By repeating the classical considerations this definition can easily be extended to functions of several variables and to line integrals. 1 It is easy to verify that the Riemann integral defined in the way described above has similar arithmetical properties to those it has in the calculus. Namely, if x(t) and y(t) are integrable functions (i.e. such that the Riemann integrals exist), then the sum x(t)+y(t) is integrable and b
b
b
J(x(t)+y(t)) dt = Jx(t) dt+ Jy(t) dt, a 1
a
a
For arbitrary Radon measure it was done by Klein and Rolewicz (1984).
(3.5.1)
Locally Pseudoconvex and Locally Bounded Spaces
121
if a function x(t) is integrable, then for any scalar C the function Cx(t) is integrable and b
JCx(t) dt
b
= C
a
Jx(t) dt;
(3.5.2)
a
if a function x(t) is integrable on an interval [a,b], then for each c, a< c < b, the function x(t) is integrable on intervals [a,c] and [c,b] and c
b
b
Jx(t) dt Jx(t)dt+ Jx(t)dt. =
a
a
c
3.5.1 (Mazur and Orlicz, 1948). An Fspace X is locally convex if and only if each continuous function x(t) defined on interval [0, 1] with values in X is integrable. Proof. Necessity. If the space X is locally convex, then the proof that each continuous fuction is integrable is exactly the same as in the calculus, but we replace the estimation of moduli by the estimation of the pseudonorms II Ilk determining the topology. Sufficiency. Suppose that the space X is not locally convex. Without loss of generality we may assume that the norm II II in the space X is nondecreasing (i.e. lltxll is a nondecreasing function for positive· t). Since X is not locally convex, there is a positive number a such that for arbitrary n there are elements x~, ... , x;n such that THEOREM
llxlli'
< ;n i= 1,2, ... ,kn)
and
llx + ·k~+xk"il >a. 1
Without loss of generality we can assume that kn+l is divisible by 2kn. Suppose that
xn(t) =
(i ~)
[xi
fort=
~0
for t = :n, i = 0, 1, ... , kn,
I. •.
;n, i = 1,2, ... , kn,
[j
. Is  kn1 , kn j . 1 2 2k] lis mear on t h e mterva 2 2 •.J  , , · · ·, n •
Obviously each of the functions Xn(t) is integrable since its values are in the finite dimensional subspace Xn spa~ned by elements x~, ... x;n.
Chapter 3
; 122
Therefore, there is a number ~n > 0 such that, for any division 0 = 1 < 11 < ...
11.2 [xn(st)Xn(s;)] (ttfi1)/1 < 3~ i=l
for all Stands; such that ft_ 1 < s1,s;
<
~t,
i
=
1,2,
... ,m1. a>
Let x(t)
= 2; Xnm(t). The function x(t) is continuous as a uniform m=l
limit of a sequence of continuous functions. We shall show that it is not Riemann integrable. Indeed, let It = i/kn.,, i = 0, 1, ... , kn"' and 2i1 let St = ft, s~ = 2kn., , i = 1,2, ... , knv· Then kn
li k~., ~ [x(st)x(s;)]ll
=II k~., ~ 6 =II k~., ~ ~ kn'P
knp
cc
[Xnm(St)Xnm(s;)]ll
p
[Xnm(st)Xnm(s;)]ll
k ...,
~II k~., ~ [xn.,(s,)xnv(s;)]ll
II k~.,6 ~[Xnm(St)Xnm(s;)]ll ~II x~+ ~~p+x~: 611 p
kn.,
p1
.. ll
p1
~a
~
2
m=l
a 3m·
a 2 ·
~:?'
1
k ...,
'
kn.,6Xnm(s,)Xnm(s;)ll
Locally Pseudoconvex and Locally Bounded Spaces
123
Thus the function x(t) is not integrable. 0 If the Fspace X is locally convex, we can also define by analogy to the Lebesgue integral, an integral called the BochnerLebesgue integral. Accordingly we aproximate a function x(t) by simple functions k..
Xn(t) =}; atXE., i=l
where Gi EX and Et are Lebesgue measurable sets of the interval [a,b]. The BochnerLebesgue integral of the simple function Xn(t) is defined in a natural way as b
len
JXn(t)dt = 2 at /Et/,
a
i=l
where /E/ denotes the Lebesgue measure of the set E. We assume now that, for every homogeneous continuous pseudonorm 1/x/1, the scalar function 1/x(t)/1 is integrable and that the sequence {xn(t)} tends to x(t) almost everywhere and 1/xn(t)/1 ~ 1/x(t)/1. Then we define b
b
Jx(t)dt = lim JXn(t)dt.
a
n~co
a
This definition is correct since one may show that this integral does not depend on the choice of the functions Xn(t). If the space X is not locally convex, we are not able to use this definition, because there are sequences of simple functions {xn(t)} uniformly b
J
tending to 0 such that the sequences of integrals{ Xn(t)dt} do not tend a
to 0. In fact, let Xn(t) =xi
for
i1
~
kn
t
i
<  (i= 1,2, ... ,kn), kn
where x~ and kn and a are defined as in the proof of Theorem 3.5.1. It is obvious that the sequence {xn(t)} tends uniformly to 0 and that
i!
j
Xn(t)dtll
=II
xi+
k~+xk,. ~a> 0. fl
a
Since, if a space X is not locally convex, there are continuous functions which are not integrable, it is reasonable to ask what classes of functions are integrabl• in what classes of spaces.
Chapter 3
124
An important class of functions is exemplified by the class of analytic functions. We say that a function x(t) with values in an Fspace X is analytic if, for any t 0 , there is a neighbourhood U of the point t0 such that, forte U, the function x(t) can be represented by a power series 00
x(t)
=;}; (tt )nxn, 0
(3.5.3)
ii,xn EX.
·n=O
If the space X is locally pseudoconvex, then the topology can be determined by a sequence of Pnhomogeneous pseudonorms II lin· This implies that if the series (3.5.3) is convergent at a point t 1 , then it is convergent for all t such that !tt0 !
hold. For example, if X= S[O, 1], then there is a power series
2 tnxn n=O
convergent only at a finite number of points (see Arnold, 1966; Zelazko, 1972). 3.5.2 (Gramsch, 1965; PrzeworskaRolewicz and Rolewicz 1966). Let X be a locally pseudoconvex Fspace. Then all analytic functions x(t) defined on an interval [a,b] with values in the space X are integrable. Proof. Since the space is locally pseudoconvex, the topology in X can be determined by a sequence of Pkhomogeneous pseudonorms II Ilk· Let x(t) be an analytic function. Then for each s 0 e [a,b] there is a neighTHEOREM
co
2 (ts )nxn is n=O convergent. This implies that there is a neighbourhood V! C U such that for t e V! bourhood U80 such that for t E U80 the series x(t) =
0
0
80
0
k = 1,2, ... ,
s1 ,
n = 0,1, ...
Since the interval [a,b] is compact, there is a finite number of points ••• , Sr such that
[a,b] C
,
U V!.
.
Let
.
V! =
.
(ai> bJ)· N,ow. we ~hall consider the Rie
j=l
mann integral of the func:ion x(t) on the interval [a1,bj]. Let e be an
Locally Pseudoconvex and Locally Bounded Spaces
arbitrary positive number. Let n0 be such an index that Mk/2no Let x(t) = Yn 0 (t)+zn 0 (t), ai ~ t ~ b,, where
...
Yno (t)
125
< e/2.
00
=}; (tSj)n Xn,
Zn0 (t)= }; (tSJ)nxn. n=n0 +1
n=O
The function Yno(t) is integrable, because it has values in the finitedimensional space spanned by elements x 0 , ••• , Xno· For any subdivision of the interval [aJ>bJ] a1
=
< '11 < ... <
t0
tp
=
b1
and arbitrary u 1, t,_ 1 ~ u 1 ~ t1, the Riemann sums of Zn0(t) can be estimated as follows: p
II}; Zn (uz) (tztzt) //k ~ 0
l=l
p
00
};
[/_l, (uzSJ)n Xn(tztzt Ilk
n=n0 +1 l=l 00
00
The arbitrariness of e and the integrability of Yn 0 (t) imply that the Riemann sums_ of x(t) converge for any normal sequence of subdivisions of the interval [aJ>bi] with respect to the pseudonorm II Ilk· Since there is only a finite number of the sets V8~, those sums converge on the whole interval [a,b]. This holds foreverypseudonorm II Ilk, whence the completeness of the space X implies that the function x(t) is integrable. 0 Let C be a complex plane. Let D be an open set in C. Let x(z) be a function defined on D with values in a locally pseudoconvex Fspace X. In the same way as for the function of the real argument, we say that x(z) is analytic in D, if for each z 0 ED, there is a neighbourhood Vz0 of the point z 0 such that the function x(z) can be represented in Vz0 as a sum of power series 00
x(z)
=}; (zz )nxn, 0
n=O
XnEX.
Chapter 3
126
As in the realargument case, we can define the Riemann integral of a function x(t) on a curve contained in D. In the same way as in the proof of Theorem 3.5.2 we can show that, if the curve Tis smooth and the function x(z) is analytic, then the integral x(z)dz exists. r Using the same arguments as in the classical theory of analytic functions, we can obtain
J
THEOREM 3.5.3. Let Then
r
be a smooth closed curve. Let x(z) be analytic.
Jx(z)dz = 0, _
(3.5.4)
J
r
x(z)  dZ
1
( 0)  XZ 21t
r
zz 0
Or each Zo inside the domain SUrrounded by
x Where Zo
(n) (
) _
Zo 
1
21t
J
x(z)
(zzo)n+l
d
r,
z,
(3.5.5) (3.5.6)
r and Fare as in (3.5.5).
As a consequence of (3.5.6) we obtain the Liouville theorem: THEOREM 3.5.4. Let x(z) be an analytic function defined on the whole complex plane C, with values in a locally pseudoconvex space X. If x(z) is bounded on the whole plane, then it is constant. Turpin and Waelbreock (1968, 1968b) generalized Theorem 3.5.2 to m dimensions and to a more general class of functions. We shall present their results here without proofs. Let U be an open domain in an mdimensional Euclidean space Rm. Let Nm denote the set of all vectors k = (k1 , ••• , km), where kt are nonnegative integers, (i = 1 ,2, ... , m). We adopt the folowing notation
Jxl = Jkl =
sup
l.,;t.;;m m
!xtl,
.L lktl' i=l
Locally Pseudoconvex and Locally Bounded Spaces
n
127
m
xk =
x~·.
i=l m
k! =
flkd, i=l
where x = (x1, .•. , Xm) E Rm, k = (kl> ... , km) E Nm. We say that a functionf(x) mapping U into an Fspace X is of class Cr(U,X), where r is a positive number, if, for any k E Nm such that lkl ~ r, there are continuous mappings Dk f of U into X and Pr,k f of Ux U into X such that D0 f=J, Pr,kf(x,x) = 0 and
Dd(y)=
~
.L.J
Dk+hf(x)•
lhl~rlkl
(yx)h 'h! +iyxirlklpr,kf(x,y) •
(see Turpin and Waelbroeck, 1968). THEOREM 3.5.5 (Turpin and Waelbroeck, 1968b). Let X be a locally pconvex space. Let U be an open set in Rm. If f(x) E C,(U,X) for r > m(lp)fp is a boundedfunction, then it is Riemann integrable. We say that a functionf(x) is of class C00 (U, R) if it is of class Cr(U, X) for all positive r .
. THEOREM 3.5.6 (Turpin and Waelbroeck, 1968b). If the space X is locally pseudoconvex, then each bounded function f(x) E Coo (U,X) is Riemann integrable. Indeed, Turpin and Waelbroeck obtained an even stronger result, namely that the integration is a continuous operator from the space C,( U, X) (Coo ( U, X)) with topology of uniform convergence on compact sets all Dkf, lkl ~r, andpr,kf(all Dkfand allpr,kf) into the space X.
3.6. VECTOR VALUED MEASURES Let (X, II II) be an Fspace. Let Q be a set, and let .E be a aalgebra of subsets of Q. We consider a function M(E) defined for E E .E with
128
Chapter 3
values in X such that, for any sequence of disjoint sets E1o ... , En, ... belonging to E M(E1 u
... uEnu ... ) = M(EJ+ ... +M(En)+ ...
The function M (E) is called a vector valued measure (briefly a measure). By the variation of the measure Mona set A C Q we shall mean the number M(A) =sup {liM(H)Ii: He A, HeE}.
THEOREM 3.6.1 (Drewnowski, 1972). The variation M of a vector valued measure has the following properties : M(e) = 0,
(3.6.1.i)
if A C B, then M(A) M(Q) <+oo,
~
(3.6.1.ii)
M(B),
(3.6.1.iii)
.M(UHn).;;;:;}; M(Hn) for any sequence{Hn} of sets m=l n=l belonging to E, (3.6.1.iv) M(H) = lim M(Hn),
if His either a union of an increasing
n+ro
sequence {Hn} C E of sets or sing sequence {Hn} of sets.
if it is an intersection of a decrea(3.6.1.v)
Proof Formulae (3.6.1.i) and (3.6.l.ii) trivially follows from the definition of the variation. (3.6.l.iii). Suppose that (3.6.1.iii) does not hold. Then there is a sequence of sets Hn E E such that IIM(Hn)li+oo. Without loss of generality we may assume that IIM(Hn)li
> 1+IIM(Hl)ll+ ··· + IIM(Hnt)li.
The sets Hn = Hn"'(H1 u
... UHnt)
A
are disjoint and IIM(Hn)li
ro
> 1. Thus the series };
n=l
M(Hn) is not con
vergent, and this leads to a contradiction. ro
(3.6.1.iv). Let A be an arbitrary set contained in
U Hn.
n=l
Let An
Locally Pseudoconvex and Locally Bounded Spaces
129 00
= (A" (H1 u
... u HnJ) () Hn The sets
An are disjoint and A =
U An. n1
Then 00
jjM(A)IJ
00
00
II}; M(An) II~}; IJM(A,.)IJ ~}; M(Hn).
=
n=l
n=l
n=l
00
The set A is an arbitrary subset of the set
U Hn.
n=l
Thus 00
M{H1 UH2u ... ) = sup{IJM(A)IJ: AC
U Hn} n=l
00
~}; M(Hn). n=l
(3.6.1.v). Let Hn E l: be an increasing sequence of sets. Thus Hn C H and by (3.6.l.ii) M(Hn) ~ M(H), n = 1,2, ... This implies that lim M(Hn) ~ M(H). Let e be an arbitrary positive number. Let A C H, A e 1: be such that M(H)e
<
IJM(A)IJ ~ M(H).
As in the classical measure theory we can prove lim M(A()Hn) = M(A). Hence for sufficiently large n M(H)2 e
<
.
IJM(A()Hn)IJ ~ M(Hn).
The arbitrariness of e implies lim M(Hn) ~ M(H). Now we shall consider the case where Hn is a decreasing sequence
n H,.. Since Hn c H, n = 1,2, ... , we have 00
of sets, Hn
E
1:. Let H
=
n1
by {3.6.1.ii) M(H) ~ M(Hn)· Thus lim M(Hn) ~ M(H) n+00
Chapter 3
130
Observe that, if An C Hn ""Hn1 , then the series
2"' M(An) converges tl=l
uniformly with respect to the choice of An. Then for an arbitrary positive· e there is an index n0 such that
II
i
n=n,+l
M(~n)ll <e.
Let A C Hn 0 , A E E, be such a set that [[M(A)/1 C()
A
Let A =A""(
U An (Hn ""HnJ). n=n +1
> M(Hn 0 )e. A
A
Observe that A C H and [[M(A)
0
M(A)fl <e. Hence M(A) pletes proof.
~
M(H)2e. The arbitrariness of e comD
A measure M is called bounded if the set { M(E): E E E} is bounded. As a trivial consequence of (3.6.l.iii), we obtain 3.6.2. Let X be a locally bounded space. Then each measure with values in X is bounded. Proof Without loss of generality we may assume that the norm in the space X is phomogeneous. Thus, by (3.6.l.iii), the set CoROLLARY
{M(E): EEE}
is bounded.
D
CoROLLARY 3.6.3. Let X be a locally pseudoconvex space. Then every measure M with values in X is bounded. Proof The topology in X can be determined by a sequence II lin of Pn·homogeneous pseudonorms. Let X~= {x: ffxlln = 0}. Let Xn be the quotient space Xn =X/X~. The pseudonorm II lin induces a Pn·homogeneous norm in Xn, and the measure M induces the measure Mn with values in the completion of Xn,Xn. Thus by (3.6.l.iii)
sup{f[M(E)[[n: EEE} <+oo, Therefore the measure M is bounded.
n = 1,2, ... D
Locally Pseudoconvex and Locally Bounded Spaces
131•
Recently Talagrand has proved that each measure with values in the space L 0(Q,E,p,) is bounded. However, there are an Fspace Hand an unbounded measure M with values in H, as follows from Example 3.6.4 (Turpin, 1975c) Let Q = [0, 1], let E be the algebra of Lebesgue measurable sets, and let .ll denote the Lebesgue measure. For fe L 0 [0, 1] we write
a(f) = .ll({t ED: f (t) # 0}, and Var {f)= inf {Var (/1): / 1 =/almost everywhere}. Let H be the space of all measurable functions f such that for each >0 there are g,he£0 [0,1] such that f=g+h and Var(g) <+oo and a(h) < 8. Let 11/11 = inf {lfgffco+Var(g)+a(h): f= g+h}, where [[g[[co = ess sup [g(t)[. 8
It is easy to verify that for fi,f2 E H we have
flii+/211::::;;; 11/111+11/211, lis/ II ::::;;;·11/11 for [sf ::::;;; 1, [[sfll+0 if s+0 11/11 = 0 if and only if/is equal to 0 almost everywhere. Thus II/II is an Fnorm on H. It is not difficult to prove that the space H is complete, but it is not essential in the construction of the example, ~ince in the opposite case we can simply take the completion of H. Let SEE. The characteristic function Xs E H. Indeed, for each 8 > 0 there is a finite union of intervals D such that a(xsxn) < 8. Since Xn is of finite variation, Xs E H. . Let {Sn}, Sn E .E, be a sequence of disjoint sets. Thus ./l(Sn)+0. From the definition of the norm II I~ we infer that IIXs.ll::::;;; .ll(Sn) for sufficiently large n. This trivially implies that the measure M(S) = XsE H,
SEE,
Chapter3
132
is a vector valued measure. We shall show that the measure M is not bounded. Indeed, let n, h be nonnegative integers. Let Inn=[.!!__, h+ 1]. ' n n
Let fn
= _!_
2
n 2h
I211,n
= _!_ M( n
U I2h, n)·
2h
We shall show that {fn} does not tend to 0 and this will complete the proof. Let fn = gn +hn. Let A be the union of those intervals Jn,n = I 211 ,n u I211 +1,n for which hn is equal to 0 on subsets of positive measures on both intervals I 211 ,n and I2h+I,n· This implies that the variation of gn on J 211 ,n is not smaller than 1/n = ~ A.(Jn,n) and 2 Var (gn) ;::?: A.(A). On the other hand by the definition of A 1A.(A) = x(D"'A)::::;; 2a(hn and finally 2Var (gn)+2a(hn);::?: 1. Therefore
/Ifni/
;::?: 1/2 and this completes the example.
A measure M is called compact if the set {M (E): E e L'} is compact. By Corollary 3.6.2 we trivially obtain CoROLLARY
3.6.5. If X is finitedimensional, then each measure is compact.
If the space X is infinitedimensional Corollary 3.6.5 does not hold, as follows from Example 3.6.6 Let D = [0, 1] and let E be the algebra of Lebesgue measurable sets. Let X= 12• We define a vector measure Mas follows M(E) = {
Jsin2nntd~}.
E
Locally Pseudoconvex and Locally Bounded Spaces
133
It is easy to verify, that M is a measure. By Corollary 3.6.3 the measure M is bounded. We shall show that it is not compact. Indeed, let
En= {t: sin2nnt
> 0}.
Observe that the nth coordinate of M(En) = 1/TC. This implies that the sequence M(En) does not contain any convergent subsequence. Thus the measure M(E) is not compact. Let f(t) be a scalar valued function defined on Q. We say that the· functionf(t) is measurable if, for any open set U contained in the field of scalars, the setf1(U) belongs to .E. A sequence of measurable functions {fn(t)} is said to tend to a measurable function f(t) almost everywhere if lim fn(t) = f(t) for all t except a set E such that M(E) = 0. n>oo
PROPOSITION 3.6.7 (Drewnowski, 1972).
If fn
tends almost everywhere to
J, then for each e > 0 00
lim k>oo
M(U
En (e)) = 0,
n=k
where En(e) = {x: //fn(x)f(x)// ~ e}. Proof The sequence
is a decreasing sequence of measurable sets. Then by (3.6.1.v) lim M(An) =
M (lim
An) = M (lim supEn(e)). n>oo
Observe that if x
E
limsupEn(e), then the sequence fn(x) is not convern
gent to f(x). Thus lim supEn(e) C A= {x: fn(x) is not convergent tof(x)}. By our hypothesis M(A) = 0. Thus, by (3.6.l.ii),
M(lim sup En(e))~ M(A) =
0.
D
11>CO
As a consequence of Proposition 3.6.7, we obtain extentions of the classical theorems of Lebesgue and Egorov
Chapter 3
134
3.6.8 (Lebesgue). If {fn(x)} tends to f(x) almost everywhere, then, for each e > 0,
THEOREM
lim M({x: 1/fn(x)f(x)// ~ e})
=
0
n+ro
3.6.9 (Egorov). If {/n(x)} tends to f(x) almost everywhere, then for each e > 0 there is an FE .E such that {fn(x)} tends to f(x) uniformly on F. Proof By Proposition 3.6.7, for each k there is an index Nk such that THEOREM
(3.6.2)
(3.6.3)
M(D"'F) =
M((J
0 En(k ))~t M( 0 En(k 1
kl nN•
k=l
nN•
1 ))
OC>
~2
;k =e.
k=l
We shall show that {fn(x)} tends uniformly to f on th set F.· Indeed, let 'YJ be an arbitrary positive number, let k be such that 1/k < 'YJ· If x E F, then
and //fn(x) f(x)//
or n
~
1
< k <
'YJ
0
Nk.
For simple measurable functions there is a natural definition of the integral with respect to the measure M. Namely, if g(t)
=
l
N 1
n=l
hnXE.,
En
E
.E, bnscalars.
Locally Pseudoconvex and Locally Bounded Spaces
135
then we define N
j g(t)dM(t) = !1
}; bnM(En). n~J.
We say that a measur:; M is L"" bounded if ll1.:: set
{J gdM:
O
is boundecl (Turpin, 1975). Of course, every L ""bounded measure is bounded. The converse is not true, as will be shown later. However, for a large class of spaces (containing locally pseudoconvex spaces) every bounded measure is L"" bounded. Let X be an Fspace. Let {An} be a sequence of nonnegative numbers. We say that the space X has property P({A.n)} if, for any neighbourhood of zero U, there is a neighbourhood of zero V such that n = 1,2, ...
PROPOSITION 3.6.10. If a space X is locally pseudoconvex, then it has property P ( { ;n }) . Proof Let topology in X be determined by a sequence mogeneous pseudonorms. Let
llxnll < e,
n
=
{II Ilk}
of Pkho
1,2, ...
Then D
THEOREM 3.6.11 (Rolewicz and RyllNardzewski, 1967; Turpin, 1975). Let an Fspace X have property P( {1/2n}). Then each bounded measure with values in X is L"" bounded. Proof Let N
g(t)
=}; hiXE•• i=l
Chapter 3
136
0 ~ bt ~ 1, be a simple measurable function. Let U be an arbitrary neighbourhood of zero. Since X has property P({1/2n}), there is a balanced neighbourhood of zero V such that 1 1 TV+ ... + 2n VC U,
n = 1,2, ...
By our hypothesis the measure M is bounded, i.e. there is an s such that M(E) c sVfor all Ee 2:. We write each bt in the dyadic form
>0
co
~bt,lc
bt = ~ 2/c ' k=l
where bt,lc is equal either to 0 or to 1. Thus N
J
g(t)dM =
.2
btM(Et) =
i=l
D
co
N
k=l
n=l
N
co
i=l
k=l
22
b~~n
M(Et)
=}; 2 b~: M(Et N
co
=
.2
;lc M(
iy bt,lcEt) C sU.
k=l
The arbitrariness of U implies that the set A= {
Jg(t)dM: 0 ~ g(t)~ 1, gsimple mesurable}
D
is bounded.
0
3.6.12 (Maurey and Pisier, 1973, 1976). Each bounded measure M with values in a space L 0(Q,E,p) is L 00 bounded.
THEOREM
The proof is based on several notions and lemmas. To begin with we shall recall some classical results from probability theory. Let (Q 0 ,E0 ,P) be a probability space (i.e. a measure space such that P(Q0) = 1). A real valued E0measurable function is called a random variable.
Locally Pseudoconvex and Locally Bounded Spaces
137
Let X(w) be a random variable. We write E(X)
j X(w)dP
=
and V(X) =
JIX(w)E(X)j dP. 2
o.
LEMMA 3.6.13 (Tchebyscheff inequality). P({w: !X(w)l;;?; e})
~4E(X'). e
Proof E(X2)
=
JIXI dP 2
o.
j
j
X 2 dP+
{co: IX(co)l~}
X 2 dP
{co: IX(o)l<s}
;;?; s2P({w: IX(w)i;;?; e}). LEMMA 3.6.14. Let X(w)
E
~
j
X 2 dP
{co: IX(co)l~s}
0
L 2(Q0) and let 0
Then . zE2(X) P({w. X(w);;?; A.E(X)});;?; (1..1.) E(X 2)"
Proof Let X'(w)
= { :(w)
ifX(w);;?; A.E(X), ifX(w)
<
A.E(X).
By the Schwarz inequality we have
E 2(X') ~ E(X' 2)P({w: X'(w) =/= 0}) ~ E(X 2 )P({w: X(w);;?; A.E(X)}). Since E(X) ~ E(X')+A.E(X), we have E(X') ;;?; (1..I.)E(X). Thus
0
Chapter 3
138
Let (fJ,l:,P) be a probality space. Random variables X 1(co), ... , X 11(co) are called independent, if P({co: X 1(co) e Bt> ... , Xn(co) e Bn}) = P({co: X 1(co) e B1})· ••• ·P({co: Xn(co) e Bn})
(3.6.4)
for all sets BI> ... , Bn E: 1:. A sequence {Xn} of random variables is called a sequence ofindpendent random variables if, for each finite set of indices i 1 , ••• , i11 , the random variables Xfl, ... , Xtn are independent. By (3.6.4) we conclude that, if {Xn(co)} is a sequence of independent random variables, then E(X11 · ... · Xtn) = E(XtJ · ... · E(Xtn)
for i 1 < i 2 < ... in. By a Rademacher sequence we shall mean a sequence of independent random variables r11(co) taking the values +I or 1 with the same probability 1/2. 3.6.15 (PaleyZygmund inequality; see Paley and Zygmund, 1932). Let {r11(co)} be a Rademacher sequence. For any sequence of numbers {a11 } LEMMA
N
P ({co:
N
I~ rn(co)anl ~A~ lanl 2}) 2
n=l
n=l
Proof We shall write N
X(co) = j ~ rn(co)an 12 n=l
and we shall calculate N
E(X) =
JI~ rn(co)anr dP J ~ rt(co) r1(co) atCii dP u U
n=l N
·=
i,j=l
=
N
N
i1
i=l
J~ latlldP = ~ latl
n
1•
;?:
+
(1.1)2 •
Locally Pseudoconvex and Locally Bounded Spaces
139
Using the formula
and observing that
Jrn (w)rn (w)rn (w)rn (w)dP = 0 1
3
2
4
n
if one index is defferent from the others, we obtain N
E(X 2)
=
JI}; rn(w)an 1dP 4
a
n=l
N
=
,2; [an[ +6
};
4
[an[ 2 [am[ 2
l~n~m~N
n=l
2: [an[ r. N
~ 3(
2
n=l
Using Lemma 3.6.14 we obtain the required inequality.
0
Let Q = {1, 1}N be a countable product of two point sets. The elements of Q are sequences a= {an}, where an take the values either +1 or 1. Let rn(a) be the following random variables rn(a) =On. Let A= {An} be a sequence ofreals such that [An[~ 1. On the two point set { 1, + 1} we define measures P1n as follows
1+A
P;.n({ +1}) =  2 , P;.n({ 1 }) =

1A 2 .
Let P1 be the product measure, P1 = P1 , X ... X P1n X .•. It is easy to verify that the random variables {rn(t)} constitute a sequence of independent random variables and
Jfn(a)dP;. =An.
a
Let A= {0,0, ... }. The measure P1 will be denoted by P0 •
(3.6.5)
Chapter 3
140
3.6.16 {Musial, RyllNardzewski and Woyczynski, 1974). Let P;. and P 0 be as described above. Let LEMMA
t
p =
(Po+P;.).
Then, for each sequence of numbers {x 1 , x 2 , integer n, n
••• }
and for each positive
n
P({a: I~ rt(a)xt I~+ I~ AtXt I})~+· i=l i=l
m =
Proof Without loss of generality we may assume that
n
If L;ixtl 2
>
i,we shall use the PaleyZygmund inequality
i=l
1 1 ( 1 )2 1 ~23 l16 ~g· n
If 2
lxtl 2 < t
we shall use the Tchebyscheff inequality (Lemma
i=l
3.6.13). By simple calculation we obtain n
JI~ (Aift (a))xtl dP.t 2
u
i=l
n
=
~ i=l
Ji(1tft(a))xtl dP.t 2
u
n
= ~ A.f i=l
n
lxtl 2 2
~
n
1tlxtl 2
i=l
n
=
l\tA.n!xtl 2 <{. i=l
Jft(a)dP.t+ ~ lxtl
u
i=l
2
Locally Pseudoconvex and Locally Bounded Sraces
141
In the calcution we have used the fact that ;, are independent random variables and formula (3.6.5). Now
PA
({a: It fc(a)xt I~+}) •=1
= PA
({a: It (ltft(a))xtl ~ +}) •=1
=
1PA({a: It (ltrt(a))xc I~ f}). t=1
n
82 ~ 82 1 I 72 L.J (IAi)lxtl 2 ~ Ip 4· i1
Hence m
>f
also in this case.
0
Proof of Theorem 3.6.I2 (modified proof given by RyllNardzewski and Woyczynski (I975). Let M be a bounded measure. Thus, by definition, the set n
K = {x: x
=}; etM(At), Et ~qual either  I or+ I, 1=1
=4t e J:, Ac disjoint sets} is bounded. We have to show that the set n
K1 = {x: x
=}; ltM(At), JkJ ~I,
At el:, At disjoint}
i=1
is also bounded. M(At) e L 0(D,J:,p). We shall write M(At) = fi(t). Let [) = {I, + 1}N, let P be a measure with the property described in Lemma 3.6.16, and let rt be a sequence of independent random variables described in that lemma. Thus by Lemma 3.6.I6 (3.6;6)
Chapter 3
142
for all t. By the arguments given in: Example 1.3.5 we may assume that
fl.(Q) <+oo. Let
n
{t: I~ At/i{t) I> sc}. that fJ.{Tc) > 0. Let fl.c(A) = fl.(A n Tc)/fJ.(Tc).
Tc =
i=I
Suppose fl.c(Tc)
Of course
= 1. For the product measure fJ.cXP we have by (3.6.6) fl.c
xP({(t,a): /4 ft(a)fi(t)l? +14 n
n
t=l
•=11
lctfi(t)
I})?+·
Then, by the Fubini theorem,
and, by the definition of Tc, n
.~a~\ fl.c ({t ETc: I~ Bt/i{t)
I> c})? +·
Hence, by the definition of fl.c, max fl. ••= ±l
({t
E
T:
" I~ Bt/i(t) t=l
?
.~~\fl.({tETc:
=
+fl.({t e T:
I> c})
n
16stfi(t)l >c})?+fJ.(Tc)
It I? sc}). ?ctfi(t)
(3.6.7)
•=I
Let Uc
= {x:
fl.({t:
lx(t)i
> c}) < c}
be a basis of neighbourhoods of zero in L 0(Q, E, fl.). Since the set K is bounded, then for each c > 0 there is an s > 0 such that sK C Uc. Thus, by (3.6.7), sK1 C U 8c. The arbitrariness of c implies that the set 0 K1 is bounded.
Locally Pseudoconvex and Locally Bounded Spaces
143
Now we shall define integration with respect to an L 00 bounded measure of scalar valued functions. Let (X, II II) be an Fspace. Let M be an L 00 bounded measure taking its values in X. Let f be a scalar valued function. Let
M·(f) =sup{ II
J
gdMII: g being simple measurable u functions, lgl ~ 111}.
3.6.17 (Turpin, 1975). M ·(f) has the following properties. If l.hl ~.hi, then M ·(/1) ~ M ·(/2) ~+oo, (3.6.8.i) Iff is bounded, then limM·(if) = 0, (3.6.8.ii)
PROPOSITION
t+0
Iff is measurable and M ·(f) = 0,
then f = 0 almost everywhere (3.6.8.iii) (i.e. except a set A such that M(A) = 0). If {In} is a sequence of measurable functions tending almost everywhere to J, then M ·(f)~ lim inf M · (fn). (3.6~8.iv) n+oo
(3.6.8.v) M·(.h+.h) ~ M·(ft)+M·(_h), providedft,.h are measurable. Proof (3.6.8.i), (3.6.8.ii), (3.6.8.iii) are trivial. (3.6.8.iv). Let {fn} be a sequence of measurable functions tending to f almost everywhere. Let g be a simple measurable function such that lg(t)l ~ lf(t)l. Let 8 be an arbitrary positive number. Lets be an arbitrary number such that 0 < s < 1. Let I fn(t)l if g(t) =I= 0, gn(t) = olg(t)l if g(t) = 0.
I
Observe that {gn(t)} converges almost everywhere. Thus, by the Egorov theorem (Theorem 3.6.9) for each 'YJ > 0 there is a set A such that M(D"'A) < 'YJ and {gn} converges uniformly on A. Since g is a simple function, there is an 'YJ > 0 such that for each setH such that M(H) < 'YJ we have I\
JsgdMII <
H
8.
144
Chapter 3
Taking 1J and A as described above, we have II
JsgdMII ~II AJsgdMII +II JsgdM[I ~lim inf M · (/n)+e,
n
n~A
n~~
because, for sufficiently large n, lfn(t)l > lsg(t)l for tEA. The arbitrariness of e and simply (3.6.8.iv). (3.6.8.v). To begin with, we shall assume that ft and / 2 are simple. Let g(t) be an arbitrary simple measurable function such that lg(t)l ~ l.h(t)+};(t)l. Let gt(t)
=
J
l
g,(t) lft(t)j
~ft(t)l+l/2(t)l
if lft(t)l+lf2(t)l =fo 0, i = 1,2. if lfl(t)l+lf2(t)l = 0.
The functions g 1(t) and g 2(t) are simple and g(t) = gl(t)+g2(t).
Hence II
JgdMII~II nJgldMII+II nJg2dMII~M·(fl)+M·(.t;).
n
The arbitrariness of g implies
Suppose now that.ft and,/; are not simple. Then there are two sequences of simple mesurable functions {.h,n} and {h,n} tending almost everywhere to / 1 and}; and such that lfi.nl ~I AI, i = 1,2. Therefore, by (3.6.8.iv),
Let B(f) = (
~
limsup M· (ft.n+An)
~
supM ·(ft,n)+supM ·(h,n)
~
M ·(h)+M·(JJ.
n
n
JgdM: g simple measurable functions, n
0 igl
~Ill}
Locally Pseudoconvex and Locally Bounded Spaces PROPOSITION
145
3.6.18. The set B(f) is bounded if and only if
1imM ·(if) = 0.
(3.6.9)
~0
Proof If the set B(f) is bounded, then limM· (if)= lim (sup{llxll: x t+0
E
B(tf)})
t>0
= lim(sup{llxll: x
E
tB(f)}) = 0.
(3.6.10)
t>0
Conversely, if (3.6.9) holds, then by (3.6.10) the set B(f) is bounded. 0 Let X denote the set of those /for which B(f) is bounded. By (3.6.8.v) and (3.6.9), M· (f) is an Fpseudonorm on X. We shall now use the standard procedure. We take the quotient space X/{f: M·(/)=0}. In this quotient space M ·(f) induces an Fnorm. We shall take completion of the set induced by simple functions. The space obtained in this way will be denoted by D(D,E, M). Since, for each simple function g,
II
JgdM[J ~ M·(g).
{l
We can extend the integration of simple functions to a linear continuous operator mapping D(D,E,M) into X.
3.7.
INTEGRATION WITH RESPECT TO AN INDEPENDENT RANDOM MEASURE
Let (D0,E0,P) be a probability space. Let Q be another set and let E be a aalgebra of subsets of D. We shall consider a vector valued measure M(A), A E E, whose values are real random variables, i.e. belong to X= L 0(D0 1:0 P). We say that M(A) is an independent random measure if, for any disjoint system of sets {AI> ... , An}, the random variables M(At) EX, i = 1,2, ... , n are independent. We recall that a vector measure M is called nonatomic if, for each A E D such that M(A) =1= 0, there is a subset A0 C A such that M(A 0) =1= M(A).
Chapter 3
146
Let X(w) be a random variable, i.e. X(w) we denote Fx(t) = P({w: X(w) ~ t}). The function Fx(t) is nondecreasing, lim
E
L 0(£2 0 ,.E0 ,P). By Fx(t)
Fx(t)
= 0, limFx(t) = I.
t+ ro
I>+ ro
It is called the distribution of the random variable X(w).
Let Q = [0, 1], and let .E be the algebra of Borel sets. We say that a random measure M is homogeneous if for any congruent sets A 1 A 2 C C [0, 1] (i.e. such that there is an a such that a+A 1 = A2 (mod 1)) the random variables M(A 1) and M(A 2) have identical distribution. It is not difficult to show that, if M is a homogeneous independent random measure, then for each n we can represent M(A) as a sum M(A) = M(A 1)+ ... +M(An), where the random variables M(Ai), i = 1,2, ... , n, are independent and have the same distribution (Prekopa, 1956). Let X(w) be a random variable. The function +ro
fx(t) =
J
eits dF(s).
(3.7.3)
ro
is called the characteristic function of the random variable X(w ). If the random variables X1 , ••• , Xn are independent, then fxi+ ... +Xn(t)
= fxit)· ··· fxn(t).
(3.7.4)
Directly from the definition of the characteristic function !ax(t) =fx(at).
(3.7.5)
We say that a random variable X(w) is infinitely divisible if for each positive integer n there is a random variable xn such that X= Xn+ ... +Xn,
(3.7.6)
in other words, by (3.7.4). fx = (/xn)n.
(3.7.7)
If X is an infirutedivisible random variable, then its characteristic function/x can be represented in the form
fx(t) =
exp(iyt+
]"'(eitx_1 1 ~xx2 ) 1 ~2x2 dG(x)),
00
(3.7.8)
Locally Pseudoconvex and Locally Bounded Spaces
147
where y is a real constant, G(x) is a nondecreasing bounded function and we assume that at x = 0 the function under integration is equal to  t 2/2. This is called the LevyKchintchin formula (see for example Petrov (1975)). If X is a symmetric random variable (which means that X and X have this same distribution), by (3.7.8) we trivially obtain co
fx(t)
= exp
J
l+u2 (costu1)zt2 dG(u).
(3.7.9)
0
Of course, without loss of generality we may assume that G(O) = 0. Suppose that M(A) is a nonatomic independent random measure with symmetric values. Then by (3.7.9) co
fM(t)
= exp
J
1+u2 (cos tu1)uzdGA(u).
(3.7.10)
0
For an arbitrary real number a we obtain by (3.7.5) and (3.7.10) co
/aM(A)
=
exp
J
1+u2 (cos tu1) zt2 dGA(u).
(3.7.11)
0
Let A 1 ,A 2 E I be two disjoint sets. By (3.7.4) (3.7.12) Formulae (3.7.11) and (3.7.12) imply that for a simple realvalued function h(s) fjh(s)dM(t) u
= exp(
JTM(th(s))ds),
(3.7.13)
0
where (3.7.14) In the sequel we assume Q = [0, 1]. Now se shall prove some technical lemmas.
Chapter 3
148
Let co
UM(x) =
J
min ( xll,
~ll) {1 +ull)dGM(u)
(3.7.15)
0
and for x
> 0,
(3.7.16)
for x =0. It is easy to see that the two functions Uy(x) and 'Py (x) are equal
to 0 at 0, continuous and increasing. LEMMA 3.7.1. For all x;;;: 0, a;;;: 0, there are positive numbers c1(a) and ell such that max TM(v)::::;;;c 1(a)UM(x) (3.7.17) O~v~ax
and 1
f TM(xt)dt ;;;: c UM(x).
(3.7.18)
2
0
Proof. Observe that
1cosuv = 2sinll
z;;:::::;;;} u v
2 2
and 1cosuv:::::;;; 2. Hence, for c1{a)
=
max(2,} all), we have
max (1cosuv):::::;;; c1 (a)min(1,x2u2). O~v~ax
Therefore ( ) . ( ll 1 ) max 1cosuv :::::;;; c1 a mm x ,  2 • 2
o,;;;v,;;;ax
U
(3.7.19)
U
Integrating (3.7.19) with respect to 1+ull dGy we obtain {3.7.17). ull Observe that sinz . (z, 2 1) 1;;;: cllmm ,
z
Locally Pseudoconvex and Locally Bounded Spaces
149
where · ( mm . ( 1· sinz) . z 2( 1 sinz)) c2 =mm  , mm  . Z
Z>1
0
Z
Integrating TM(tx) with respect to t, we obtain 1 Joo( 1 sinxu) ~ 1+u2 dGM(u) J TM(tx)dt =
xu
0
0
0 Observe that
Integrating by parts the second integral on the righthand side, we obtain
J
1+u2 dGM(u)
1/z
J oc
00
~
= ~
GM(u) 2+u21oo +2 ~
1IX
1/X
GM(u) du ~
2'l'M(X).
Thus (3.7.20) On the other hand, integrating by parts both terms, we obtain
(3.7.21) Formulae (3.7.20) and (3.7.21) imply that the functions 'l'.lll(x) are equivalent at infinity.
u.~~~(x)
and
150
Chapter 3
3.7.2 (Urbanik and Woyczyiiski, 1967). Let G{u) be a continuous nondecreasing function such that G(O) = 0 and lim G(u) = 1. Let LEMMA
uo+oo
J 00
([>(x)
=
(3.7.22)
G;;) du.
1/X
Then the function ([>(yx) is concave. Proof We apply a change of variable v = l/u 2 in formula (3.7.22), and obtain ([> (yx) =
f G~~) jG( ylv) du =
1/y;
dv.
0
Thus the derivate of ([>(yx) is equal to G(I/yx). It is nonincreasing, since G is nondecreasing. Therefore the function ([>(yx) is concave. D 3.7.3 (Urbanik and Woyczyiiski, 1967). Let {In} be a sequence of simple measurable functions belonging to L 0 [0, 1]. The sequence
LEMMA
1
Jfn(s)dM(s) tends to 0 in L [0, 1] 0
0
if and only if 1
J'l'M(/fn(s)l)ds = 0.
lim
n+oo 0
Proof We shall use the classical statement that a sequence {hn} of functions belonigng to L 0 [0, I] tends to zero in L 0[0, I] if and only if the logarithms of its characteristic functions tend uniformly to 0 on 1
each compact interval. Thus by (3.7.13) {
Jfn(s)dM(s)} tends to zero in 0
1
L 0 [0, 1] if and only if { jTM(fn(s))ds} tends to Ouniformly on each como
pact interval [0, T]. 1
If~
JTM(tfn(s))ds} tends uniformly on each compact interval, then 0
1
lim
1
J JTM(ifn(s))dtds =
noo 0
0
1
lim ~oo
1
J JTM(ifn(s))dsdt = 0
0
0.
Locally Pseudoconvex and Locally Bounded Spaces
151
Hence by (3.7.18) 1
lim
JUM(Ifn(s)l)ds = 0.
(3.7.23)
n+oo 0
Since UM(x) and IJfM(x) are equivalent at infinity, 1
lim
J'PM(Ifn(s)l)ds = 0.
(3.7.24)
x+co 0
Conversely, if (3.7.24) holds, then (3.7.23) holds. By (3.7.17) we find 1
that { JrM(tfn(s))ds} tends to Ouniformly on each compact interval. 0 0
THEOREM 3.7.4 (Urbanik and Woyczyiiski, 1967). Le M be a symmetric homogeneous independent random measure with values in the space L 0[0, 1]. Then there exists a continuous function N(x) defined on [O,+oo) such that N(O) = 0, N(J./i) is concave and such that the set of all real 1
functions for which the integral J fdM exists is the space N(L). 0
Conversely, if N(yx) is a concave continuous function such that N(O) = 0, then there is a symmetric independent homogeneous random measure M with values in the space L 0 [0, 1], such that the set of real val1
ued functions f for which the integral J fdM exists is precisely the space 0
N(L). Proof The first part of the theorem follows immediately from Lemmas 3.7.2 and 3.7.3. Conversely, if N(yx) is concave, then it can be represented in the integral form
N(yx) = J q(u)du, 0
where q(u) is a nonincreasing function. We put G(O) = 0 and G(x) = min(l,q(1/x2)) for x > 0. Let M be a symmetric independent homogeneous random measure such that
152
Chapter 3
(3.7.10) holds. To complete the proof it is enough to show that the functions N(x) and :~:,
00
NG(x)
J G~~)
=
du
=
1/Z
J
min (I ,q(u))du
0
are equivalent at infinity. Since q(u) is nonincreasing, q(u) ~ q(1) for u > 1. Thus z
1
<1:
N(y'X)
Jq(u)du = Jq(u)du+ Jq(u)du = Jq(u)du+q(1) J min(I,q(u))du
=
0
0
1
z
1
0
1
~ K+CNG(yx). 1
where K =
Jq(u)du and C =
q(1).
0
On the other hand, trivially NG({x) ~ N(y'x).
0
Woyczynski (1970) extended this result to the case where M(A) take values in the product space L 0 (D 0 ,E0 ,P) X ... XL0 (!J0 ,E0 ,P) and f(s) is n..times
a matrix function.
3.8. UNCONDITIONAL CONVERGENCE OF SERIES
Let Q = N be the set of all positive integers. Let ~ be the algebra of all subsets of N. Let M be a vector measure defined on A E E taking its values in an Fspace (X, II II). Let Xn = M({n}). Since M is a vector 00
measure the series
2
Xn
has the property that for each sequence
en
n=1 00
taking values ·either 0 or 1 the series
2 en x~ is convergent. The series n=1
with this property are called unconditionally convergent series. Thus
Locally Pseudoconvex and Locally Bounded Spaces
153
we have shown that each vector measure on N induces unconditionally convergent series. But the converse is also true. Namely, each unconditionally convergent series }; Xn induces a vector measure by the formula n=l
M(E)=};
Xn.
neE
PROPOSITION
3.8.1. The measure M induces by an unconditionally concc
vergent series };
Xn
is compact (and thus bounded).
n=l 00
Proof Since the series };
Xn
is unconditionally convergent, for each
n=l
e > 0 there is an index N such that for each sequence either 0 or 1 00
l
2;
BnXn
n=.N+l
I<
{en}
taking values
(3.8.1)
e.
.N
Thus the finite set };
e11 Xn,
where {e1 ,
••• ,
en}
runs over all finite systems
n=l
taking values either 0 or 1, constitutes an enet in the range of the meas
D
~M 00
THEOREM
3.8.2 (Orlicz, 1933). A series};
Xn
is unconditionally convergent
n=l
jf and only if for each permutation p(n) of positive integers the series 00
2 Xp(n) is convergent. ft=l 00
Proof Suppose that the series .for each e
>
2
Xn
is unconditionally convergent. Then
n=l
0 there is an index N such that
k+m
IIJ:
enXn
.n=k
fork> N, m
>
I<
e
0 and Bn taking the value either 0 or I.
Chapter 3
154
Let K be such a positive integer that, for n arbitrary r > K and s ;;::;: 0 r+s
II~ Xp(n) I! = n=r
>
K, p(n)
>
N. Then for
q
114 I < 8 tXt
(3.8.2)
8•
•=p
where p = inf {p(n): r ~ n ~ r+s}, q =sup {p(n): r ~ n ~ r+s}, t = {1 if i = p(n) (r ~ n ~ r+s), 8 0
otherwise.
The arbitrariness of 8 implies that the series
2"'
Xp(n)
is convergent.
n=l
Suppose now that the series 2"'
Xn
is not unconditionally convergent.
n=l
This means that there are a positive number tJ and a sequence {8n}, taking the value either 1 or 0 and a sequence of indices {rk} such that r•+l
il ~
8 nXn n=r•+l
8
I > J.
Now we shall define a permutation p(n). Let m be the number of those 8n, n = rk+ 1 , ... , rk+I• which are equal to 1. Let p(rk+v) = n(v), where n(v) is such an index that 8n(v) is a vth 8i equal to 1, rk < i ~ rk+I• Q < v ~ m. The remaining indices rk < n ~ rk+1 we order arbitrarily. Then r.+m
r•+l
n=r•
n=r.+l
II~ Xp(n)ll =II~
This implies that the series
8
nXnll >
J.
2"'
is not convergent.
xp(n)
D
n=l
A measure M induced by a series "'
2
Xn
is L"' bounded if and only if
n=l
for each bounded sequence of scalars {an} the series 2"'
anXn
is con
n=l
vergent. The series with this property will be called bounded multiplier convergent.
Locally Pseudoconvex and Locally Bounded Spaces
155
THEOREM 3.8.3 (Rolewicz and RyllNardzewski, 1967). There exist an 00
Fspace (X,
II II) and an uncoditionally convergent series}; Xn
of elements
n=1
of X which is not bounded multiplier convergent.
The proof is based on the following lemmas. LEMMA 3.8.4. Let X beakdimensional real space. There exists an open symmetric starlike set A in X which contains all points PI> ... , p 3k of the type (e~> ... , ek), where ei equals 1 or 0 or 1, such that the set Ak1 =A+ ... +A (k1)fold
does not contain the unit cube C=
{(a1,
... ,
ak):
lail <
1, i = 1,2, ... , k}.
Proof Let A 0 be the union of all line intervals connecting the point 0 with the points PI> ... , Ps•· Obviously the set A~ 1 is (k1)dimensional. Therefore there is a positive number e such that the set (A 0 +A.)k\ where A. denotes the ball of radius e (in the Euclidean sense), has a volume less than 1. Thus the set A= A 0 +A. has the required property. D
LEMMA 3.8.5. There is a kdimensional Fspace (X, II II) such that IIPill ~I, i = 1,2, ... , 3k and there is a point p of the cube C such that IIPII;;:, k1. Proof We construct a norm II II in X in the way described in the proof of Theorem 1.1.1, putting U(l) =A. Since PtE A = U(I), IIPill ~ 1. Furthermore, since Ak 1 = U(kI) does not contain the cube C, there is a point p E C such that p E U(k1). This implies that II PII ;;:, k1. D Proof of Theorem 3.8.3. We denote by (Xk, space constructed in Lemma 3.8.4. Let 1
II
II~) the 2k dimensional
'
llxllk = 2k llxllk· Let X be the space of all sequences a
= {an} such that
00
lllalll
=}; ll(a2••+t• ... , a2~)/lk1+la1l <+oo. k=l
Chapter 3
156
It is easy to verify that X is an Fspace with the norm
Ill Ill. Let
(0,0, ... , 0, 1,0, ... )
Xn =
nth place
and let {en}' be an arbitrary sequence of numbers equal either to 0 or to 1.
,2; e
The series
11
x 11 is convergent. Indeed,
n=l r
IIIL
Ill~ Ill Yo 111+111 Yk+llll+ ... +Ill Yk'III+IIIY~III,
1
BnXn n=m
where k is the smallest integer nongreater than log 2 m, k' is the largest less than log2 r, and 2•
,2;
Yo =
BnXn,
n=m 2i
YJ =
,2;
j=k+1, ... ,k',
BnXn, n=2J'+l r
Yo= . : : ,_; I
\'
BnXn.
n=2•'+1
In virtue of Lemma 3.8.5 1
IIIYolll ~ 2kl ' 1
IIIYJIII ~ 2il ' '
1
II!Yolll ~lk'.
j = k+1, ... , k'
.
Therefore r
~~~~ n=m
k'
BnXnlll
~2
i=k1
;i < 2L2 ~ ! ' w
00
and the series
,2;
BnXn
n=l
ditionally convergent.
is convergent. Thus the series
,2; n=l
Xn
is uncon
Locally Pseudoconvex and Locally Bounded Spaces
151
On the other hand, as follows from Lemma 3.8.5, for each k there is a point ak E Xk, ak = (a2k+1• ... , a2k+1), SUCh that latl < 1, i = 2k+ 1, •.. , 2k+1 and
Let bn =an, n
= 1,2, ... The sequence {bn} is bounded and
2t'+l
Illn=2k+1 }; bnxnlll =
llakllk
>
+·
00
Therefore the series }; bnxn is not convergent. This implies that the n=l 00
series };
Xn
is not bounded multiplier convergent.
D
n=l
Observe that the measure M induced by an absolutely convergent 00
series };
Xn
which is not bounded multiplier convergent is bounded, but
n=l
is not L 00 bounded. Turpin (I 972) constructed a nonatomic measure with this property.
3.9.
INVARIANT A(X)
We shall prove the results of Turpin (1975), however restricting ourselves to the metric case. Let (X, II II) be an Fspace. By A(X) we denote the set of such sequences {An} that the space X has property P({A.n}), i.e. for any neighbourhood of zero U there is a neighbourhood of zero V such that (3.9.1)
Of course, by definition, A (X) always contains all sequences with finite support, i.e. sequences of the form A= {A.1 , A2 , ... , A.n, 0, 0, ... }. If A (X) contains a sequence with an infinite support, in other words : if there is a sequence A= {An}, An> 0 belonging to A(X), we say that the space X is strictly galbed.
Chapter 3
158
3.9.1. Let (X, II II) be an F*space. Then the set A(X) is equal to the set of all such sequences {A.n} that, for any bounded sequence {xn} C X, the sequence
PROPOSITION
N
{YN}
= {}; AnXn}
(3.9.2)
i=l
is bounded. Proof Necessity. Let {A.n} e A(X). Let U be an arbitrary neighbourhood of zero. By the definition of A(X) there is a neighbourhood zero V such that (3.9.1) holds.
Let {xn} be an arbitrary bounded sequence. Then there is a b > 0 such that Xn e bV, n = 1,2, ... Thus, by (3.9.1) YN E bU, N = 1,2, ... Since U is arbitrary, the sequence YN is bounded. Sufficiency. Suppose that {A.n} ¢ A(X). This means that there is a neighbourhood of zero U such that (3.9.1) does not hold for any neighbourhood of zero V. This implies that for an arbitrary neighbourhood of zero V and an arbitrary positive integer n (3.9.3)
A.nV+A.n+ 1 V+ ... ¢ U.
Now we shall choose a sequence of elements {xn} and an increasing sequence of indices {nm} in such a way that i
=
(3.9.4)
nm+l, ... ,nm+l,
(3.9.5)
Such a choice is possible since (3.9.3) holds. The sequence {xn} tends to 0 by (3.9.4), thus it is bounded. On the other hand by (3.9.5) the sequence
{zm}
=
{A.nm+!Xnm+l+ ... +A.nm+IXnm+I}
is not bounded. This implies that the sequence {YN} is not bounded either. 0 PROPOSITION
3.9.2. lf{A.n}eA(X)
and0~p. 11 ~An,
n= 1,2, ... , then
{P.n} E A(X). Proof It follows immediately from the definition of A(X).
0
Locally Pseudoconvex and Locally Bounded Spaces
159
PROPOSITION 3.9.3. Let {A.i} e A(X). Let!., be a sequence of nonempty finite disjoint subsets of the set of positive integers. Let
Then {.Un} e A(X). Proof. Let {x.,} be an arbitrary bounded sequence. Let
, {x., Xi =
for i e /.,, elsewhere.
0
The sequence {rn} is of course also bounded. Then the sequence N
{YN}
= {};
N•J
,Unxn} =
n=l
{}; }; n=l
Atx;}
iEln
is bounded.
D
3.9.4. Let {1.,}, {.Un} e A(X) and let (nt,m;) be a double sequence with terms not equal to one another. Then the sequence {vi} = {An,·.U,..} E A(X). Proof. Let {xi} be an arbitrary bounded sequence. Then PROPOSITION
N
YN
N
= }; VtXi = }; An,,llm, X~ t=l
t=l
p
~};An (2; ,llmXin,m)' n=l
where p
=
(3.9.6)
m
max nt and the summation with respect to m is taken over IEO;i~N
all m for which there is an index i = i.,, m ~ N such that (n;, mt) = (n, m). = }; ,UmXtn,m, is bounded. Therefore the sequence The set {y:},
y:
m
{YN} is also bounded.
D
COROLLARY 3.9.5. If an F*space (X, II II) has property P({qn}) for a certain q, 0 < q < 1, then in the space X bounded multiplier convergence and unconditional convergence are equivalent.
+t}
Proof. We shall show that if {qn} e A(X), then {(
e A(X). Indeed
let (nt,mt) be such a double sequence that (nt+mt) is a nondecreasing
Chapter 3
160
sequence and the sequence (nt,mt) contains all pairs (n,m). Let An
=
qn
=
fln·
It is easy to verify that An.flm, = qn•+m• has the same order of growth as
qv'n. Then {qv'n} E A(X). Since lim (_!_)n qvr. = 0, by Proposition 3.9.2 '11>00
{(~f}eA(X).
2
Theorem 3.6.11 implies the corollary.
D
PROPOSITION 3.9.6. A(X) is an invariant of isomorphisms. Proof An isomorphism maps bounded sets onto bounded sets, and the inverse also maps bounded sets onto bounded sets. Thus by Proposition 3.9.1 A(X) is an invariant of an isomorphism. D PROPOSITION
3.9.7. Let Y be a subspace of an F*space X. Then
A(Y):) A(X).
Proof Let {.An} E A(X). Let {xn} be an arbitrary bounded sequence in Y. Then it is also a bounded sequence in X. Therefore the sequence N
{YN} = {
.J.: AnXn} n=l
is bounded. This implies that {.An}
E
A(Y).
D
3.9.8. A(X) c [1. Proof. Each space X contains a one dimensional subspace X 0 • Since A(X0 ) = [1, by Proposition 3.9.7 the corollary holds. D CoROLLARY
3.9.9. Let a continuous linear operator T map an Fspace X onto an Fspace Y. Then PROPOSITION
A(Y):) A(X).
Proof Let {.An} E A(X). Let U be an arbitrary neighbourhood ofzero in Y. The set T 1( U) is open. Then there is a .neighbourhood of zero V in the space X such that
A1V+A.2 V+ ... C T 1(U). then
Locally Pseudoconvex and Locally Bounded Spaces
161
By the Banach theorem (Theorem 2.3.1) the sets T(V) are open. The arbitrariness of Uimplies that{A.n} e A(Y). D As an immediate consequence of Propositions 3.9.6, 3.9.7 and 3.9.9 we obtain THEOREM
3.9.10. If
dimz
X~
dimz Y
(codimz X~ codimz Y),
then A(X)~A(Y).
PROPOSITION 3.9.11. If X is a locally pconvex space, then
A(X) ~ lP. Proof Let {llxllm} be a sequence of phomogeneous pseudonorms determining the topology. Let {xn} be a bounded sequence in X. Let
llxnllm·
Mn =sup n
Let co
C=
};
IA.niP ·
n=l
Let N
YN
=}; AtXt. i=l
Then co
IIYNiim ~}; IA.tiP llxtllm ~ CMm.
(3.9.7)
D
i=l
CoROLLARY 3.9.12 (compare Mazur and Orlicz, 1948). If an F*space X is locally convex, then A(X) = [1, PROPOSITION 3.9.13. Let X be a locally pseudoconvex space. Then A(X)~ /P.
n
p>O
Chapter 3
162
Proof Let {A.J
n
E
[P.
Let {llxJim} be a sequence of Pmhomogeneous
p>O
pseudonorms determining the topology in X. Let w
Cm = } ; IA.tJPm • i=l
Taking Xn and YN as in the proof of Proposition 3.9.11 we obtain an estimation similar to estimation (3.9.7)
IIYNIJm ::=;:;: CmMm. 0
This implies the proposition.
Now we shall prove propositions converse to Propositions 3.9.11, 3.9.13. PROPOSITION 3.9.14. (cf. Metzler, 1967). If A(X)) [P, then the space X is locally pconvex. Proof Suppose that an F*space (X, II JJ) is not locally pconvex. Then there is a positive number e such that for any m there are points {Xm, 1 , ••• ... , Xm, km} such that i
= 1,2, ... , km,
(3.9.8)
and (3.9.9) Let for n, k 1 + ... +km1
1 < n ::=;:;: k 1 + ... +km, let An===.2mkl/Pm m
2mxm,t, where i = n(k1 + ... +km1). Since (3.9.8), the sequence {xn} tends to 0, and thus it is bounded. On the other hand, by (3.9.9) the sequence Xn =
N
{YN}
= {
.2; Anxn} n=l
is not bounded. Therefore {..1.n} does not belong to A(X). Since {..1.n} E /11, A(X) does not contain /P. 0
Locally Pseudoconvex and Locally Bounded Spaces
n
PROPOSITION 3.9.15. If A(X) :::>
163
/P, then the space X is locally pseudo
P>O
convex. Proof The proof follows the same line as the proof of Proposition 3.9.14 if we replace, in the definition of A.n and in formula (3.9.9), p by
Pm tending to 0.
D
As follows from Proposition 3.9.15, locally pseudoconvex spaces are stricly galbed. There are also nonlocally pseudoconvex spaces which are strictly galbed, as follows from PROPOSITION 3.9.16 (Turpin, 1975). For any sequence {..1n} belonging to all spaces lP (in a particular case for any geometric sequence qn, 0 < q < 1), there is a nonlocally pseudoconvex space X such that {A.n} e A(X). co
Proof Let Cn
= }; IA.tl 1/n. Let X be the space of all sequences x = {xn}. i=l
such that Xn e L 1fn[o, 1] and for all positive integers k c~
lim
llxnllvm[o,l] = 0.
We introduce a topology in X by a sequence of Fpseudonorms llxllk = s~pc!llxnllvrn 10, 11 • It is easy to verify that X is an Fspace. The space X is not locally pseudoconvex. Indeed, let Xn = {x eX: Xi= 0 for i =F n }. The space Xn is isomorphic with L lfn[o, 1]. Take any neighbourhood of zero Uin X such that U::PXn, n = 1,2, ... By the definition of topology in X, this is possible. Then the modulus of concavity of U can be estimated as follows c(U)
~
c(UnXn)
~
c(Xn) = 2n,
n
= 1,2, ...
Thus c( U) = +oo. This implies that X is not locally pseudoconvex. On the other hand, let U be an arbitrary neighbourhood of zero in X. U contains a neighbourhood of zero U0 of the form U0
= {x: llxllk < e}.
Let V = {x: llxiiHI
< e}. co
Let JI, Y2• .•. E V, Yi = Yi, n• Let X = •
1: A.,y,. i=l
164
Chapter 3
Then 00
1\x\\ =
supc~ //}; AtYt, n//v,• n
i~1
~ supc~}; \A.t\ 11n 1\Yi, n\\L•I•[O,J] n
i~1
~ supc~+ 1 sup\IY«,n\\L•I•[o,1J n
Therefore x
E
i
U and {An}
E
<e.
A(X).
There are non strictly galbed spaces, as follows from 3.9.17. Let (D,E,p) be a measure space. Suppose that the measure f1 is not purely atomic (i.e. there is a set D 0 such that the measure f1 restricted to the set D 0 is a nonatomic measure which does not vanish identically). Then the space L 0(D,E,p) is not strictly galbed. Proof According to our hypothesis about the measure, we can find D 0 , O
+
11+00
and 00
D0 C
U En
(3.9.11)
forK= 1,2, ...
n~K
Let {An} be an arbitrary sequence of positive numbers. Let n Xn
= An XEn•
By (3.9.10), {xn} tends to 0, and thus it is bounded. On the other hand, • the seque6ce N
YN(t) = };..tnxn(t) "~1
tends to infinity for
t E
D 0 • Thus it is not bounded.
3.9.18. Let X be the space l{P•}, 0 x = {xn} such that PROPOSITION
00
1\x\1 = }; \xn\P• n~1
< +oo,
D
< Pn ~ 1, of all sequences
Locally Pseudoconvex and Locally Bounded Spaces
165
with a topology given by the Fnorm //xll. If p,+0, then the space X is not strictly galbed. Proof Let {A.n} be an arbitrary sequence of positive numbers. Since p 11 ~0, for each n we can find an index k(n) such that A.~Jc
n=l
divergent. It is obvious that the sequence {x11 } = {~11 e11 }, where {e11 } denotes the standard basis in l{Pn}, tends to 0. Thus it is bounded. On the other hand, N
1/YN/1
N
=112: Anxn// =}; 1/A.nXn/1 n=l n=l
+,}; N
~
n=l
~ 11Pk(n) ~00. N
Therefore the sequence YN =}; A.11 x 11 is not bounded. This implies
n=l
that {A.n} ¢ A(X).
D
3.10. CSEQUENCES AND CSPACES
The following .notion of Csequences is related to bounded multiplier convergence. We say that a sequence {xn} of elements of an Fspace X is a Csequence if for any sequence of numbers {t11 } tending to 0 the series 00
}; t 11 x 11
is convergent. Of course, a subsequence of a Csequence is a
n=l
Csequence, and the image of a Csequence by a continuous linear operator is also a Csequence. PROPOSITION 3.10.1. A sequence {x11 } of elements of an Fspace (X, is a Csequence if and only if the set 00
A= {}; anxn: /an/~ 1, n = 1, 2, ... } n=l
is bounded.
•
1/ //)
Chapter 3
166
Proof. Sufficiency. Suppose that the set A is bounded. Let {en} be an arbitrary sequence of numbers tending to 0. Let Cn =sup JctJ. Of n~i
course, {C11 } also tends to 0. Let e be an arbitrary positive number. Since the set A is bounded, there is an index N 0 such that for n > N 0
JJCnxJJ <
(3.10.1)
B
for all x EA. Thus for n, m, N 0 < n < m,
since
Necessity. Suppose that the set A is not bounded. Then there are a sequence of numbers {tn} tending to 0, a sequence of numbers {an}, lanl ~ 1, and a sequence of indices {nk} such that the sequence n•+l { fk
};
GnXn}
does not tend to 0.
n=n•+l Let Cn
=
{~an
for nk < n ~ nk+I• elsewhere. 00
The sequence {en} tends to 0, but the series Therefore, the sequence
{xn}
2
CnXn
is not convergent.
n=l
is not a Csequence.
D
An Fspace X is called a Cspace if for any Csequence {Xn} the series 00
2
Xn n=l
is convergent. As an example of an Fspace which is not a Cspace
we can consider the space c0 (see Example 1.3.4.a). Indeed, the standard 00
basis {e11 } in c0 is a Csequence, but the series
2 en is not convergent. n=l
A subspace of a Cspace is also a Cspace. Therefore, if an Fspace X contains a subspace isomorphic to c0 , then X is not a Cspace.
Locally Pseudoconvex and Locally Bounded Spaces
167
Now we shall prove the following THEOREM 3.I0.2 (cf. Bessaga and Pelczynski, I958). Let X be an Fspace with a basis {en}· If X is not a Cspace, then X contains a subspace isomorphic to the space c0 • The proof is based on the following PROPOSITION 3.10.3 (Schwartz, 1969). Let (X, II II) be an Fspace such that each Csequence tends to 0. Then X is a Cspace. Proof Suppose that X is not a Cspace. Let {xn} be a Csequence such 00
that the series
2
Xn is not convergent. Then there are an increasing
n~l
sequence of indices nk and a positive number /J such that IIYkll
> IJ, where
nk+t
Yk
=
.2.)
Xn.
n~n•+l
The sequence {Yk} is also a Csequence, and it does not tend to 0. Proof of Theorem 3.10.2. If (X,
II II) is not a Cspace, then there is a
0 Cse
CD
quence {xn} such that the series
2 Xn
is not convergent. Basing our
n~I
selves on Proposition 3.10.3, we can assume without loss of generality that the sequence {xn} does not tend to 0. Let {fk} be basis functionals corresponding to the basis {ek}. Since {xn} is a Csequence, {fk(Xn)} are Csequences for k = I ,2, ... Hence lim fk(Xn) = 0,
k =I, 2, ...
(3.I0.2)
.,_..00
By Theorem 1.2.2 we may assume that the norm II II is nondecreasing. Since {xn} does not tend to 0 and (3.10.2) holds~ we can find a subsequence {xflp} = {zp}, a positive number/Jand an increasing sequenceofindices {np} such that · '
1
llzpZpll < 2p IJ.
(3.I0.3)
Chapter 3
168
where n,+,
2,;
z~ =
fn(zp)en.
n=np+l
Corollary 2.6.6 implies that {z~} is a basic sequence. We sp.all show that it is equivalent to the standard basis of c0 • Let {tp} be a sequence of coefficients of x belonging to X 0 =lin {z~} with respect to the basis {z~}. Since //z~ll > 6/2, the sequence {tn} tends to 0. On the other hand, (zp} is a Csequence. Therefore, by (3.10.3), {z~} is also a Csequence. co
This means that if tn+0, then the series}; tpz~ is convergent. Therefore, p=l
the basis z~ is equivalent to the standard basis of c0 • Thus, by Theorem 2.6.3, the space X 0 is isomorphic to the space c0 • D PROPOSITION 3.10.4. Let (X, II II) be a locally bounded space. with a basis {en}. Let Y be a subspace of the space X. Suppose Y is not a Cspace. Then the space Y contains a subspace Y0 isomorphic to the space c0 • Proof Without loss of generality we may assume that I II is phomogeneous. Using the same construction as in the proof of Theorem 3.10.2, we find elements {zp} and {z~}. By formula (3.10.3) and Theorem 3.2.16 the sequence {zp} is a basic sequence. The rest of the proof is the same as in Theorem 3.10.2. D Let X be a nonseparable Fspace. If Xis not a Cspace, then it contains an infinite dimensional separable subspace X which is not a c~space. Indeed, if X is not a Cspace, then there is a Csequence {xn} such that 00
the series };
Xn
is not convergent. Let
X= lin{xn}. The space X has the
n=l
required properties. By the classical theorem of Banach and Mazur (1933), space C[O, I] is universal for all separable Banach spaces. The space C[O, 1] has a basis (Example 2.6.10). Thus we have PROPOSITION 3.10.5 (Bessaga and Pelczynski, 1958). If X is a Banach .space which is not a Cspace, then X contains a subspace isomorphic to c0 • Now we shall prove
Locally Pseudoconvex and Locally Bounded Spaces
THEOREM 3.10.6 (Schwartz, 1969). The spaces LP(D,E,p), 0 ~ p are Cspaces.
169
< +oo,
The proof is based on the following propositions. PRoPOSITION 3.10. 7 (KolmogorovKchintchin inequality ; see also Orlicz, 1933b, 1951, 1955). Suppose that in the space L 0(D,E,p,) a Csequence {x,.(t)} is given. Then on each set D 0 of finite measure the series 00
}; lx,.(t)l 2 convergent almost everywhere. n=l
Proof(Kwapien, 1968). Let {r,.(s)} = {sign(sin2n211s)} be a Rademacher system on the interval [0, 1]. Since under our hypothesis {x11 (t)} is a Csequence, the set of elements
"
{ Yn,s(t)
=}; rt(s) Xt(t):
n = 1, 2, ... , 0
~ s ~ 1}
i=l
is, by Proposition 3.10.1, bounded in the space L 0 (D,E,p). This implies that for any positive B there is a constant C such that, for all n = 1,2, ... and all s, 0 ~ s ~ 1, ~t({te
D 0 : IYn,s(t)i ~ C}) ~ ~t(D0)B.
(3.10.4)
Let n be fixed al!d let
At= {s: IYn,s(t)i ~ C}.
(3.10.5)
By the Fubini theorem and (3.10.4) we find that the set E of those t for which the Lebesgue measure IAel of the set At is greater than lye has a measure greater than ~t(D0)ye, i.e. if E
= {t e D0 : jAel
~
2y'e},
then ~t(E) ~ ~t(D 0)ye.
Lett e E. Formula (3.10.5) implies n
JI}; rt(s) Xt(t)r ds ~ C
Ao
i=l
2•
Chapter 3
170
Thus n
J _2/xt(t)/ 2 2 J ( _2 A, i=l
A,
Re(xt(t)x,(t))rt(s) r;(s)) ds
~ C2 •
(3.10.6)
l~i<j~n
Applying the Schwarz inequality to (3.10.6), we obtain n
C ;;?: (1y'B)}; /xt(t)/ 2 2
i=l
2 (
_2
/xt(t)x1(t)J2) 112 (
l~i<j~n
_2
( Jrt(s)r1(s)dsrf' 2.(3.10.7)
l~i<j~n
A,
Since rt is a Rademacher system, J rt(s) r;(s)ds = A•
(3.10.8)
J ri(s) r,(s)ds, 1'\..A•
where I denotes the interval [0, 1]. The system {ri(s)rJ(s)}, 1 ~ i <j, is an orthonormal system in the space £2[0, 1]. The Fourier coefficients of the characteristic function X['\..A• of the set /~At are equal to fri(s)r,(s)ds, thus equalto Jri(s)rJ(s) 1'\..A•
A,
ds, by (3.10.8). Therefore, by the Parseval inequality. _2
(
l~i<j~n
f rt(s) r,(s)dsr ~/!"'At/ <{e.
(3.10.9)
A•
Observe that
( l,
/xt(t)x,(t)/ 2
l~i<j~n
n
= ( };Jxt(t)/ 2 i1
tt =
f'
2
~
(
_2 l~i,j~n
/xt(t)x1(t)J 2
f'
2
n
(3.10.10)
};Jxt(t)/ 2 • i~l
Formulae (3.10.7), (3.10.9), (3.10.10) imply n
(1y'e 2tJe) _2/xt(t)/ 2 ~ C 2 •
(3.10.11)
i=l 00
Formula (3.10.11) implies that the series 2;/xi(t)/ 2 is convergent on the i=l
171
Locally Pseudoconvex and Locally Bounded Spaces
set E. Since p(E) > p(.Q0)yc, the arbitrariness of 'e implies that the D series is convergent almost everywhere on .Q0 • 3.10.8. Let Then {xn} tends to 0.
CoROLLARY
{x11 }
be a Csequence in the space LO(.Q,E,p).
LEMMA 3.10.9 (Schwartz, 1969). Let {x11 } be a Csequence in a space LP(.Q,E,p), 0
represent the set .Q in the form .Q
=
U
Dt u N, where, on the set N,
i~I
all functions x 11 (t) are equal to 0 almost everywhere, .Q~, C .Qi+I• and for each i the sequence of functions {x 11(t)} tends uniformly to 0 on Dt. Suppose that the sequence {x 11 (t)} does not tend to 0 in LP(.Q,E,p). Then there is a positive number b' such that, for infinitely many n, J/xn(t)/Pdp
> b'.
Q
Let 0 < b < b'. We shall construct by induction a sequence of disjoint sets {Ak}, k = 1,2, ... , and increasing sequences of indices {nk}, {mk} such that Ak C Dm. and
J/x .(t)/P dp > b, 11
A•
(3.10.12)
forj=/=k.
Let n0 be an arbitrary index such that
J /x D
is an index m 0 and a set A 0 C Dm0 such that
110
(t)/ Pdp > b. Then there
J /x
110
(t)/Pdp >b. Suppose
A,
that Ak, nk, mk are defined for k
=
0,1, ... , r1. Then there is an index
m~ and a set Br C Dm'T such that Ak C Br for k
k
= 0, 1, ... , r1 and
= 0, 1, ... , r1.
Since Br C !Jm'T and the sequence {x11(t)} tends uniformly to 0 on !Jm',T
172
Chapter 3
there is an index n,.
> nr1 such that
Jlxnlt)l dp ~ 2~,
Br
and
J lxnr(t)IP dp >b.
(3.10.13)
t:J"Br
Formula (3.10.13) implies that there is an index m, and a set A, C Qmr disjoint with the sets A 0 , ••• , Ar1 such that
Jlxnr(t)IPdp >b.
(3.10.14)
Ar
In this way we have constructed by induction the required sequence. Let {en} be a sequence of positive numbers tending to 0 such that co
,2;c: = +oo.
(3.10.15)
n=1
Then r
r
r
f\2: CkXn.(t)\P dp ~ 2 J12: CkXn.(t)\p dp (J
k=O
j=O A1
r
k=O
r
J(cJixnlt)IP L cflxn.(t)IP)dp
~};
j=OA1
k=O k#j
r
~b
2
j=O
r
cJ C
2 2;~k
j,k=O
r
~ b}; cJ4C,
(3.10.16)
;~o
where C =sup c!. n co
Then, by (3.10.15), the series}; Ck xn. is not convergent in LP(Q,E,p). k=1
Therefore, {xn} is not a Csequence and we obtain a contradiction.
0
LEMMA 3.10.10 (Schwartz, 1969). If {xn} is a Csequence in a space LP(Q,E,p), 1 ~p <+oo, then Xn+0.
Locally Pseudoconvex and Locally Bounded Spaces
173
Proof The proof is the same as the proof ofLemma3.10.9, but in (3.10.16) the Minkowski inequality is used. D Proof of Theorem 3.10.6. The proof is a trivial consequence of Proposition 3.10.2 and Lemmas 3.10.9, 3.10.10. 0 3.10.11. The space LD[O, 1] is not universal for separable Fspaces. Proof The space L 0 [0, 1] is a Cspace. Hence it does not contain a subspace isomorphic to c0 • D CoROLLARY
The notions of Csequences and Cspaces introduced by Schwartz (1969) are similar to what Matuszewska and Orlicz (1968) called condition (0). . A sequence {Xn} of elements of an Fspace X is called perfectly bounded if the set N
A0 = {
.2;
Xpn:
Pn runs over all finite increasing systems of
n=l
positive integers} is bounded. An Fspace X satisifies condition (0) if each perfectly bounded sequence is unconditionally convergent. It is easy to verify that for Fspaces satisfying condition (0) Theorem 3.10.2 holds. Therefore, if an Fspace X can be imbedded into an Fspace with a basis, then X is a Cspace if and only if it satisifies condition (0). Matuszewska and Orlicz (1968) showed that a large class of modular spaces (in particular, spaces N(L(Q,E,p))) satisfy condition (0).
3.11.
LOCALLY BOUNDED ALGEBRAS
Let X be a linear space. We say that X is an algebra if there is an operation·: xxx~x, called multiplication, which is associative (x·y)·z
=
x·{y·z),
x,y, zeX
(3.11.1)
174
Chapter 3
and bilinear (x1+x2)·y = X1·y+x2·y,
(3.11.2)
X·(JI+Y~
= X·Y1+X·Y2
(3.11.3)
= t(x·y) = x·(ty)
(3.11.4)
(tx)·y
for x,y E X, t being a scalar. For brevity we shall write x·y = xy. We say that an algebra X is commutative if xy
=
yx.
(3.11.5)
We say that an algebra Xhas a unit ifthere is anelemente EX such that ex= xe
= x.
(3.11.6)
It is easy to verify that the unit is unique. If an algebra X has a unit, then (3.11.4) automatically holds. In the sequel only commutative algebras with a unit will be considered. For brevity we shall call them simply algebras. We say that an algebra X is an F*algebra if it is an F*space and the operation of multiplication is continuous. We say that an F*algebra X is locally bounded if it is a locally bounded space. The theory of locally bounded algebras was developed by Zelazko (1960, 1962, 1962b, 1963, 1965). It is a generalization of the classical theory of Banach algebras. THEOREM 3.11.1 (:Zelazko, 1960). Let X be a locally bounded algebra. Then the topology in X can be determined by a phomogeneous norm which is submultiplicative, i.e.
llxyll ~ llxiiiiYII·
llxll
(3.11.7)
If X is complete, then it is also complete with respect to the norm II 11. Proof By Theorem 3.2.1 the topology in X can be determined by a pho
mogeneous norm
llxll
=
Observe that
llxll'. Let llxyll'
~1o TYIT''
llxll is finite, since the multiplication is a continuous oper
Locally Pseudoconvex and Locally Bounded Spaces
ator (cf. Theorem 3.2.13). It is easy to verify that submultiplicative and [fell = 1. Since
175
llxll is phomogeneous,
Jlxll ?: Jjxjj' lle!l' the norm llxll is stronger than the norm llxJ['.
(3.11.8)
Now we shall identify x e X with a continuous linear operator L.e mapping X into itself by formula Lxy = xy. Observe that the norm operator IILxll of the operator Lx is equal to llxll. Thus we can consider (X, II JJ) as a subspace of the space (B(X), II JJ). If { Xn} tends to x in the space (X, II II'), then, by the continuity of multiplication, Lx,. tends to Lx in (X, II II). Thus the norms II II and II II' are equivalent. D Let X be an algebra. An element x e X is called invertible if there is an element x 1 e X, called inverse to x, such that (3.11.8) It is easy to verify that x 1 is uniquely determined (provided it exists) and that (x 1) 1 = x. PROPOSITION 3.11.2. Let (X, II II) be a complete locally bounded algebra. Let x be an invertible element in X. Then there is a neighbourhood of zero U such that, for y e U, the element xy is invertible. Proof Without loss of generality we may assume that II II is phomogeneous and submultiplicative (see Theorem 3.11.1). We can write co
(xy)1
= x1(e.x1y)1 = x1), (xIy)n, ..:....; n=O
where the series is convergent provided
IIYII <  11 . llx 11
D
Let X be an algebra. A set M C X, {0} =I= M =I= X is called ideal if it is a linear subset of X and for all x e X, xM C M. Let X be an Falgebra, by the continuity of multiplication we conclude that for any ideal M its closure M is also an ideal provided that M =I= X. We say that an ideal M C X is maximal, if for any ideal M1 such that M C M 1 we have M = M 1 •
Chapter 3
176
PROPOSITION 3.11.3. Let X be a complete locally bounded algebra. Then every maximal ideal is closed. Proof Let M be a maximal ideal in X. By Proposition 3.11.2, e ¢ M. Thus M is an ideal. Since M is maximal and M C M, M = M. D PROPOSITION 3.11.4. Let X be a complete locally bounded algebra over complexes. The function (x.A.e) 1 is analytic on its domain. Proof Let A.0 belong to the domain of the function (x.A.e) 1 • This means that the element xA.0e is invertible. Let y = (A.A. 0 )e. Then (xA. 0 e)y = xA.e. Therefore (xA.e) 1
=
((xA. 0 e)y)1 CXl
= (xA. 0 e)1 } ; (xA. 0 e)n(A.A.0)n, n=1
where the series on the right is convergent provided
JA.A.oJ ~ JJ(xA.oe)1 111 • Thus (xA.e) 1 is analytic.
D
Proposition 3.11.4 implies an extension of the theorem of Mazur and Gelfand (Mazur, 1938; Gelfand, 1941). THEOREM 3.11.5 (Zelazko, 1960). Let X be a complete locally bounded field over complex numbers. Then X= {A.e: A. being a complex number}. Proof Suppose that there is an x E X which is not of the form .A.e, i.e. x =1= A.e for all A.. Since X is a field (xA.e) 1 is well defined on the whole complex plane. By Proposition 3.11.4 it is an analytic function. Observe that lim (xA.e)1 = 0 . .t...oo
Thus, by the Liouville theorem (Theorem 3.5.4), (xA.e) 1 = 0 and we obtain a contradiction. D Let X be a complete locally bounded algebra over the complex numbers. Let f be a linear functional defined on X (not necessarily continuous). We say that the functional/is a multiplicativelinear functional if f(xy) = f(x)f(y).
(3.11.9)
Locally Pseudoconvex and Locally Bounded Spaces
177
There is a onetoone correspondence between multiplicativelinear functionals and maximal ideals. Namely, for a given multiplicativelinear functional f the set M1 = {x: f(x) = 0}
is an ideal. It is a maximal ideal since it has codimension 1. Since X is locally bounded, it is closed by Proposition 3.11.3. Thus the functional f is continuous. Thus we have proved PROPOSITION 3.11.6 (Zelazko, 1960). In complete locally bounded algebras all multiplicativelinear functionals are continuous.
Let M be a maximal ideal in the algebra X. Then X/ M is a complete locally bounded field. Therefore it is isomorphic to the field of complex numbers (see Theorem 3.11.5) and this isomorphism induces the required multiplicativelinear functional. A locally bounded algebra X is called semisimple if {x: F(x) = 0 for all multiplicative linear functionals} = {0}. If a complete locally bounded algebra X over complex numbers is semisimple, then x E X is invertible if and only if F(x) # 0 for all multiplicativelinear functionals F. Indeed, x is invertible if and only if x does not belong to any maximal ideal. In the case of complete locally bounded algebras this implies that F(x) # 0 for all multiplicative functionals. Now we shall present an application of locally bounded algebras to the theory of analytic functions. For this purpose we shall present the following example of a locally bounded algebra. Example 3.11.7 Let N(u) be a nondecreasing continuous function, defined for u?: 0, such that N(u) = 0 if and only if u = 0. We shall assume that for sufficiently small v, u ~
N(u+v)
and there is a C N(uv)
> 0 such that for sufficiently small u,
~
(3.11.10)
N(u)+N(v)
CN(u)N(v)
v
(3.11.11)
Chapter 3
178
> 0 such that
and there is a p
N(u) = N 0(uP),
(3.11.12)
where N 0 is a convex function in a neighbourhood of zero. Let X= N(l) be the space of all sequences x = {x0 ,x1 , that
••• }
such
co
llxll = PN(X) =
J: N(lxni) < +oo.
n=O
The space (X, II II) is an Fspace (see Proposition 1.5.1). By Theorem 3.4.3 it is locally bounded. Now we shall introduce multiplication in X by convolution, i.e. if x = {xn}, y = {Yn} then we define n
X·y
=
{
.2: XkYnk}·
k=O
By (3.11.10) and (3.11.11) we conclude that for sufficiently small xandy (3.11.13) llxyll ~ C llxiiiiYII. Formula (3.11.13) implies that the multiplication is continuous. Thus X is a complete locally bounded algebra. Now we shall give examples of functions satisfying conditions (3.11.10)(3.11.12). The simplest are functions N(u) =uP, 0
0 (positive homogeneity). Let X 0 be a subspace of the space X. Let fo(x) be a linear functional defined on X 0 such that fo(x)
l
0 uP logu 2p 1 e2
foru = 0,
for 0 < u ~ e 2/P, for e 2/P < u.
We shall show that N(u) satisfies (3.10.10)(3.10.12) By the definition, N(O) = 0. The function N(u) is continuous at point 0 since limN(u) = 0, U+0
and at point e 2/P since N(e 21P) = 2p 1 e 2• In the interval (0, e 2/P) it is continuous as an elementary function. The function N(u) is nondecreasing. Indeed, on the interval (0, e 2/P) dN du
= puP1 loguuP 1 = uP1 (p logu+1) > 0,
because logu < 2/p.
Locally Pseudoconvex and Locally Bounded Spaces
179
. Now we shall calculate the second derivative d2 N du
= (pl)uP2(plogu+l)puP 2 2 = uP2(p(pI)logu+2pl) = uP2((pI)(plogu+ I)+ I) < 0 for 0 < u ~ e 2/P. Thus N(u) is concave on the interval (0, e 21P). Now formula (3.Il.1I) will be shown. Suppose we are given u, v 0 < u, v ~ e 2/P. Then (uv)Piloguvl = (uv)P llogu+ log vi
~
uP llogul vP llogvl,
because llogu, logvl ~ 2/p ~ 2. Let N 0(x) = x 2 logx. It is easy to see that if x = uPI 2 , then N 0 (uP1 2) =
2p
N(u).
(3.Il.14)
We shall show that N 0 is convex in a neighbourhod of zero. Indeed, dN a:x= 0
2xlogxx,
dN~
dx 2 =. 2logx21 = 2logx3
and the second ?erivative is greater than 0 on the interval (0, e 3/ 2). Therefore the function N 0 is convex in a neighbourhood ofO and (3.Il.12) holds. Example 3.11.8 Let N be as in Example 3.11. 7. By N~ (I) we shall denote the space of all sequences of complex numbers x = {xn}, n = ... , 2,1,0,1,2, ... such that 00
llxll
= PN(x) =
.2;
N(lxni)
< +oo.
(3.11.15)
n=c:o
In a similar way as in Example 3.11. 7 we can show that (N~(l),
II II) is
Chapter 3
180
a complete locally bounded space. We introduce multiplication a by the convolution +oo
xy
= { };
XnkYk}.
k=00
In a similar way as in Example 3.11.7 we can show that N::_(t) is a complete locally bounded algebra. By NF we shall denote the algebra of measurable periodic functions, with period 2n, such that the coefficients of the Fourier expansions +oo
x(t)
= };
Xneint
n=oo
belong to the space N::_(1). The operations of addition and multiplication are determined as pointwise addition and multiplication. It is easy to see that the pointwise multiplication of functions in NF induces the convolution multiplication in the space N::_(1). Thus the space NF can be regarded as a complete locally bounded algebra with topology defined by the norm (3.11.15). We shall show that each multiplicative linear functional defined on algebra NF is of the form F(x)
=
x(t0 )
(3.11.16)
Indeed, let z = eit. The element z is invertible in NF. We shall show that JF(z)J = 1. Suppose that JF(z)J > 1. Then there is a, 0 1 and we can repeat the preceding considerations. Thus JF(z)J = 1 and F(z) = eti, for a certain t 0 , 0 < t 0 ~ 2n. Since F is a multiplicative linear functional, we obtain for every polynomial n
p(z) =
.2 a,zi 1
i=k
(here n, k are integers not necessarily positive), F(p(z))
= p(F(z)) = p(eil•).
Locally Pseudoconvex and Locally Bounded Spaces
181
The polynomials are dense in algebra NF and hence the functional F is of the form (3.11.16). This implies THEOREM 3.11.9. Let N be a function satisfying the condition described in Example 3.11. 7. Let x(t) be a measurable periodic function, with period 2n, such that the coefficients {xn} of the Fourier expansion
+oo
x(t) =
L Xnetnt n=oo
form a sequence belonging to the space N~(l). If x(t) # 0 for all t, then the function Ifx(t) can also be expanded in a Fourier series +oo
.J:
_I= Ynetnt x(t} n~oo
such that {Yn} eN~(/).
For N(u) = u this is the classical result of Wiener. For N(u) =uP, 0 < p ~ 1 it was proved by Zelazko (1960). Let X be a locally bounded complete algebra over complex numbers. Let x eX. By the spectrum a(x) of x we mean the set of such complex numbers A. that (x A.e) is not invertible. By Proposition 3.11.2, the set of such complex numbers A. that (xA.e) is invertible is open. Moreover, if A.> llxll, the element (xA.e) is also invertible. This implies that the set a(x) is bounded and closed. Hence it is compact. Let A. 0 e a(x). Then by the definition of a spectrum, the element (xA.0e} is not invertible. Then there is a multiplicative linear functional F such that F(x A.0e) = 0, i.e. F(x) = A. 0 • Conversely, if A.0 ¢ a(x), then, for each multiplicative linear functional F, F(x) # A.0 • Thus a(x) = {F(x): Fruns over all multiplicative functionals}.
Let (1}(A.) be an analytic function defined on a domain U containing the spectrum a(x). Let FeU be an oriented closed smooth curve containing a(x) inside the domain surrounded by r. We shall define I
(1}(x) = 2ni
J(1}(A.)(xA.e)ldA.. r
Chapter 3
182
The integral on the right exists since r f"'\ a(x) = that, for any multiplicative linear functional F, F((/J(x)) = {/}(F(x)).
0.
It is easy to verify (3.11.17)
Applying (3.11.17) to the algebra NF, we obtain 3.11.10. Let x(t)ENF. Let (/J(A.) be an analytic functions defined on an open set U containing THEOREM
a(x) = {z: z = x(t),O
<
t::s;;2n}.
Then the function (/J(x(t)) also belongs to NF.
For N(u) = u we obtain the classical theorem of Levi. For N(u) = uP, 0 < p:::;;;; 1, Theorem 3.11.10 was proved by Zelazko (1960). Let N satisfy all the conditions described in Example 3.11. 7. By N H we denote the space of all analytic functions x(z) defined on the open unit disc D such that the coefficients {xn}, n = 0, I, ... of the power expansion 00
x(z) =
.2; Xnzn n=O
form a sequence {xn} belonging toN(!). There is a onetoone corespondence between pointwise multiplication in N H and the convolution in N(l). Thus we can identify NH with N(l). Now we shall show that every multiplicative linear functional F defined on NH is of the form F(x) = x(z0), lzol:::;;;; 1. To begin with we shall show this for x(z) = z. Suppose that F(z) = a, lal > I. Then F(zfa) = I. Therefore F(anzn) = 1. On the other hand, anzn*O, and this leads to a contradiction with the continuity of F. Observe that N(l) C 1. Therefore each function x(z) E NH can be extended to a continuous function defined on the closed unit disc D. Thus we have 3.11.11. Let x(z) E NH. Let (/J be an analytic function defined on an open set U containing x(D). Then (/J(x(z)) E NH. THEOREM
Locally Pseudoconvex and Locally Bounded Spaces
183
Theorems 3.11.10 and 3.11.11 can be extended to the case of many variables in the following way THEOREM 3.11.12 (Gramsch, 1967; PrzeworskaRolewicz and Rolewicz, 1966). Let x 1, •.• , Xn E NF (or NH). Let
=
{(x1(t), ... ,Xn(t)): 0
~
t ~ 2n}
a(x) = {x1(z), ... , Xn(z)): lzl ~ 1}).
Then the function <J>(xl> ... , Xn) belongs to NF (or respectively, to NH).
We shall not give here an exact proof. The idea is the following. Replacing the Cauchy integral formula by the Weyl integral formula, we can define analytic functions of many variables on complete locally bounded algebras. ·
3.12.
LAW OF LARGE NUMBERS IN LOCALLY BOUNDED SPACES
Let (Q, .E, P) be a probability space. Let (X, [[ I[) be a locally bounded space. Let the norm II II be phomogeneous. As in the scalar case, a measurable function X(t) with values in X will be called a random variable. We say· that two random variables X(t), Y(t) are identically distributed if, for any open set A C X P({t: X(t)
E
A})= P({t: Y(t)
E
A}).
A random variable X(t) is called symmetric if X(t) and X(t) are identically distributed. Random variables X 1(t ), ... , Xn(t) are called independent if, for arbitrary open sets AI> ... , An
n n
P({t: Xt(t) EAt, i
=
1, ... , n})
=
P({t: Xt(t) EAt}).
i=l
A sequence of random variables {Xn(t)} is called a sequence of independent random variables if, for each system of indices n1o ... , nk, the random variables Xn 1 (t), ... , Xn.(t) are independent.
Chapter 3
184
3.12.1 (Sundaresan and Woyczynski, 1980). Let X be a locally bounded space. Let II II be a phomogeneous norm determining the topology in X. Let {Xn(t)} be a sequence of independent, symmetric, identically distributed random variables. Then
THEOREM
E(IIX1ID
=
JIIXl(t)lldP < +oo
(3.12.1)
a
if and only if (3.12.2)
almost everywhere. Proof. Necessity. To begin with we shall show it under an additional hypothesis that X1 takes only a countable number of values x1,x2 , ... Since Xn are identically distributed, all Xn admit values x 1, ... For each positive integer m we shall define new random variables Xk'(t)
= { ~k(t)
if Xk(t) = x 1 , elsewhere.
... , Xm,
By Rk'(t) we shall denote Xk(t)xr(t). For each fixed m, {IIXrll} constitutes a sequence of independent identically distributed symmetric random variables taking real values. Moreover, (3.12.3)
The random variables {xr} takes values in a finitedimensional space. Thus we can use the strong law of large numbers for the onedimensional case (see for example Petrov, 1975, Theorem IX.3.17). Let an = n11P. Then CX)
.}; ak2 n=k
CX)
= .}; n2/p n=k = O(n 2/P+l) =
O(na;: 2 ).
(3.12.4)
Having (3.12.3) and (3.12.4), we can use the strong law of large numbers by coordinates (here we use the fact that xr takes values in a finite dimensional space).
Locally Pseudoconvex and Locally Bounded Spaces
185
Thus (3.12.5) almost everywhere. At the same time, IIR::'II is a sequence of independent identically distributed real random variables with finite expectation, so that, by the classical strong law of large numbers,
IIR~II+ ... +IIR::'II +E(IIR~II) n
(3.12.6)
almost everywhere. Since IIR~II tends pointwise to 0 as m tends to infinity and IIR~II ~ !lXIII, by the Lebesgue dominated convergence theorem E(IIR~II) tends to 0. The set D 0 of those t for which (3.12.5) and (3.12.6) converge at t is of full measure, i.e. P(D0 ) = P(D) = 1 Let 8 > 0. Choose an m > 0 such that m
8
E(IIR1 ID < 4· For any t E D 0, we can find anN= N(8,t,m) such that
llnliP(X~(t)+ ... X~(t))ll < ; for n > N. Thus, for t E D 0 and n
(3.12.6)
> N,
This completes the proof under the condition that Xn are countable valued. To complete the proof of necessity we shall use the standard approximation procedure. For each 8 > 0, there is a symmetric Borel function T, taking values in a countable set in X such that
IITs(x)xll <
8.
(3.12.7}
Chapter 3
186
Hence lln 11P(X1+ ... +Xn)ll ~ 1 n +lln 11P(T,(X1)+ ... +T.(Xn))ll.
~ CIIX1 T.(X1) II+ ... + IIXn T.(Xn)ll)
Since the first term is less than e and the second one tends to 0 we obtain (3.12.2). Sufficiency. Observe that (3.12.2) implies that n1
n 1/PXn =n 11P(X1+ ... +Xn) ( n
)1/p(n1) 1P(X 1
1
+ ... +Xn1) . IIXnll tends to 0 almost everywhere. Thts means that 70 almost everyn where. The random variables {IIXnll} are independent. Then, using the classical result from probability theory (see for example Petrov, 1975, Theorem IX.3.18), we obtain that co
}; P({t: IIX1(t)ll ;;::o n}) < +oo. n=O
Renee co
E(IIX111) ~}; nP({t: n1 ~ IIX1(t)ll n=1
< n})
co
=}; P({t: IIX1(t)ll ;;::on}) < +oo.
0
n=O
Other results concerning convergence of random variables in nonlocally convex spaces the reader can find in Woyczynski (1969, 1974), RyllNardzewski and Woyczynski (1974); and Marcus and Woyczynski (1977, 1978, 1979).
Chapter 4
Existence and NonExistence of Continuous Linear Functionals and Continuous Linear Operators
4.1. CONTINUOUS
LINEAR
F UNCTIONALS
AND
OPEN CONVEX
SETS
Let X be an F*space. Let f be a continuous linear functional defined on X. We say that functional f is nontrivial iff =I= 0. If there is a nontrivial linear continuous functional defined on X, we say that X has a nontrivial dual space X* (briefly X has a nontrivial dual). If each linear continuous functional defined on X is equal to 0, we say that X has a trivial dual. Let U =
{x: Jf(x)l < 1}.
The set U is open as an inverse image of the open set {z: lzl < 1} under a continuous tra,nsformation. Moreover, the set U is convex, since if x,y e U, a,b ~ 0, a+b = 1, then if(ax+by)i < aif(x)i+blf(y)i < 1. This implies that if an F*space X has a nontrivial dual, then there is an open convex set U C X different from the whole space X. We shall show that the converse fact is also true. Namely, if in an F*space X there exists an open convex set different from the whole space X, then there is a nontrivial continuous linear functional. To begin with we shall prove this for real F*spaces. Let X be a real F*space. Let us suppose that there is a convex open subset U of X different from the whole space X, U =I= X. Since a translation maps open sets on open sets, we may assume without loss of generality that 0 e U. 187
Chapter 4
188
Let
= inf{t > 0:
llxllu
~ E u}.
Evidently, (4.1.1)
llxllu:?::0 and U
= {x: llxllu < 1}.
Moreover lltxllu
= tllxllu for t > 0
(positive homogeneity)
(4.1.2)
and llx+yllu ~ llxllu+IIYIIu.
(4.1.3)
Formula (4.1.2) is trivial. We shall prove formula (4.1.3). Let e be an arbitrary positive number. The definition of llxllu implies and The set U is convex; therefore llxllu x IIYIIu Y (1e) llxllu+II.YII~ llxllu +(1e) llxllu+II.Ylk7" IIYIIu
x+y
= (1e) llxllu+IIYIIu
E
U.
Since e is an arbitrary positive number, we obtain llx+yllu =:;::: 1 llxllu+IIYIIu ""' ' and this implies (4.1.3). A functional satisfying conditions (4.1.1)(4.1.3) is called a Minkowski functional. If we replace (4.1.2) by llxllu
= ltlllxllu
for all scalars t,
(4.1.2')
then a Minkowski functional becomes a homogeneous pseudonorm (see Section 3.1). Let us remark that llxllu is a homogeneous pseudonorm if and only if the set U is balanced.
Existence of Continuous Linear Functionals and Operators
189
Let X be an F*space. Let f(x) be a linear functional defined on X. If there is an open convex set U containing 0 such that Jf(x)J
< llxllu,
then the functionalf(x) is continuous. On the other hand, if f(x) is a continuous linear functional and
{x: Jf(x)J < 1 }.
U=
then lf(x)J
< llxllu.
THEOREM 4.1.1 (Hahn 1927, Banach 1929). Let X be a rea/linear space. Let p(x) be a realvalued functional (generally nonlinear) such that: (1) p(x+y)
< p(x).
Then there is a linear functionalf(x) defined on the whole space X such •that f(x) =fo(x)
for x
f(x) <,.p(x)
for xe X.
E
X0
and (4.1.4)
Proof We say that a iinear functional h(x) is an extension of a linear functional g(x) if the domain of the functional h(x) contains the domain of the functional g(x) and both functionals coincide on the domain of the functional g(x). Let m: be the set of all extensions of the functional fo such that g(x)
for x belonging to the domain of g.
We can introduce a relation of partial order in m: in the following way. We say that h 3 g if g is an extension of h. By the KuratowskiZorn lemma there is a maximal element of the family m:. Let {f(x) be such a maximal element. To complete the proof it is sufficient to show that the domain of the functionalf(x) is the whole space X.
Chapter 4
190
Suppose that the above does not hold, i.e. that the domain Y of the functionalf(x) is different from the whole space X. Let y 0 e Y and let us denote by Y0 the space spanned by Y and y 0 • Obviously each element x e Y0 can be written uniquely in the form x = y+ay 0 , where y e Yanda is a scalar. Let x',x" e Y. Then f(x')+f(x")
= f(x' +x") ~ p[(x' +Yo)+(x" ~p(x' +yo)+p(x" y0 ).
Yo)]
Hence f(x")p(x" y0 )
~
f(x')+p(x'+y0 ).
Since x' and x" are arbitrary elements of Y, we obtain A= sup (f(x)p(xy 0 ))
~
XEY
B
=
inf(f(x)+p(x+y0 )). XEY
Let c be an arbitrary number such that A tional/1 in Y 0 in the following way: f1(y+ayo)
=
~
c ~ B. We define a func
f(y)+ca.
Obviously, ];_ is an extension of the functional
f We shall show that
f~e~.
Let a be a positive number. Then f1(y+ayo)
=
ca+f(y)
+J(y) =
=
~ aB+f(y) ~a(!(~ )+P(Yo+ ~))
f(y)+p(ayo+Y)+f(y)
p(y+ayo).
In the case where a is negative, the proof is similar, but the inequality c ~ A is applied instead of c ~ B. Hence f is not a maximal element of A and we obtain a contradiction. D 4.1.2. Let X be a real F*space. Let U be an open convex set in the space X different from X. Then there is a nontrivial continuous linear junctional in X. Proof Without loss of generality we can assume that 0 e U. Since U is different from the whole space X, there is an element x 0 such that CoROLLARY
Existence of Continuous Linear Functionals and Operators
191
llxollu =F 0. Let X 0 be the space spanned by x 0 • Let fo be a functional defined on the space X, in the following way: fo(x) = fo(axo) = allxollu.
Obviously,f0(x) ~ llxllu for x E X 0 • By Theorem 4.1.1 we can extend the functionalfo to a linear functional f(x) defined on the whole space X and satisfying the condition f(x) ~ llxllu for x EX. Therefore, the functional f(x) is continuous. D CoROLLARY 4.1.3. Let X be a complex F*space. Let us suppose that there is an open convex subset U C X different from the whole space X. Then there is in the space X nontrivial functional, continuous and linear (with respect to complex numbers). Proof Obviously, the space X can also be regarded as a linear space over reals. Therefore, by Corollary 4.1.2. there exists a nontrivial functional g(x), continuous and linear (with respect to reals). Let f(x) = g(x)ig(ix). Obviously, f(x) is a nontrivial functional, continuous and linear (with respect to reals). We shall show that f(x) is also linear with respect to complex numbers. Indeed, letz = a+ ib ; then f((a+ib)x) = g((a+ib)x)ig(i(a+ib)x) = ag(x)+bg(ix)iag(ix)+ibg(x) = (a+ib)(g(x)ig(ix)) = (a+ib)f(x).
D
THEOREM 4.1.4. Let X be a B~space. Thenfor any x 0 =F 0 there is a continuous linear functional f such that f(x 0 ) is different from 0. Proof Since the space X is locally convex, there is a convex neighbourhood of zero U such that x 0 rt U. Then, of course, llxollu =F 0. The rest of the proof follows the same line as the proof of Corollaries 4.1.2 and 4.1.3. D THEOREM 4.1.5 (Bohnenblust and Sobczyk, 1938). Let X be a normed space and let Y be a subspace of the space X. Let f 0(x) be a continuous linear functional defined on Y. Then there is an extension f(x) of the functionalfo(x) to the whole space X such that IIIII = llfoiiProof In the real case the theorem is a trivial consequence of Theorem
192
Chapter 4
4.1.1. Indeed, if we put p(x) such that
=
lllollllxll, then there is an extension
I
l(x) ~ p(x) = 11/ollllxll· This implies that II/II~ 11/oll· On the other hand, since I is an extension offo, we have 11/11 ;:::, ll.foll· Let us now consider the complex case. We shall define two linear real valued functionalsl1 andl2 on Y by the equality
fo(x) = };_(x)+ih_(x). Of course, II!III ~ ll.foll and lll2ll ~ 11/oll· Moreover, for x
E
Y,
l1(ix)+if2(ix) = lo(ix) = ifo(x) = i.h(x)12(x). Hence j;(x) =  };_(ix). Therefore fo(x) = };_(x)if1(ix). Let F 1(x) be a realvalued norm preserving extension};_(x) to the whole space X. Of course, by definition, IIF1II = ll.hll. Letl(x) = F1(x)iF1(ix). The functional f(x) is a continuous functional linear with respect to complex numbers (compare Corollary 4.1.3). Since F 1is an extension of};_, I is an extension of the functional fo. To complete the proof it is enough to show that 11/11 ~ ll.foll. Let x be an arbitrary element of X. Let 8 = argf(x). Then lf(x)l = lf(ei0x)l = IFI(eiOx)l ~ 11Flllllei0xll ~ ~ ll/1llllxll ~ lllollllxll· Hence 11/11 ~ llloll· COROLLARY
D
4.1.6. Let X be a normed space. Then
llxll = sup lf(x) I, 11/11~1
where the supremum is taken over all continuous linear lunctionalsle X*. Proof Let x 0 be an arbitrary element of X. Let X 0 be the space spanned by x 0 , i.e. the space of all elements of the type ax0 • Let us put fo(x) = allxoll for x E X 0 • The functional fo is of norm one. Basing ourselves on Theorem 4.1.5, we can extend it to a continuous linear function F 0 of norm one. Then llxoll = fo(x 0) = F0(x0 ). Hence D
Existence of Continuous Linear Functionals and Operators
193
Duren, Romberg and Shields (1969) gave an example of an Fspace X with a total family of continuous linear functionals such that X possesses a subspace N such that the quotient space X/N has trivial dual. Shapiro (1969) showed that /P, 0 < p < 1, also contain such subspaces N. Kalton (1978) showed that every separable nonlocally convex Fspace has this property.
4.2.
EXISTENCE
AND
NONEXISTENCE OF CONTINUOUS
LINEAR
FUNCTIONALS
4.2.1 (Rolewicz, 1959). Let
THEOREM
. . f N(t) 1tmtn 
> 0.
t Then in the space N(L(Q,E,p.)) there are nontrivial continuous linear functionals. Proof Corollaries 4.1.2 and 4.1.3 imply that it is enough to show that. in the space N(L(D,E,p.)) there is an open convex set U different from the whole space. Let E E E, where p.(E) = a, 0 < a < +oo. Since 1>00
lim inf N(t) 1>00
> 0, there are a positive constant a and a posii:ive number
t
T such that, for t > T, N(t) >at. Let U be a convex hull of the set {x: PN(x) < 1}. The set U is an open convex set. We shall show that it is different from the whole space N(L(Q,E,p.)). Let x1 , ••• , Xn be arbitrary elements such thatpN(xt) ~ 1, i = 1,2, ... , n Let Bt = {s: lxi(s)l > T}. Let Xt(s) xi(s) = { 0
for SE Bt, elsewhere,
and
Let
,
xi+ ...
+x~
,
xi'+ ...
+x~'
Xo = =and xo = =''n n Since lx;'(s)l~ T(i= l,2, ... ,n),lx"(s)I~T. It implies that (lx~'(s)ldp. il ~
Ta. Moreover a
Jlxi(s)ldp. ~JE N(lxi(s)ldJi.~PN(Xt) < 1
E
(i= 1,2, ... ,n).
Chapter 4
194
Hence
J/x~(s) /dp < 1/a. Therefore, the function z(s) = (T+ 1/a)xE does E
D
not belong to the set U.
If the measure fl has an atom E 0 of finite measure, then there is a nontrivial continuous linear functiomil in the space N(L(Q,E,p). Indeed, by the definition of measurable function, x(t) is constant on E0 palmost everywhere, x(t) =c. Let us put f(x) = c. It is obvious that f(x) is a continuous linear functional. If the measure Jl is atomless, the following theorem, converse to Theorem 4.2.1, holds :
THEOREM 4.2.2 (Rolewicz, I959). Let Jl be an atomless measure. N(t) 0 . . fI1m1n = ,
(4.2.1)
t
t>00
If
then there are no nontrivial continuous linear functionals in the space N(L(Q,E,p)). Proof Lets be an arbitrary positive number. Suppose that (4.2.1) holds.
Then there is a sequence {tm} tending to infinity such that N~:m) J>0. Let km be the smallest integer greater that N(tm)/s. Let E be an arbitrary set belonging to E of the finite measure. Since the measure Jl is atomless, there are measurable disjoint sets E 1 , ••. , Ek., such that km
E=
U Et i=l
(i
and
= I, 2, ... , km).
Let xi,.(s)
for s E Et (i = I, 2, ... , km), elsewhere.
tm
= {0
Obviously, for sufficiently large m, PN(x:J:::::::;; s. On the other hand, for seE Ym (s)
I
km
\1
t
tm
= k...::.... Xm(s) = k' m
i=l
m
Existence of Continuous Linear Functionals and Operators
Since
~= +oo, every function
195
of the type axE belongs to the convex hull
U of the set {x: PN(x) <e}. The setEis an arbitrary setoffinitemeasure. Hence all simple functions with suports of finite measures belong to U. The set U is open and convex. Therefore U = N(L(Q,E,J1)).
4.2.S (Day, 1940). In the spaces LP(Q,E,p), 0 <,p < 1, there is no nontrivial continuous linear functional provided the measure p is atomless. COROLLARY
Let X be an Fspace. We say that a family X' of continuous linear functional is total if f(x) = 0 for all f EX' implies that x = 0. Theorem 4.1.4 implies that, if X is locally convex, then the family of all continuous linear functionals is total. Let Y be a subspace of the space X. If there is a total family X' of continuous linear functionals on X, then the family X' is a total family of functionals on the space Y. Problem 4.2.4. Does every infinitedimensional F*space X contain an infinite dimensional subspace Y with a total family of continuous linear functionals?
We do not kpow the solution of the even simpler Problem 4.2.5. Does every infinitedimensional F*space X contain an infinitedimensional subspace Y with a nontrivial continuous linear functional?
We do not know whether Problems 4.2.4 and 4.2.5 are equivalent. There is also a sequence of weaker questions, for example : Problem 4.2.6. Does every infinitedimensional locally bounded space contain an infinitedimensional space with a nontrivial continuous linear functional?
We know only the following sufficient condition.
Chapter 4
196
We say that an Fspace Y contains arbitrarily short lines if, for each positive s, there is an element x E Y, x =ft. 0, such that
< s.
sup lltxll tscalars
PROPOSITION 4.2.7 (Bessaga, Pelczynski and Rolewicz, 1957). If an Fspace X contains arbitrarily short lines, then it contains a subspace X 0 isomorphic to the space (s) of all sequences (see Example 3.l.a"). Proof Let r(x) = sup lltxll· Let {en} be a sequence of elements of X /scalars
such that r(en+J)
<} r(en).
(4.2.2)
Formula (4.2.2) implies that for each sequence of scalars {tn} the 00
series~ tnen
is convergent.
n=1 00
We shall say that ~ tnen = 0 implies tn = 0 for n = 1, 2, ... n=1
Indeed, suppose that the above does not hold. Let tc be the first scalar different from 0. From the definition of r(x) it follows that there is a number a such that (4.2.3.) 00
On the other hand,
2
atnen
= 0. Hence
n=1 00
00
llateill :::;; }; llatnenll :::;; }; r(en):::;; n=i+1
n=i+l
00
(14 )ntr(et) = 31 r(ei),
~
:::;; ~
n=i+l
and this contradicts (4.2.2). By a simple calculation we find that if 00
a sequence {xn} =
{2 tfet} is convergent i=1
00
to x =
2
ttet, then tj+tt,
i=1
n = 1,2, ... and conversely. Therefore, the set of all elements of type 00
2
n=l
tnen constitutes a subspace X 0 isomorphic to (s).
D
Existence of Continuous Linear Functionals and Operators
197
4.2.8. If an Fspace X contains arbitrarily short lines, then X contains an infinitedimensional subspace X 0 with a total family of continuous linear functionals. Proof Proposition 4.2.7 implies that the space X contains a subspace isomorphic to (s) and in the space (s) there is a total family of continuous linear functionals, because the space (s) is locally convex. D COROLLARY
Kalton (1979) has shown that any strictly galbed space that is not locally bounded contains an infinite dimensional locally convex space. Let us observe that there is an Fspace X with a total family of continuous linear functionals and a subspace Y of the space X such that in the quotient space X/Y there are no nontrivial continuous linear functionals. Indeed, let X= lP, 0
0 be chosen so that the set B = {t E Q: lz(t)l > c} has a positive measure, p(B) > 0.
J
Example 4.3.1a' Each continuous linear functional defined on the space [P, 1 < p < +oo, is of the form F(x)
"'
= }; YnXn,
(4.3.2)
n=l
where {Yn}
[a. The norm of the functional F is given by the formula
E
IIFII
= ll{Yn}llzq.
Example 4.3.2 Each continuous linear functional F(x) defined on the space L(!J,.E,f.l) with the norm llxll = J Jx(t)J df.l is of the form n F(x)
=
Jy(t)x(t)df.l.
(4.3.3)
n
where y(t) E M(D,l:,f.l) and the norm of the functional F is given by the formula
IJFII = (i.e.
IJFII =
esssupJy(t)J ten,
IIYIIM
Existence of Continuous Linear Functionals and Operators
201
Example 4.3.2a
Each continuous linear functional F defined on the space 1is of the form (4.3.2), where {Yn} em. The norm of the functional F is given by the formula
IIF/1 =max IYnl· n
Example 4.3.3
Each continuous linear functional F(x) defined on the space C(.Q) is of the form F(x) =
Jx(t)d,u,
(4.3.4)
u
where ,u is a countable additive scalar valued function defined on Borel subsets of .Q. Moreover, ,u is regular (i.e. for any Borel set E and for any positives there are an open set G and a closed set F such that F C E C G, and J,u(G""F)J < s) and the total variation is bounded, V(,u, .Q) =
sup E,vE,=U E, Borel sets
J,u(E1)J+J.u(E2)J < +oo.
If we take in C(.Q) the standard norm
JJxJJ
=
sup Jx(t) I, lEO
then the norm of the functional F is equal to the total variation of the measure. Example 4.3.3a In the space C[O, 1] each continuous linear functional is of the form 1
F(x) =
J x(t)dy(t),
(4.3.5)
0
where y(t) is a function of the bounded variation. The norm of the functional F is given by the formula 1
IIFII = Var y(t). 0
Chapter 4
202
Example 4.3.3b In the space c each continuous linear functional is of the form co
F(x)
=
1 YnXn+Yo lim Xn, 1
. (4.3.6)
n+co
n=l
where {yp} Eland the norm of the functional F is given by the formula co
IIFII = IYol+
L IYnl.
n=l
Example 4.3.3b' Each continuous linear functional defined on the space c0 is of the form (4.3.6), where of course Yo= 0. Example 4.3.4 In the Hilbert space H each continuous linear functional is of the form F(x)
=
(4.3.7)
(x, y),
and the norm of the functional F is equal to the norm of y. Formulae (4.3.2)(4.3.7) define continuous linear functionals in the respective spaces. The proof of formulae (4.3.2)(4.3.7) can be found for example, in Dunford and Schwartz (1958). The proof of formula (4.3.1) is given in Krasnosielski and Ruticki (1958).
4.4. CONTINUOUS LINEAR
F
UNTIONALS IN E 0SPACES
Let X be a E 0space with the topology determined by a sequence of homogeneous pseudonorms {llxlln}. As a simple consequence of the considerations of Section 4.1 we obtain THEOREM 4.4.1 (Mazur and Orlicz, 1948). Let X be a E0 space with the topology determined by a sequence of pseudonorms {jjxJJ,.}. A linear functional f(x) defined on the space X is continuous if and only if there are a pseudonorm llxlln, and a constant Kt such that
I*J(x)l < Ktllxlln,.
Existence of Continuous Linear Functionals and Operators
203
4.4.2. Let X be a B0 space with the topology determined by a sequence of pseudonorms //x//n. Let COROLLARY
X!= {x:
Jlxlln = 0}
and
Xn= X/X~.
The pseudonorm Jlxlln induces the norm in the space X. Then the conjugate space X* is the union of the space conjugate to Xn :
x:
ux:. 00
X*=
n=l
Let us remark that if
then
ll.f//1 ~ ll.fl/2 ~/Ill/a~···
and
for norms of functionals. The topology of bounded convergence in X is equivalent to the topology given by the following basis of neighbourhoods of zero: U
= {.f: /E
xz, /l.fl/k <
8
for a positive 8 and an index k}.
If X is not a 8*space, then the space X is not metrizable; in fact, if X is not a B*space, then there is a system of pseudonorms JlxJ/ 1 < Jlxl/ 2 < ... determining the topology equivalent to the original one such that X~::;t=x;_ 1 • Let .fn be an arbitrary sequence of functionals such that In Ex; and In¢ 1. Of course, for any sequence of scalars {tn}, tn =I= 0, n = 1, 2, ...
x;_
ln.fn Ex:
and
Let U be an arbitrary neighbourhood of zero in X*. It is easy to verify that only a finite number of elements tnfn belong to U. Since {tn} is an arbitrary sequence of scalars different from 0, the space X* is not a linear metric space. Theorem 4.4.1 and the knowledge of the general form of continuous linear functionals in Banach spaces permit us to give the general form of continuous linear functionals in B0spaces
Chapter 4
204
Example 4.4.3
Each continuous linear functional defined on the space Example 1.3.6) is of the form F(x) =
Jx(t)d,u,
e (.Q) 0
(see
(4.4.1)
flk
for a certain k and some measure in Example 4.3.3.
.u satisfying the conditions described
Example 4.4.4 Each continuous linear functional defined on the space C""(.Q) (see Example 1.3.7) is of the form
(4.4.2)
for a certain positive integer m and some measure .Uk satisfying the conditions described in Example 4.3.3. Example 4.4.5 Each continuous linear functional on the space LP(am, k), 1 < p < +oo, is of the form 00
F(x) =
.2; YnXn,
(4.4.3)
n=l
where {Yn} is such a sequence that, for a certain index m, 00
~ IYn1lq < +oo am,n
n=l
and 1/p+ 1/q
=
1.
Example 4.4.6
Each continuous linear functional defined on the space L(am,n) is of the form (4.4.3), where {Yn} is such a sequence that, for a certain index m,
1 < +oo.
1 suplynam,n
Existence of Continuous Linear Functionals and Operators
205
TitEOREM 4.4.7 (Eidelheit, 1936). Let X be a B 0 space. Let {in} be a . sequence of linearly independent continuous linear functionals defined on X. Let us suppose that for any continuous pseudonorm llxll there is only a finite number of functionals fn,, ... ,fnm continuous with respect to the pseudonorm llxll (i.e. for i =/= nk sup {fi(x): llxll < 1} = +oo). Then for any sequence ofscalars {an} there is in X such an element x 0 that fn(x 0) = an. Proof. Let llxllk be an increasing sequence of pseudonorms determining the topology
llxll1 ~ llxll2 ~ llxlla ~ ··· The hypothesis implies that for any k there is an index nk such that for n > nk the functional fn is not continuous with respect to the pseudonorm II Ilk· Now we construct by induction a sequence of elements {xn} such that for
fi(xn)
=
0
fn(Xn)
=
anfn(xl+ ... +xn1),
if n
i= 1,2, ... ,n1,
(4.4.5)
1
llxnllk < 2n·
> nk,
(4.4.4)
(4.4.6)
Since the functionals fn are linearly independent and only a finite number of them are continuous with respect to II Ilk such a construction is possible. · 00
Condition (4.4.6) implies that the series}; Xn is convergent. Let us ~
n=l
denote its sum by x 0 • Conditions (4.4.4) and (4.4.5) imply that x 0 has the required property. D COROLLARY 4.4.8. Let {a 11 } be an arbitrary sequence of numbers and let {tn} be an arbitrary sequence of points of the interval (0, 1). Then there is an infinitely differentiable function x(t) defined on the interval [0, 1], vanishing together with all its derivatives at point 0 and 1 and such that x
206
Chapter 4
with all their derivatives at 0 and I, satisfy the hypotheses of Theorem 4.4.7. [] 4.4.9. For any B0 space X which is not a Banach space there is a continuous linear operator T mapping X onto (s).
COROLLARY
4.5.
NONEXISTENCE OF NONTRIVIAL COMPACT OPERATORS
Let X, Y be two Fspaces. An operator T mapping X into Y is called compact if there is an open set U such that the closure of its image,
T( U), is compact. An operator T mapping X into Y is called finitedimensional if dim T(X) foo. Of course, each finitedimensional operator is compact. If there is a nontrivial linear continuous functional f defined on X, then there are finitedimensional operators mapping X into Y, namely operators of the form T(x) = f(x)y, y E Y, y :;i= 0. Kalton and Shapiro (1975) showed that there is an Fspace X with a trivial dual, such that there is a compact operator K :;i= 0 mapping X into itself. Pallaschke (1973) proved that in certain spaces N(L(Q, l:,Jl) there is no compact operator different from 0 acting in those spaces.
<
4.5.1 (Pallaschke, 1973). If J1 is nonatomic and finite, then each compact operator T acting in the space L 0 (Q,l:,J1) is equal to 0. Proof Sppose that Tis a compact operator mapping L 0(Q,l:,J1) into itself and that T :;i= 0. The~ there is a compact set K and a positive integer k such that THEOREM
where, as usual,
B(O,r)={x: llxlf
=
1, 2,
000
••. ,
Ak
Existence of Continuous Linear Functionals and Operators
207
Then, for each positive integer nand fori= 1,2, ... p(Q) llnxXA,II ~  k  ,
and therefore k
n(T(x))
= T(nx) =}; T(nxxA,) E K, i=l
which contradicts to the fact that K is compact.
0
There are also spaces N(L(Q,E,p)) with unbounded N such that each compact operator mapping N(L(Q,E,p)) into itself is equal to 0. The proof of this fact is based on the following lemma: LEMMA
each s have
4.5.2. Let K be a compact set in a space N(L(Q,E,p)). Then, for 0, there is a IJ > 0 such that for each A such that p(A) < IJ we
>
SUPPN(XXA)
<
S.
XEK
Proof Suppose that the lemma does not hold. Then there are s0 > 0, and a sequence {Xn} C K and a sequence of sets {An} such that p(An)+0 and P.v(XnXA.) :;;:: s0 • Now we shall choose by induction subsequences {xn.} and {An) in the following way. As Xn 1 we shall take an arbitrary element of the sequence. Suppose that the elements {xn 1 , ••• , xn.} are chosen. Of course, there is a number IJ > 0 such that for each set A, p(A) < IJ, we have
i = l, 2, ... , k.
(4.5.1)
Now take as An.+> a set from the sequence {An} such that p(An.+) < IJ and as xn.+, the corresponding element. By the property of the sequence {xn} and (4.5.1) for i::j::j. Formula (4.5.2) implies that the set K is not compact.
(4.5.2)
0
Chapter 4
208
THEOREM 4.5.3 (Pallaschke, 1973). If a measure J1 is nonatomic and . . f 1 Itmtnn+co
n
N(Nl(n)) =a> n
0
,
(4.5.3)
then each compact operator T mapping N(L(Q,E,p)) into itself is equal to 0. Proof Let T =F 0 be a compact operator mapping N(L(Q,E,p)) into itself. Let r > 0 be such a number that T(B(O, r)), where B(O, r) = {x: PN(x) < r}, is compact. Since simple functions are dense in N(L(Q,E,p)), there exists a set A, 0
Let . {sgnz(t) J(t) = c 1 sgnz(t)
for for
tE
Q"". B,
tE B,
where sgnz = 0 for z = 0 and sgnz =
z JZT for z =F 0 and a denotes the
number conjugate to a. Suppose that T1(x) = j(t) T(x)t, i.e. that T1 is a composition of the compact operator T and the operator of multiplication by the function j(t). Of course, T1 also maps B(O, r) into a compact set K1 • Observe that (4.5.4) Let oc = p(A). We take a partition of the set A, An, 1 , ••• , An,n, such that oc p(An, i) =
n.
Let
Then N 1(n)
1
XA = (Yn,l+ ··· lyn,n). n n
(4.5.5)
Existence of Continuous Linear Functionals and Operators
Let
In,k
{ te Q; JT1(yn,k)Jtl
=
N 1
209
(n)} ·
> n
Then by (4.5.4) and (4.5.5) N 1(n)
( N 1(n)
)I
nXn(t)~ T 1 nXA t =
~ Tl(yn,t)lt n1 61_
and (4.5.6)
Let
M =sup
J N(JxJ)dJl.
xEK 1 0
Since the set K 1 is compact, M is finite. By the choice of the set A, PNCYn,i) < r, i = 1,2, ... , n. Hence
Nl(n)) N (  n  Jl(Jn,k) ~ M and
M
p(ln,>),;;
N( N~(n)) .
(4.5.7)
Let e < { a(l(B), where a is defined by (4.5.3). The set K 1 is compact, hence by Lemma 4.5.2 there is a (J > 0 such that
PN(ITI(Yn,k)IXA)
<e.
provided Jl(A) < b. By (4.5.7) and (4.5.3) we can choose an m0 such that for m Jl(lm,k) < b, k = 1,2, ... , m and _I N(Nl(m))
m Thus by (4.5.6)
m
>}:_a. 3
> m0
Chapter 4
210
Hence
2 1 (Nm(m)) ~
3a~mN
1
s p(B)
a
<2,
and we obtain a contradiction.
D
Now we shall give an example of a function N(u) satisfying (4.5.3). Let N(u) = log(1 +u). Then
en)
en
1 1 (N1 l o(n1 1 N  (n))  =g   +  ;:?:log
n
n
n
n
n
n
n
= 1 _ logn + 1. n Kalton (1977b) showed that for any Fspace Y every continuous linear operator T=/= 0 mapping LP(O, 1]), 0 < p < 1, into Y is not compact. ~6.EXISTENCE
OF RIGID SPACES
In the preceding section we showed that there are Fspaces in which each compact endomorphism is identically equal to 0. In this section we shall show that there are Fspaces in which each continuous endomorphism is of the form al, where a is a scalar and I denotes the identity operator. The F*spaces with this property will be called rigid spaces. The first example of a rigid F*space was given by Waelbroeck (1977) however, his space is not complete. Roberts (1976) gave an example of a rigid Fspace, but his construction was only published in 1981 in a paper written jointly with Kalton (see Kalton and Roberts, 1981). We shall present this example, following the paper mentioned above. The construction of a rigid Fspace is based on several notions and lemmas. Let (X, II I[) be a locally bounded space and let I II be aphomogeneous norm. A function
[x]
=
llxll1iP
Existence of Cof'tinuous Linear Functionals and Operators
211
is called a quasinorm (see Hyers, I939). It is easy to observe that if and only if x = 0,
[x] = 0 [ax]
= IaI[x] for
(4.6.1)
a being a scalar,
x E X,
(4.6.2)
:s;; 211P([x]+[y])
[x+y]
(4.6.3)
and [x] = inf{t
where B =
> 0:
~ E B},
(4.6.4)
{x: llxll :s;; I}.
4.6.I (Peck; see Kalton and Roberts, I98I). Let (X, II II) be an ndimensionallocally bounded space and let II II be phomogeneous. Then LEMMA
(4.6.5) Proof Without loss of generality we may assume that = I, ... , m. Since X is finitedimensional, the set B = {x:
# 0, i llxll :s;; I}
Xi
is compact. Let
Of course, u E conv B. Since X is ndimensional, by the classical Caratheodory result we obtain u = c1 v1 +
where Vi
... +cnVn,
= I, ... , n,
E
B, i
c1
+ ... +en= I.
Ci ~
(4.6.6) 0, i = I, ... , n and (4.6.7)
Therefore n
u=
lluii 11P:s;; (_r lctiPIIvtiiY'P i=l
n
:s;; sup {(_r lctiP)1/P: c1 + ...
+en
:s;; I} =
n11P 1.
i=1
This trivially implies (4.6.5).
D
Chapter 4
212
The next lemmas concern the spaces LP, 0 < p standard norm in LP. We shall write
<
1. Let /I!IIP be the
1
[f]P
iifii~P =
=
(f if(t)jPdtY'P·
(4.6.8)
0
4.6.2. Let 0 < q ~ p ~ 1. Let f E Lq. Then there is a continuous linear operator T: LP +Lq such that T I (here I denotes a constant function equal to one) and LEMMA
[Tx]q
~
for x
[f]q [x]p
E
LP.
f
(4.6.9)
Proof. To begin with we shall prove the theorem for p that inflf(t)i > 0. Let ~ fo
=
=
q. Suppose
f
[.f]q ·
=
Thus [fo]q = 1. Let T0(x) = x(F(t))fo(t), where t
F(t)
=
Jlfo(t) qdt. 0
The operator T0 is linear and 1
IITo(x)iiq
=
1
Jix(F(t)).f (t)/q dt Jix(F(t))iq dF 0
0
=
=
llxllq.
0
It is easy to verify that Tl = [f]qT0 l = f(t).
We recall that, in general, Lq :::> LP and, for yELP, [y]q
~
[y]p.
Thus [Tx]q
~
[x]q[.f]q
~
[x]p[f]q.
(4.6.9)
holds provided infl.f(t)i > 0. Since the functions f with this property are dense in Lq, by continuity arguments we find that (4.6.9) holds for all functions belonging to Lq. D
Existence of Continuous Linear Functionals and Operators
213
Let C be a onedimensional subspace of the space LP, 0 :::::;;; p < 1 consisting of constant functions. Let p0 : LP+LP/0 be the quotient map associating with each x E LP the coset containing x. Of course by the definition of the quotient space IIPofll~
= inf {llfaiiP:
a being scalar}
where II II denotes the quotient norm. Since there is no danger of confusion, we shall denote II II~ also by II liP· The space LP/0, 0 < p < I is not isomorphic to the space LP (see Kalton and Peck, 1979); nevertheless, it embeds into LP, as follows from LEMMA
4.6.3. There is alinear operator S: LP+LP, 0 :::::;;p
<
I, such
that (4.6.II)
IIPofiiP:::::;;; JIS(f)IIP:::::;;; 2IIPofllv·
Proof By the classical results of measure theory the space LP is isometric to the space LP([O, I] X [0, I]). We define S: LP+LP([O, 1] X [0, I]) by Sf)(x,Y) = f(x)f(y). Then 1
liS/liP~
1
1
JJJf(x)f(y)IPdxdy:;? JIIPo!IIPdy = IIPofllv·
0
0
0
On the other hand, 1
IISfiJp:::::;;;
1
f f (Jf(x)IP+ lf(y)IP)dxdy:::::;;; 2JIJIIP·
(4.6.I2)
0 0
Observe that S(f)
= S(fa)
for all costant functions a. Thus by (4.6.I2) IIS(f)IIP = infiiS(fa)IIP:::::;;; 2inf[jja1Jp = 2IIPofiJp. a
a
D
Chapter 4
214 LEMMA
4.6.4. Let {en}, n = 0, 1,2, ... , be a sequence such that 00
.J:
Cn1f2
< }.
(4.6.13)
n=1
Then we may select sequences {Pn}, {en}, n = 0, 1, ... , and a sequence of finitedimensional subspaces V11 C LPn such that
en> 0, Pn is increasing, Po= 1/2, limpn = 1,
(4.6.14)
n+oo
(4.6.15)
1 E Vn, nMnen
<
n = 1, ... ,
Cn,
where n1
Mn =
.J: dim Vi,
(4.6.16)
i=O
for each n ~ 0 there exists {vn,k: I~ k ~ r(n)} such that Vn,k e Vn, k =I, ... , r(n) and r(n)
.J: Vn,k = 1,
(4.6.17)
k=1 r(n)
.J: [vn,k]pa <en,
(4.6.18)
k=1 r(n)
~ [v ]Pn+t L.,; n,k Pn
< BnPn+t •
(4.6.19)
k=1
Proof We select the sequences by induction. We begin by takingp0 = 1/2, eo= 2, V0 = C, V0 C Ljll 2 , r(O) = 1 and v0 , 1 = 1. Observe that (4.6.15) and (4.6.17)(4.6.19) hold. Condition (4.6.16) ought to be valid for n ~ 1. Suppose that (p 0 , ... , Pm 1), (e0 , ... , sm_ 1), ( V0 , ... , Vm_ 1) are chosen so that (4.6.15)(4.6.18) hold for n = m1 and (4.6.19) holds for n = m2. Since r(m1)
.J: [Vm1,k]Pm1 <
k=1
Bm1,
Existence of Continuous Linear Functionals and Operators
215
we can find Pm sufficiently close to 1, such that Pm > Pm~> Pm > 11 fm and (4.6.19) holds for n = m1. Now take em such that (4.6.16) holds Since 1 belongs to the convex hull of every neighbourhood of zero in LPm, we can find Vm, 1 , ••• , Vm,r(m) E LPm such that (4.6.18) holds for n = m. Finally, we put Vm = lim(vm, 1 , ••• , Vm,r(m))· 0 Let Z be a space of real valued functions f defined on (0, oo) such that oo
A(j)
n±l
=};I
< +oo.
/J(t)/Pn dt
n=On
A is an Fnorm on Z and (Z,A) is an Fspace. It is easy to verify that the space Z is locally bounded and that the unit ball B = {x: A (x) ~ 1} is convex. Thus there is a homogeneous norm II II such that
t
4
B = {x:
llxll ~ 1}. Write [x] = llx// 2 •
Let Z(a,b) be a subspace of Z supported on the interval (a,b). Let Pn, En, Qn be the natural projection of Z onto Z(O,n), Z(n,n+1) and Z(n, oo) respectively. Then I= Pn+Qn = Pn+En+Qn+I
and 1/En/1 = 1/Pn/1 = 1/Qn/1 = 1.
Note that iff, g Z(n, oo) then [f+g]Pn
~
[f]Pn+[g]Pn.
(4.6.20)
There is a natural isomorphism rn: LPn+Z(n,n+1) given by f(tn) 7:nf= { O
forn
Let Un = 't'nVn, en= 't'n1, en, k = 't'n Vn,k, n = 0, 1, ... k = 1,2, ... , r(n) 00
Let Y =
U Un,
M =lin {en}. Let p be the natural quotient map
a=l
p: Z+Z/ M. Thus 1/pf/1 = infl/fg/1, gEM
Chapter 4
216
where we denote the norm in the quotient space induced dy the norm II 1/ also by// 1/. Note that iffe Z(n,n+I), then 1/pf/1 = inf//faen/1 = min//faen/1. aER
aER
LEMMA 4.6.5. Suppose that /E Z(O,n) and 1/fli  . I. Then there exists a linear operator A: Z(n,n+ I))Z(O,n) such that A(en) = f and //A//~ I. Proof Suppose thatf = h0 + ... + hn1 , where hiE Z(i,i+ I), i = 0, I, ... . . . , n1. Since 1/f/1 = I implies A(f) = I, we have n1
1
= 1/f/1 = A(f) =}; [hi]P•. i=O
By Lemma 4.6.2 there exist linear operators Fi: Z(n,n+1))Z(i,i+I) such that [Fi(/)] ~[hi][/] and Fien =hi, i = 0, I, ... , n1. Let A= F 0 +... + Fn_ 1 • Then, of course, Aen = f If g E Z(n,n+1) and 1/g/1 = 1 = A(g), then n1
n1
A(Ag) = }; [Fi(g)]P• ~}; [hi]Pt ~ 1. i=O
This implies that 1/A/1
i=O
~
I.
D
Now let {B7,}, n = 0, I, ... , be a partitioning of the set N of nonnegative integers into infinite disjoint subsets with the property that n ~ minBn. For each n, let {yk: k E Bn} be a dense subset of the set {!: fe U0 +... + Un, 1/f/1 =I}, with the property that Yk =en infinitely often. By Lemma 4.6.5 we can find operators Ak: Z(k, k+I))Z(O, k) with 1/Ak/1 ~ 1 so that Akek = Yk, k = I, 2, ... Define T: Z+Z so that 00
T
=
'),
~
CkAkEk
k=O
with the convention A0 = 0. By simple calculations we obtain 00
1/TI/ ~}; Ck1/2 ~~. k=1
(4.6.2I)
Existence of Continuous Linear Functionals and Operators
217
Moreover T(Z(O, k+ 1)) C Z(O, k)
(4.6.22)
T(Z(O, 1)) = 0.
(4.6.23)
and By (4.6.21) the operators S =/Tis invertible. Observe that T(M) C Y. Let M 1 = S(M) C Y. Let 1t: Z'?Z/M be the quotient map and let X be equal to n(Y) (isomorphic to YfM1 ). We shall show that X is a rigid space. LEMMA
4.6.6. For je Z(O,n+l).
/lpEnf/1 ~ 2llnfll.
(4.6.24)
where the both norms in quotient spaces Z/M and Z/M1 are denoted in the same way by II 11. Proof Take an arbitrary number b > 1. By the definition of the norm in the quotient space there is a g E M such that
11/Sgll ~ lln/11. Then
IIPEnfpEnSgll ~ blln/11·
(4.6.25)
By the definition of M, pEng = 0 forgE M. Thus by the .definition of S
IIPEnf+pEnTgll ~ b/lnf/1. By (4.6.22), EnTg = EnTQn+ 1 g, and so
IIPEnf/1 ~ b/lnJ/1+/IEnTQn+lg/1
~ b/lnf/1+} /IQn+Ig/1. Since Qn+ 1 f= 0, Qn+ 1 (fSg) = Qn+I S g and
/IQn+ISg/1 ~ b/lnf/1. Hence, by the definition of S,
(4.6.26)
Chapter4
218
so that IIQn+lgll ~ 2ll1t/ll. Thus, by (4.6.26), IIPEn/11 ~ 215ll1t/ll. Since 15 is an arbitrary number greater than 1, we obtain (4.6.24).
0
LEMMA 4.6.7. The space X is infinitedimensional. Proof By (4.6.16) en+0, and by (4.6.17) and (4.6.18) dim Vn+oo. Thus dim Un+oo. For fe Un, by Lemma 4.6.6, ll1t/ll
> ~ IIP/11 =
Hence dim 1t(Un)
~
~ minl!faenll· a
dim Un1 and dimX =
+oo.
D
LEMMA 4.6.8. The set {a1t(en): a e R, n eN} is dense in the space X. Proof Suppose thatfe U0 Un and A(f) = 1 = IJ/11· Then by the definition of {Yk}, there is a subsequence ~ C N such that
+ ... +
limy;= f. j~
By the definition ofT, T(c; 1e;) =
Yf,
and so
n(c; 1 e,) = n(yJ) + n(f). The multiples off are dense in Y and this completes the proof.
D
THEOREM 4.6.9. The space X is rigid. Proof Suppose that a linear continuous operator A maps X into itself and that IIAII < 1. We shall show that, for each n, 1t(en) is an eigenvector of A. Fix n eN and let B~ = {j e Bn: YJ =en}. For j e B~, Te1 = CJen. Hence 1t(e,) = c11t(en). Now
and r(j)
~ [eJ,k] < BJ. k=l
Existence of Continuous Linear FunctiOnals and Operators
Since [[A[[
< 1, there are gi,k E
n(gi,k)
219
Y such that
= An(ei,k)
and
k= 1,2, ... ,r(j). Let
Then Now Pigi,kE U0 +
... +Uj1,
and so, by Lemma 4.6.1 (for p = 2), r(j)
~Mi _27 [Pigi,k] ~ MteJ ~c~.
[P1h1]
k=1
}
Similarly Qj+Jgj,kE Z(j+ 1, oo),
and by (4.6.19) and (4.6.20) 1
r(j)
[QJ+lhi]
~ (,27 [gi,k]PJ+~ )vst1 < ej ~ _c4 . J
k=l
Thus ·
[hiEihi]
4c·
= [Pthi+Q;t1hi] < !.
This implies that, for j
E
B:,
[A:n(en)ci 1 :n:Eihi]
< ~.
[c£ 1 :n:Eihtcj 1 :n:Eihi]
<
s(! +;).
Observe that Cj 1 :n:EthJci 1:n:Etht E
Z(O,j+ 1),
and by Lemma 4.6.6
<
(4.6.28)
}
By (4.6.28) we trivially obtain for i,j E B:, i
[ci 1PEihi]
(4.6.27)
}
32(! +;).
< j,
Chapter 4
220
Thus, by the definition of p, there is a A = A(i,j) such that [c; 1 EJhJAeJ] < 32 ( This implies [c; 1 n(E1h1Ae;)]
++;). ++ ; ).
< 32 (
Since, by the definition of B~ cj 1n(e;) = n(en), [An(en)n(en)]
< 64 (
++;).
As i,j E B~ can be chosen arbitrarily large, we deduce that there is a real f1 such that An(en) = pn(en). Thus, by Lemma 4.6.8, there is a dense set of eigenvectors. Using the continuity of A, we can deduce that each element x EX is an eigenvector corresponding to an eigenvalue Ax. Taking two eigenvectors x, y, we infer, by simple considerations in the twodimensional space lin({x,y}), that x+y is an eigenvector if an only if the eigenvalues are D equal Ax = Ay = A. Thus A = AI. Kalton and Roberts (1981) proved also that the constructed rigid space is isomorphic to a subspace of LP for all 0 ~ p < 1. Refining the construction described above, they constructed a rigid space such that each quotient is rigid, and formulated the following problems: Problem 4.6.11 (Kalton and Roberts, 1981). Does LP(O
< 1) have
a rigid quotient? Or, more generally, does every Fspace with a trivial dual have a rigid quotient? Problem 4.6.12 (Kalton and Roberts, 1981). Does every Fspace with a trivial dual have a closed rigid subspace?
Chapter 5
Weak Topologies
5.1.
CONVEX SETS AND LOCALLY CONVEX TOPOLOGICAL SPACES
Let X be a linear space over real or complex numbers. A set A C X is said to be convex if, for arbitrary nonnegative, a, b such that a+b = 1, x,y E A implies ax+by E A (see Section 3.1). The intersection of an arbitrary family of convex sets is a convex set. Let A be a convex set and let a 1 , ••• , an be nonnegative numbers such that n
Then
l
n 1
ai.XiEA
i=l
for an arbitrary system of elements x 1, ••• , Xn E A. Let A be an arbitrary set. By conv(A) we denote the intersection of all convex sets containing A. The set conv(A) is called the convex hull of the set A. It is easy to verify that the convex hull of the set A may be characterized in the following way : m
conv(A) =
{2: anxn:
m
XnE
n=l
A,
an~ 0, .2; an= 1}.
(5.1.1)
n=l
The algebraic sum of two convex sets is a convex set. The image of a convex set A under a linear operator Tis a convex set. A counterimage of a convex set under a linear operator is a convex set. Let M be an arbitrary subset of X. We say that a point p E M is 221
Chapter 5
222
a Cinternal point of the set M if, for each x e X, there is a positive numbers such thatp+tx eM for [t[ <e. A point p e X is called a Cbounding point of the set M if it is neither a Cinternal point of the set M nor a Cinternal point of the complement ofM. Let K be a convex set containing 0 as a Cinternal point. Then we can define a functional p(x) (generally nonlinear) in the following way: p(x)
=
inf{t
> 0:
~ e K}.
(5.1.2)
Let us remark that the functional p(x) has the following properties (compare Section 4.1) : (1) p(x) ~ 0, (2) p(x) <+=for xe linK (3) p(x) = tp(x) for positive t, (4) p(x+y) ~p(x)+p(y), (5) if x e K, then p(x) ~ 1, (6) if xis a Cinternal point of the set K, then p(x) < 1, (7) if xis a Cbounding point of K, then p(x) = 1. We say that a linear functional f(x) separates two sets M and N if there is a constant c such that ~
c
for xeM
Ref(x)<: c
forxeN.
Ref(x)
and Here by Re z we denote the real part of a complex number z. Obviously, if X is a linear space over reals, then Ref(x) = f(x). Of course, a functional f(x) separates the sets M and N if and only if it separates the sets M Nand {0}. PROPOSITION 5.1.1. Let M and N be two disjoint convex sets in a linear space X. Suppose that M has a Cinternal point. Then there is a linear functionalf(x) separating the sets M and N. Proof Without loss of generality we can assume that 0 is a Cinternal point of the set M. To begin with, we shall consider the case where X is a linear space over reals. Let y be an arbitrary Cinternal point of the set MN.
Weak Topologies in Banach Spaces
223
Since the sets M and N are disjoint, the point 0 does not belong to the set MN, Let K = MN+y. Obviously, 0 is a Cinternal point of the set K and the point + y does not belong to K. Let p(x) be the functional defined by formula (5.1.2). Obviously, p( +y) ~ 1. Let X0 be the space spanned by the element y. We define on the space X0 a linear functionalfo in the following way fo(ay) = ap(y). Of course, for x E X0, fo(x) ~p(x). By the HahnBanach theorem (Theorem 4.1.1) the functional fo(x) can be extended to the functional f(x) defined on the whole space X such thatf(x) ~p(x). This implies thatf(x) ~ 1 for x E K andf(y) ~ 1. Hence the functional f separates the sets MN and {0}. Therefore, it separates the sets M and N. Now we shall consider the complex case, i.e. the case where X is a linear space over complex numbers. The space X may obviously be considered as a linear space over reals, too. Then there is a realvalued linear functional f(x) separating the sets M and N. Let g(x) = f(x) if(ix) (cf. Corollary 4.1.3). The functional g(x) is linear with respect to complex numbers and it separates the sets M and N. 0 We recall that X is a linear topological space if it is a linear space with a topology and if the operations of addition and multiplication by scalars are continuous. Let A be a subset of a linear topological space X, If A is a convex set, then the closure A of the set A is also a convex set and the interior Int(A) of the set A is also convex. The continuity of multiplication by scalars implies that each internal point of the set A is a Cinternal point of this set. If a convex set A contains internal points, then the necessary and sufficient condition for the point x E A to be Cinternal is that x be an internal point. By conv(A) we denote the intersection of all closed convex sets containing A. The set conv(A) is called the closed convex hull of the set A. It is easy to verify that conv(A) = conv(A), conv (aA) = a conv (A) for all scalars a.
(5.1.3) (5.1.4)
Chapter 5
224
PROPOSITION 5.1.2. If conv(A) is a compact set, then conv (A+ B) = conv (A)+ conv (B). The above formula is a consequence of the following lemma: LEMMA 5.1.3 (Leray, 1950). Suppose we are given a topological space Y, a compact space K, and a continuous mapping f(t,k) of the product Xx K into a topological space X. Let F be a closed subset of the space X disjoint with the set f(t 0 , K). Then there is a neighbourhood V of the point t0 such that the set F does not have common points with the set f( V, K). Proof Let k E K. The continuity of the function f implies that there are a neighbourhood V(k) of the point t0 and a neighbourhood W(k) of the point k such that Fnf(V(k), W(k)) =¢.The set K is compact, hence we can choose a finite cover of the set K by the sets W(k1 ), ••• , m
... , W(km), Kc
U W(kt). Let i=l
n V(kt). m
V=
•=1
The set Vis an open set with the required properties.
D
LEMMA 5.1.4 (Leray, 1950). Let X be a linear topological space. Let F be a closed set in X and let K be a compact set in X. Then the set F+K is closed. Proof If x ¢= F+K, then the set F does not have common points with the set x K. Therefore, by Lemma 5.1.3, there is a neighbourhood V of the point x such that Fn (VK) =¢.Hence vn(F+K)=¢.
D
The proof of Proposition 5.1.2 is a trivial consequence of Lemma 5.1.4 and of the fact that the algebraic sum of two convex sets is con~L
D
Let us note some further properties of convex sets. By similar considerations to those in Corollary 4.1.2 we can prove that if a functional f(x) separates two sets M and N and one of them contains an internal point, then the functionalf(x) is continuous.
Weak Topologies in Banach Spaces
225
Conversely, Proposition 5.1.1 implies that if M and N are two disjoint convex sets and one of them contains an internal point, then the sets M and N can be separated by a continuous linear functional f(x). We say that a linear topological space is locally convex if each neighbourhood of zero contains a convex neighbourhood of zero (compare Section 3.1). THEOREM 5.1.5. Let X be a locally convex topological space. Let M and N be two disjoint convex closed sets. If, in addition, M is compact, then there are constant c, and e > 0 and a continuous linear functional f(x) defined on the space X such that Ref(x)~ce
Ref(x)
~
c
for xEN, for XEM.
(5.1.6)
Proof Lemma 5.1.4 implies that the set M N is closed. Since M and N are disjoint, the set M N does not contain the point 0. The space X
is locally convex, and therefore there is a convex balanced neighbourhood of zero U such that U n (M N) = 0. Proposition 5.1.1 implies that there is a continuous linear functional f(x) which separates the sets U and M N. The set U is open, whence sup{Ref(x): x E U} = e > 0. Therefore, Ref(x) osition.
~
e for x
E
(5.1.7)
MN. This trivially implies the prop
D
5.1.6. Suppose that in a linear space X there are two locally convex topologies (X, r1) and (X, r 2). If both topologies designate the same continuous linear functionals, then a convex set A is closed in the first topology r1 if and only if it is closed in the second topology "l'2 • Proof Let K be a convex set closed in the topology r1 . Then for each point p f. K there is a linear functional f(x) which is continuous in the topology r1 and there are numbers c,e > 0, such that Ref(x) ~ ce for x E K and Ref(p) ~ c. But the hypothesis implies that the functional f(x) is continuous in COROLLARY
the topology r2 • Hence p could not belong to the closure K 2 of the set K in the topology r2 • This implies that K 2 = K.
226
Chapter 5
If K is a convex set closed in the topology r2 , then using the same 0 arguments we infer that it is closed in the topology r1 •
5.2.
WEAK TOPOLOGIES. BASIC PROPER TIES
Let X be a linear space over real or complex numbers. Let X' denote the set of all linear functionals defined on the space X. A subset of the set X' is called total if f(x) = 0 for all f E F implies that X = 0 (compare Section 4.2). Let F be a total linear set of functionals. By the Ftopology of the space X we shall mean a topology determined by the neighbourhoods of the type
r
N(p,fi, ... ,Jk; a 1 , ... , ak) = {x: Jfi(xp)J < ai (i = 1, 2, ... , k)}, where at> 0 andfi E F. Obviously, the space X with a Ftopology is a locally convex space. We say that a set A C X is Fclosed (Fcompact) if it is closed (compact) in the Ftopology. The closure of a set A in the Ftopology will be called the Fclosure. A functional f(x) is said to be Fcontinuous if it is continuous in the rtopology. Let X be a locally convex topological space. Let X* be the set of all continuous linear functionals defined on X (conjugate space). The X*topology is called the weak topology. Obviously, the weak topology is not stronger than the original one. Let X be the space conjugate to a locally convex space X. Let us recall that in the conjugate space we have the topology of bounded convergence. Each element XEX induces a continuous linear functional F on the space X by the formula F(x) = x(x_).
(5.2.1)
We shall indentify the set of functionals defined by formula (5.2.1) with the space X. The Xtopology in X is called the weak topology of functionals or the weak*topology. Since we always have X*:::> X, the weak topology offunctionals is not stronger than the weak topology.
Weak Topologies in Banach Spaces
227
5.2.1. Let X be a linear space. Let r be a total linear set of functionals. A linear functional f(x) is continuous in the Ftopology if and only iffe PROPOSITION
r.
The proof is based on the following lemma : LEMMA 5.2.2. Let X be a linear space. Let g, ];_, ... .fn be linear functionals defined on X. If
ji(x) = 0,
i = 1, 2, ... , n,
implies g(x)
=
0,
then g(x) is a linear combination of the functionals ft, ... .fn. Proof Without loss of generality we can assume that the functionals ft, ... .fn are linearly independent. Let X0
=
{xe X: Ji(x)
=
0, i
=
1, 2, ... , n}.
(5.2.2)
Let X be the quotient space XJX0 • The functionals ];_, ... ,fn induce linearly independent functionals ];_, ... ,fn on X. The assumption about g(x) implies that the functional g (x) also induces a linear functional g(x) defined on X. Then space X is ndimensiona1, therefore, g (x) is a linear combination of];_, ... ,f,,


i=


ad1 + ... +an/n.
It is easy to verify that g
=
ad1 + ... +anfn·
D
Proof of Proposition 5.2.1. Sufficiency. From the definition of neighbourhoods in the rtopology it trivially follows that each functional fer is rcontinuous (i.e. continuous in the rtopology). Necessity. Let g(x) # 0 be a functional continuous in the Ttopology. Then there is a neighbourhood of zero U in that topology such that sup jg(x)l < 1. But the neighbourhood U is of the type XEU
U= {xeX: i.fi(x)i
=
1,2, .... , n}.
228
Chapter 5
This implies that g(x) = 0 for x E X 0 • Therefore, by Lemma 5.2.2, it follows that g(x) is a linear combination of f 1 , ••• Jn. Since is linear, gET. D
r
PROPOSITION 5.2.3. Let X be a linear space and c(x) a nonnegative valued function defined on X. Let X' be the set of all functionals defined on X. Let K={JEX': /f(x))::o:;;c(x)}. Then the set K is compact in the Xtopology of the space X'. Proof Let
I(x) ={a: a scalar, /a/::::;; c(x)}
and
/ I= " /"I(x). xEX
It is well known (the Tichonov theorem) that the set I is compact in
the product topology. We define a mapping T of the set K into the set I by the formula
"/
(5.2.3)
Tf= /"f(x). xEX
It is easy to verify that T is a homeomorphism. In order to finish the proof it is sufficient to show that the image T(K) of the set K is closed. Obviously, I may be considered as a set of functions defined on X Let A(x, y) = {gE I: g(x+y) = g(x)+g(y)} (5.2.4) and B(a, x) = {gE I: g(ax) = ag(x)}. (5.2.5)
The sets A(x,y) and B(a,x) are closed. Hence the set T(K) = (
n
x_yEX
A(x, y)J n [
n
B(a, x)]
xEX
a scalars
is also closed.
D
TliEOREM 5.2.4 (Alaoglu, 1940). The closed unit ball of the space X* conjugate to a Banach space X is compact in the weak*topology. Proof By definition, the closed unit ball of the space X is the set S* = {fE X': /f(x)/::::;; 1/x/1} and by Proposition 5.2.3 we trivially obtain the theorem.
0
Weak Topologies in Banach Spaces
229
COROLLARY 5.2.5. A set A of the space X* conjugate to a Banach space X is compact in the weak*topology if and on/ y if it is closed in that topology and bounded in the norm topology. Let X be a Banach space. Let X** denote the space conjugate to the space X*. The space X** is called the second conjugate. As we have seen before, each element x E X can be regarded as a continuous linear functional on the space X* (see formula (5.2.1)). This means that there is a natural embedding n(X) of the space X into the space X**. Corollary 4.1.6 implies that the embedding n is norm preserving, i.e. that lln(x)ll = =llxll. THEOREM 5.2.6 (Goldstine, 1938). Let X be a Banach space and let x** be its second conjugate. Let SandS** denote the unit balls in the spaces X and X** respectively. Then the set n(S) is dense in the set S** in the X*topology. Proof By S1 we denote the %*closure of the set n(S). Theorem 5.2.4 implies that S** is closed in the X*topology, thus S 1 C S**. Since S 1 is a closure of a convex set, it is a convex set. We shall show that S1 = S**. Suppose that this does not hold. Then there is an element X** E S** which does not belong to S1 • Applying Theorem 5.1.5, we can find an %*continuous functional f(x), a constant c and a positive number s such that Ref(x) ~ c for xE S 1 . Rej(x**) ): c+s.
(5.2.6)
Since the functional f(x) is %*continuous, by Proposition 5.2.1 we find that there is an element x* EX* such thatf(x) = x(x*) for all XEX**. Since n(S) C S 1 , Rex*(x) ~ c for xES. The set S is balanced, which implies lx*(x)l ~ c for all xES. Therefore llx*ll ~c. Hence 1/(x**)l = lx**(x*)l ~ cllx**ll ~ c and we obtain a contradiction of formula (5.2.6).
D
As an obvious consequence of Theorem 5.2.6 we find that the set n(X) is dense in X** in the %*topology. A Banach space X is said to be reflexive if n(X) =X**.
Chapter 5
230
5.2. 7. A Banach space X is reflexive if and only if the closed unit ball is compact in the weak topology (i.e. the %*topology). Proof Necessity. If X is a reflexive space, then n(S) = S** and (X**)* =X*. Therefore the weak topology in S** is equivalent to the %*topology. Hence, by the Alaoglu theorem (Theorem 5.2.4) S** is compact in the %*topology. Since S = S**, the set Sis compact in the weak topology. Sufficiency. Let us now suppose that the unit ballS is weakly compact. The natural embedding n is obviously a homeomorphism between S and n(S) in the %*topology of both sets. Therefore n(S) is compact, and hence closed, in the %*topology. Theorem 5.2.6 states that n(S) is dense in S in the %*topology. Therefore, n(S) = S** and n(X) =X**. D THEOREM
5.2.8. A Banach space X is reflexive if and only if each bounded weakly closed set is weakly compact.
CoROLLARY
5.2.9. Let X be a reflexive Banach space. Let A be a bounded closed convex set in X. Then the set A is weakly compact. Proof Let p ¢ A. By Theorem 5.1.5 there are a continuous linear functional f(x), a constant c, and a positive number e such that PROPOSITION
Ref(p)~cs
and
Ref(x) ;;;::, c for x EA.
This implies that the point p does not belong to the weak closure of the set A. Therefore, the set A is weakly closed and, by Corollary 5.2.8, D it is weakly compact.
5.3.
WEAK CONVERGENCE
Let X be a Banach space. We say that a sequence {xn} of elements of X is weakly convergent to x E X if, for each continuous linear functional f(x), limf(xn) =f(x). nHO
Weak Topologies in Banach Spaces
231
THEOREM 5.3.1 (Eberlein, 1947; Smulian, 1940). Let A be a subset of a Banach space X. Then the following three conditions are equivalent: A. The weak closure of the set A is weakly compact. B. Each sequence of elements of A contains a subsequence weakly convergent to an element of X. C. Each countable subset A 0 of the set A has a cluster point x 0 E X in the weak topology (this means that, for each neighbourhood of zero U, A 0 n (x 0 + U) ::;i: 0). The proof included here was given by Whitley (1967) and it is based on the following lemmas : LEMMA 5.3.2. Let X be a Banach space and let the conjugate space X* contain a total sequence {x:} of functionals. Then the weak topology on a weakly compact subset of the space X is metrizable. Proof Without loss of generality we may assume that llx!/1 = 1, n = 1,2, ... Define a metric for X by 00
d(
X,
Y
)=
~ ixn(xy)i
L.J
2n
.
n=l
Let A be a weakly compact set. The identity mapping of A with the weak topology onto A with the metric topology induced by d is clearly continuous and thus it is a homeomorphism, since A is weakly compact. D LEMMA 5.3.3. Let F be a finitedimensional subspace of the space Y* conjugate to a Banach space Y. Then there is a finite system of elements Y1, ... , Ym E Y such that for ally* E F max /ly*(y,)/1 ~
l~i~m
1
2 /ly*ll.
Proof The surface of the unit sphere of the space F is compact. Therefore, there is a fnet on this surface, i.e., there is a system of points y,! C F such that, for any element y* E F of norm one, inf lly*y:11 < f. Let y1 , ••• , Ym be elements of Y, each of norm one,
Yi, ... , l~i~m
Chapter 5
232
such that IY7{yt)l
>
i· Then
max ly*{yt)l
max ly*(y,)yt(Yt)+y7{yt)l
=
l~i~m
r=:;;;i~m
for ally* E F of norm one.
D
Proof of Theorem 5.3.1. A_,.B. Let {an} be a sequence of elements of A. Let X 0 denote the space spanned by {an}· The set X0 is weakly
closed as a linear subspace. Therefore, the intersection of the weak closure w(A) of the set A with the space X0 is a weakly compact set in X. By Lemma 5.3.2 the weak topology on w(A) n X 0 is metrizable, because the space X 0 is separable. Therefore, there is a subsequence {an.} weakly convergent to an element a E X0 , i.e. lim f(an.)
=
f(a)
(5.3.1)
kHt:J
xr
for all f E Hence this is also true for f E X*. B_,.C. This is clear. C_,.A. Suppose that a set A satisfies condition C. Then, for each continuous linear functional f(x), the set of scalars f(A) = {f(x): x E A} also satisfies this condition. Therefore it is bounded. Hence the BanachSteinhaus theorem implies that the set A is also bounded. Let.n denote the natural embedding X into X**. Let w*(n(A)) be the closure of the .set n(A) in the X*topology. Now we shall show (5.3.2)
w*(n(A)) C n(X).
Let X** be an arbitrary element of the set w*(n(A)). Let xt be an .arbitrary element of the space X* of norm one. Since x** belongs to the set w*(n(A)), there is a point a1 E A such that l(x**n(ai))(xi)l
< 1.
(5.3.3)
The space spanned by x** and x**n(a1) is twodimensional. Hence by Lemma 5.3.3 there are points x:, ... , x: <2> each of norm one, xt EX*,
Weak Topologies in Banach Spaces
i
233
= 2, 3, ... , n (2), such that max jy**(x*)l
>
!..lly**ll
(5.3.4)
2
2
for ally** belonging to the space spanned by x** and x**n(a1). Again using the fact that x** is in w*(n(A)), we can find a point a 2 E A such that max
j(x**n(a2))(x!)l
< {.
l<m.,;n(2}
Then find
x! <2>+1 , ••• , x! (a) of norm one in X* so that
max
jy**(x!)l
> {lly**ll
n(2)<m..;n(3)
for ally** belonging to the space spanned by the elements x**, n(a1 ), n(a2). Once more using the fact that x** is in w*(n(A)), choose a 2 E A so that max I(x**n(a3))(x!)l < 1/3 and continue the construction. I.,;m..;n(3)
Then we obtain : (1) an increasing sequence of positive integers {n(k)}, (2) a sequence of elements X, {xi> x 2, ••• } , (3) a sequence of elements of A, {a1oa 2, •• •} such that max
jy**(x;)j
>
(5.3.5)
{lly**ll
I.,;m.,;n(k)
for ally** belonging to the space xk spanned by the elements x, n(al), ... . . . , n(ak_ 1). And max
jx**n(ak))(xm)l
1
< k.
(5.3.6)
I.,;m..;n(k)
Let X 0 be the space spanned by the elements x, n(a1), ... By hypothesis there is a point x 0 EX which is a cluster of the sequence {an} in the weak topology of X. Since X 0 is weakly closed as a closed subspace of X, x 0 EX. Formula (5.3.5) implies that max jy**(xm)l ~ flly**ll for y E X 0 • m
In particular, for x**n(x0) we have max I(x**n(x0)) (x!)j
>
m
fllx**n(x 0)11.
On the other hand, formula (5.3.6) implies that for m j(x**n(ak))(x!)i
<
lfp.
< n(p) <
k
Chapter 5
234
Thus
The point x 0 is a cluster point of the sequence {an} in the weak topology. Therefore, there is a number k such that
* /xm(akx)/
1
form= 1, 2, ... , n(p).
Obviously, we can assume that k imply that for all p l/x**n(x0 )1/
>
n(p). Therefore, (5.3.6) and (5.3.7)
4 p
~ .
(5.3.8)
Hence x** = n(x0). This means that w*(n(A)) C n(X). Since n is a homeomorphism between X and n(X), both with X*topology, the weak closure of the set A is exactly w*(n(A)). The Alaoglu theorem (Theorem 5.2.4) implies that the set w*(n(A)) is compact. 0
5.4.
EXAMPLE
OF
AN
INFINITEDIMENSIONAL
BANACH
SPACE
WHICH IS NOT ISOMORPHIC TO ITS SQUARE
In the majority of known examples of infinitedimensional Banach spaces, those spaces are isomorphic to their Cartesian squares. Now we shall give an example which shows that this is not true in general. The example is based on the following 5.4.1 (James, 1951). Let X be Banach space with a basis {xn}· Let Xn denote the space spanned on the elements en+~> en+ 2 , ••• If for any functional f belonging to X* LEMMA
lim 1/f/xn/1 = 0,
(5.4.1)
where f/y denotes the restriction of the functional f to a subspace Y, then the basis functionals {In} (see Corollary 2.5.3) const~tute a basis in X*.
Weak Topologies in Banach Spaces
235
00
Proof Letfe X*. Let x EX, x
=.};
fn(x)en. Then
n=l 00
f(x)
00
00
=J(}; fn(x)en) =}; fn(x)f(en) = n=l
n=l
[ ~ f(en)fn](x). n=l
00
Formula (5.4.1) implies that the series }; f(en)fn is convergent to fin n=l
the norm of the functionals and that this expansion is unique.
D
Example 5.4.2 (James, 1951) Let x = {x1 ,x2 , .. • } be a sequence of real numbers. Let us write n
JJxll =
Sup(
.2 (Xp
2;_1
r
Xp2;)2 +(Xp2nH)2
i=l
2
,
where the supremum is taken over all positive integers n and finite increasing sequences of positive integers p 1 , ••• , p 2n+ 1 • Let B be a Banach space of all x such that llxll is finite and lim Xn = 0.
llxll is a norm. Indeed, llxll = 0 if and only if x = 0, lltxll = it lllxll for all scalars t. Now we shall show the triangle inequality. Let x = {x~>x2 , ... }, y = {Y~>Y2• ... }.From the definition of the norm llx+yll it follows that for any positive 8 there is an increasing sequence of indices PI> ... , p 2n+l such that n
Jlx+yJJ ~ ( .2; (xp,i1+YP•i1Xp,;yp,i) 2 +(xp,n+>+ i=l +YP•nH)2]1/2+8 n
~ [}; (xp.t1xp.;)2+(xp•n+>)2Y' 2 i=l n
+(
L (ypai1 yp,;)2+(YP•n+>)2r' +8 ~ llxll + JJyJJ+8 · 2
i=l
Thus the arbitrariness of 8 implies that
llx+yli ~ llxii+IIYII. Let zn
= {0, ... , 0, 1, 0, ... }. ~
Chapter 5
236
It is easy to verify that the linear combinations of the elements zn are dense in the whole space B, because lim Xn = 0. Moreover, for all posn>oo
itive integers n and p,
Then Theorem 3.2.15 implies that {zn} is a basis in B. We shall show now that the space B is not reflexive. For this purpose we shall prove that the closed unit ball in B is not weakly compact. Let Yn = z1 + ... +zn. Of course, IIYnll = 1. If the closed unit ball is weakly compact, then {Yn} converges to a y E B. Since {zn} is a basis, y ought to be of the form (1, 1, ... ). This is impossible, because, for all x €: B, lim Xn = 0. n>00
Now we shall describe all functionals f belonging to the second conjugate space B**. Let {gn} be the basis functionals with respect to the basis {en}. According to Lemma 5.4.1, {gn} constitute a basis in the conjugate space B*. Let F be a functional from B**. Then the functional F is of the following form: there is a sequence of real numbers 00
00
{Fi} such that F(f) = 2 Fi}i for any /E B, f = 2 figi. Let us calcui=I
i=l
late the norm of the functional F. We have n
n
I}; Fi/il
=
n
If(}; Fizi)l ~ 11/11[12 Fizill
i=l
i=l
i=l
n'
 11/11 sup 
[
"\' L..J (Fp,i1 Fp,;) 2 +(Fq,n'+>)2]1/2 ,
i=l
where the supremum is taken over all positive integers n' and increasing sequences of indices PI> ... , p 2n'+l with Fpk replaced by 0 if Pk > n. The arbitrariness of n implies n
IIF II~ sup [}; (Fp,i1Fp.t)2 +(Fp,n'+t)2 i=l
r. (5.4.2)
Weak Topologies in Banach Spaces
237
n
Let us now fix nand let un =
1: F;zi. Let us define a linear functional i=l
f on the space Yn spanned by the elements un: zn+l, zn+2, ... in such a way that and
f(zi) = 0
fori= n+l, n+2, ...
Then 00
L
jJ(aun+
00
atzi)j=JiaunJI~jjaun+};
i=n+l ~
Thus
11/11 =
atzijj.
i=n+l CD
f = }; figi be an extension of the functional f to
1. Let
i=l
whole space B of norm one. Thenfi = 0 fori> nand oo
F(f) =
n
j}; Ftfi\j = j}; Ftfij = /f(un)/ = i=l
//unJI
~ //F//.
i=l
Hence, calculating the norm of un, we obtain (5.4.3) for all positive integers n and all finite increasing sequences of integers .. • ,p 2n+l· C~mbining (5.4.2) and (5.4.3), we obtain
p1,
(5.4.4) where the supremum is taken over all positive integers n and finite increasing sequences p 1 , ••• , p 2n+l· The norm IIF II is finite if and only if there is a limit lim Fn. Since the space B is not reflexive, B** contains n=co
an element which does not belong to n(B). Then the only possibility is B** that n(B) is a subspace of codimension 1, i.e. that dim n(B) = 1. There
(BXB)** X** fore, dim n(B X B) = 2, and since dim fl(X) is an invariant of an isomorphism, the space B is not isomorphic to its Cartesian square (see Bessaga and Pelczynski, 1960b).
Chapter 5
238
Pelczyiiski and Semadeni (1960) showed another example of a space which is not isomorphic to its square. Their example is of the type C(.Q). An example of a reflexive Banach space nonisomorphic to its square was given by Figiel (1972). Problem 5.4.3. Does there exist a Banach space nonisomorphic to its Cartesian product by the real line?
The answer is positive for locally convex spaces, as will be shown in Corollary 6.6.12. Rolewicz (1971) gave an example of normed (noncomplete) space X nonisomorphic to its product by the real line. Dubinsky (1971) proved that each B 0space contains a linear subset X which is not isomorphic to its product by the real line. Bessaga (1981) gave an example of a normed space which is not Lipschitz homeomorphic to its product by the real line.
5.5.
EXTREME POINTS
Let X be a linear space over the real or the complex numbers. Let K be an arbitrary subset of X. We say that a point k e K is an extreme point of the set Kif there are no two points k1 , k 2 e K and no real number a, 0 < a < 1 such that k = ak1 +(1a)k2 •
(5.5.1)
The set of all extreme points belonging to K is denoted by E(K). A subset A of the set K is called an extreme subset if, for each k e A the existence of k 1 ,k2 , 0 A. Since the set K is compact, the intersection of a decreasing family of closed sets is a closed nonvoid set, and obviously it is also an extreme set, provided the members of the family belong to m.
Weak Topologies in Banach Spaces
239
Then, by the KuratowskiZorn lemma, there is a minimal element A 0 of the family m. We shall show that the set A 0 contains only one point. Indeed, let us suppose that there are two different points p, q e A 0 • Then there is a functional x* e X* such that Rex*(p) o:F Rex*(q). (5.5.2) Let (5.5.3) A1 = {xe A0 : Rex*(x) = inf Rex*(y)}. yEA,
Since the set A 0 is compact, the set A1 is not empty, Moreover, formula (5.5.2) implies that the set A1 is a proper part of the set A 0 • Let k1ok 2 be points of K such that there is an a, 0 < a < 1, such that ak1 +(1a)k2 eA 1 •
(5.5.4)
Since A 0 is an extreme subset, k1 and k 2 belong to A 0 • Since (5.5.4) aRex*(k1)+(1a)Rex*(k2) = inf Rex*(y). yE.Ao
This is possible if and only if Rex*(k1) = Rex*(k 2) = inf Rex*(y). yE.A,
This implies that k1 ,k2 e A1 • Hence A1 is an extreme set. Thus we obtain a contradiction, because A 0 is a minimal extreme subset. Therefore, A 0 is a onepoint set, A 0 = { x 0 } and, from the definition, x 0 is an extreme point. D THEOREM 5.5.2 (Krein and Milman, 1940). Let X be a locally convex topological space. Let K be a compact set in X. Then conv E(K) ) K.
(5.5.5)
Proof Suppose that (5.5.5) does not hold. This means that there is an element k e K such that k ¢ conv E(K). Then there are a continuous linear functional x* and a constant c and positive e such that
Rex*(k) ~ c
(5.5.6)
and Rex*(x);;, c+e
for x e conv E(K.).
(5.5.7)
Let K 1 = {xe K: Rex*(x)
= infRex*(y)}. yEK
(5.5.8)
Chapter 5
240
Since the set K is compact, the set K1 is not empty. By a similar argument to that used in the proof of Proposition 5.5.1, we can show that K1 is an extreme set. By formula (5.5.7) the set K1 is disjoint with the set E(K). This leads to a contradiction, because, by Proposition 5.5.1, K1 contains an extremal point. 0 5.5.3.
COROLLARY
If a set K
is compact, then
conv K = conv E(K), 5.5.4. For every compact convex set K,
COROLLARY
K
= conv E(K).
5.5.5. Let X be a locally convex topological space. Let Q be a compact set in X such that the set convQ is also compact. Then the extreme points of the set conv Q belong to Q. Proof Letp be an extreme point of the set convQ. Suppose thatp does not belong to the set Q. The set Q is closed. Therefore, there is a neighbourhood of zero U such that the sets p+ U and Q are disjoint. Let V be a convex neighbourhood of zero such that PROPOSITION
vvc u. Then the sets p+ V and Q+ V are disjoint. This implies that p E Q+ V. The family {q+ V: q E Q} is a cover of the set Q. Since the set Q is compact, there exists a finite system of neighbourhoods of type qi+ V, 11
i = 1,2, ... , n, covering Q, QC
U (qi+V).
i=l
Let Kt
= conv ((qi+ V) n Q).
The sets Ki are compact and convex; therefore conv(K1 u ... u Kn) = conv (K1 u ... u Kn) = convQ. Hence n
p =
.J; atki, i=l
n
at~ 0,
.J; at =
1, kt E K t.
i=l
Since p is an extreme point of conv Q, all at except one are equal to 0.
Weak Topologies in Banach Spaces
241
This means that there is such an index i that pEKt C Q+V, which leads to a contradiction.
0
REMARK 5.5.6. In the previous considerations the assumption that the space X is locally convex can be replaced by the assumption that there is a total family of linear continuous functionals T defined on X. Indeed, the identity mapping of X equipped with the original topology into X equipped with the Ttopology is continuous. Thus it maps compact sets onto compact sets. Therefore, considering all the results given before in the space X equipped with the Ttopology we obtain the validity of the remark.
5.6. EXISTENCE OF A CONVEX COMPACT SET WITHOUT EXTREME POINTS Roberts (1976, 1977) constructed an Fspace (X, II ID and a convex compact set A C X, such that A does not have extreme points. The fundamental role in the construction of the example play a notion of needle points (Roberts, 1976). Let (X, II ID. be an Fspace. We say that a point x 0 EX, x 0 ::j::. 0, is a needle point if for each c; > 0, there is a finite set F C X such that x 0 EconvF,
(5.6.1)
sup{llxll: xEF}<s,
(5.6.2)
conv {0, F} c conv {0, x 0 }+B.,
(5.6.3)
where, as usual, we denote by B. the ball of radius c;, B.= {x: llxll < s}. A point x 0 is called an approximative needle point if, for each c; > 0, there is a finite set F such that (5.6.2) and (5.6.3) hold, and moreover x 0 E conv F+B•.
(5.6.4)
Since c; is arbitrary, it is easy to observe that x 0 is a needle point if and only if it is an approximative needle point. Let E denote the set of all needle points. The set Eu {0} is closed.
242
Chapter 5
From the definition of needle points and the properties of continuous linear operators we obtain PROPOSITION 5.6.1. Let X, Y be two Fspaces. Let T be a continuous linear operator mapping X into Y. If x 0 EX is a needle point and T(x0) # 0, then T(x0 ) is a needle point. We say that an Fspace (X, II II) is a needle point space if each x 0 EX, x 0 # 0 is a needle point. The construction of the example is carried out in two steps. In the first step we shall show that in each needle point space there is a convex compact set without extreme points, in the second step we shall show that a large class of spaces (in particular, spaces LP, 0 < p < 1) are needle point spaces. THEOREM 5.6.2 (Roberts, 1976). Let (X, II II) be a needle point Fspace. Then there is a convex compact set E C X without extreme points. Proof Without loss of generality we may assume that the norm II II is nondecreasing, i.e. that lltxll is nondecreasing for t > 0 and all x EX. Let {sn} be sequence of positive numbers such that co
,2;
Sn
< +oo.
(5.6.5)
n~o
Let x 0 # 0 be an arbitrary point of the space X. We write £ 0 = conv({O,x0}). Since X is a needle point space, there is a finite set F = E 1 = {x~, ... , x~} such that (5.6.1)(5.6.3) holds for s = s0 • For each xt, i = 1, ... , n~> we can find a finite set Ft such that x}Econv({O} u Fi), sup{llxll: XE Fl}
(5.6.6)1
< ~,
(5.6.7)1
n1
conv({O} u Ft) C conv{O, xl}+B~.
(5.6.8)1
n,
Observe that (5.6.8)1 implies conv({O} u E 2) C conv({O} u E1 )+Be,,
(5.6.9)1
Weak Topologies in Banach Spaces
243
where (5.6.10\ The set E 2 is finite, and thus we can repeat our construction. Finally, we obtain a family of finite sets En such that for each x E En we have XE
conv({O} u En+ 1),
(5.6.6)n
sup{JJxJJ: XEEn} <en, conv({O}
U
(5.6.7)n
En+ 1 ) C conv({O} U En)+Ben·
(5.6.9)n
Let <X>
K 0 = conv(
U En U n=O
{0}).
The set K 0 is compact, since it is closed and, for each e a finite enet in K0 • Indeed, take n0 such that
>
0, there is
00
.2; en< e. n=n0 no
By (5.6.9)n the set
U En constitute an enet in the set K
0•
n=O
Observe that no x =I= 0 can be an extremal point of K0 , since 0 is the 00
unique point o£ accumulation of the set
U En,
and, by construction,
n=O
no x E En is an extremal point of K 0 • Thus the set K 0  K 0 does not have extremal points. D Now we shall construct a needle point space. Let N(u) be a positive, concave, increasing function defined on the interval [O,+oo) such that N(O) = 0 and lim N(u) = 0. u
11+00
(in particular, N(u) could be uP, 0 < p < 1). Let Q = [0, 1]"' be a countable product of the interval [0, 1] with the measure p, as the product Lebesgue measure. Let E be a ualgebra induced
Chapter 5
244
by the Lebesgue mesurable sets in the interval by the process of taking product. 1
Take now any function f(t)
E
L co[O, I] such that
J f(t)dt = I. We 0
shall associate with the function fa function St(j) defined on· [0, I]w by the formula St(f)lt = f(tt),
where t = {tn}. Observe that the norm of Si(/) in the space N(L(Q,E,Jl.)) is equal to 1
IISt(f)JI =
f N(f(t))dt,
i~
(5.6.11)
I, 2, ...
0
Of course Si(f) can be treated as an independent random variable. Thus, using the classical formula n
£2
n
2; (Xt E(Xt)) = 2; E (Xt E(Xt)) 2
i=l
i=l
n
we find that, for at ;:?o 0 such that }; ai = I, i=l
n
n
f [2; at(St(f)I)r dJi. = 2; af a~(St(f)1)2dJ1
a
i=l
i=I n
:::=;::;a
2; at J (St(/)1) dJ1. 2
a
i=I 1
=
J(f(t)1) dt,
a
(5.6.12)
2
0
where a = max {aI>
••• ,
an}.
By the Schwartz inequality we have n
n
( j 12; a,S(fi)IJdJl.r::::;::; f 2; at(St(f)1) dJ1. 2
a
i=1
a
i=1
(5.6.13)
Weak Topologies in Banach Spaces
245
The function N(u) is concave, hence the following inequality results directly from the definition (compare the Jensen inequality for convex functions) n
n
(5.6.14)
N(); atut} ) ' ) ; atN(ut). i=1
i=1
As an intermediate consequence of formula (5.6.14), we infer that for each g E N(L(Q,E,p)) n L(Q,E,p), we have
JlgJJ ~ N(
j
(5.6.15)
JgJ dp).
!1
By (5.6.12), (5.6.13) and (5.6.15) we obtain n
1
J[); atSt(f)1[[ ~N( a( J(f(t)1) dtr' 2
i=1
2
(5.6.16)
).
0
Now we shall introduce the notion of IJdivergent zone. Letfe L 00 [0, 1] 1
be such that
I
f(t)dt = 1 and let IJ
> 0. An interval [a,b], 0
0
is called a IJdivergent zone for thefunctionfiffor arbitrary real numbers, a 1 , ••• ,an, ai )' 0 such that a 1 + ... +an= 1 we have (5.6.17) and sup[[
2: atSt(f)[[ < IJ.
(5.6.18)
at:::;a
LEMMA 5.6.3 (Roberts, 1976). Let N(u) be as above. Then for each b .
there are a nonnegative function fe L ""[0, 1] such that
1
I
N(Jf(t)J)dt
>0 < IJ
0
1
and
I f(t)dt =
1 and a number a, 0
0
is a IJdivergent zone for the function f Proof In view of the properties of the function N it is easy to to find
Chapter 5
246
a functionfe L"'[O, 1] such that
J f(t)dt =
1 and
0
b
1
j
N(jj(t)j )dt
o
< .
(5.6.19)
m
where m ~ 1/b. Take al> ... , an such that a 1 + ... +an ~ 1 and at ~ b, i = 1,2, ... , n. Thus n ~ m and, by (5.6.11) and the triangle inequality, we obtain m
1
II}; atSt(/)11 ~m j N(if(t)i)dt
(5.6.20)
0
For chosen/, by (5.6.16) there is an a such that (5.6.18) holds.
D
5.6.4 (Roberts, 1976). A function equal to 1 everywhere is a needle point in the space N(L(Q,E,p)). Proof. Let s be an arbitrary positive number. Let a positive integer k be chosen so that N(Ifk) < s/3. Let b = s/3k. By Lemma 5.6.3 we can PROPOSITION
1
choose functions h,
... ,/k F. L "'[0, I]
such that .fi(t) ~ 0,
J.fi(t)dt = I, 0
1
JN(ifi(t)i)dt < b i =
I,2, ... , k and .fi have disjoint bdivergent zones
0
[at,hd. Let
By (5.6.I6) there is ann such that I/n
< min ai and
I _n!_ i; St(f)III < s. i=1
I
Thus, for the set F = {S1(f), ... , Sn(f)}, (5.6.4) holds. By the choice of .fi we trivially obtain k
IIIII =
II
k
! 4 Jill~ 4 11./ill < ~=1
and this implies (5.6.2).
~=1
k
3~ .
= ;'
Weak Topologies in Banach Spaces
247
Now we shall show (5.6.3). Take cx1 , •.• , cxn ~ 0 such that cx1 1, ... , k, we shall write
+ ... +
+an= 1. For eachj,j =
Lj = } ; CXt St(li)' «t~bJ
Mj =
};
CXtSt(fi),
lli~«c
Rt =
}; CXtSt(fi). «c
Of course, l
n
}; cxtSt{f) =
k
k}; Lt+ Mt+ Rt.
i=l
j=l
By the definition of the €5divergence zone we have
< €5, JIRJCJJI < €5,
(5.6.21)
JILt!!
(5.6.22)
where
Thus (5.6.23) and
1
1
k
k
Ilk~ Rjkj~ Cjl
<;.
The intervals [a1, b1] are disjoint. Thus each ext can belong to at most one interval [a1,b1]. Therefore
1
k
k
Jk _27 Mtdfl. = aj ~ }; };
a
j=l
cxtSt(fi) dp.
J=l «<E[tli,bJ]
1
k
=k};
2:
j=l «cE[tli,bJ]
1
CXF';;;,k
n
1
2: CXt~k'
i=l
Chapter 5
248
and by (5.6.15) 11
~
t. ~ N( n M111
< ;.
k
Finally, for c = }
,2;
CJ,
j=l
n
!I}; <XtSt(/)cll <
1
(5.6.24)
i=l
and (5.6.3) holds. Thus 1 is an approximative needle point and it is also a needle point. D THEOREM 5.6.5 (Roberts, 1976). For N, (Q,E,p) as above, the space N(L(D,E,p)) is a needle point space. Proof Take agEL 00 (Q,E,p), g::/= 0. The function g induces a continuous linear operator Tg, Tgf = gf mapping N(L(Q,E,p)) into itself. Observe that Tu(l) =g. Thus, by Propositions 5.6.1 and 5.6.4, g is a needle point. Since each element of the space N(L(D,E,p)) can be approximated by bounded functions and the set of needle points with added 0 is closed, each element different from 0 is a needle point. D. By the classical measure theory, there is a onetoone mapping of the interval [0, 1] onto [0, I]<" preserving the measure. This implies that the spaces N(L(E, D,p)) and N(L) are isomorphic. In this way we obtain CoROLLARY 5.6.6. If the function N has the properties described above, then there is in N(L) a convex compact set without extreme points. Shapiro (1977) formulated the problem of the existence of an Fspace X with a trivial dual in which each convex compact set has extreme
points. Kalton (1980) showed that certain spaces N(L) have these properties. Some further information can be found in Kalton and Peck (1980).
Chapter 6
Montel and Schwartz Spaces
6.1.
COMPACT SETS IN FSPACES
The definition of compact sets implies that if K is a compact set in an Fspace X, then for any neigbourhood of zero U there is a finite system of points x1 , •.• , Xn such that n
K C
U (xt+U). i=l
(6.1.1)
PROPOSITION 6.1.1. Let X be an Fspace. Let K be a closed subset of the space X such that, for every neighbourhood of zero U, there is a finite system ofpoints x1 , ••• , Xn such that (6.1.1)holds. Then the set Kis compact. Proof Let
Un
= (x: llxll <
!),
where, as usual, llxll denotes the Fnorm in X. Let {Yn} be an arbitrary sequence of elements of the set K. The property of the set K implies that there are an infinite subsequence {y~} of the sequence {Yn} and a point x1 such that Y! E K1 = (x1 + ~). The set K1 is a closed subset of K, therefore, there are an infinite subsequence {y!} of the sequence {y!} and an element {x2} such that Y! E K2 = (x2 + UJ. Repeating this argument, we find by induction that there are a family of sequences {Y!} is a subsequence of the sequence {Y! 1} and a sequence of points {xk} such that n, k =I, 2, ... 249
(6.1.2)
Chapter 6
250
Let {Yn} = {y:}. The sequence {Yn} is fundamental. Indeed, if n,m > k, then Yn, Ym are elements of these sequences {y!}, thus by (6.1.2) llYnYmll < < 2/k. The space X is complete, hence there is a limit y of the sequence {Yn}· Since the set K is closed, y E K. This means that each sequence of elements K contains a convergent subsequence. Then the set K is com
0
pKL
Proposition 6.1.1 holds also for complete linear topological spaces but here we shall restrict ourselves to the metric case. Let X be a finite dimensional space. It is easy to verify that each closed bounded set in X is compact. Since a finitedimensional space is locally bounded, this means that there are neighbourhoods of zero such that their closures are compact sets. We say that an Fspace is locally compact if there is a neighbourhood of zero U such that the closure U of the set U is a compact set. THEOREM
6.1.2 (Eidelheit and Mazur, 1938). Each locally compact space
X is finitedimensional. Proof Let V be such a neighbourhood of zero that the closure V of V is a compact set. Let Y be an arbitrary finite dimensional subspace different from the whole space X. Obviously, X# Y + V. Suppose that VC Y+ V. Then Y+ V = Y+ V. Lemma 5.1.4 implies that the set Y + V is closed. On the other hand, the set Y + V is open. and we obtain a contradiction. Therefore, there is an a E V such that a¢ Y+V. Now we shall construct by induction the following sequence. y 1 is an arbitrary element. Suppose that the elements y 1 , •.• , Yn are defined. Let Yn be the space spanned by those elements. Let Yn+I be such an element that Yn+1 E V and Yn+I ¢ Yn+ V. It is easy to verify that if the space X is infinitedimensional, we could construct such an infinite sequence {Yn}. But this sequence would not contain any convergent subsequence, because, fork# n, ykYn ¢ V. This leads to a contradiction since the set Vis compact. Therefore, the space X is finite dimensional. 0
Monte) and Schwartz Spaces
251
PROPOSITION 6.1.3 (Mazur, 1930). Let X be a B0 space. If a set A C X is compact, then the set conv A is also compact. Proof Let U be an arbitrary convex neighbourhood of zero. Since the set A is compact, there is a finite system of elements xi> ... , xn such that n
A C
U (xt+U). i=l
K
conv({x1, •.. , xn}).
{6.1.3)
Let =
The set K is bounded and finitedimensional, therefore, there is a finite system of points y 1 , ... , Ym such that m
K C
U (Yi+U). i=l
Thus, by (6.1.3), conv A c conv(K+ U) Therefore m
convA C
=
(6.1.4)
K+ U. m
U (ydU)+U = i=l U (Yt+2U). i=l
The arbitrariness of U and Proposition 6.1.1 implies the proposi~.
D
Since Proposition 6.1.1 holds for complete topological spaces, Proposition 6.1.3 holds for complete locally convex spaces. Therefore, in the case of complete spaces, we can omit in Corollary 5.5.5 the assumption that the set conv Q is compact.
6.2. MONTEL SPACES In the preceding section we proved that each locally compact space is finitedimensional. In finite dimensional spaces each bounded closed set is compact. There are also infinitedimensional spaces with this property. Examples of such spaces will be given further on. Fspaces in which each closed bounded set is compact are called Monte/ spaces.
Chapter 6
252
PROPOSITION 6.2.1. Locally bounded Monte/ spaces are finitedimensional. Proof Let X be a locally bounded Montel space. Since X is locally bounded, there is a bounded neighbourhood of zero U. The space X is a Montel space, hence the closure U of the set U is compact. Therefore, the space X is locally compact and, by Proposition 6.1.2, finitedimensional. D PROPOSITION 6.2.2 (Dieudonne, 1949; Bessaga and Rolewicz, 1962). Every Monte/ space is separable.
The proof of the theorem is based on the notion of quasinorm (Hyers, 1939; Bourgin, 1943), similar to the notion of pseudonorm. Let X be an Fspace. By ~ we denote the class of all open balanced set. Let A E ~. The number [x]A=inf{t>O:
~ EA}
is called the quasinorm of an element x with respect to the set A. Quasinorms have the following obvious properties: (a) [tx]A = It I[x]A,, (b) if A) B, then [x]A :::( [xB], (c) the quasinorm [x]A is a homogeneous pseudonorm (i.e., satisfies the triangle inequality) if and only if the set A is convex. PROPOSITION 6.2.3. [x+y]A+B :::( max((x]A, [y]B). Proof Let us write r = max([x]A, (y]h). Let e be an arbitrary positive number. By the definition of the quasinorm,
XE(1+e)rA
and
yE(1+e)rB.
Hence
x+yE(1+e)r(A+B). Therefore [x+y]A+B :::( (1 +e)r. The arbitrariness of e implies the proposition. D PROPOSITION 6.2.4. Let a sequence {An} C ~ contitute a basis of neighbourhoods of zero. Then a sequence {xm} tends to 0 if and only if lim [xm]An = 0 ~0
(n = 1, 2, ... ).
(6.2.1)
Monte! and Schwartz Spaces
253
Proof. Necessity. Suppose that Xm+0; then, for arbitrary s > 0, and n, there is an m 0 dependent on n and s such that for m > m 0 , Xm E sAn, whence [xm]A,. :::;;;; s.
Sufficiency. Let us suppose that (6.2.1) holds; then for every n there is an m 0 dependent on n such that, for m > m 0 , [xm]A,. :::;;;; 1. This means that Xm E An. Since {An} constitutes a basis of neighbourhoods of zero,
D
~~
PROPOSITION 6.2.5. Let {An} C m: constitute a basis of neighbourhoods of zero. A set K is bounded if and only if there is a sequence of numbers {Nm} such that sup[x]A .. :::;;;; N,. XEK
Proof. Sufficiency. Suppose that a sequence {Nn} with the property
described above exists. Let {xm} be an arbitrary sequence of elements of K and let {tm} be an arbitrary sequence of scalars tending to 0. Then [tmXm]A,.:::;;;; ltmiNn+ 0.
Hence, by Proposition 6.2.4, the sequence {tmxm} tends to zero. Therefore, the set K is bounded. Necessity. Suppose that there is an index n such that sup[x]A. = +=. xEK
Then there is a sequence {xm} C K such that 0 < [xm]A .. +oo. Let tm = 1/[xm]A ..· The sequence {tm} tends to 0. On the other hand [tmxm]A .. = 1, l).ence, by Proposition 6.2.4, the sequence {tmxm} does not tend to 0. This implies that the set K is not bounded. D Proof of Proposition 6.2.2. Let X be a nonseparable Fspace with the Fnorm llxll· Since the space X is nonseparable, there are a constant ~ > 0 and an uncountable set Z such that liz z'll > ~ for z, z' E Z, z i= z'. Let Kn = {x: llxll < 1/n}, and le~ us write briefly [x]n = [x]x..·
Since the set Z is uncountable, there is a constant M 1 such that the set Z 1 = Z n {x EX: [x] 1 :::;;;; M 1 } is also uncountable. Then there is a constant M 2 such that the set Z 2 = Z 1 n {x EX: [x] 2 :::;;;; M 2 } is uncountable. Repeating this argumentation, we can find by induction a sequence of uncountable sets {Zn} and a sequence of positive numbers {Mn} such that the set Zn is a subset of the set Zn1 and sup[x]t:::;;;; Mn for XEZn
i= 1,2, ... ,n.
Chapter 6
254
Let us choose a sequence {zn} such that Zn E Zn and Zi =I= Zk fori =I= k. Then sup [zn]k ,;;; max (Mk, [z1]k, ... , [zkt]k) n
<
+oo
for k = 1, 2, ...
Therefore, by Proposition 6.2.5, the sequence {zn} is bounded. On the other hand, Zn E Z, hence llziZkil > t5 for i =F k. This implies that the D set {.Zn} is not compact. Therefore, X is not a Montel space. The following question has arised: is it sufficient for separability if we assume that each bounded set is separable? The answer is negative. Basing on the continuum hypothesis Dieudonne (1955) gave an example of a nonseparable space in which each bounded set is separable. For Fspaces such an example was .given by Bessaga and Rolewicz (1962). For D0 spaces it was given by RyllNardzewski.
D:
PROPOSITION 6.2.6 (RyllNardzewski, 1962). There is a nonseparable D0 space in which all bounded sets are separable. Proof Let S denote the class of all sequences of positive numbers. We introduce in S the following relation of order ~. We write that {In} ~ {gn} ifln < gn for sufficiently large n. A subclass S1 of class Sis called limited if there is a sequence {hn} E S such that {fn} ~ {hn} for all sequences {In} E S1 • Let us order class Sin a transfinite sequence {I:} of type w 1 (here we make use of the continuum hypothesis). Now we define another sequence of type w 1 as follows: {g~} is the first sequence (in the previous order) which is greater in the sense of relation ~ than all {I~} for oc < {3. It is easy to see that no noncountable subclass of this sequence is limited. Let us consider the space X of all transfinite sequences {x"} (oc < w1) of real numbers such that x" vanishes except for a countable number of indices and ,, p
ll{xp}lln = L.J gnjxpj < +oo,
n
=
1, 2, ...
P<w,
Let us introduce a topology in the space X by the sequence of pseudonorms ll{xp}lln· The space X with this topology is a Dri space. We shall
Monte! and Schwartz Spaces
255
show that the space X is complete. Let {xn} be a fundamental sequence of elements of the space X. Let A= {a: x::;t=o for certain n}.
The definition of the space X implies that the set A is countable. Therefore, the subspace X 0 spanned by the elements {x!}, {x!}, ... is of the type V(am,n). Thus it is complete. Therefore, the sequence {x:} has the limit {xa} E X 0 C X. To complete the proof it is enough to show that each bounded set Z C X is separable. If the set Z is bounded, then there is a sequence of positive numbers {Mn} such that sup ll{xa}/ln ~ Mn,
n = 1, 2, ...
{x..)EZ
Let k be a positive integer. By h we denote the set of all such indices that there is an x ={xu} E Z and such anindex p such that [xp] > 1/k. Then we 1
have kg~< ll{xa}lln ~ Mn. Hence, for
P E h, we have {g!} 3
{Mn} and 00
by the property of the class {g~} the set h is countable. Let I= ,._..
U Ik. k=l
Then the set I is, of course, also countable. Let y be the smallest ordinal greater than all the terms of the set I. Then from the definition of the set I it follows that if {xa} E Z, then X 0 = 0 forb > y. Thus the set Z is separable. 0
6.3.
SCHWARTZ SPACES
Let X be an F*space. We say that a set K is totally bounded with respect to a neighbourhood of zero U, if, for any positive c:, there is a finite system of points xi> ... , Xn. such that ~ C
00
U (xi+c:U). A set which is totally i=l
bounded with respect to all neighbourhoods of 0 is called totally bounded or precompact. Proposition 6.1.1 implies that if a set K is closed and totally bounded with respect to all neighbourhoods of zero, then it is compact.
Chapter 6
256
An Fspace X is called a Schwartz space if, for any neighbourhood of zero U, there is a neighbourhood of zero V totally bounded with respect to U. 6.3.1. Let X be a Schwartz space. Then its completion also a Schwartz space. PROPOSITION
X
Proof Let U0 be a neighbourhood of 0 in X. Let U = U0 n X. Since X is a Schwartz space, there is a neighbourhood of zero V C X such that for any s > 0 there is a system of points x1 , ••• , Xn such that n
V C
U (xi+sU). i=l •
Thus n
V C
n
U (xi+sU) c U (xi+2sU). i=l
D
i~l
6.3.2. Every Schwartz Fspace is a Monte/ space. Proof Let K be a closed bounded set. Let U be an arbitrary neighbourhood of zero. Since we consider a Schwartz space, there is a neighbourhood of zero V totally bounded with respect to U. The set K is bounded. Then there is a positive a such that K C a V. The neighbourhood Vis totally bounded with respect to U, and so there is a finite PROPOSITION
system of points Then Kc aV c
Y~> ... , Yn
such that V C
0 1
(Yi+
~). Let
Xi
= ayi.
co
U (xi+U). i=l
Since the set K is closed and U is arbitrary, the set K is compact (see Proposition 6.1.1). D There are also Mantel spaces which are not Schwartz spaces. An example will be based on PROPOSITION 6.3.3. Let am, n ~ am+I,n. The space M(am, n) (or LP(am, n)) (see Example 1.3.9) is a Schwartz space if and only. if, for any m, there
Monte! and Schwartz Spaces
257
is an index m' such that
. am.n O 1Im = '
(6.3.1)
am•,n
lnl~oo
where lnl = lnii+In2l+ ... +lnkl· Proof. Sufficiency. Let U be an arbitrary neighbourhood of zero. Let U0 be such a neighbourhood of zero that U0 + U0 C U. Let m be such
an index that the set
~
{x: llxllm <
Um =
(resp. Pm(x) <
~)}
is contained in U0 • Let m' be such an index that (6.3.1) holds. Let 8 beanarbitrarypositive number. Since (6.3.1) holds, there is a finite set A of indices such that, for n ¢ A, am.n
<
8.
(6.3.2)
am',n
Let L be a subspace of M(am,n) (resp. LP(am,n)) such that {xn} E L if and only if x 71 = 0 for n ¢A. The space L is finitedimensional. Let KL = {x E L: am'.n lxnl < I}. The set KL is compact. Then there is n
a finite system_ of points y 1 ,
••• ,
Yn such that KL C
U (Yt+ U
0 ).
i=l
Let V = {x: llxllm• < 1/m}. Let x be an arbitrary element od V. By (6.3.2) there is an x 0 E VnL such that xx0 EBUmC8U0 • Since VnLcKL, we have n
V C V n L+8U0 C KL+8U0 C
U (yi+eU +8U 0
0)
i=l
n
C
U (Yi+8U).
i=l
Thus the set Vis totally bounded with respect to U. Necessity. Let us suppose that there is such an m that for all m'
. sup'amn I1m lnl+oo
am•,n
..:
= um
> O.
>m
(6.3.3)
Chapter 6
258
Let U = {x: Jlx!Jm < 1 (resp. Pm(x) < 1)}. Let V be an arbitrary neighbourhood of zero. Then there are a positive number b and an index m' such that V:> {x: l!xm•ll < b}. Let A be the set of such indices n that amn bm' '  >2 . am•,n Since (6.3.3) holds, the set A is infinite. Let yn =
{y~},
where
fork= n, fork =I= n. It is obvious that yn E V (n = 1,2, ... ). On the other hand, if n,n' E A, n=/= n', then llynyn'll > Mm.f2 (resp. Pm(ynyn') > [Mm,f2]P). Since the set A is finite, this implies that Vis not totally bounded with respect to U. The arbitrariness of V implies the proposition. D Example 6.3.4 (Slowikowski, 1957) Example of a Monte! space which is not a Schwartz space. Let k, m, n~> n2 be positive integers. Let
ak,m,n,,n, = n~max(l,n~n·).
Let X denote the space of double sequences x = {xn1,n 2} such that l!xllk.m = sup ak,m,n,,n, Jxn,.n,l < nhna
+ 00
with the topology determined by the pseudonorms l!xl!k,m· X is a B0 space of the type M(am, n). The space X is not a Schwartz space. Indeed, let us take two arbitrary pseudonorms l!xl!k,m and l!xllk',m'· Let n~ > m,m'. Then lim n 2~co
ak,m,n~.n. = (n?)kk'
Therefore lim sup ak,m,n,,n, lnl+o:>
> 0.
ak',m',nY,na
ak','11'tnl,nl
> 0,
and from Proposition 6.3.2 it follows that the space X is not a Schwartz space.
Monte) and Schwartz Spaces
259
Now we shall show that the space X is a Montel spac~. Let A be a bounded set in X. Since X is a space of the type M(am, n), it is enough to show that lim ak,m,n.,n,sup lxn,,n,l = 0.
lnl+oo
(6.3.4)
XEA
Let us take any sequence {(n~,nD}, such that lim ln~l+ln~l
= +oo.
We have two possibilities : (1) n~+oo,
(2) nr is bounded. Let us consider the first case. Let x = {Xn 1, n 2} E A and let k' > k, m' > m. Since the set A is bounded, there is a constant Mk',m' such that
Then for sufficiently large
n~
(6.3.5) Let us consider the second case. Let m' Then
> m and m' >
n~,
k'
>k
Therefore, by (6.. 3.5) and (6.3.6) formula (6.3.4) holds. This implies that X is a Monte! space. D
6.4.
CHARACTERIZATION OF SCHWARTZ SPACES BY A PROPERTY OFFNORMS
In the previous section we introduced the notion of Schwartz spaces. Now we shall give a characterization of those spaces by a property of Fnorms. Let Y be an arbitrary F*space with the Fnorm llxll and let 8 be an arbitrary positive number. We write c(Y, 8, t)
= inf {lltxll : x E Y, llxll = 8}
260
Chapter 6
if there is such an element x c(Y, e, t)
if sup
llxll <
=
{~
E
llxll =
Y that
e and
for t#O, fort=O,
e.
xeY
THEOREM 6.4.1 (Rolewicz, 1961). Let X be a Schwartz space. Then, for every increasing sequence of finitedimensional subspaces {Xn} such that 00
the set X*=
U Xn is dense in the whole space X, thefunctionsc(X/Xn,e,t) n=l
are not equicontinuous at 0 for any e. Proof Let us write K 11 = {xeX:
llxll <1]}.
Suppose that the theorem does not hold. Then there are a positive e and a sequence of finitedimensional spaces {Xn} such that Xn C Xn+I> 00
the set X* =
U Xn is dense in the space X and the functions c(X/ Xn,e,t) n=l
are equicontinuous at 0. Let 15 be an arbitrary positive number. By A.0 we denote such a positive number that c(X/ Xn, e, t) <
15
T
for 0
< t < A.0 •
Since the set X* is dense in X, this implies that c(X*/Xn,e, t)
< ~
for 0
< t < A.0 •
We shall choose by induction sequences of positive integers {kn} and of elements {xn} such that (1) llxnll < 15, (2) (3)
Xn E
Xkn>
Xn  Xi ¢ AoKe/2" As k 1 we take 1 and as x 1 we take an arbitrary element of X 1 • Let us suppose that for some n we have chosen the elements xi> ... , Xn and the integers lei> •.. , kn. By hypothesis, there is in th~ space X/Xkn a coset Z
Monte! and Schwartz Spaces
261
such that IIZII = E and llxoZII ~ b/2. By xn+I we denote an arbitrary element of the coset XoZ such that llxn+ 1 ll < CJ, and by Xk•+• we denote a subspace containing the element xn+I· Since for every x E Xk.• (xn+1x) E EAoZ,
II
Xn+).1 x 0
!I ~ IJZII =
E
}we have
fori= 1,2, ... ,n1,n. Thus, by (3), the set K 0 is not totally bounded with respect to the set K, 12 • The arbitrariness of (J implies that X is not a Schwartz space. 0 It is not known whether the inverse theorem to Theorem 6.4.1 holds in the general case. THEOREM 6.4.2 (Rolewicz, 1961). Let X be an Fspace for which there is a positive Eo such that for every x E X, x =I= 0, sup lltxll > E0 • If there is t>O
a sequence of finitedemensional spaces {Xn} such that Xn C Xn+l and the functions c(X/Xn,E, t) are not equicontinuous at 0 for any E, then the space X is a Schwartz space. Proof Let us write
K~=
U tKrr.
ttl
Let E be an arbitrary positive number such that (6.4.1) Suppose that there is a sequence of finitedimensional spaces {Xn} such that Xn C Xn+ 1 and the functions c(X/Xn,E, t)are not equicontinuous at 0. This means that there are a sequence {nk} of positive integers and a positive number (J such that for each tt > 0 and for sufficiently large k sup c(X/Xn.,E,t) > CJ,
(6.4.2)
O~t~Jl
i.e. there is a Jlk, 0
<
< tt such that > CJ.
Jlk
c(X/Xn., E, Jlk)
(6.4.3)
This implies that in the quotient space X/ Xn. Kli C ttkK; C ttK;.
(6.4.4)
262
Chapter 6
Thus (6.4.5) Let x be an arbitrary element of a norm less than b, llxll < b. Let Z denote the coset containing x. Since (6.4.5), IIZ/.ull <e. Let x0 E Z be such an element that llx0 /.ull <e. This implies that (6.4.6) Let us write
x = x0 +(xx0 ). Then
(6.4.7)
(xx0) E Xn. and xx0 eK/5+.uK;
c
,U(K;+K;).
(6.4.8)
Since the space Xn. is finitedimensional, by (6.4.1) the set XnJI (K; + K;) is bounded. Thus it is totally bounded with respect to the set .uK;. Therefore, by Proposition 6.2.3 and by (6.4.8), there is a finite system of points y 1 , •.• , Yn such that m
Ko
c
U (Yi+.u(K;+K;+KD).
i~l
Hence the set K8 is totally bounded with respect to the set K = K; +K; +K;. Since the set K.. constitutes a basis of neighbourhoods of zero, the space X is a Schwartz space. D CoROLLARY 6.4.3 (Rolewicz, 1961). Let X be an Fspace with a basis {en}· Suppose that there is a positive e0 such that sup O
lltxll >eo
for every x E X, x i= 0. Then the space X is a Schwartz space if and only if the functions c(X~,e, t), where X~ are the spaces generated by the elements en+r. en+ 2 , ••• , are not equicontinuous at 0 for any e. Basing ourselves on Corollary 6.4.3, we shall give an example of a Schwartz space which is not locally pseudoconyex.
Monte! and Schwartz Spaces
Example 6.4.4 Let {Pn} be a sequence of real numbers 0
263
< Pn <
1. By
[(p•)
we deriote
the space of all sequences x = {xn} such that co
[[x[[ = }; [xn[P• < +oo n=l
with the Fnorm [[x[[. It is easy to verify that [(p•) is an Fspace. The sequence {en}, en= {0, ... , 0,2,0, ... }, constitutes a basis in the space nth place
[CPn).
By simple calculation we find that
c(X~, s, t) = s[tJP., where X~ is the space spanned by the elements en+I,en+ 2 , Pn = inf{pi: i
•••
and
> n}.
Therefore, by Corollary 6.4.3, the space !(p•) is a Schwartz space if and only if Pn~O. Let us remark that if Pk~o, then the space l(p.) is not locally pseudoconvex. Indeed, let() be an arbitrary positive number. Let Xn = {0, ... , 0, (b) 11P•, 0, ...}. Then [[xnll = b. Let p be an arbitrary positive number
nth place
and let np be such an index that, for n > np, Pn ~p/2. Let us take the pconvex combination of the elements xn»+l, ... , Xn,+k· Then
I XnP+;+ ..• +xnP+k II= klfp
n\.±,k . .:::..
CJ(1)Pn
\=np+l
;;:;, (Jk ( k;fp
klfp
t'2 =
(Jkl/2
The arbitrariness of p and b implies that the space pseudoconvex.
6.5.
~ 00 .
z(pn)
is not locally
APPROXIMATIVE DIMENSION
Let X be an F*space. Let A and B be two subsets of the space X. Let ·B be a starlike set. Let M(A,B,s) = sup{n: there exist n elements xi> ... , Xn e: A such that XiXk ¢ sB for i =1= k}.
Chapter 6
264
The quantity M(A,B,s) is called the scapacity of the set A with respect to the set B. Obviously, M(A,B,s) is a nonincreasing function ofs. Let M (A, B)= {qJ(e): qJ(e) real positive function, .
qJ(e)
}
hm M(A, B, s) = +oo . Let us denote by 0 the family of all open sets and by c:f the family of all compact sets. The family of real functions
nn
M(X) =
M(A, B)
.AE() Be:J
is called the approximative dimension of the space X (see Kolmogorov, 1958). PROPOSITION
6.5.1.
If two
F*spaces X and Yare isomorphic, then
M(X)= M(Y). Proof Let T be an isomorphism mapping X onto Y. Then a set A is open (compact) if and only if the set T(A) is open (compact). Hence M(X)=
nn
M(A,B)
.Ae:J BE() .AcX BEX
=
n n M(T(A), T(B)) = M(Y).
.AE':J BE() .A eX BcX
PROPOSITION
6.5.2. Let X be a subspace of an F*space Y. Then
M(X) :::> M(Y). Proof Let U be an open set in X. Let x
E
rx=inf{Jixyjj: y¢U,yeX}. Let us put V=
U {ze Y:
XEU
jjzxjj
<+rx}.
U and let
D
265
Monte! and Schwartz Spaces
The set V C Y is open as a union of open sets. On the other hand, it is easy to verify that U = V tl X. Then M(Y)=
nn
AE~
BEQ
Ac:Y Bc:Y
= =
nn
M(A,B) C
AE~
BEQ
M(AnX,B)
Ac:Y Bc:Y
n n M(A n X, B n X) n n M(A, B)= M(X).
A~ BeY Ac:Y Bc:Y
AE~
BEQ
D
Ac:X Bc:X
CoROLLARY
6.5.3. If
dimzX ~ dimzY(dimzX = dimzY), then
'
(resp. M(X) = M(Y)).
M(X) :::> M(Y)
In order to prove a similar fact for linear codimension, we shall use the following 6.5.4. Let Y be an Fspace and let X be a subspace of the space Y. Let us consider the quotient space Y/X. Let K be a compact set in Y/X. Then there is a compact set K0 C Y such that PROPOSITION
K= {fx]: xEK0}, where, as usual, we denote by [x] the coset containing x. Proof Since no confusion will result, we denote by the same symbol II 1/ the Fnorm in Yand the norm induced by it in the quotient space Y/X. We shall say that a finite system of elements Zr. ••. , Zn constitutes a finite snet in a set A if for any x E A there is an index i such that 1/ziX/1
<s. Let r0 be an arbitrary positive number. Since the set K is compact, there is a finite ronet z~, ... , Z!, inK. Let Xi, i =I, 2, ... , no, be arbitrary elements such that Xi E z?. Let n
A 0=
U {xE Y: i=l
1/xxt/1
From the definitions of an r0net and of a quotient space it follows that
Chapter 6
266
A0}. Let a~= sup {p: there {x: !!xxzl/
K C {[x]: x
E
2r1
IS
an Xz
E
Z
such that
= inf a~ > 0. ZEK
Indeed, let us suppose that there is a sequence {Zn}, Zn E K, such that a~n +0. Since the set K is compact we can assume without loss of generality that {Zn} tends to Z 0 e K. Therefore, for sufficiently large n,
IIZnZoll <}a~,. This implies that ~n :;;>fa~,, and we obtai11 a contradiction, because a~n +0. Let us take a finite r1net inK, Z~, ... , z;,. The definition of r 1 implies that there are points xf E Zf, i = 1, 2, ... , ni> such t~at n
A1 =
.U {x: llxxfll < r 1}
C
Ao.
t=l
Repeating this argument, we can choose by induction a sequence of systems of cosets {Z~, ... , z~.} and a sequence of positive numbers {nc} tending to 0 such that (1) Z~, .. . , z~. constitute an r~cnet inK, (2) A~c C Ak1> where Ai =
rn
U {x:
llx}xll < ri}.
i=l
Let n•
oo
Kri
=
U U {xn.
k=l i=l
The set K0 = Kri is compact. Indeed, let 8 be an arbitrary positive number. Let us take an'" such that r~c. The finite set
k
nt
U U {xj} constitutes i=l j=l
an r~cnet in K 0 , because Am C A~c for m > k. Since the set K0 is closed, the arbitrariness of 8 implies that the set K is compact. Let K 1 = {[x]: xe K 0}. Since the set K0 is compact, the set K1 is also compact. Moreover, [~] oo
= Z! and the set
n•
U U {Zf} is dense inK. Hence K = k=l i=l
'
K1•
D
Monte) and Schwartz Spaces
267
Proposition 6.5.4 implies the following fact. Let The a continuous linear operator mapping an Fspace Y onto an Fspace X. Then for every compact set Kin X there is a compact set K 0 in Y such that T(K0) = K. Indeed, let Z = {x: Tx = 0}. Then the space X is isomorphic to the quotient Y/Z and the operator T induces the operator T' mapping y E Y into the coset [y] E Y/Z. 6.5.5. Let X and Y be two Fspaces. linear operator T mapping Y onto X, then
PROPOSITION
If there
is a continuous
M(X) C M(Y). Proof To begin with, les us remark that if Tis a linear operator, then M(A, B,
e)~
M(T(A), T(B), e).
Hence M(A, B) C M(T(A), T(B)).
Since the inverse image of an open set is always open and in our case, by the Banach theorem (Theorem 2.3.1 ), the image of an open set is open,
n M(A, B) c n M(T(A), B).
BE(j
BEQ
BcX
BeY
On the other hand, for any compact set K c X there is a compact set K 0 C Y such that T(K0) = K. Therefore M (X)= COROLLARY
nn M(A, B) c nn
AE':J BEO
AE':f BEO
AcXBcX
AcYBcY
6.5.6. If
codim1 X~ codim1 Y (codiml X= codim1 Y), then, M(X) ::> M(Y)
(resp. M(X) = M(Y)).
By a simple calculation we obtain PROPOSITION
6.5. 7. If X is anndimensional space, then
M(X) = {tp(e): lim entp(e) = n~oo
+=}.
_
M(A, B) M(Y).
268
Chapter 6
PROPOSITION 6.5.8. Let X be an F*space. If X is not a Schwartz space, then the set M(X) is empty. Proof If X is not a Schwartz space, then there is a neighbourhood of zero U such that, for any neighbourhood of zero V, there is a positive number ev such that
M(V, U, ev) = +oo.
(6.5.1)
Let Vn = {x: llxll < 1/2n}, where llxll denotes, as usual, the Fnorm in X. Let us write, for brevity, en = evn· Let rp(e) be an arbitrary positive function. Formula (6.5.1) implies that there is a system of points x~, ... , x~n such that mn > rp (en), x£ E Vn, i = I, mn
oo
2, ... , mn, xjx~¢enUforj=f= k. LetK =
U U {x£}. The set Khas a n=l i=l
unique cluster point 0. Then the set K is compact. On the other hand, M(K, U, en)> mn > rp(en). Therefore rp(e) does not belong to M(K, U). The arbitrariness of rp(e) implies that the set M(X) is empty. D The Kolmogorov definition of approximative dimension has some disadvantages. Simply there are "too many" open and compact sets. This is the reason why in many cases it is more convenient to consider a definition of approximative dimension introduced by Pelczyri.ski (1957). The Pelczynski approximative dimension of a space X can be defined (after certain modifications) as the set of functions M'(X)=
Do lJOnpOive M(nV,
U).
integers
In other words, M'(X) = {rp(e): for each open set U there is an open . 'P(e) set Vsuch that, for all n, hm M( V U ) = +oo}. £* 0 n , ,e Let { Ui} be a basis of neighbourhoods of zero. Let M~(X) = {rp(e): for each i, for almost allj, for all n, lim M( e>0
~(e) U
n i+i•
)
i, e
=
+=}.
6.5.9. For all F*spaces, M'(X) = M~(X). Proof The definitions of M'(X) and M~(Y) imply that M'(Y) C M~(Y). Suppose that rp(e) ¢ M'(Y). This means that there is a.neighbourhood of
PROPOSITION
Monte! and Schwartz Spaces
269
zero U such that for all neighbourhoods of zero V there is a positive integer n such that lim M( '1'~8)U ) < +=. Let Ut be an arbitrary e~o n ' ,8 neighbourhood contained in U. Putting V = Ut+i(j = 1, 2, ... ),we find that rp(8) does not belong to M 0(X) C M'(X). D In applications the most useful class is the class M~(X), because it is described by a countable family of functions. PROPOSITION
6.5.10. Let X be a Schwartz space. Then the set M'(X) is not
empty. Proof Let us choose a decreasing basis of neighbourhoods of zero { Ut} in such a manner that M(nUt+~o Ui, 8) < for all positive integers n. Such a choice is possible, because X is a Schwartz space.
+=
Let
n n
rp(8)
=
_.!._
M(nUi+I, Ut, 8)
8 i=l
1 1 for<8<. nl n
Let us take an arbitrary index i and an arbitrary n trary j and 8 < 1/n,
> i. Then, for arbi
0 s M(nUi+j• Ut, 8) sM(nUi+l, Ui,8) s ~ rp(8) ~ rp(8) ~ 8·
This implies that rp(8) E M~(X). Therefore, by Proposition 6.5.10, rp(8) E M'(X). D PROPOSITION
6.5.11. Let X be an Fspace. Then
M(X) :) M'(X). Proof Let K be a compact set and let V be an arbitrary neighbourhood of zero. Then there is a positive integer n such that K C n V. Therefore, M(nV, U,8)
~
M(K, U, 8).
Thus M(n V, U) C M(K, U). This implies the proposition. From Propositions 6.5.8, 6.5.10, 6.5.11 follows
D
Chapter 6
270 CoROLLARY
6.5.12. X is a Schwartz space if and only if the set M(X) is not
empty. CoROLLARY
6.5.12'. X is a Schwartz space
if and only if the set M'(X)
is
not empty.
Proposition 6.5.11 shows that M(X)) M'(X). We do not know whether the converse inclusion is true. It is so in the case where X is a finitedimensional space. For infinitedimensional spaces only the following partial answer is known. PROPOSITION 6.5.13. Let X be an infinitedimensional Schwartz space. Suppose that, for each neighbourhoods of zero U and V, there is a neighbourhood of zero W such that for, all n,
M(nW, U, s) ~ M(V, U, s)
(6.5.2)
for sufficiently smalls. Then M(X) = M'(X). Proof Formula (6.5.2) implies that M(n W, U) J M( V, U). Then the
n n
set M'(X) is equal to the set Mv (X)=
nn
M(V, U).
UEQ VE(j
By Proposition 6.5.11 Mv (X) C M(X).
Now we shall show the converse inclusion. Suppose that a function tp(s) does not belong to Mv (X). This means that there is a neighbourhood of zero U such that for all neighbourhoods of zero V . . f 1Imin eo
tp(s) M(V,2U,s)
<
+oo.
Since the space X is infinite dimensional, lim M(V, 2U, s) M(V, U, s)
•o
=
O .
Thus 1.
· f
Imin e~o
tp(s) 0 M(V, U, B) = .
Monte! and Schwartz Spaces
This means that there is a sequence {s
271
Dtending to 0 such that
v
rp(sk) _ . I liD v  0. k>oo M(V, U,sk)
Let Vn
=
{x: j[x[[
< 21n
},
(6.5.3)
where, as usual, [[x[[ denotes the Fnorm in
the space X. Basing ourselves on formula (6.5.3) we can choose by the diagonal method a sequence {en} tending to 0 and such that lim rp(en) = 0. n.ro M(Vn, U, en) This implies that for sufficiently large n there are systems of points {x~, ... , x~ln}, mn > rp(en), such that xf E Vn (i = 1, 2, ... , mn) and xjoo
xj f# snU fori =I= j. Let K
=
mn
U U {xf}. The set K has a unique clusn=l i=l
ter point 0. Therefore, the set Kis compact. Moreover M(K, U, en))' mn
>
rp(en).
This implies t4at the function rp(s) does not belong to M(X). Hence M'(X) = Mv (X)) M(X).
0
Since it is not known whether the equality M'(X) = M(X) holds in general, we shall prove for M'(X) propositions and corollaries similar to Propositions and Corollaries 6.5.16.5.6. PROPOSITION 6.5.14. Let X andY be two Schwartz spaces. If the spaces X and Yare isomorphic, then
M'(X) = M'(Y).
Proof The above follows immediately from the definition of M'(X) and · the fact that the image (the inverse image) of an open set under an isomorphism is an open set. D
Chapter 6
272
6.5.15. Let X be an F*space and let Y be a subspace of the space X. Then PROPOSITION
M'(X) C M'(Y). Proof The proof is the same as the proof of Proposition 6.5.2. CoROLLARY
0
6.5.16. If
diml X:;:;;; dimlY
(dimlX =dimlY),
then M'(X)
~
M'(Y)
(resp. M'(X) = M'(Y)).
PROPOSITION 6.5.17. Let X andY be two Fspaces. LetT be a continuous linear operator mapping X onto Y. Then
M'(X) C M'(Y). Proof The Banach theorem (Theorem 2.3.1) implies that the image of an open set is an open set. In general, the inverse image of an open set is an open set. The rest of the proof is the same as the proof of Proposition 6.5.6. 0 COROLLARY
6.5.18. If
codiml X:;:;;; codiml Y
(codimlX = codimlY),
then M'(X)
~
M'(Y)
(resp. M'(X) = M'(Y)).
It is easy to calculate approximative dimensions of spaces M(am,n). For the calculation of M'(X) it is enough to know the functions M(Ui+1• Ui, s) for a basis of neighbourhoods of zero {Un} (see Proposition 6.5.9). PROPOSITION
6.5.19. Let X= M(am,n) be a real Schwartz space. Let
Ut = {x: JJxllt < 1}. Then (6.5.4) where E'(r) is the greatest integer less than r.
Monte! and Schwartz Spaces
273
Proof Letx = {xn} andy= {Yn} be two elements of U;+i· Letxy E eU;. Then there is an index n such that a;,nlxnYni >e. On the other hand, ai+i•nXn and ai+i•nYn are less than 1. Since there may exist no E'(1+l:__at.n )numbersbksuch thata;+f,nlbkl < 1 and e ai+j,n . Jbkbk,Ja;,n > e fork=!= k', and systems with this number of elements
more than
exist, the proposition holds.
D
Proposition 6.5.13 and formulas (6.5.4) and (6.3.1) imply that M'(M(am,n))
=
M(M(am,n)).
If X= M(am,n) is a complex space, then tp(e) E M(X) if and only if
y cp(e) belongs to M(Xr), where Xris a real space M(am,n).
Propositions 6.5.19 and Corollaries 6.5.12', 6.5.16, 6.5.18 imply 6.5.20 (Pelczynski, 1957). There is no Schwartz space universal (or couniversal) for all Schwartz spaces. Proof Since for every Schwartz space X the set M'(X) is nonempty, it is sufficient to show that for any functionf(e) there is a Schwartz space X1 such thatf(e) ¢; M'(X,). Let a!= 2/k for nkl < n ~ nk, where nk = logd(l/k)+l. Let X1 = M(am,n), where am,n = (a!) 11m. By proposition 6.3.3 the PROPOSITION
space X1 is a Scl].wartz space. On the other hand,
M( U;+i, U;, !)= ~
i1 E'(1+k(a~) ~ i:i) [1 E'(l+ka~) ~ > 1(!).
Therefore,J(e) <j M'(XJ).
2n•
D
Let us observe that, for any sequence {Xn} of Schwartz space, there is a Schwartz space X universal for the sequence {Xn}. Indeed, let X be the space of all sequences x = {xn}, Xn E Xn with the Fnorm 00
~ 1
llxll = ;21
2n
llxnlln l+JJxnlln'
Chapter 6
274
where JjxJJn denotes the Fnorm in Xn. It is easy to verify that X is a Schwartzspace and that it is universal and couniversal for all spaces Xn.
6.6.
DIAMETRAL DIMENSION
In this section we shall consider another definition of approximative dimension, socalled diametral approximative dimension or briefly diametral dimension (see Mityagin; 1960, Tichomirov, 1960; Bessaga, Pelczynski and Rolewicz, 1961, 1963). Let A, B be arbitrary sets in a linear space X. Let B be balanced. Let L be a subspace of X. We write c5(A, B, L)
= inf(e > 0: L+eB >A).
Let us write c5n(A, B)= infc5(A, B, L),
where the infimum is taken over all ndimensional subspace L. Let c5(A, B) denote the class of all sequences t = {tn} of scalars such that lim
tn c5n(A, B)
n+co
=
0.
The following properties of the class c5(A, B) are ob':ious: if A' C A and B c5(aA, bB)
~
B', then c5(A', B') C c5(A, B);
= c5(A, B) for all scalars a, b different from 0.
(6.6.1) (6.6.2)
Let X be an Fspace. Let 0 denote the class of open sets and c;J the class of compact sets. Let c5(X) =
U U c5(B, U).
UE() BE':f
The class c5(X) is called the diametral approximative dimension (briefly diametral dimension) of the space X. PROPOSITION
6.6.1. Let X andY be two isomorphic Fspaces. Then
c5(X)
~
c5(Y).
Monte! and Schwartz Spaces
275
Proof The proposition immediately follows from the fact that the classes of open sets and compact sets are preserved by an isomorphism. D In many cases diametral dimension is easier to calculate than approximative dimension. Unfortunately we do not know the answer to the following question: do we have b(X) C o(Y) is X is a subspace of an Fspace Y? As we shall show later, the answer is affirmative under certain additional assumptions. PROPOSITION 6.6.2 (Mityagin, 1961). Let X andY be two Fspaces. LetT be a continuous linear operator mapping X onto Y. Then
b(X) :::> b(Y). Proof The definition trivially implies that, for arbitrary A, B C X and an arbitrary subspace L, b(A, B, L) ?= o(T(A), T(B), T(L)). Since dim T(L)
~
dimL, this implies
bn(A, B)?= On(T(A), T(B))
and
o(A, B) :::> o(T(A), T(B)).
The inverse image of an open set is an open set. For any compact set K C Y there is a compact set K 0 C X such that T(K0 ) = K (cf. Proposition 6.5.4). Then
o(X) =.
U BEO U o(A, B) :::> U BEO U b(T(A),
AE~
~
:::>
AE~
T(B))
AE~
A eX BeX
AeX Be X
U b(A, B)= o(Y).
BEO
AeY BeY
COROLLARY 6.6.3.
/f
codim1 X~ codim1 Y
(codim 1X= codim1 Y),
then (resp.b(X)
b(X) C o(Y)
=
o(Y)).
PROPOSITION 6.6.4 (Bessaga, Pelczynski and Rolewicz, 1963). Let 3(X) =
U
n
UEO VEO
UeX VeX
b(V, U).
D
Chapter 6
276
Then b(X) = b(X). Proof Let B be an arbitrary compact set and V, U arbitrary neighbourhoods of zero. Since the set Bis compact, there is a positive number a such that BcaV. Then by (6.6.1) and (6.6.2) we obtain r5(B, U)Cb(V, U) for all B and V. Hence
U b(B, U) C
BE~
BcX
n
b(V, U).
VEO VeX
Therefore
u ub(B, u) c u n b(v, u),
UEO BE~
UEO VEO
i.e. b(X) C '"" b(X). Now we shall show the converse inclusion. Let U0 be a neighbourhood of zero and let
{tn}E
n
b(V, Uo).
VEQ
Let { V1c} be a decreasing countable basis of neighbourhoods ofzero. Fort mula (6.6.3) implies that~~ bn(V:, Uo) = 0 (k = 1, 2, ... ).Then we can choose an increasing sequence of indices kb k 2 , tn lim =0 11+CO bn( vkn, Uo) . For every V1cn there is a finite Zn C V1cn such that
•••
such that
1
bn(Zn, Uo) ~ 2b(VTcn, Uo). co
Let B =
U Zn. Since the set B has a unique cluster point 0, it is compact. n=l
On the other hand, JtnJ ~ JtnJ ~ 2JtnJ bn(B, Uo) ""' bn(Zn, Uo) ""' bn(Vkn, Uo)
This means that {tn}
E
b(B, U0 ). Then {tn}
E
O +
b(X).
·
D
As an obvious consequence of Proposition 6.6.4 and (6.6.1) we obtain
Montel and Schwartz Spaces
277
PROPOSITION 6.6.5 (Bessaga, Pelczynski and Rolewicz, 1961). Let { Un} be a basis of neighbourhoods ofzero in an Fspace X. Then
n b(Um, Un). co
00
b(X) =
U
n~l m~l
CoROLLARY 6.6.6 (Bessaga, Pelczynski and Rolewicz, 1961). If X is a Schwartz space, then there is a sequence {t;} tending to 0 such that
{t:} ¢b(X). Proof If X is a Schwartz space, then there is a basis of neighbourhoods of zero { Un} such that lim bn( Ui+I• UJ) = 0 (j = 1, 2, ... ). Since we have
only a countable number of sequences bn( Ui+I• UJ), it is not difficult to construct a sequence {t;} tending to 0 such that . I1m
X
In
n~oo bn(Ui+l• Uj)
= +oo,
j= 1, 2, ... ,
Proposition 6.6.5 implies that {t;} ¢ b(X).
D
PROPOSITION 6.6.7 (Mityagin, 1961). Let X be a B 0 space with the topology given by a sequence of Hilbertian pseudonorms, i.e. the pseudonorms [[x[[n = (x,x)n, where (x,x)n are inner products. Let Y be a subspace of the space X. Then b(Y) C b(X). Proof Let Un = {x: [[x[[n
By similar arguments to those used in the proof of Proposition 6.6.3 we obtain b(Up n Y, Uq n Y) = b(Tp,q(Up), Tp,q(Uq)) C IJ(Up, Uq). Since the sets Vp = Up n Y constitute a basis of neighbourhoods of zero in Y,
U n IJ(Vp, Vq) C q=lp=l U n b(Up, Uq) = IJ(X). q=lp=l 0000
IJ(Y) =
0000
D
278
Chapter 6
We do not know whether the converse fact to Corollary 6.6.6 holds, i.e. the following question is open : Problem 6.6.8. Let X be an Fspace. Suppose that there is a sequence ·
{t} ten ping to 0 such that {t;} ¢ (J (X). Is X a Schwartz space? 6.6.8 is strictly connected with
Problem 6.6.9. Let (X, II II) be an Fspace and let A be a closed bounded set in X. Suppose that, for each starklike open set U, the sequence {bn(A, U)} tends to 0. Is the set A compact?
The answer is positive for locally pseudoconvex spaces and spaces N(L(Q, E, /1)). LEMMA 6.6.10 (Turpin, 1973). Let (X, II II) be an F*space. Suppose that for each neighburhood of zero U there is a neighbourhood of zero V such that for each finite dimensional subspace Landfor all a > 0 there is a finite set H C V n L such that (6.6.3)
Vn L C H+aU.
Then each bounded set B such that lim bn(B, U) = 0 for all open balanced enighbourhoods of zero U is totally bounded. Proof Let W be an arbitrary neighbourhood of zero. Let U be a balanced neighbourhood of zero such that
U+Uc W. By our hypothesis there is a finitedimensional space L such that B C L+ +U. Hence Bc(B+U)nL+U. Without loss of generality we may assume that U nL is bounded. Hence for sufficiently small a> 0 1
Bc(B U)nL+UcVn L+U. a
(6.6.4)
Thus by (6.6.3) and (6.6.4) there is a finite set H such that 1
1
Be H+U+UC H+W. a a
(6.6.5)
Monte! and Schwartz Spaces
279
Therefore, for all a' > 0, there is a finite setH such that B C H+a'W.
(6.6.6)
0
REMARK 6.6.11. In Lemma 6.6.9 the hypothesis that H is finite can be replaced by the hypothesis that it is totally bounded.
Indeed, if His totally bounded, then for each a' > 0 and a neighbourhood of zero W there is a finite set H 0 such that H C H 0 +a'W.
(6.6.7)
Thus by (6.6.6) and (6.6.7) we obtain B C H 0 +a'W+a'W.
6.6.12. Let X be an F*space without arbitrarily short lines. Then each bounded set B such that lim (Jn(B, U) = Ofor an arbitrary hal
PROPOSITION
anced neighbourhood of zero U is totally bounded. Proof The proposition follows immediately from Remark 6.6.11 and Lemma 2.4.6. D
As an immediate consequence of Proposition 6.6.12 we obtain 6.6.13. Let X be a locally bounded space. Let B C X be a bounded set such that lim (Jn(B, U) = 0 for each open balanced neigh
PROPOSITION
bourhood of zero U. Then the set B is totally bounded. PROPOSITION 6.6.14. Let X be an F*space with a topology given by a sequence of Fpseudonorms {II lin} (see Section 1.3). Suppose that for each n there is an an > 0 such that ,for all x such that' llxlln # 0,
sup lltxlln > an.
(6.6.8)
tER
Let B be a bounded set such that, for each open balanced set V, lim (Jn(B, V) = 0. Then the set B is totally bounded. n+ro
Proof Let U be an arbitrary neighbourhood of zero. Then there are n and
280
Chapter 6
a number a'
> 0 such that
{x: llxlln
(6.6.9)
C U.
Let X~= {x EX: llxlln = 0}. Let Xn =X/X~ denote the quotient space. The Fpseudonorm II lin induces an Fnorm in the space Xn. By (6.6.9) Xn is without arbitrarily short lines. The bounded set B induces a bounded set Bn = {[x]: x E B} in Xn. Similarly, U induces a neighbourhood of zero in Xn. Thus, by Proposition 6.6.12, for each a > 0 there is a finite setH such that
0 By Propositions 6.6.13 and 6.6.14 we trivially obtain PROPOSITION 6.6.15. Let X be a locally pseudoconvex space. Let B be a bounded set such that, for eac_h open balanced set U, lim bn(B, U) = 0. 11+00
Then the set B is totally bounded.
LEMMA 6.6.16 (Turpin, 1973). Let X be an F*space. Suppose that there is a basis of neighbourhoods of zero {Um}, such that, for m = 1, 2, ... , aUm is a linear space and there is a neighbourhood V m such that
n
a>O
Then each bounded set B such that lim bn(B, U) = 0 for each balanced neighbourhood of zero U is totally bounded. Proof Let {Vm,n} be a sequence of neighbourhoods of zero such that Vm,o = Vm and Vm,n+l+ Vm,n+l C Vm,n•
Let
Of course,
n aWm,n = n aUm
a>O
a>O
(6.6.10)
Monte) and Schwartz Spaces
281
and W m,n+I + W m,n+I C W m,n.
By the Kakutani construction (see Theorem 1.1.1), the sequence II lim and, by (6.6.10), formula (6.6.8) holds. Of course, the topology determined by the sequence of Fpseudonorms {II lim} is equivalent to the original one. Therefore Proposition 6.6.14 implies the Lemma. 0 {Wm,n} induces an Fpseudonorm
LEMMA 6.6.17 (Turpin, 1973). Let X be an F*space such that for each neighbourhood of zero U there is a neighbourhood of zero V with the following property : if there are sequences {sn,i}, i
= 1, 2, ... , k
such that Sn,i;:, 0,
Sn,il · 1, 2, .... , k and . Sn,i= +oo, 1"1m= 11m oo, z= n~co
n+oo
Sn,i
k
}.; Sn,ie,e V, thenlin(e1 ,
•••
,ek) CU.
(6.6.11)
i=l
Then the hypotheses of Lemma 6.6.10 hold. Proof Let U and V satisfy condition (6.6.11). Let Wbe a balanced neigh
bourhood of zero such that W + W C V. Let L be a finitedimensional subspace. We shaJl show that there is a bounded setH C L such that W n L C H+
n
aU.
(6.6.12)
a>O
Suppose that (6.6.12) does not hold. Then there is an unbounded sequence {xn}C WnL such that, for each subsequence {Yn} of thesequence {xn} and for each bounded setH C L.
{Yn} ¢ H+
n
aU.
(6.6.13)
a>O
The existence of such a sequence follows from the fact that L is finitedimensional. We shall show that (6.6.13) does not hold. Namely, we shall show that each unbounded sequence {xn} C W n L contains a subsequence {Yn} C WnL such that there is a bounded sequence {zn} such that YnEZn+ aU.
n
a>O
282
Chapter 6
Since {xn} is unbounded and the space L is finitedimensional, we can find e1 E L, e1 =I= 0 and a subsequence {y~} of the sequence {xn} such that
z
where sn, 1 +oo and _n +0 and, moreover {zn} belongs to a subspace Sn,l
L 1 of the space L such that e1 ¢ L 1 • Either the sequence {z~} is bounded or it is unbounded. In the second case we repeat our process. Finally we can choose a subsequence {Yn} of the sequence {xn} which can be represented in the form k'
. Yn
=.}; Sn,iet+zn,
(6.6.14)
i=l
where k' ~ k, Sn,i ;? 0, Sn,t+oo, Sn.il +oo and {zn} is a bounded seSn,i
quence. Since {zn} is bounded, there is a number b, 0 < b < 1, such that {b zn} C WnL. The sequence {Yn} is a subsequence of the sequence {xn}; thus {b Yn} C WnL. Therefore, by (6.6.14), k'
b.}; Sn,iei E V i=l
and, by (6.6.11), lin (et. ... , ek.) C U. This implies that lin (e 1 , C aU and (6.6.13) does not hold.
n
... ,
ek.)
a>O
Thus we have (6.6.12) and since Lis finitedimensional the hypotheses of Lemma 6.6.10 hold. D THEOREM 6.6.18 (Turpin, 1973). In the space N(L(Q, E, fl.)) each bounded set B such that lim bn(B, U) = 0 for each balanced neighbourhood of zero ,__,.co U is totally bounded. Proof We shall show that the hypothesis of Lemma 6.6.17 holds. Lets be an arbitrary positive number. Let ,h, ... ,fk be measurable functions. Suppose that there are sequences {sn,i}, i = 1, ... , k such that sn,i >0,
Monte! and Schwartz Spaces
283
Sn i1 sn,i+oo,' +oo and for all n Sn,i n
k
PN(l' Sn,ifi)
=
i=l
f N(ll' Sn,Jil)d,u <e.
(6.6.15)
i=l
Q
Let k
A=
U {t: .fi(t)::;i:O}. i=l k
For each tEA,
I};
S11 ,i./i(t)
I tends to infinity.
Thus, by (6.6.15) and
i=l
the Fatou lemma,
J
A
N(u)d',u ~ e,
sup
O
i.e., sup
O
N(u) ~
,U
e (A) .
(6.6.16) k
By (6.6.16), for each linear combination f = }; cifi, i=l
PN(/)~e
< 2e.
The hypothesis of Lemma 6.6.17 and the above formula imply the theorem. D Turpin (1973rshowed in fact a stronger result, namely that Theorem 6.6.18 is valid for some generalizations of spaces N(L(Q, E, Suppose we are given a space M(arn,n), where mare positive integers and n are nonnegative integers (see Example 1.3.9). We say that the space M(arn,n) is regular if, for m < m', the sequence arn,n/am,,n is nonincreasing.
,u)).
PROPOSITION 6.6.19. If a space M(am,n) is regular, then b(M(am,n))
=
{{tn}: lim tn aq,n n+oo
ap,n
=
Ofor some p and all
q}.
Proof Let Up= {x: llxll < 1}. Since the space M(am,n) is regular, we have, for q > p, bn_ 1(Uq, Up)= ap,nfaq,n and Proposition 6.6.5 trivially implies the proposition. 0
Chapter 6
284 COROLLARY
6.6.20. Let {an}, {bn} be two sequences of reals tending to in
finity. Then b(M(a~)) = M(a~),
b(M(b~l/m))
= M+(b;; 11m) = {{tn}: tnb;; 11m +Oforsomem}.
6.6.21 (Bessaga, Pelczynski and Rolewicz, 1961). There is an infinitedimensional B 0 space X which is not isomorphic to its product by the onedimensional space. Proof Let X= M(expm2 2"+} The sequence {exp(22"+"')} belongs to b(M(expm2 2")), but it does not belong to b(M(expm22"+')). 0 COROLLARY
Now we shall introduce a class of sequences in a certain sense dual to the class b(X). We denote by b'(A, B) the set of all sequences {tn} such that lim tnbn(A,B) = 0. 11>00
The class b'(A, B) has the following properties: if A' C A, B':) B, then b'(A', B') :) b'(A, B), (6.6.1') b' (aA, bB) = b' (A, B) for all scalars a, b different from 0. (6.6.2') Let b'(X) =
n n b'(K, U).
KE':JUeO
By a similar argument to that used for b(X) we find that b'(X) is an invariant of linear co dimension, which means that if codimzX ~ codimzY
(codimzX = codimzY),
then b' (X) :) b' (Y)
(resp. b' (X)= b' (Y)).
Let J'(X) =
U U b'(V, U).
UeO VeO
By a similar argument as to that used in the proof of Proposition 6.6.4 we obtain '
PROPOSITION

6.6.22. b (X) = b'(X).
As a consequence of Proposition 6.6.22 we obtain the following proposition, in a certain sense dual to Proposition 6.6.20.
Monte] and Schwartz Spaces PROPOSITION
6.6.23./f a space M(am,n) is regular, then
()' (M(am,n))
=
{
{tn}: for all p there is a q such that tn
CoROLLARY
285
ap,n aq,n
+
o}.
6.6.24. Let {an}, {bn} be sequences ofreals tending to infinity.
Then (j'(M(b;;I/m)) ()'(M(a~))
=
M(b;;l/m),
= M(a?;:) = {{tn}: tna;;m +0 for some m }.
In the same way as in Corollary 6.6.7 we can prove 6.6.25. Let X be a B0 space with the topology given by a sequence of Hilbertian pseudonorms. Let Y be a subspace of the space X. Then ()'(X):::> ()'(Y).
COROLLARY
The following, natural question arises : when does the equality of diametral approximative dimensions imply isomorphism? We shall show that this holds for an important class of spaces called Kothe power spaces. Let an be a sequence of positive numbers tending to infinity. The space M(a?;:) is called a Kothe power space of the infinite type, the space M (a~Ifm) is called a Kothe power space of the finite type. To begin with, we shall show that two Kothe power spaces of infinite type (of the finite type) induced by sequences {an} and {~n} are equal as the sets if and only if there are two positive constant A, B such that
A< logan
(6.6.17)
logan
Indeed, x = {xn}
e: M(a~), (M(a~Ifm)) if and only if
logixnl+mlogan+ oo,
m= 1,2, ...
(6.6.18)
(resp. 1
1ogixnl1ogan+ oo, m
m= 1,2, ... ).
(6.6.18')
286
Chapter 6
If (6.6.17) holds, then (6.6.18) (resp. (6.6.18') holds if and only if logJxnl+mlogan+ oo,
m= 1,2, ...
(resp. logJxnJ1logan+ oo, m
m = 1, 2, ... ).
Thus (6.6.17) implies that the spaces as the sets are equal. In a similar way we can show that the sets M(a";:) and M(a";:) (resp. M+(a~Ifm) and M+(a~Ifm) , where M(a:) is defined in Corollary 6.6.24 and M+(a~Ifm) is defined in Corollary 6.6.20, are equal if and only if (6.6.17) holds. Thus, by Corollaries 6.6.20 and 6.6.24, we obtain COROLLARY 6.6.26. Let X, Y be two Kothe power spaces of the finite type (of the infinite type). Then the following three conditions are equivalent. X and Yare isomorphic,
(6.6.19.i)
the diametral approximative dimensions of X and Yare equal, = b(Y), (6.6.19.ii)
b(X)
o'(X)
=
b'(Y),
(6.6.19.iii)
Corollary 6.6.26 shows that Kothe P
ap,n
n+cc
aq,n
=
0.
Let N denote the set of all subsequences v = {nk} of the sequence of positive integers. Let
and Jet
Monte] and Schwartz Spaces
Let NI(L(am,n))
=
and N 2 (L(am,n)) =
287
U n n N~.q,s p
q
8
n U n N!.q.s· p
q
8
The following facts hold NI(L(am,n)) () N 2 (L(am,n))
= 0,
NI(L(am,n)) U N 2 (L(am,n)) =f::. 0.
Moreover, NI(L(am,n)) and N 2(L(am,n)) are invariants of isomorphisms inside the class of all spaces L(am,n), which means that if two spaces L(am,n) and L(bm,n) are isomorphic, then i= 1, 2.
We shall introduce an order inN in the following way vi ~ v2 if almost all elements of vi belong to v2 • The class Ni(L(am,n)), i = 1, 2, satisfies one of the following three conditions : there is a maximal element in Ni(L(am,n)),
(6.6.20.1)
there is no maximal element in Ni(L(am,n)),
(6.6.20.2)
the class Ni(L(am,n)) is empty.
(6.6.20.3)
We say that a space L(am,n) belongs to the class_ Ei,J, i,j = 1, 2, 3, if NI(L(am,n) satisfies condition (6.6.20.1) and N 2(L(am,n)) satisfies condition (6.6.20.2). The class E 3 , 3 is empty. Dragilev (1970b) showed that no other class is empty. Let E 0 = EI, 3 u E 3 ,I u EI,I· Dragilev (1970b) showed that if two spaces belonging to E 0 have the same approximative diametral dimension, then they are isomorphic. On the other hand, if Et,i ¢ E 0 , then there are two spaces belonging to the union Et, 1 u E 0 such that they are not isomorphic but they have the same approximative diametral dimension. It is not known whether two spaces L(am,n) and L(bm,n) having the same approximative dimension and equal sets NI and N 2 are necessarily isomorphic.
288
6.7.
Chapter 6 ISOMORPHISM AND NEARISOMORPHISM OF THE CARTESIAN PRODUCTS
The results of this sectionforlocally convexlinear topological spaces was obtained by Zahariuta ( 1973). We shall formulate it for metric linear spaces not necessarily locally convex. Let (X, II llx) and (YII lly) be two F*spaces. We recall that a linear continuous operator T mapping X into Y is called compact if there is a neighbourhood of zero U in X such that the set T( U) is totally bounded. We say that an ordered pair ofFspaces (X, Y) satisfies condition R (briefly (X, Y) E R) if every linear continuous operator T mapping X into Y is compact. 6.7.1. If Y is a Monte! space, then (X, Y) E Rfor each locally bounded space X. Proof Let T be a linear continuous operator mapping X into Y. Let U C X be a bounded neighbourhood of zero. By Theorem 2.1.1 the set . T(U) is bounded. Since Y is a Monte! space, the set T(U) is totally bounded. D PROPOSITION
6.7.2. Let Y be a locally pconvex space. Jf(X, Y) E Rfor all locally bounded spaces X with phomogeneous norms, then Y is a Monte/ space. Proof The topology in Y is given by a sequence {II l!n} of phomogeneous pseudonorms. Let A be an arbitrary bounded set in Y. Then there is a sequence of numbers {mn} such that PROPOSITION
P (y)
= sup mniiYIIn::::;; I n
for ally EA. Let (X,p(y)) be the space of such y that p(y) < +oo. Of coursep(y)is apnomogeneous norm on X. The operator Tequal to identity maps X into Yin a continuous way. Observe that A C U = {y: p(y) ::::;; 1}. Thus A C T(U). Since T(U) is totally bounded, the set A is also totally bounded. D
Monte! and Schwartz Spaces
289
6.7.3. Let X be a Schwartz space. Then, for each locally bounded space Y, (X, Y) E R. PROPOSITION
Proof Let T be a continuous linear operator mapping X into Y. Since the space Y is locally bounded there is a neighbourhood of zero U c X such that T(U) is bounded. Take any neighbourhood of zero We Y. Since the set T(U) is bounded, there is a positive number a such that aT(U) c W. The space X is a Schwartz space, thus there is a neighbourhood of zero V totally bounded with respect to U, i.e. for each s > 0 there is a finite system of points {x 1, ... , Xn} such that (6.7.1)
Thus
The arbitrariness of s and W implies that the set T( V) is totally bounded. D PROPOSITION 6.7.4. Let X be a locally pseudoconvex (locally pconvex) space. If (X, Y) E R for each locally bounded space Y (locally bounded space Y with a p:homogeneous norm,) then X is a Schwartz space. Proof Take an arbitrary absolutely pconvex neighbourhood of zero U. The set U induces a phomogeneous pseudonorm II 11. Take as X 0 the quotient space X 0 = Xf{x: llxll = 0}. The space X 0 is a locally bounded space with the norm induced by II 11. The canonical mapping nu: x~X0 is a continuous linear operator. Thus, by our hypothesis, there is a neighbourhood of zero V such that nu(V) is totally bounded with respect to nu(U). This implies that Vis totally bounded with respect to U. Since there is a basis of absolutely Pnconvex (pconvex) neighbourhoods of zero in X, Vis totally bounded. D
6.7.5. (X, Y) E R and(Y, Z) E: R does not imply (X, Z)E R. Proof Let X be a Schwartz space. Let Y be an arbitrary Banach space. Then (X, Y) E Rand (Y, X) E R, whereas (Y, Y) ¢ R. 0 CoROLLARY
Chapter 6
290
6.7.6. Let (X, Y) E R. Let X 0 be a subspace of X and let Y 0 be a subspace of Y. We assume that X 0 is complemented, i.e. that there is a continuous projection of X onto X 0 • Then (X0 , Y 0) E R. Proof Let T 0 be an arbitrary linear continuous operator mapping X 0 into Y 0 • Since (X, Y) E R, the operator T 0 P mapping X into Y is compact. Thus there is a neighbourhood of zero U in X such that T0(P( U)) is totally bounded. Therefore the set T 0(X0 n U) C T0 P(U) is also totally bounded. D LEMMA
6.7.7. If (X, Y) E R and X is isomorphic to Y, then X is finitedimensional. Proof The lemma is a trivial consequence of the EidelheitMazur theorem (Theorem 6.1.2). D LEMMA
Combining Lemmas 6.7.6 and 6.7.7., we obtain PROPOSITION 6.7.8. Let (X, Y) E R. Let X 0 be an infinitedimensional subspace of X. Let X 0 be complemented. Then X 0 is not isomorphic to any subspace Y 0 of Y.
Let (X, I llx) and (Y, II IIY) be two Fspaces. A linear continuous operator T mapping X into Yis called a C/Joperator (or a near isomorphism) if T(X) is closed and dim{x: T(x) = 0} = a(T) <
+=
(6.7.2)
and dim Y/T(X)
=
{J(T)
< +oo.
(6.7.3)
By the index of a C/Joperator Twe shall mean the difference "(T) = rt(T){J(T).
(6.7.4)
If there is a C/Joperator mapping X into Y we say that the spaces X and Yare nearly isomorpic. The theory of C/Joperators in Banach spaces is presented in a fundamental paper by Gohberg and Krein, 1957. The results concerning nonlocally convex spaces can be found in PrzeworskaRolewicz and Rolewicz (1968). We shall formulate those results without proofs.
Monte! and Schwartz Spaces
291
Let X, Y, Z be three Fspaces. Let T: x_,.y and S: y_,.z be two 4>operators. Then the superposition ST is a 4>operator and we have the following equality for indexes
x(ST) = x(T)+x(S).
(6.7.5)
Formula (6.7.5) for Banach spaces was proved by Atkinson (1951, 1953). Basing ourselves on the Riesz theory of compact operators on nonlocally convexspaces(Wiliamson, 1954), we find that if T: X__,.. Yis a 4>operator and S: x_,.y is a compact operator then T+Sis a 4>operator and
x(T+S) = x(T).
(6.7.6)
Let T: X__,.. Y be a 4>operator, then there is a 4>operator T such that
TT=I+B,
(6.7.7)
TT= I+C,
(6.7.8)
where B and C are compact operators. Let X = X1 X X 2 , Y = Y1 X Y 2 be Fspaces. Let T be a continuous linear operator mapping X into Y. Of course T can be represented in the matrix form Tl,l T1,2l T= [ r..2,1 r.2,2 ' where Tt, 1 : Xt_,.Y,, i,j = I, 2. Now we shall formulate a lemma proved by Douady (1965) for Banach spaces and by Zahariuta (1973) for locally convex spaces. LEMMA
=
6. 7.8/f Tis a 4>operator and (XI> Y2)
[T1'1 T1'2l is a 4>operator. 0 T2,2
E
If moreover (X1,
R, then the operator T 0
Y2) E R then
x(T) = x(TI,I)+x(T2,2).
(6.7.9)
1
0 is compact. Hence R, the operator S = [ 0 T2,1 0 the operator TS = [T1'1 T1'2l is a 4>operator. Proof Since (X1, Y 2)
E
0
T2,2
292
Chapter 6
If(X2, Yl) E R, then the operator sl u(TSS1) =
=
[~ ~1 ' 2] is compact and
u([~1 ' 1 ~2 . 2 ]) = u(TI,1)+u(T2,2).
0
As a consequence of Lemma 6.7.8 we obtain PROPOSITION 6.7.9. Let X= X 1 XX2 andY= Y1 X Y2 • Let (X1 , X 2 ) E R and ( YI> X 2) E R. Then X is nearly isomorphic to Y if and only if X 1 is nearly isomorphic to Y1 and X 2 is nearly isomorphic to Y 2 •
Let X be an Fspace. By X(i) we shall denote an arbitrary subspace of codimension i if i;?: 0 (observe that all such subspaces are isomorphic) and an arbitrary space Xx Z, dimZ = Iii, if i < 0. THEOREM 6.7.10. Under the hypotheses of Proposition 6.7.9 the space X is isomorphic to the space Y if and only if there is an integer s such that Y 1 is isomorphic to x~> and y2 is isomorphic to x~·>. Proof Sufficiency. If Y1 is isomorphic to x~•> and Y2 is isomorphic to X~s), then Y1 X Y2 is isomorphic to X~> X X~s) and it is isomorphic to
X 1 xX2•
Necessity. LetT: X1 X X2+ Y1 X Y2 be an isomorphism. Then, by Lemma 6.7.8, T1 , 1: X 1 +Y1 and T2, 2: X2+Y2 are (/)operators. Then Y1 is isomorphic to X~,> and Y 2 is is isomorphic to X~·>. By formula (6.7.9) 0 = u(T) = u(T1,1)+u(T2,2) = s1+s2 and
0 Now we shall apply the results given above to a certain class of locally convex spaces. Let X be a B0space with a topology defined by a sequence of pseudonorms {II Ilk}. Suppose that {en} is a basis in X such that, for each x ro
ro
= }; Xnen, the series }; lxnlllenllk are convergent for k = 1, 2, ... We n=l
n=l
shall call a basis with this property an absolute basis. We say that X E d1 (is of type d1) if there are an absolute basis {en} in X and an indexp such that for each index q there are an index r and N
Monte! and Schwartz Spaces
293
= N(p,q) such that for n >N. (6.7.10) llenll: ~ lleniiP llenllr We say that X e d2 (is of type d2) if there is an absolute basis {en} in X such that for each p there is a q such that for each r there is an N = N(p, r) such that lien II~?= llenllpllenllr
for n
> N.
(6.7.11)
Example 6. 7.11. Let {an} be a sequence of positive numbers tending to infinity. The spaces LP(a'::), 0 < p < and M(a'::) are of the type d1 • The spaces LP(a~ 1 1m), 0 < p < +oo, and the spaces M(a~Ifm) are of type d2. By Proposition 6.3.3 the spaces in Example 6. 7.11 are Schwartz spaces.
+=
THEOREM 6.7.12 (Zahariuta, 1973). Let X, Y be two B 0 spaces. Let X E d2 and Y e d1 • If Y is a Monte! space, then each continuous linear operator mapping X into Y is compact. Proof By definition there are absolute bases {en} in X and {fn} in Y such that (6.7.11) and (6.7.10) hold. Since the bases are absolute, we may assume without loss of generality that the topology in X (in Y) is given by a sequence of pseudonorms co
{llllm}(resp.{ll ll~})suchthatforx=
co
2 XneneX(resp.y= n=l 2 Ynfne Y) n=l
co
llxllm =
J; lxnlllenllm,
m= 1,2, ... ,
n=l
(resp. 00
IIYII~ =
J; IYnlllenllm,
m = 1,2, ... )
n=l
Let T be a continuous linear operator mapping X into Y. Let hn co
= T(en) =
2
co
ti,nfi. Of course for each x =
i=l
2
n=l co
T(x) =
co
J; Xn T(en) = J; Xn J; t;,nfi n=l co
=
00
n=l co
.2; (.2; t;,nXn)fi. i=l
n=l
i=l
Xnen
294
=
Chapter 6
The continuity of the operator T implies that for each p there is a q q(p) such that co
jjT(ek))jp ;~ )t;,k 111./ill~ II ek II q = sup II ek II q < k
C(p) = sup k
+=
(6.7.12)
We have assumed that Y is a Montel space. Thus to prove the theorem it is enough to show that the operator T maps some neighbourhood of zero Uq, = { x: jjx))q, < I} into a bounded set. Since Y E d1 , there is a p 1 such that for each p there are i0 (p) and p 2(p) such that
(JJ./iiJ~)2 ~ IJ./iii~,JJ./ill~.
for i > io(P)
On the other hand, X E d2. Take q for each q2 there is a k 0 (q2) such that
j)ekii!, ~ j)ek))q,J)ek))q,
=
(6. 7.13)
q(p1). Thus there is a q0 such that
fork> k(q 2).
(6.7.14)
Take q 2 = q(p 2). Of course, k (q 2) depends implicitly on p. By (6.7.13) and (6.7.14) there is a constant L(p) such that 2 ( JJ./iJJ~ ) ~ L(p) ( JJ./iJJ~, ilftil~,) l)ek))q, ~ Jlek)lq,llekl)q,
(6.7.15)
fori, k = 1, 2, ... Hence, by the Cauchy inequality, we obtain 00
'
}; it· I 11./illp i=l
j)ek))q,
•.k
~ L(p) ~ L(p)
, , (Jt;,k)
6;{
00
(
JJ./iJJ~, )1/2 (lt;,k) 11./iJJ~. )1/2
Jlek)lq, '
};
jt;,k) JJ./illp,
i=l
J)ekllq,
~L(p)C(p 1 )t/2C(p 2 )t/2
Thus sup
Jlekllq, )
1/2 (
00
'
}; Jt;,k) JJ./i))p, i=t
< +oo.
)
1/2
llek)lq,
11~(~~)11~ ~ M(p),
where M(p) = L(p)C(p1)112C(p 2) 112, p ek q, = 1, 2, ... By the form of the pseudonorms, this implies that k
II T(x)JI~ ~ M (p) Jjxj)q,. Therefore the set T(U) is bounded.
D
Monte! and Schwartz Spaces
295
COROLLARY 6. 7 .13. If X is a Kothe power space of the finite type and Y is a Kothe power space of infinite type, then each continuous linear operator mapping X into Y is compact. Later we shall give an example of a Kothe power space of infinite type, which is a subspace of a Kothe power space of the finite type (see Section 8.3). Now following Zahariuta (1973) we shall introduce a new topological invariant. Let 8 1 , 8 2 be two classes of spaces such that (XI> X 2) E R for XI' E 81 and x2 E 82. With each X which is a product X= XI X x2 we shall associate the set T(X) of all pairs (t5(xr)), t5(X~s))), s = 0, ± 1, ±2, ... , where t5(Y) denotes the diametral approximative dimension of Y. As a consequence of Theorem 6.7.10 we imediately obtain THEOREM 6.7.14. T(X) is an invariant of an isomorphism, i.e. if X= X1 X X 2 , Y = Y1 X Y 2 , XI> X 2 E 81> X 2 , Y 2 E 8 2 and the spaces X and Y are


isomorphic, then T(X) = T(Y). If, moreover, the classes 8 1 and 8 2 are such that zl, z2 E 8i, i = 1, 2, and t5(Zl) = t5(Z2) implies the isomorphism of

Z 1 and Z 2 , then T(X)
 implies that X is isomorphic to Y. = T(Y)
As a consequence of Theorem 6.7.14 and Corollary 6.6.26 we obtain PROPOSITION 6.7.15 (Zahariuta, 1973). Let X, X1 be two Kothe power spaces offinite type and let Y, Y1 be two Kothe power spaces ofinfnite type.
If T(Xx Y)
X1 X
Y1.
=

T(X1 X Y 1) then the space Xx Y is isomorphic to the space
Chapter 7
Nuclear Spaces. Theory
7.1.
DEFINITION AND BASIC PROPERTIES OF
NUCLEAR
SPACES
Let X be an F*space. The space X is called nuclear if, for any neighbourhood of zero U, there is a neighbourhood of zero V such that lim m5n(V, U) = 0,
(7.1.1)
n>co
the definition of bn(V, U) being given in Section 6.6. 7.1.1. If a space X is nuclear, then for any neighbourhood of zero U and for any positive integer k there is a neighbourhood of zero V such that lim nkbn(V, U) = 0. (7.1.2) PROPOSITION
n>oo
The proof of Proposition 7.1.1 is based on LEMMA 7.1.2 (Mityagin and Henkin, 1963). For arbitrary U, V, W C X
bn+m(W,
U)~bn(W,
V)·bm(V, U)
(7.1.3)
provided bn(W, V) and bm(V, U) are finite. Proof Let a= bm(V, U),
b = bn(W, V).
Lets be an arbitrary positive number. Then, by the definition of bm(V, U) and bn(W, V), there are subspaces L 1 and L 2 , dimL1 = m, dimL2 = n, such that
Nuclear Spaces. Theory
297
Therefore W C L 1 +L2 +(a+e)(b+e) U. Since dimL1 +L 2 ~ n+m, bn+m(W, U) ~ (a+e)(b+e). The arbitrariness of e implies the lemma. D Proof of Proposition 7.1.1. Let U be an arbitrary neighbourhood of zero. Since X is a nuclear space, there is a system U = V0 , V1 , ••. , Vk of balanced neighbourhoods of zero such that Iimnbn(Vt, Vi_ 1)=0
(i= 1,2, ... ,k).
Then, by formula (7.1.3), lim nkbkn(Vk, U) = 0.
(7.1.4)
Since bm( Vk, U) is monotonic,
0~ m'~m(V, U) ~ £(~)' [£( :p"'(iJ(V,, U)]~o. where, as usual, E(a) denotes the greatest integer not greater than a. This implies the proposition. D Further on, other equivalent definitions of nuclear spaces will be given. 7.1.3 (Ligaud, 1973). Let X be a nuclear F*space. Then its completion Xis also nuclear. Proof Let U be a neighbourhood of zero in X. Let U1 be a neighbourhood of zero in X such that U1 + U1 C U. Let V = U1 n X. V is a neighbourhood of zero in X and V C U1o where V is a completion of V. Since X is nuclear, there is a neighbourhood of zero W, W C X, such that <'lm( W, PROPOSITION
A
V)
A
1
< m+ 1 . By the definition of bm, there is a subspace L, dim L = m, 1
A
such that W C L+ m+ 1 V. Hence, for the completion W of W we have
1 1 1 W C L+ m+ 1 V+ m+ 1 U1 C m+ 1 U+L. A
A
Thus bm(W, U)
A
1
< m+ 1 and this implies the nuclearity of X.
D
298
Chapter 7
THEOREM 7.1.4 (Ligaud, 1971). Every locally pseudoconvex nuclear space is locally convex. The proof is based on several lemmas. The set n
l'p(A)
n
{x:x=}; atXt, XtE A,}; latiP ::( 1, n = 1, 2, ... }. i=l i=l is called the absolute pconvex hull of the set A. Observe that Fp(Fp(A)) = Fp(A). If II II is a phomogeneous norm, then =
Fp({x: llxll
< r}) =
< r}.
{x: llxll
LEMMA 7.1.5. Let X be anndimensional space. For any set A C X, 1
F 1 (A) C n"P 1Fp(A). n
n
Proof Let X E rl(A), i.e. X= ); aiXi, where); latl ::;;;; 1. Observe that i=l i=l
Hence n
F 1 (A) C max
{(.I; latiP)l/p: i=l
=
n~ 1Fp(A).
n
}; latl ::( i=l
1} Fp(A) D
LEMMA 7.1.6 (Auerbach, 1935). Let (X, II II) be anndimensional real Banach space. Then there are elements e1 , ••• , en of norm one and functionals !t, ... , fn of norm one such that for i = j,
(7.1.5)
for i::j=.j.
Proof Let x 1 , ... , Xn be elements of norm one and let vol (x 1 , ... , Xn) denote the volume in the Euclidean sense of a parallelepiped with the vertices (e1 x 1 , ... , enxn), where et = 1 or 1 (i = 1, .. .). Let el> ... , en be such elements of norm one that vol(e1, ... ,·en)= supvol(x1 ,
... ,
Xn).
Nuclear Spaces. Theory
299
Since the unit sphere in anndimensional space is compact and the volume is a continuous function of x 1 , ••. , Xn, such e1 , ••• , en exist. Let Hi be a hyperplane passing through the point ei and parallel to vectors e1 , ••• , ei1> ei+ 1, •••,; en. The hyperplane Hi does not have common points with the interior of the unit ball in X. Indeed, suppose that Hi has common points with the interior of the unit ball. Then there is a point e~ of norm one such that Hi separates 0 and e; and moreover e~ ¢ Hi. Therefore
which contradicts the definition of eto ... , en. Let.fi be such a functional that {x: .fi(x) = 1} =Hi. Since Hi do not have common points with the interior of the unit ball, the functionals .fi have norm equal to one. The definition of Hi implies (7.1.5). D 7.1.7. Let (X, I II) be an ndimensional complex Banach space. Then there are a system of elements e1 , •.. , en and a system offunctionals j;_, ... .fn such that lleill = II./ill = 1 (i = 1, 2, ... , n) and (7.1.5) holds. Proof. Let x 1 , ••• , Xn be arbitrary linearly independent elements of X. Let X 0 be a real space spanned by x 1 , •.. , Xn. By Lemma 7.1.6 there are in X 0 elements e1 , ••• , en and realvalued linear functionals g 1 , ••. , gn such that lleill = llgi!l = 1 (i = 1, 2, ... , n), and (7.1.5) holds. Let h1 be a realvalued normpreserving extension of g1 on the whole space X considered as a real 2ndimensional space. Let jj(x) = hJ(x)ihJ(ix), j = 1, 2, ... , n. The functionals f1 are linear (see Corollary 4.1.3). In the same way in the proof of Theorem 4.1.5 we can show that llfJII = 1 (j = 1, 2, ... , n). By the definition off1, formula (7.1.5) holds. D LEMMA
LEMMA 7.1.8. Let X be anndimensional space. Let A be a balanced closed convex set. Then there are points e1, .•• , en E A such that
Proof. Without loss of generality we may assume that X= lin A. Since A is convex and balanced, it induces the norm llxll = inf {t: xft E A}. By Proposition 7.1.6 (or 7.1.7) there are elements e1 , •.• , en and functionals h, ... .fn of norm one such that (7.1.5) holds.
Chapter 7
300
Observe that
lfi(x)! ::s;; 1, i = 1, 2, ... , n} C
A C {x:
n
n
i=l
i=l
l, l.ft(x)l ::s;; n} = n{x: }; l.ft(x)l ::s;; 1}
C{x:
D LEMMA
are e1,
7.1.9. Let X be anndimensional space. For any set A C X, there en E Fp(A) such that
••• ,
•
A C n"PFp(e1 ,
,en).
•••
Proof. Of course, A C F 1(A). Thus, by Lemma 7.1.8, there are e~, ... , e~ E F 1(A) such that
E
A C nFp({el, ... ,en}). ,
By Lemma 7.1.5, e1,
1
,
•.. ,

1
en E nP Fp(A). Let i=1,2, ... ,n.
Then et
E
Fp(A), i = 1, 2, ... , nand, by Lemma 7.1.5, 1
A C n·n"P 1Fp({ei, ... , e~}) = n1 1PFp({el, ... , e~}) =
n~ 1Fv({e 1 ,
•••
n~Fp({e1 ,
,en}) C
•••
D
,en}).
LEMMA 7.1.10. Let (X, II ID be a locally bounded space with a phomogeneous norm II 11. Let A be a balanced bounded set in X. If there is an ndimensional space L such that
(7.1.6)
A C L+tJU, where U is a unit ball in X, U = {x: llxll ei> ... , en E Fp(A) such that
• •
•
A C 2"Pn"PtJ'U+n"PFp({e1 , Proof. Let A'= {x
E
...
L: dist(x,A)
::s;; 1}, then for all (J' >
(J
there are
,en}).
<
b'}, where dist(x,A) = infllxyll. yEA
The set A' is bounded and, by (7.1.6), A C A' +tJ'U. By Lemma 7.1.9 there are e~, ... , e~ E Fp(.i') such that A' C
Nuclear Spaces. Theory
301
C n 21PF(p{ e~, ... , e~}. From the definition of an absolute pconvex hull we can represent e~ in the form n
n
ei = .}; Ot,j Xi,j,
Xi,j E
.}; [at,J[P ~ I.
A',
i=l
J=l
Let Yi, 1 E A be chosen so that [[Yt,i Xi,J[[ < 15'. Let n
et
= .}; at,iYi,J. j=l
Then n
[[eiei[[ ~.}; [ai,i[P[[xt,tYd[ ~ 15'.
(7.1. 7)
j=l
r
r
Let x= ,l'aie~EFp({ei>····e~}). Then for y= ,l'aieiEFP({e1 , t=l
•••
!=1
... ,en}) by (7.1.7), llxyl[ ~ 15'. This implies Fp({el, ... , e~}) C Fp({e1 ,
... ,
en})+15'U.
Finally, A C n 21P(Fp({e1 ,
c
n 21PFp({e1 ,
en})+I5'U)+I5'U
... , .. • ,
en})+2 2 1Pn21PI5'U.
LEMMA 7.1.11. Let (X, II I[) be a locally bounded space with a phomogeneous norm II 11. Let U = {x: llxll < I} denote the unit ball in X. Let A be a balanced set in X. If thereis a > 0 such that
(7.1.8) then there is a sequence {xn} such that
lim nal[xn[l = 0 n>co
and A C F 1 ({x1 ,
X2, ••. }).
Proof Let {sn} be such a sequence of positive numbers that limnksn n>co
= 0,
k =I, 2, ...
Chapter 7
302
Let
dn(P, U)
=
bn(P, U)+sn
for every set P. Let mn = 2 2n. We shall construct by induction a sequence of finite systems {xn,i, 1:::;:; i :::;:;mn} and a sequence of sets {Bn}in the following way, We put B1 = A. Suppose that the set Bn is defined. Then by Lemm'a 7.1.10 we can choose a system of points {xn,i, 1:::;:; i:::;:; mn} such that Xn,i E Tp(Bn)and
Bn
c m~PTp({xn,l, ... , Xn,mn})+2 21pm~pdm"(Bn, U)U.
(7.1.9)n
Let
Bn+l
=
[Bnm~PTp({xn,t, ... , Xn,mn})J n 221Pm~P dm"(Bn, U) U.
(7.1.10)n
Of course, by (7 .1. 9)n and (7 .1.1 O)n,
Bn C Bn+I+m~PTp({xn,l, ... , Xn,mn}). Since bk(Tp(A), U)
=
bk(A, U), we have, by (7.1.10)n,
bk(Bn+I, U):::;:; bk(Bn+n21PBn, U):::;:; 221Pn 21Pbk(Bn, U). Then by induction
n
m~ 1Pbk(A, U).
(7.1.11)
n
m~ 1Pdk(A, U).
(7.1.12)
nl nl
bk(Bn, U):::;:; 2 2p
i=~l
Hence nl nl
dk(Bn, U):::;:; 22_P_
i=l
Now take any positive integer r. We can represent r in the form r = m0 + ... +mnl+i,
where we put m0 Zr =
=
0 and 0 :::;:; i
< mn. Let
2nmn2/p Xn,i.
We shall show that A C F 1 ({z1 , z2 , ••• }).Let x EA. Basing ourselves on formulas (7.1.lO)n, we can represent x by induction in the form
Nuclear Spaces. Theory
303
where li E 2im~IPFp( {Xi,l, and Yn E Bn. Observe that 1
.•. ,
Xi,mi})
1
2tn+ .. . + 2nl fn1 E Fp({z1 , z 2, ... }) c F 1({z1 ,
... })
and Yn
E
dm.(Bn, U)U
and by (7.1.12)
n m; n
Yn
E
22nfp
1pdm.(A,
U)U,
i=l
n
But
/J
mt = 22 +···+ 2• = 22n+ 1 1 < 22n+I = m!,
t=1
Hence Yn
E
2 2/p mn4/pdmn( A'
u) u c mn5fpdmn(A ' u) .
and, by (7.1.8) and the definition of Sn, Yn~O. Therefore xEF1 ({z1 ,z2 , .•. }).Now we ought to investigate the converge nee of razr. By definition razr = (m 0 + ... +mn1+it2nm~PXn,i E (mo+ · · .+mn1 +i)a2nm~PFp(Bn).
By (7.1.10)n and (7.1.12), putting k = mn_ 1 we obtain
n
n1
r aZr
E
n amna 22
m,21vdmn·• (A , U)U
i=l 4a~
C mn! 1 dm._,(A,U)U
and, by (7.1.8), raz" tends to zero.
D
Proof of Theorem 7.1.4. Let X be a nuclear locally pseudoconvex space,
with topology determined by a sequence of Prhomogeneous pseudonorms II IIi· Let At= {x: llxlli < 1}. We shall show that each At contains an open convex set. Observe that, by the nuclearity of the space X, for each j there is an index k such that lim n 131P1dn(AJ+k, AJ) = 0. n> 00
Chapter 7
304
Let XJ = {x: llxllt = 0} and X1 = X/XJ. The pseudonorm a Prhomogeneous norm on Xi. Observe that
II IIi induces
i = 1, 2, ...
Putting A = At+k+ XJ and U = A,, by Lemma 7.1.11 we find that there is a sequence {zn} of elements of A1 such that lim n21P1IIznlli
0
=
and (7.1.13)
Since
II 111 is Prhomogeneous by (7.1.13), there is an M > 0 such that
F 1({zl> z 2 , Therefore Ai+k C
••• })
C MAJ.
1
MF1({z1, z2 ,
1 and the set M IntF1({z1 ,z2 ,
••• })
••• })
c A1
is the required open convex set.
D
The existance of nonlocally convex nuclear spaces follows from PROPOSITION
7.1.12. A space [fP•} (see example 6.4. 7) is nuclear if and only
if
lim supp~logn < +oo,
(7.1.14)
where {p~} is the sequence obtained from the sequence {Pn} by ordering it in a nonincreasing sequence. Proof Let Kr
=
{x: llxll
~
r}, co
where
llxll is the norm in the space [fP•l, llxll = }; n=l
By a simple calculation we obtain for r < s CJn(Kr, Ks) = (:
F~.
lxniP•.
Nuclear Spaces. Theory
305
( p~logn+log~) p~ s .
(7.1.15)
Hence n6n(Kr, Ks) = exp
If (7.1.14) holds, for each s there is an r such that
r
p~logn+logs +00
p~
(7.1.16)
and by (7.1.15) the space X is nuclear. Conversely, if X is nuclear by (7.1.15), formula (7.1.16) holds for a certain r0 < s. Thus (7.1.I4) holds. D Another example of a nonlocally convex nuclear space was given by Fenske and Schock (I970). The abovementioned examples of nonlocally convex nuclear spaces have the property that the dual spaces are nontrivial. Ligaud (I973) constructed an example of a nuclear space without nontrivial linear continuous functionals.
Example 7.1.13 (Ligaud, I973) Let E be the linear subset of the space L0 [0, I] spanned by characteristic functions of intervals with rational ends. The algebraic dimension of the space E is countable. Let V be an arbitrary neighbourhood of zero in E in the topology of L0 [0, I]. Observe that
Indeed, for each e > 0, every function belonging to L 0 [0, I] can be represented as a finite sum of functions with supports contained in intervals of measure less than e. Now we shall introduce a new topology on E. We shall construct it in the following way. Let {ei>e 2 , ... } be a Hamel basis in E. Let Vn be a sequence of balanced neighbourhoods of zero in the primal topology such that
Chapter 7
306
We define by induction new neighbourhoods W~ = Vn and let
w;.+I =
W~
as follows: let
Do(m~l Wi.+Lm),
where Lm = lin({ei> ... , em}). The neighbourhoods W~ define a new topology on E. By induction we trivially infer that the sets W~ are balanced,
w;.+l + w;.+l c w;., w;., c w~ w;.' c w;.
ifn
~
n',
ifp~p'.
Hence W np n wp' n'
U/max(p,p')
::::> rr max(n,n')
Now we shall show by induction that the sets W~ are absorbing. Of course, w~ = Vn is absorbing. Suppose that w~ is absorbing. Let X E E. Then there is an m0 such that, form ?:o m0 , x ELm. Since W~ is absorbing form < m 0 , there are A.m. > 0 such that 1 p AmXE m+fWn.
Let A= min(l, A1 ,
... ,
Am, ). Then
1 1 p A X E 1  Wn+Lm
m+
wr
1 and W~+ 1 is absorbing. Therefore the family for all m. Thus AXE { W~, n = 1, 2, ... , p = 0, 1, ... } defines a linear topology on E stronger than the primal topology. The family W~ is countable, and thus the topology in question is metrizable. Since for all m ?:o 0
Wnp+l C
m+l 1
TUP
L
rrn+ m,
15n(W~+l, Wi.) < m~ l
.
Therefore the space E with this new topology is nuclear. By the defini
Nuclear Spaces. Theory
307
tion of WP
n lcVnC W~, n=l,2, ... ,p=O,I, ... F1(~)) F 1(n leV)= E . '>0
Hence
.:1>0
Thus there are no nontrivial linear continuous functionals on E. The ... space E is not complete, but its completion E is, by Proposition 7.1.3, also nuclear, and of course there are no nontrivial linear continuous functionals on E. The example described above has been specially constructed to show the existence of nuclear spaces without nontrivial linear continuous functionals. Burzyk (1980) investigating the completeness of the Mikusinski operator field, has introduced in a natural way an algebra which is a Monte! space without nontrivial linear continuous functionals. It is not clear whether the space constructed by Burzyk is also nuclear. Kalton (1979) has proved that, if a strictly galbed infinitedimensional Fspace X is not locally bounded, then it contains an infinitedimensional nuclear locally convex space. For a locallyconvex X this has been proved in Bessaga, Pelczynski and Rolewicz (1961). The notion ofnuclearity can be refined by the notion of /cnuclearity. Let }. be a linear space of sequences of numbers. We assume that A is normal (i.e. {c;n} E A, I'IJnl ~en imply {'IJn} E A) additive (i.e. if {c;n} E A {1Jn} E /c, for C2, _ 1 = c;,, C2n = 'l}n the sequence Cn(n)belongs to A. for all bijections 1t(n) of positive integers onto themselves) and decreasing rearangement invariant (i.e. if {c;n} E A then the sequence g~} obtained from the sequence {c;n} by a rearangement such that {l~nl} is non increasing, also belongs to /c). We say that an F*space X is Anuclear if, for each neighbourhood of zero U, there is a neighbourhood of zero V such that {bn(V,U)} EA. Taking as A.= {{c;n}: suplc;nlnk < +oo, k = I, 2, ... } we obtain nuclear spaces. Investigations of Anuclear locally convex spaces have been carried out in Ramanujan (1970), Dubinsky and Ramanujan (1972), Dubinsky and Robinson (1978), Moscatelli (1978), Dubinsky (1980).
Chapter 7
308
7.2.
NUCLEAR
OPERATORS
AND
NUCLEAR
LOCALLY
CONVEX
SPACES
In this section we shall begin investigations of nuclear operators in Banach spaces. Let (X, II llx) and (Y, II IIY) be two Banach spaces. By Bx,By we shall denote the unit balls in X and Y, Bx = {xE X: llxllx
< 1},By = {yE
Y:
IIYII < 1}.
Write dn(T) = ~n(T(Bx), By)
for any linear continuous operator Tmapping X into Y. It is obvious that an operator Tis compact if and only if dn(T)+0. Let three Banach spaces X, Y, Z be given. Let B be a continuous linear operator mapping X into Y and let A be a continuous linear operator mapping Yinto Z. Then by Lemma 7.1.2 dn+m(AB) = dn(A)dm(B).
Let X be a B0space and let {llxllt} be an increasing sequence of homogeneous pseudonorms determining the topology. Let
X~= {x: llxlli = 0} and let Xt be the quotient space X/X~. The pseudonorm llxllt induces a homogeneous norm in the space Xt. Since no confusion will result, we shall denote this norm by the same symbolllxlli· Of course, Xi with the norm llxlli is a normed space. Since the sequence of pseudonorms {llxllt} is increasing, there is a natural continous embedding of the space Xi+I into the space Xt. We shall denote it byh The definition of nuclear spaces implies PROPOSITION 7.2.1. Let X be a B 0space. The space X is nuclear if and only if there is an increasing sequence of pseudonorms {llxlli} determining a to
pology such that limndn(Tt) = 0
(i
= 1, 2, ... ).
(7.2.1)
A continuous linear operator T mapping a normed space X into a normed space Y is called nuclear if it can be represented as the sum of
Nuclear Spaces. Theory
309
onedimensional operators Pn(x) = .fn(x)en (en an element,.fn a continous linear functional)
""
T=}; Pn n=l
such that IITIIA =
l
00
1
11Pnll.
n=l
The number II Til Ais called the nuclear norm of the operator T. This definition trivially implies that the sum of two nuclear operators is a nuclear operator and that a superposition of a nuclear operator with a continuous linear operator (also a superposition of a continuous linear operator with a nuclear operator) is a nuclear operator. PROPOSITION
If
7.2.2.
I
lim n4dn(T) = 0,
n_,.ro
then the operator Tis nuclear. The proof is based on the following LEMMA 7.2.3. Let Y be anndimensional subspace of a Banach space X. Then there is a_linear projection P, with the norm not greater than n, of the whole space X onto Y. Proof. Basing ourselves on Lemma 7.1.6 and 7.1.7, we can find in Y such elements e~> ... , en and functionals ];_, ... , fn that lieill = IIfill = 1 (i = 1, 2, ... , n) and (7.1.5) holds. By the HahnBanach theorem we can extend each functional jj to a functional F1 of norm one defined on the whole space X. Let n
P(x)
=}; Ft(x)eJ.
(7.2.2)
i=l
The operator Pis a continuous linear projection of X onto Y. Moreover, n
IIPII
~}; IIFilllleill = n. j=l
0
Chapter 7
310
Proof of Proposition 7.2.2. Let Ln be anndimensional subspace such that Ln+2d11 (T)By :J T(B)x and let Pn be a projection, with the norm not greater than n, onto L 11 • The existence of such a projection follows from Lemma 7.2.5. Let x E Bx. Then T(x) = y+z, where yE Ln an9 liz// ~ 2d11 (T). Thus I!Pn(T(x))T(x)l! = I!Pn(z)zl! ~ (I!Pnl!+l)l!zl! ~
2(n+1)dn(T).
(7.2.3)
Now let T(x) = (T(x)P1 (T(x)))+(P 1(T(x))P 2(T(x)))+...
(7.2.4)
On the other hand, dim(P11 TPn+IT) (X)~ 2n+2.
Therefore, by Lemma 7.1.6 and 7.1.7 and by the HahnBanach theorem, there is a system of one dimensional operators Kf, ... , K~"+ 1 , I!Ki/1 = 1, i = 1, 2, ... , 2n + 1, such that the operator 2n+l
P'=}; Kj j=l
is a projection of the space X onto the space (Pn TPn+ 1 T)(X). By (7.2.4) oo 2n+l
T=};}; (P11 TPn+IT)Kj'. n=l'j=l
The operators (PnTPn+ 1T)Kj are onedimensional and, moreover, by (7.2.5) oo 2n+l
oo
}; }; 1/(PnTPn+IT)Kj/1 ~}; (2n+2)1!(PnTPn+IT)// n=lj=l
n=l
00
~
l, (2n+2) (//PnTT/1+1/Pn+ITT//) n=l 00
~}; (2n+2) (4nt3)dn(T) < +oo. n=l
D
Nuclear Spaces. Theory
311
7.2.4. Let T be a nuclear operator mapping a Hilbert space H into itself. Then PROPOSITION
limndn(T) = 0.
(7.2.5)
n~oo
Proof The operator Tis of course compact. Let {A.!} be eigenvalues of the selfadjoint operator T*T, and let {en}, llenll = 1 be the eigenvectors corresponding to {A.~}. Let us choose en in such a way that they are orthogonal even if they correspond to the same eigenvalue. The operator T can be written in the form 00
Tx =
.2; A.n(X, en)fn, n=l
where A.nfn = Ten and sequence
11/nll =
I. Let us order all An in a nonincreasing
Then dn(T) = A.n+I· On the other hand, the operator Tis nuclear, and thus, by definition, it can be written in the form 00
T(x) =
00
.2; (x, Xn)Yn,
where
n=l
.2; llxnlliiYnll < +oo. n=l
Hence 00
).n = (Ten,/n) =
00
(.J; (en, Xj)YJJn) = .2; (en, Xj)(YJ,Jn). j=l
j=l
Thus 00
00
00
co
l, dn(T) = .2; An= .2; .2; (en, Xj)(YJ,Jn) n=O
n=l 00
n=lj=l 00
j=l n=l
00
r' (.J; [(YJ ,/n)[ r'
~ .2; (.J; [(en, XJ)i 2
2
2
2
n=l
00
=
.2; llxill IIYJII < +oo. j=l
and, since {dn(T)} is a nonincreasing sequence, (7.2.5) holds.
D
•
312
Chapter 7
PROPOSITION 7.2.5. Let X andY be two Banach spaces. Any nuclear operator T mapping X into Y can be factorized as follows :
T
where His the the space 12. Proof Let us write Tin the form 00
(Tx)
=}; Anfn(x)en:• n~l
00
where /Ifni I =
llenll
=
2
I, An~ 0 and
An
< +oo! Let us put
n~l
XEX,
Sl{hi}) E Y. It is easy to verify that S 2 S1 = T.
0
7.2.6. Suppose we are given four Banach spaces xl> x2, Xa, x4 and letT., be nuclear operators mapping Xi into Xi+ 1 (i = I, 2, 3). Then
PROPOSITION
limndn(TaT2T1)
=
0.
(7.2.6)
ti+CO
Proof Let us factorize T1 and T 2as follows: ~
xl
~
'',
sl
'~
/'
/ /
'\./ HI
~
_..x2+X2 ~
~
x4 /
',
'\./
/ / s2
>H2
where H 1 = H 2 = P, The operator T 0 is a nuclear operator mapping H 1 into H 2 as a super
Nuclear Spaces. Theory
313
position of the nuclear operator T2 with continuous operators. Hence, by Proposition 7.2.4, lim ndn(T0 ) = 0. n>exo
0 ~ ndn(TsT2Tt) ~ nJJS2JJ dn(To) JIStJJ+0. This implies (7.2.6). THEOREM 7.2.7 (Dynin and Mityagin, 1960). A E0space X is nuclear if and only if there is an increasing sequence of pseudonorms {Jixlli} determining a topology such that the canonical mappings Ti are nuclear operators. Proof Sufficiency. Suppose that a sequence of pseudonorms {Jixlli} satisfies the properties described above. Let JJxJJ~ = Jlxlls, and let denote the canonical mappings with respect to the pseudonorms llxll~. Then, by Proposition 7.2.6,
r;
1imndn(T[)
..co
= 0,
i = 1, 2, ...
Thus, by Proposition 7.2.1, X is a nuclear space. Necessity. Suppose that X is a nuclear space. Then, by Proposition 7.2.1, there is an increasing sequence ofpseudonorms {llxlli} such that the canonical embeddings satisfy (7.2.1). Let us put JJxJJ~ = JJxJJ 4,. Then, by Proposition 7.2.2, the canonical mappings T~ with respect to the pseudoD norms JJxJJ; are nuclear. PROPOSITION 7.2.8 (Mityagin, 1961). Let X be a E 0 nuclear space. Then there is a sequence of Hilbertian pseudonorms JlxJJi = (x, x)i determining a topology equivalent to the original one. Proof Theorem 7.2.7 implies that there is a sequence of pseudonorms {llxlli} such that the canonical mappings Ti of Xi+ 1 into Xi are nuclear. This means that Ti can be written in the form
y
Tt(X) =
2:"' A~Ytn(X)Yi,n, · n~l
where Yi,n
E
Xi, Y;,n E
x;+l> IIYi,nll =
IIY;,nll = I,
..1f > 0 and the series
Chapter 7
314 00
}; A.~ is convergent. Let us introduce now inner products in X by the n=l following formula 00
(x, Y)t
=}; A.~Ytn(X)Ytn(y). n=l
On the one hand, we have 00
(x, x)i
~}; IA.~IIY~n(x)! 2 ~ A~l!xii~+I, n=O
00
where At = (}; IA.~IY 12 • On the other hand, n=l 00
l!xl!t
=
j}; A~Y~n(X)Yi,nj ~}; A.j IYtn(x)j n=l 00
n=l
00
~ (}; A.?r' 2 ( } ; A.~IY~n(x)l 2 ) 112 = n=l
At(X, X)t,
n=l
Then Ai 112 1!xl!i ~ (x, x)~12 ~ Ail!xl!i+t·
Hence the Hilbertian pseudonorms (x,x)f 12 yield a topology equivalent to the original one. D As an obvious consequence of Proposition 7.2.8 and Proposition 6.6.7, we obtain PROPOSITION 7.2.9 (Mityagin, 1961). Let X be a nuclear B 0space and let Y be a subspace of the space X. Then
(J(Y)C (J(X).
~3.UNCONDITIONAL
AND ABSOLUTE CONVERGENCE
In Section 3.6 we have considered unconditional convergence in Fspaces. Now we shall consider other kinds of convergence of series in Fspaces.
Nuclear Spaces. Theo~y
315
00
We shall say that a series
2
Xn of elements of an Fspace X is absolutely
n=l 00
convergent if for each continuous quasinorm [x] the series
2 [xn] is conn=l
vergent. The classical Riemann theorem shows that if X is a finitedimensional 00
space, then a series
2
Xn is unconditionally convergent if and only if it
n=l
is absolutely convergent. Let {Um} be a basis of balanced neighbourhood of zero and let [X]m
=
inf{t
> 0:
~ E Um}
be the quasinorm with respect to Um. It is easy to verify that a series oo
2
ro
Xn absolutely convergent if and only if the series
n=l
2 [Xn]m are
vergent form= I, 2, ... If X is a locally bounded space with a phomogeneous norm
2
llxll, then ro
~
a series
con
n=l
Xn is absolutely convergent if and only if the series }; [xnr!P
n=l
n=l
is convergent. PROPOSITION 7:3.1. An Fspace X is locally convex if and only if each absolutely convergent series is unconditionally convergent. Proof Necessity. Let X be a locally convex space and let {Um} be a basis of balanced convex neighbourhoods of zero. Let us denote by llxllm the 00
pseudonorm generated by Um. Let
2
Xn be an absolutely convergent
n=l 00
series in X. Then the series }; llxnllm are convergent for m = I, 2, ... Let n=l
{en} be a sequence of numbers equal either to 1 or to 1. Then ro
0~
ro
~~~ enXnllm ~ ~ llxnllm70 n=k
n=l
fork tending to infinity and form = 1, 2, ...
Chapter 7
316
ro
Therefore, by definition, the series }; Xn is unconditionally convergent. n=l
Sufficiency. Let X be a nonlocally convex £space. Let { Um} be a basis of balanced neighbourhood of zero such that Um+l C { Um. Since the space X is not locally convex, there is a neighbourhood of zero V such that cony Um cj: V (m = 1, 2, ... ). This means that there are elements Xm, 1 , ..• , Xm,nm of Um and nonnegative reals am, 1 , ••• , am,nm such that nm
~ am,i =
(7.3.1)
1
i=l
and

~ am,iXm,i ¢ V.
(7.3.2)
i=l
co
Let us order the elements am, iXm, i in the sequence {Yn}. The series
2; Yn n=1
is absolutely convergent. Indeed, let us denote by [x]k the quasinorm with respect to the set Uk. Then for j, / > k we have j"
nm
j'
nm
} ; ~ [am,iXm,i]k = ~}; [am,iXm,i]k m=j i=l m=j i=l j'
j'
1
<}; sup[x}k<}; 2mk < 2ik1. m=j xEUm
m=j
00
On the other hand, formula (7.3.2) implies that the series
2; Yn is not n=Jt
unconditionally convergent.
0
If a space X is infinitedimensional, then unconditional convergence does not imply absolute convergence. Dvoretzky and Rogers (1958) have shown that in each infinitedimensional Banach space there is an unconditionally convergent series which is not absolutely convergent. This theorem has been extended to locally bounded spaces by Dvoretzky (1963). In general, the problem when unconditional convergence implies absolute convergence is open. For locally convex spaces such characterization is due to Grothendieck.
Nuclear Spaces. Theory
317
THEOREM 7.3.2 (Grothendieck, 1951, 1954, 1955). Let X be a B 0 space. The space X is nuclear if and only if each unconditionally covergent series in X is absolutely convergent. The proof of this theorem, the main theorem of the present section is based on several notions, lemmas and propositions. We say that a linear continuous operator T mapping a Banach space X into a Banach space Y is absolutely summing if there is a positive constant C such that, for arbitrary x 1, .•• , Xn EX, n
n
2; IIT(xt)ll ~ cll2; xtll·
i=1
(7.3.3)
i=1
PROPOSITION 7.3.3. An operator T satisfies (7.3.3) if and only if n
n
2; II T(xt)ll ~ C sup 112; t:txtl/·
i=1
Proof Necessity. Let
(7.3.4)
••=±1 i=1
t:.g =
1. Then
Thus (7.3.4) implies (7.3.3). . Sufficiency. Let x 1 , .•. , Xn be arbitrary elements of X. Let t:~, ... , t:~ be arbitrary numbers equal to + 1 or 1. Then putting Yi = t:?xt, i = 1, ... ... , n, and applying (7.3.3) to Yt, we obtain n
n
n
n
2; IIT(xt)ll = 2; IIT(y.,)ll ~ c/12; t:hi\\ ~ C sup l/2; t:txt/1· i=1
i=1
i=1
8 t=±1
CJ
i=1
PROPOSITION 7.3.4. A linear operator T satisfies (7.3.4) if and only if n
n
2; IIT(xt)ll ~ C II!sup 2; lf(xt)l. II i=1
=1 i=1 jEX*
(7.3.5)
Chapter 7
318 co
=
Proof Sufficiency. Lety
2
B~Xt be such an element that
i=l
n
IIYII
=
sup
~~~etXill·
•t=±l i=l
Letf'be a functional of norm one such thatf'(y) = n
sup
IIYII. Then
n
~~~ etXtll
=
IIYII
=
f'(y)
=
Bf=±l i=l
~ f'(shi)
i=l
n
n
~ ~ if'(xt)i ~ sup ~ if(xt)i. i=l
llfll=li=l /EX*
Therefore (7.3.4) implies (7.3.5). Necessity. Let s0 = signf(xi) for afunctionaljE X* of norm one. Then n
n
n
~ if(xt)i = ~ f(shi) = !(~ s?xi) i=l
i=l
i=l
n
n
~ ~~~ s?xill ~ sup ~~~ BiXill· i=l t<=±l i=l Therefore, (7.3.5) implies (7.3.4).
D
Formulae (7.3.3)(7.3.5) give us three equivalent definitions of absolutely summing operators. The infimum of those T which satisfy (7.3.3) will be denoted ~y a(T). Let T be an absolutely summing operator belonging to B(X+ Y). Let A EB(Y+Z) (or A EB(Z+X)). Then the operator AT(resp. TA) is absolutely summing. PROPOSITION 7.3.5. A linear operator T mapping a Banach space X into a Banach space Y is absolutely summing if and only if it maps unconditionally convergent series into absolutely convergent series. co
Proof Let
2
Xn
be an unconditionally convergent series. Then
n=l k'
l~m sup~~~ BnXnll =
k,k+WSn=±l n=k
0.
(7.3.6)
Nuclear Spaces. Theory
319
Let Tbe an absolutely summing operator. Then by (7.3.4) and (7.3.6) k'
lim }; IIT(xn)ll = 0, k,k'HtJn=" 00
.J;
and the series
T(xn) is absolutely convergent.
n=l
On the other hand, if we suppose that an operator T E B(X+ Y) is not absolutely summing, then, by definition, for any k there are elements {xk, 1, ••• , Xk,nJ of X such that (7.3.7) and 11k
} ; IIT(xk,t)li
> 1.
(7.3.8)
i=l
Let us order all Xk,i into a sequence {y 11 }. Formula (7.3.7)implies that the 00
series
2
Yn is unconditionally convergent. Formula (7.3.8) implies that
n= 00
the series
2
z:'(Yn) is not absolutely convergent.
0
n=l
PROPOSITION 7.3.6. Each nuclear operator is absolutely summing. Proof Let TE B(X+Y) be a nuclear operator. This means that the operator T can be written in the form 00
T(x)
=}; Angn(x)yn, n=l 00
where An> 0,
c= 2
n=l
An<
+oo,
gn E X*, Yn E Y,
llgnll = IIYnll = 1
= 1, 2, ... ). Let xi> ... , xN be arbitrary elements of X. Letfi, i = 1, ... , N be a continuous linear functional of norm one defined on Y such that fi(T(xt)) (n
Chapter 7
320 =
IIT(xt)ll· Then N
N
N
N
}; IIT(xt)ll = };fi(T(Xt)) = };fi(L Angn(Xt)Yn) i=l
i=l oo
i=l
i=l
N
N
~};An}; [gn(Xt)[[/t(yn)! ~ C sup n=l
i=l
l, [g(xt)[.
llull=li=l
ueX*
Hence, by (7.3.5), the operator Tis absolutely summing.
D
We say a continuous linear operator T mapping a Hilbert space H 1 into a Hilbert space H 2 is a HilbertSchmidt operator if, for any orthonormal sequence {en} in the space H 1, 00
}; [[T(en)[[ 2 < +oo. n=l
This definition is clearly equivalent to the following one. An operator T E B(Hc~H2) is called a HilbertSchmidt operator if, for an arbitrary orthonormal sequence {en} in H 1 and an arbitrary orthonormal sequence {In} in H 2 , 00
}; !(T(et),/t)[ 2 < +oo. j,i=l
This implies that an operator conjugate to a also a HilbertSchmidt operator.
Hilb~rtSchimdt
operator is
7.3.7. If an operator TE B(H1 +H2), where H 1 and H 2 are Hilbert spaces, is absolutely summing, then it is a HilbertSchmidt operator. Proof Let {et} be an arbitrary orthonormal set in H 1 and let {at} be an arbitrary sequence belonging to 12• Let Xi= atet. Then, by (7.3.5),
PROPOSITION
00
Thus, by the arbitrariness of n, we find that the series
J; at[!T(et)ll is coni=!
Nuclear Spaces. Theory
321 CJ)
vergent. Since this holds for all sequences {an} E f2, the series}; IIT(et)W is 
i=l
convergent. This means that Tis a HilbertSchmidt operator.
0
7.3.8. The superposition of two HilbertSchmidt operators is
PROPOSITION
a nuclear operator. Proof Let H 1 , H 2 , H 3 be Hilbert spaces. Let T E B(Hc~H2), and let S E B(H2 ?>H3) be HilbertSchmidt operators. Let {en} be an arbitrary orthonormal set in H 2 • Then co
co
= }; (T(x), en)S(en) = }; (x, T*(en))S(en),
ST(x)
n=l
n=l
where T* E B(H2 ?>H1 ) denotes the operator conjugate to the operator T. The operator T* is also a HilbertSchmidt operator. Thus 00
}; IIT*(en)IIIIS(en)ll n=l
co
~ (}; IIT*(en)ll 2 n=l
r
12
J)
(};
IIS(en)ll2
n=l
r'
2
<
+=.
Hence ST is a nuclear operator.
0
7.3.9 (Pietsch, 1963). Let TE B(X?>Y) be an absolutely summing oprator. ·Then there is a probability measure (i.e. a regular positive Borel measure with total mass 1) f1 on the unit ball S* of the conjugate space X* such that THEOREM
IIT(x)ll
~ a(T)
J lx*(x)l dfl(x*).
s•
Proof (Linden strauss and Pelczytiski, 1968). Let n
n
W = {g E C(S*):g = a(T)}; lfx,(x*)l with}; IIT(xi)ll = 1}, i=l
wherefx(x*)
i=l
= x*(x) for x* E S* and x EX.
We shall show that the set W is convex. Let m
n
gl
=
a(T) .2;1fx,,Jx*)l, i=l
g2
= a(T)}; 1/x,,,(x*)l, i1
Chapter 7
322
where n
m
.2;11T(xi,l)ll = .2;11T(x;,2)11 = i=l
(7.3.9)
1.
i=l
Let a,b?;;O,
a+b = 1.
(7.3.10)
Let forj= 1,2, ... ,n, for j = n+1, ... , n+m. Then, by (7.3.9) and (7.3.10) n+m
n
m
};IIT(yJ)II =a}; IIT(xi,l)ll+b}; IIT(x~,2)1i = 1. · j=l
i=l
i=l
Moreover, n+m
g(t) = a(T)}; lfuix*)l j=l
n
m
= a(T) [}; 1/ax,,,(x*)l+.2;1/bx,,,(x*)J] i=l
i=l
n
m
= a(T) [a}; 1/x,,,(x*)J+b}; J/x,,,(x*)J] = ag1+bg2. i=l
i=l
Thus the set W is convex. n
The definition of a(T) implies that if}; IIT(xi)ll = 1, then i=l
n
n
sup }; Jx*(xi)l = sup }; x•es• i=l x•es• i=l
J.fx.(x*)J
?;; 1
(see Proposition 7.3.4). Therefore, the set W is disjoint from the set N
= {/E C(S*): f(x*) <
1}.
The set N is open and convex. Therefore, there is a continuous linear functional F defined on the space C(S*) such that F(f)?;; 1
forjE W
(7.3.11)
Nuclear Spaces. Theory
323
and
F(f)
<
1
forfE N.
(7.3.12)
The general form of continuous linear functionals on the space of continuous functions implies that there is a regular Borel measure flo defined on S* with its weak*topology such that F(f)
=
Jf(x*)dfl (x*). 0
s•
Since the set N contains the cone of negative functions in C(S*), by (7.3.12) the measure flo is positive. Thus it is of the form flo= afl, where fl is a probability measure and a = IIFII· The set N contains the unit ball in C(S*), hence, by (7.3.12), a= IIFII < 1. 1 Let x EX and T(x) #0. Then g = a(T) IIT(x)JI i.fx(x*)i E W. Therefore, by (7.3.11)
Jgdfl ~ s•Jgdflo ~ 1.
s• Thus
J
IIT(x)ii ~ a(T) lfz(x*)dfl(x*) s•
=
J
a(T) lx*(x)i dfl(x*) s•
and this completes the proof.
D
THEOREM 7.3.10 (Pietsch, 1963). LetT be an absolutely summing operator mapping a Banach space X into a Banach space Y. Then the operator T can be factorized as follows T
X+Y
t
,),
C(M1+H
where H is a Hilbert space, M is the unit ball S* in the conjugate space X* with its weak*topology, and i is the natural embedding of X into C(M). Proof Let fl be a probability measure defined in Theorem 7.3.9 on the
Chapter 7
324
set M. Let V(jt) denote the completion of C(M) with respect to the norm llxll = lx(t)i dp, and let V(jt) denote the completion of C(M) with re
J
M
spect to the norm
Let
be natural injections and let Z be the closure of {JIX i(X) in the space D(fl). The theorem follows from the diagram T X+Y
""""'y
>zc fJ/
t
C(M)
+
H
/
V(fl).
= L2(/1)
Theorem 7.3.9 implies that the operator y is continuous.
D
THEOREM 7.3.11 (Pietsch, 1963). A composition of five absolutely summing operators is a nuclear operator. Proof Let us consider the diagram
Tl
T2 +X2 /
xl
""'
·""\./ Hl
Ta
T4
+Xa
"',
/
""__,.H2
\./
IX
/
__,.x4 /
Ts +X,
""
""
+
\./
/
+Xs / j
Ha
fJ
The existence of such factorization follows from Theorem 7.3.10. The operators IX, fJ are absolutely summing as compositions of absolutely summing operators with continuous operators. Therefore, by Proposition 7.3.8, the operator {JIX is nuclear. Thus the operator T 5 T4 T3 T2 T1 = j(Jai is nuclear. D
Nuclear Spaces. Theory
325
Proof of Theorem 7.3.2. Sufficiency. Let X be a nuclear B0space and let the topology in X be given by an increasing sequence of homogeneous pseudonorms {llxllt} such that the canonical mappings Tt from Xi+ 1 into Xi are nuclear. By Proposition 7.3.6 the operators Ti are absolutely summing. 00
Xn
Let };
be an unconditionally convergent series in X. This means
n=1
that 00
lim
sup~~~ BnXnlli =
0,
i = 1, 2, ...
k>oo tn=±1 n=k
Since the canonical mappings Tt are absolutely summing, this implies that 00
the series };
llxnlli 1 are convergent fori= 2, 3, ...
n=l
Necessity. Let X be a B0space and let {llxlli} be an increasing sequence of pseudonorms determining the topology. Theorem 7.3.11 implies that it is sufficient to show that for any pseudonorm llxllt there is a pseudonorm llxlli such that the canonical embedding X1 into Xi is an absolutely summing operator. Suppose that the above does not hold. This means that there is a pseudonorm llxllio such that the canonical embedding Xi into Xio is not absolutely summing for any i > i 0 • Then, by definition, there are elements Xi, 1 , ... Xi, n, s u~h that n,
~llxdlt, =
1
(7.3.13)
j =1
and n•
sup~~~ BJXi,JIIi < ~i.
(7.3.14)
BJ=±1 j=1
Let us order Xi,n in a sequence. Formula (7.3.14) implies that this series is unconditionally convergent. On the other hand, by (7.3.13) it is not absolutely convergent. D CoROLLARY 7.3.12 (Dvoretzky and Rogers, 1950). In each infinite dimensional Banach space there is an unconditionally convergent series which is not absolutely convergent.
Chapter 7
326
Proof. If each unconditionally convergent series were absolutely convergent, then the space would be nuclear, and thus a Montel space. Since each Banach space is locally bounded, it would be locally compact. Thus it would be finite dimensional. D Rosenberger (1972, 1973) gave the following extension of the notion of nuclear spaces. Let cf> be a class of continuous, subadditive, strictly increasing functions defined on [O,+oo], such that all functions (jJ E cf> vanish at 0. An F*space X is called cf>nuclear if, for any balanced neigh· bourhood of zero U, there is a balanced neighbourhood of zero V such that 00
_2; qJ(bn(V, U)) < +oo
for all
(jJ E
W.
n=O
When cf> contains only one function (jJ, we shall mark cf>nuclear spaces as qJnuclear spaces. By Theorem 7.3.2 if X is locally convex, then it is nuclear if and only if it is tnuclear. Moscatelli (1978) gave the conditions for (jJ ensuring the existence of a universal qJnuclear space for locally convex qJnuclear spaces.
7.4.
BASES IN NUCLEAR SPACES
Let X be a locally convex space, i.e. a space in which the topology could be given by a sequence of homogeneous pseudo norms. Let {en} be a basis in X. We say that a homogeneous continuous pseudonorm llxll is admissible if r
r+s
II}; tnenll ~!I}; tnenll n=l
forr,s~
0.
(7.4.1)
n=l
Theorem 3.2.14 implies that in each locally convex space there is a sequence of homogeneous admissible pseudonorms determining a topology equivalent to the original one. PROPOSITION 7.4.1. Let X be a locally convex space with a basis {en}. Iffor each sequence of homogeneous pseudonorms {llxllm} determining a topology
Nuclear Spaces. Theory
327
equivalent to the original one for every i there is a j such that ro
~ Jlenllt
Ct,i
(7.4.2)
= L; llenlli < +oo, n=l
where we assume
0
0 =
0, then the space X is nuclear.
P··,.,ot: Let {llxllt} be an increasing sequence of homogeneous pseudonorm ,~.. ,,_ ~
n;;[; :ng the topology. Without loss of generality we can assume that t:,._ p~·udonorm llxllt are admissible. Let us take an arbitrary i. Then, by th: hypothesis, there is an indexj such that (7.4.2) holds. Let us denote by {fn} the sequence of basis functionals, Let llxlli < 1. Since the pseudonorm llxlli is admissible n
llfn(x)eniiJ<
n1
lll fi(x)etjj +lll' fi(x)et// 1
1
i=l
1
< 2llxlli < 2.
(7.4.3)
i=l
The canonical embedding T1, i of X 1 into Xt is nuclear, provided (7 .4.2) holds. Indeed. ro
Ti,t(x)
=}; Pn(x),
where Pn(x) = fn(x)en
n=l
are onedimensional operators. Moreover, by (7.4.3)
IIPnll =. sup IIPn(x)Jit = sup llfn(x)enllt ~~~Cl
~~~Cl
llenllt
sup llfn(x)enllill11 en j
1/x//JCl
llenllt < 2en 1 11

11
•
Hence, by (7 .4.2) 00
}; IIPnll < +oo.
(7.4.4)
n=l
Therefore, X is a locally convex nuclear space.
0
7.4.2. Let X be a locally convex space with a basis {en}. If there is in X an increasing sequence {llxllt} of pseudonorms determining a topology such that (7.4.2) holds, then each increasing sequence ofpseudonorms {llxiO determining the topology has the same property.
PROPOSITION
Chapter 7
328
Proof Let us take an arbitrary i: Since the pseudonorms {llxlli} yield the topology, there are an index i 1 and a constant C such that 11x11; ~ Cllxll;,·
Let A be such an index Ci,,;, < +oo. The pseudonorms llxll;, yield the topology, therefore, there are an index j and a constant K such that llxll;, ~ Kllxll;. Thus ~ KC
llenlli llenllj
llenlli, lleniiJ, ·
Therefore,
We shall now show a theorem converse in a certain sense to Proposition 7.4.1. 7.4.3 (Dynin and Mityagin, 1960; Mityagin, 1961). Let X be a locally convex space with a basis {en}. Let the topology in X be given by an increasing sequence of admissible pseudonorms {llxlli}. If the space X is nuclear, then for every i there is aj such that (7.4.2) holds. Proof Since the space X is nuclear, then for each i there is an indexj such THEOREM
that limn4bn(Bf, Bi) = 0,
(7.4.5)
n>00
where Br =
{xe X: llxllr ~ 1}.
Let n en z;=
Since (7.4.5),
llenlli lim llzj'lli =
for
llenlli =F 0.
0. Let us reorder the positive integers in a se
n...oo
quence {kn} in such a way that
llekmllt llekmll;
;:?:
llekm+•lli llekm+tlli
(m =I, 2, ... ).
(7.4.6)
Nuclear Spaces. Theory
329
Let
let 1
(we admit 1/0
an= /len/It'
= oo),
and let
A = {x Ei X: lft(x)l ~ at(i = 1, 2, ... )}, B = {x EX: /ft(x)j ~ bt(i = 1, 2, ... )}, where {.fn} are the basis functionals with respect to the basis {en}. If x EX and /lx/lt < f, then /l.fn(x)en/lt < 1 (i = 1, 2, ... ) because the pseudonorm /lx/lt is admissible. Therefore, f Bi CA. On the other hand, B C B1, because for x E B 00
00
00
/lx/li ~}; /l.fn(x)en/li = };lfn(x)l /len/li ~}; bnllen/li n=l
n=l
n=l
00
=_.!._ ~_;.=1. a L.J n n=l Therefore,
<5n(BJ, (
~ )Bt) > <5n(B, A).
By (7.4.6)
<5nl(B, A)= bk. = ak. Thus (7.4.5) imply
(7.4.7)
(1/lek.llt). an llek.lli 2
lim n2 /lek.llt = 0. /lek./li
n+oo
Therefore, (7.4.2) holds.
0
We say that a basis {en} of an Fspace X is unconditional if the series of expansions with respect to this basis are unconditionally convergent. A basis {en} of an Fspace X is called absolute if the series of expansions
Chapter 7
330
are absolutely convergent. Proposition 7.3.1 implies that in locally convex spaces each absolute basis is also unconditional. Theorem 7.4.3 implies THEOREM 7.4.4 (Dynin and Mityagin, 1960). lf X is a nuclear B 0space, then each basis {en} in X is absolute. Proof Let {llxlft} be an increasing sequence of admissible pseudonorms determining the topology. By Theorem 7.4.3, for each i there is a j such that (7.4.2) holds. Therefore 00
00
~
~
L.J 11/n(x)enllt = L.J n=l
llenllt lfn(x) llleniiJ llenlfJ
n=I
COROLLARY 7.4.5 (Dynin and Mityagin, 1960).Jn a nuclear B 0 space X all bases are unconditional. It is not known what situation there is in nonlocaUy convex spaces. There is an interesting question: does Theorem 7.4.4 and Corollary 7.4.5 characterize nuclear B0spaces? Pelczynski and Singer (1964) have proved that in each Banach space with a basis there is a nonunconditional basis. Wojtynski (1969) has proved that if, in a B0space X, the topology is given by a sequence of Hilbertian pseudonorms and each basis in X is unconditional, then the space X is nuclear. In the same paper it is shown that if in a B0space X all bases are absolute, then the space X is nuclear. Let X be a nuclear B0space with a basis {en}. The basis functionals corresponding to {en} are denoted by {.fn}. Let {llxllm} be a sequence of pseudonorms determining the topology in X. Let us assign to each element x eX a sequence {.fn(x)}. Since {en} is a basis, we have, by Theorem 2.6.1.
liml/n(x)l llenllm = 0, 11>00
m= 1,2, ... ,
Nuclear Spaces. Theory
331
i.e. {fn(x)} E M(am,n), where am,n = llenllm (compare Example 1.3.9). Moreover, the operator T(x) = {fn(x)} is a continuous operator mapping X into M(am,n)(see Theorem 2.6.1). Now we shall show that T(X) = M(am, n). To do this it is sufficient to prove that if {tn} e M(am,n) then 00
the series }; tnen is convergent in X. The space X is nuclear, therefore,
n=l
for each index m there is an index r such that co
co
Cm r = ~ am,n = ~ ' .L..J ar,n .L..J
JJenJim < +oo n=l JJenJJr
n=l
(see Theorem 7.4.3). Hence co
00
l;lltnenllm
n=l
=}; JJtnJJJJenJir ~~~:~~
where, according to Example 1.3.9
JJ{tn}IJr =
sup JtnJ ar,n = sup JtnJJJenJJr. n
n
00
Therefore, the series }; tnen is absolutely convergent, and thus conver
n=l
gent. Therefore, T(X) = M(am,n). Since both spaces are complete, by the Banach theorem (Theorem 2.3.2) the operator T 1 is continuous. Hence the following proposition holds : PROPOSITION 7.4.6 (Rolewicz, 1959c). Let X be a nuclear B0 space with a basis {en} and the topology determined by a sequence of pseudonorms {JJxJJm}. Let am,n = llenllm· The set of expansions with respect to the basis {en} constitutes the space M(am,n). The corresponyence between x EX and {fn(x)} e M(am,n) is a homeomorphism.
In many applications n and m are vectors consisting of nonnegative integers (see Example 1.3.9). There are nuclear B0spaces without bases. The first such an example was constructed by Mityagin and Zobin (1974) (see also Djakov and Mityagin, 1976; Bessaga, 1976). Even more, Dubinsky (Dubinsky, 1979,
Chapter 7
332
1981; see also Vogt, 1982) showed that there are B0spaces without the bounded approximation property1 • Proposition 7.4.6 play an important role in the theory of nuclear spaces, because the spaces M(am,n) have a very simple structure. Sometimes, however, it is more convenient to consider in M(am,n) other families of pesudonorms determining the topology. PROPOSITION
2 n
7.4.7. Let [am,n] be a matrix ofnonnegative reals. Let am,n am+I,n
Let LP(am,n) (1 <,_p sequences {tn} that 1/{tn}//m,p
=
< +oo.
(7.4.8)
< +oo) (see Example (27/tnam,n/P)lfp <
1.3.9) be the space of all such
+oo
(7.4.9)
n
with the topology determined by the pseudonorms 1/x//m. Then the space LP(am,n) is identical with the space M(am,n). Proof Since 1/{tn}//m,p ~ 1/{tn}//m,
(7.4.10)
where 1/{tn}//m =sup /tn/am,n, we have LP(am,n) C M(am,n). On the other hand, 1/{tn}//m,p
= (,2itnam,n/Prp = (_2/tnam+I,n/P( n
n
a::l:nrrp
<,_ Cm,vl/{tn}l/m+I,
where by (7.4.8) Cm,p = (
~( ..:.....J n
am,n )p)Ifp < am+I,n
+oo.
D
As an application of Proposition 7.4.7, we obtain 1 We say that an Fspace X has the bounded approximation property, if there is a uniformly continuous sequence of continuous finite dimensional operators Tn mapping X into itself such that lim Tn x = x for all x E X.
n
Nuclear Spaces. Theory
333
7.4.8. Each continuous linear functional defined on a space M(am,n), where am,n satisfies (7.4.8), is of the form PROPOSITION
f(x) = l'Jnxn, n
where {xn}
E
M(am,n) and {fn} is such a sequence of scalars that
sup Ifni < += (7.4.11) n am,n for a certain m. Proof By Proposition 7.4.7 the space M(am,n) is identical with the space V(am, n). The general form of continuous linear functionals on B0spaces implies that the functional/is continuous with respect to a certain norm Hxllm, 1 . Hence the general form of continuous linear functionals in f1 implies (7.4.11). 0 We say that a basic sequence (see Section 2.6) of elements {xn} of a nuclear B0space is represented by a matrix [am, n] (we shall denote it briefly by {xn} ,...._ [am,n]) if there is a sequence of pseudonorms {llxllm} determining the topology in X such that llxnllm = am, n·
7.5.
SPACES WITH REGULAR BASES
Let X be aB0space with a basis {en}. A basis {en} is called regular if there is a sequence of pseudonorms {II lim} determining the topology such that t he sequence·{
llenllm }·IS nonmcreasmg . . 10r aII m. llenllm+I t"
Let {an} be a nondecreasing sequence of positive numbers. The standard bases in the Kothe power spaces M(a;;:) and M(a~l/m) are regular. Let X be an Fspace. Two bases {en} and {.fn} inXarecalledsemiequivalent if there is sequence of scalars {rn}, rn > 0 such that the bases {en} and {rn.fn} are equivalent. THEOREM 7.5.1 (Djakov, 1975; Kondakov, 1974). Let X be a nuclear B 0space with a regular basis {en}. Then each regular basis {fn} in X is semiequivalent to {en}.
Chapter 7
334
Proof By the definition of regular bases there are sequences of pseudo
norms {II lim, 0 } and {II II~. 0 } both determining a topology equivalent to . . { llenllm,o } {[[fn[[~.o } the ongmal one and such that the sequences llenllm+ 1•0 and llfnll~+ 1 • 0 are nonincreasing. By Proposition 7.4.3 we may assume without loss of generality that co
llxl/m,o
=}; le~(x)l 1/en//m,o
(7.5.1)
n=1
and co
1/x//:n,o =}; lf~(x)l 1/fnll:n,o,
(7.5.2)
n=1
where {e~} ({!~}) denote the basis functionals with respect to the basis {en} (resp. to the basis {/n}). Let Um,o = {x: 1/x//m,o < 1} and Vm,o = {x: llxl/~,o < 1}. Clearly it is possible to find a subsequence {mr} and two sequence of positive scalars {ar} and {br} such that (7.5.3) Let and
1/x/1; = b; 11/xl/:n,,o. Of course Ur = {x: 1/x//r < 1} = arUm,,o and
Vr = {x: 1/x/1; < 1} = brVm,,o. Using the fact that the bases {en} and {fn} are regular, we can easily calculate the approximative diameters, .ll
(
Un1 Vr, Vs Take
t ~ s;
)
=
/Ifni/; /Ifni/; .
(7.5.4)
then, by (7.5.3), Ut ::> Vt ::> V8 ::> UB+ 1 and
~n(Us+l,
Ut)
~ ~n(Vs,
Vt).
(7.5.5)
Nuclear Spaces. Theory
335
Thus, by (7.5.4),
llenllt . 11/nll! llenlls+I ·~ llfnll; ' Take t
> s, then by (7.5.3) ~
l5n(Vt, Vs)
(7.5.6)
Vs :::> U8 +1 :::> Ut :::> V, and
l5n(Ut,
U:~J1).
7.5.7)
Thus, by (7.5.4),
llfnll; llfnlle
llenlls+l llenllt •
&
~
(7.5.8)
Therefore, by (7.5.6) and (7.5.8),
llenllt llfnll!
llenlls+l ll.fnll; '
&
~
(7.5.9)
for all t, s, n = I, 2, ... This implies that rn = s~p
llenllt llfnll! < +"9·
(7.5.10)
Thus, by (7.5.9),
llenllt ~ llrn.fnll! ~ Ilen litH· Therefore the bases {en} and {rnfn}e ar equivalent.
(7.5.11)
D
We say that two bases {en} and {.fn} are quasiequivalent (see Dragilev, 1960) if there is a permutation a of positive integers such that the bases {en} and {fa(n)} are semiequivalent. THEOREM 7.5.2 (Crone and Robinson, 1975). Let X be a nuclear B 0space with a regular basis {en}. Then each basis {fn} in X is quasiequivalent to {en}. The proof of the theorem is based on several notions and lemmas and propositions. LEMMA 7.5.3 (Kondakov, 1983), Let X= V(am,n). Assume that 1
llxllm ~ 2 m llxllm+t·
(7.5.12)
Chapter 7
336
Suppose that for an element f f'(f) = 1 and
supJif'll:r.+211fllm+l = m
X there is a functional f'
€
C
<
E
X* such that
+=,
(7.5.13)
where !I II~ denotes the norm of the functionals induced by the pseudonorm
II lim·
Then there are A > 0 and an index i such that
m= 1,2, ...
(7.5.14)
where {en} denotes the standard basis in Ll(am, 11 ). Proof To begin with, we shall show the first inequality. Let {e:} denote the basis functionals corresponding to the basis {e11 }. Then 00
00
00
~ leMf)lsup llenllm ~ ~ ~ le~(f)lllenllm
6
m
llfnllmH
= {: nJ;l Jle~(f)enllm =
6i
llfllmH 00
= r(.r e~(f)en) ~
n=l
6 ;8
llfllmH
~ llfllm ~ ~_I =
;8
IJfllmH
1
;8 2m
00
l, le'(f)llf'(en)l.
n=l
(7.5.15)
Comparing the series on the left and on the right, we find that there is an index i such that
lletllm If'( )j llfllm+l ~ et .
(7.5.16)
Putting A= f'(et), we obtain the first part of the inequality. By (7.5.13) we have
lf'(et)lllfllmH ~ llf'JI~+211etllm+21Jfllm+l ~ Clletllm+2· and we obtain the second part of the inequality.
D
PROPOSITION 7.5.4 (Kondakov, 1983; cf. Dragilev, 1965). Let {fs} be a basis in a space L 1 (am,n). Then there are a sequence of constants {as}, as > 0, a sequence of indices {ns} and a subsequence {II liP} of the sequence of standard norms such that
llen.IIP ~ asll.fsiiPH ~ l~n.IIP+2, where {en} denotes the standard basis in D(am,n).
(7.5.17)
Nuclear Spaces. Theory
337
Proof Let a:n,n = 22m sup at,n.
(7.5.18)
l~i~m
It is easy to verify that the space V(a~1 ,n) is isomorphic to the space V(am,n) and, for the space V(a~,n), (7.5.12) holds. Let X= V(a~,n).
Let {Is} be a basis in X. Let{!;} denote the basis functionals corresponding to the basis {fs}. By (7.5.12) for each pseudonorm II lim and each a> 0, there is a pseudonorm II llm1 such that aJJxllm ~ Jlxllm1• Thus for each index r there is an index m(r) such that
JJJ;(x)fsJJr ~ JJxJJm(r)
(7.5.19)
JJ/:JJm(r)JJ/sJJr ~ 1.
(7.5.20)
and Then there is a sequence of pseudonorms determining a topology such that (7 .5.13) holds for f = h and C = 1. D PROPOSITION 7.5.5 (Dragilev, 1965). Let X be a nuclear B 0 space with a regular basis {en}. Let {fn} be an arbitrary basis in X. Then {fn} can be reorderd in such a way that it becomes a regular basis. Proof By Proposition 7.4.7 and 7.4.8 the space X is isomorphic to the space V(JJenllm· Thus, by Proposition 7.5.4, for each s there are a8 and n8 such that (7.5.} 7) holds. Now we shall show that in the sequence {n8 } each index can be repeated only a finite number of times. Indeed, suppose that en., = en,,= en,. = ... Then the space X0 spanned by {i'& i'&,, fs., ... } is isomorphic to the space /. It leads to a contradiction, since X is nuclear and it does not contain any infinitedimensional Banach space. Hence we can find a permutation a of positive integers such that the sequence na(s)· is nondecreasing. Now we shall introduce new pseudonorms in X 1,
cc
JJx!Jo,p =
};
a; 1 JI.fs' (x)IIPllen,llP
s=l
Since the space X is nuclear, by (7.5.17) the sequence of pseudonorms
{II llo,P} determines a topology equivalent to the original one.
Chapter 7
338
Observe that
Jl/a(s)llo,m ll.fa<s>llo,m+l
llena<,>llo,m llena(slllo,mH .
(7.5.21)
Since a(s) is nondecreasing and the basis {en} is regular, by (7.5.21) the basis {fa(s)} is also regular. 0 Proof of Theorem 7.5.2. Let X be a nuclear E 0space with a regular basis {en}. Let {fn} be an arbitrary basis in X. By Proposition 7.5.5 there is a permutation a(n) such that the basis {fa(n)} is regular. Then, by Proposition 7.5.4 {fo.(n)} is semiequivalent to {en}, and this completes the 0 proof.
Theorem 7.5.2 was first proved by Dragilev (1960) for the space of analytic functions in the unit disc. He then extended it to nuclear spaces of the types d1 and d2 with regular bases (see Dragilev, 1965). Let d1, i = 1, 2 denote the class of spaces with regular bases belonging to di, i = 1, 2. PROPOSITION 7.5.6 (Zahariuta, 1973). Let X, Y be nuclear E0 spaces such that X E d1 and Y E d2. Then all bases in the product XX Yare quasiequivalent. Proof Let {e;}, {e;} be regular bases in X and Y. Let e2n_ 1 = (e;,o), e2n = (o,e;). In this way we obtain a basis in Xx Y. Let {In} be another basis in XX Y. By Proposition 7.5.4 there are a sequence of indices {nk} and a sequence of positive numbers {ak} and a sequence ofpseudonorms {II liP} determining a topology equivalent to the original one such that

Jlen.IIP ::;::;; llakfkllp+l ::;::;; llen.Jip+2.
(7.5.22)
Let X1 =: lin{fk: en. EX} and Y1 = lin {.fk: en.
E
Y}.
Let {h,s} be a subbasis of the basis {fk} consisting of those elements which belong to X1 • Let {,/;, 8} be a sub basis of the basis {/k} consisting of
Nuclear Spaces. Theory
339
those elements which belong to Y1 • By (7.5.I7), in the same way as in the proof of Proposition 7.·5.5, we infer that {h,s} and {As} are regular bases in X1 and in Y1 • Moreover, X1 e d2, Y1 e d1 • By Theorem 6.7.IO, this implies that there is an integers such that xl is isomorphic to x<s> and yl is isomorphic to yls>. Since we can shifts elements of the basis form X into Y if s > 0, and conversely from Y into X if s < 0, we can assume without loss of generality that s = 0. The bases {e;}, {e;} are regular; by Theorem 7.5.2 there are permutations u(s), u'(s) and sequences of positive scalars {a8 }, {a:} such that.h,s =as~ and As= a;e.;,(s)• which trivially D implies that the bases {en} and {/n} are quasiequivalent. 1
Theorem 7.5.6 for X= M(expm(n 1 + ... +nk)) and Y=M(exp
m
(n 1 + ... +nk) was proved by Dragilev (1970) and Zahariuta (1970). Zahariuta (1975) proved Theorem 7.5.2 for another important class of spaces. Let {a1 ,n}, {b1 ,n}, {a2,n}, {b 2 ,n} be four sequences of real numbers tending to infinity. We shall assume that there is a positive number c such that i =I, 2, i =I, 2.
Let n = (n1 ,n2). We shall consider two spaces, X= M(a~n, b~~~~) and Y = M(a~n, h;~~~). Zahariuta (1975) showed that if the diametral dimensions of the spaces X and Yare equal, t5 (X) = t5 ( Y), then the spaces X and Y are isomorphic and, what is more, that there is a permutation of indices u such that {fn} is equivalent to {ea(n)}, where {en} and {fn} are standard bases in X and respectively in Y. As a consequence of this fact, it is possible to obtain the following results, arrived at independently by Djakov (I974) and Zahariuta (1974). Let X= M(exp(!
nf+mn~)) andY= M(exp(! nf'+mnf.))
Then the spaces X and Yare isomorphic if and only if I
1
1
I
+=+p q p' q'"
Chapter 7
340
Moreover, if X and Yare isomorphic, there is a permutation u of indices such that the bases {fn}, {ea(n)} are equivalent, where {en} and {/n} are standard bases in X and Y respectively. There are also other classes of spaces in which all absolute bases are quasi similar. Let H be a Hilbert space and let A be a selfadjoint positively defined operator acting in H. Let
(oo
oo)
Let Ha be the completion of the domain D A of the operator A with respect to the norm llxlla = (x,x)a. The system of spaces {Ha} ( oo
[[x[[m
=
.J:
(nlfm[xn[)P,
n~l
with the topology defined by the sequence of phomogeneous £pseudonorms {II lim} (see Example 1.3.9). It is easy to verify that LP(nlfm) are Schwartz spaces, and thus Monte! spaces. This means that every closed
FNorms and lsometries in FSpaces
397
bounded set is compact. Moreover, the space LP(n 1/m) has a total family of linear continuous functionals. Thus, by Remark 5.5.6, every compact set has an extreme point. Hence the spaces LP(nIfm) have the strong KreinMilman property. The spaces LP(nlfm) are not locally convex. Indeed, let en = {0, ... . . . , 0, I, 0, ... }. The sequence {en} tends to 0, because nth place
l!enllm = (n 1 /m)P +0 form=I,2, ... On the other hand,
;;?:
n(~ n1/m )P =
nlpplm
1 provided p(I + Ifm) < 1. Therefore the sequence { e
+oo,
+ ...n +en} is
not
bounded. This implies that the spaces LP(n_ 1,m) are not locally convex. The important class of spaces, namelyLP[O,l], 0 < p < I, do not have the strong KreinMilman property. For this reason we shall prove THEOREM 9.3.12 (Rolewicz, 1968). Let (X, II llx) and (Y, II IIY) be two real locally bounded spaces. Suppose that the norms II llx and II IIY are concave, i.e., for all x EX, y E Y, the functions lltxllx and lltYIIY are concave for positive t. Then every rotation mapping X onto Y is a linear operator. Proof Let r be a positive number such that the set K2, = {x EX: llxll ~ 2r} is bounded. Such an r obviously exists, since the space X is locally bounded. Using the concavity of the norm, we shall show that sup llxllx < r.
(9.3.11)
J12x[[x.;;r
Suppose that (9.3.11) does not hold. Then there is a sequence {xn} of
I llx +r. Therefore
elements of X such that llxnllx = r and x;
r
Chapter 9
398
The concavity of the function [[tx[[x implies that [[anxn[[x ~ 2r. This leads to a contradiction, because the set K 2r is bounded. If the set K 2 r is bounded, then the set K 28 is also bounded for all s, 0 < s < r. Let n(r) = sup [[x[[x. The function n(r) is continuous and ll2xllx.;;;r
it is strictly increasing, provided r lows from (9.3.11). Let us define by induction
rn=n(rn1), Obviously, by (9.3.I), r 0 We shall show that
<
r 0 , where K 2r, is bounded as fol
n= I,2, ...
>
r1
>
r2
> ... > rn > ... (9.3.I2)
1imr11 =0. T+00
indeed, suppose that (9.3.I2) does not hold, i.e. that r' = lim r11
> 0.
(9.3.13)
11+00
Since n (r) is strictly increasing, n (r') < r'. The continuity of the function n(r) implies that there is an r > r' such that n(r) < r'. By the definition of r' there is a positive integer n such that r11 < r. Hence n(rn) < n(r) < r'. This leads to a contradiction, because n(rn) = r11 +1 ~ r'. Let x' andy' be two arbitrary elements of X such that [[x'y'llx < r 0/2. Let H 0 = {xe X: [[xx'[[x
<
n(;)
and [[xy'[[x
Obviously fl(H0) = sup [[xy[[x x,yEH,
~ 2~( ; ) < r
0•
We define by induction Hn = {xE Hn1: [[xy[[x ~ rn for all yE Hn1} (n =I, 2, ... ).
We shall prove by induction that the sets Hn are not void and are such that x'+y' 2EHn,
n
=
0, I, 2, ... ,
if x E Hn, then .X= x'+y'xe H 11 , n = 0, I, 2, ... , fl(Hn) ~ r11 , n = 0, 2, 2, ...
(9.4.I4.i) (9.3.I4.ii) (9.3.I4.iii)
FNorms and Isometries in FSpaces
399
For n = 0 this is trivial, since xx' = y'x and xy' = x'x. Suppose that the above holds for a certain k1. Then (9.3.14.iii) implies that (9.3.14.i) holds for n = k. Hence the s'et H1e is not empty. By the definition of H~e b(H~e) = sup 1/xyJJx :s;;; sup JJxyJJx :s;;; rn, x,yEHk
xEHk yEHie1
i.e. (9.3.14.iii) holds for n = k. Let x nition of H~e, llxyllx
E H~e
andy
E H~c_ 1 •
= llyxJJx:s;;;rn.
Then, by defi(9.3.15)
(9.3.15) implies that (9.3.14.ii) holds for n = k. Since rn+0, the intersection of all sets Hn is the set which consists of one element (x' +y')/2. It is a metric characterization of the centre of the points x' andy'. We can apply a similar reasoning to the space Y. Since we have defined the point (x' +y')/2 in the metric language, this implies that there is a number a > 0 such that if Jlx Yllx < a then, for each isometry U,
u( x~y)
=
U(x)~U(y)
This implies that if llxllx 2U(kx)
=
<
•
a/2 then
U((k+1)x)+U((k1)x).
(9.3.16)
Basing ourselves on formula (9.3.16), we shall show by induction that U(nx)
=
nU(x),
n
=
1, 2, ...
(9.3.17)
Putting k = 1 in formula (9.3.16), we find that (9.3.17) holds for n = 2. Let us suppose that (9.3.17) holds for n = m and let us put k =min (9.3.10). Then the induction hypothesis implies 2mU(x) = 2U(mx) = U((m+ 1)x)+ U((ml)x) = U((m+1)x)+(m1)U(x).
Hence U((m+1)x)
= (m+1) U(x),
and we have proved (9.3.17).
Chapter 9
400
Let x andy be arbitrary elements of X. Obviously, there is a positive integer n such that llxfnllx < a/2 and llyfnllx < a/2. Therefore U(x+y)
u( 2n(~~y)) = x~y) =22n(u(:)+u(~))=
=
2nu(
U(x)+U(y).
Hence the operator U is additive. Since it is continuous and the spaces are real, it is a continuous linear operator. D We do not know whether Theorem 9.3.12 is true for arbitrary norms in a locally bounded space. We can only prove THEOREM 9.3.13. Let X and Y be two locally bounded real spaces. Let U be a rotation such that lltU(x)=tU(y)IIY = lltxtyllx for all positive t. Then the operator U is linear. Proof Since no confusion will result, we shall denote the two norms II llx and II llr by the same symbol II 11. Let llxll* = sup lltxll and llxlf** O
sup
=
(bb= 1 1[ax+bl=a llbxll**) (cf. Lemma 9.1.2). llxll** is a norm
a a equivalent to the original one and the function lltxll** is concave for positive t. It is easy to verify that fiU(x) U(y)ll** = llxyll**. Therefore by TheoD rem 9.3.12, the operator U is linear. b>l>a>O
9.4.
ISOMETRICAL EMBEDDINGS IN BANACH SPACES
Let (X, I fix), (Y, II lfy) be two real Banach spaces. Let U be an isometry mapping X into Y. If U is not a surjection (i.e. U does not map X onto Y), then U need not be linear, as can be seen from the following Example 9.4.1 Let X be the space of real numbers R with norm llxll = lxl and let Y be a twodimensional real space RxR with norm lf(x,y)ll = max(lxl, lyl).
FNorms and Isometries in FSpaces
401
Let U be a mapping X into Y given by formula U(x) = (x,sinx). It is easy to verify that U is a nonlinear isometry. However, for Banach spaces the following substitute of Theorem 9.3.12 holds: THEOREM 9.4.2 (Figiel, 1968). Let X andY be two real Banach spaces. Let U be an isometry (nonnecessarily linear) mapping X into Y and such that U(O) = 0. Let the linear hull of the set U(X) be dense in the space Y. Then there is a continuous linear operator F mapping Y into X and such that the superposition FU is an identity on X. The operator F is uniquely determined and it has norm one. The proof of Theorem 9.4.2 is based on the following notions and lemmas. Let X be a Banach space with a homogeneous norm llxll Let Sr
= {x: llxll = r}.
We say that a point x E Sr is a smooth point if there is only one continuous linear functionalfx of norm one such thatfx(x) = r. PROPOSITION 9.4.3 (Mazur; see Phelps, 1966). If (X, II II) is a separable Banach space, then the set S 8 of all smooth points is a dense Gbset in X. Proof In the case of a complex Banach space X, we consider it as a linear space over reals. The set of all smooth points S 8 remains unchanged. Therefore we can restrict ourselves to the real Banach spaces. As can easily be verified, it is sufficient to show that S 8 n sl = s; is a dense G(j set in S 1 • Let {xn} be a dense subset of S 1 • For each positive integer m, let Dm,n = {xE: S1 : llf(xn)g(xn)ll < l/m for all continuous linear functionals j, g such that 11/11 = f(x) = 1 = g(x) = llgll}. Of course, if xis a smooth point, then there is only one functional/ such that 11/11 = f(x) = 1. Therefore, llf(xn)g(xn)ll
=
1
0 "'(  , m
n, m = 1, 2, ...
Chapter 9
402
This implies that
n 00
Ss c
m,n=I
(9.4.1)
Dm,n·
On the other hand, if x is not a smooth point, then there are at least two different continuous linear functionals f, g of norm one such that /(~) = g(x) = 1. Since the set {xn} is dense inS~> sup lf(xn)g(xn)l
=
llfgll.
n
Hence, there are such m and n that x ¢ Dm,n· This implies 00
Ss
:::>
n Dm,n·
(9.4.2)
n,m=l
(9.4.1) and (9.4.2) imply
n 00
Ss =
Dm,n.
n,m=l
We shall show now that for every n, m the set Dm,n is open. Let Yk E sl""Dm,n andyk +y. Sinceyk E S""'Dm,n, therearefunctionals f~cand g~c
such that k
=
1, 2,...
(9.4.3)
and k = 1, 2, ...
(9.4.4)
By the Alaoglu theorem (Theorem 5.2.4) the sequence {f~c} and {g~c} have cluster pointsjand g respectively. By (9.4.3)
IIIII = lf(y)J = 1 = Jg(y)J = llgJ!, and by (9.4.4)
Jf(y)g(y)J
>
1/m.
This implies that y ¢ Dm,n· Then the set Dm,n is open. For the completion of the proof it is enough to show that for each nand m the set Dm,n is dense in S 1 . Suppose that the above does not hold. Then there are such n, m, y e S1 and b > 0 that, if x e S1 and llxyll
FNorms and lsometries in FSpaces
403
x ¢ Dm,n· Let us write y1 = y. Then there are continuous linear functionals / 1 and g1 that and
h(x) ~ g 1(xn)+ Ifm.
Now we shall choose by induction a sequence {Yk} of functionals {/k} and {gn} of norm one such that IIYIYkll /k(Yk)
=
< I
(l2k)b,
= k
gk(Yk),
.
/k(Xn) ?= +gl(xn).
m
c
S 1 and sequences
(9.4.5.i) (9.4.5.ii) (9.4.5.iii)
For k = I conditions (9.4.5.i)(9.4.5.iii) are satisfied. Suppose that they hold for a certain k. Then
where a is chosen so small that (9.4.5.i) holds. Then Yk+1 ¢ Dm,n and there are continuous linear functionalsfk+ 1 , gk+ 1 such that (9.4.6) and . I fk+I(xn) ?= m +gk+l(xn)
We have (9.4. 7) Since (9.4.8) we have (9.4.9) By (9.4.7) and (9.4.9) /k(Xn) ~ gk+l(Xn).
(9.4.10)
Chapter 9
404
Then, by the induction hypothesis, 1 1 fk+I(xn) ?= +gk+I(xn) ?= +fk(Xn) m m k+1 ?= +gl(xn). m
(9.4.11)
Hence (9.4.5.iii) holds. Let us observe that this leads to a contradiction, because
lfk+l(Xn)l ~ ll.fic+JIIIIxnll
=
1.
D
LEMMA 9.4.4. Let U be an isometry of the space ofreals R with the standard norm lxl into a Banach space (Y, II II). Let U(O) = 0. Then there is a continuous linear functionalf E Y of norm one such that
f(U(x))
=
x.
Proof Let n be an arbitrary positive integer. The HahnBanach theorem implies that there is a continuous linear functional.fn of norm one such that fn(U(n)U(n)) = IIU(n)U(n)ii = 2n. Thus, for every t, it I < n, we have
2n = lntl+it(n)i = II U(n) U(t)ii+IIU(t) U( n)ii ?= fn(U(n) U(t))+fn(U(t) U( n)) =fn(U(n)U(n)) = 2n.
(9.4.12)
Therefore, in formula (9.4.12) the equality holds, and this implies
fn(U(t)U(n)) = IIU(t)U(n)ii = t+n.
(9.4.13)
Putting t = 0 in (9.4.13) we obtain.fn( U( n)) = n. Thus
fn(U(t)) = t.
(9.4.14)
The Alaoglu theorem (Theorem 5.2.4) implies that the sequence {.fn} has D a cluster point/ Formula (9.4.14) implies thatf(U(t)) = t. LEMMA 9.4.5. Let x be a point of the Banach space X. Let a be a smooth point of the set S 11a 11 = {x: iixii = llall}. Let fa be afunctional ofnorm one such thatfa(a) = llaii· Letfa(x) # 0. Then there is a real t such that
ila+txil
< llall.
(9.4.15)
FNorms and Isometries in FSpaces
405
Proof Suppose that (9.4.15) does not hold, i.e., for all real t, Jla+txJJ ~ llall.
(9.4.16)
Sincefa(x) ::;C 0, x ::;C 0. Formula (9.4.16) implies that a and x are linearly independent. Let X 0 denote the space spanned by a and x. The formula
g(ax+f3a) = f311all defines a continuous linear functional on X 0 • Formula (9.4.16) implies that Jlgll = 1. Since g(a) = llallg is a restriction of the functional fa into X 0 ,fa(x) = g(x) = 0 and we obtain a contradiction. 0 LEMMA 9.4.6. Let U be an isometry of a Banach space X into a Banach space Y such that U(O) = 0. Let a be a smooth point of the sphere S 11 a 11 • Let f E Y* be a continuous linear functional of norm one such that, for all real r, (9.4.17) f(U(ra)) = rJ!aJ!. Then (9.4.18) f(U(x)) = fa(x).
Proof Let x, y EX. We have lf(U(x))f(U(y))J
=
~
Suppose that for a certain p
E
lf(U(x) U(y))J IIU(x)U(y)ll = llyxJI.
(9.4.19)
X
(9.4.20)
fa(P) ::;Cf(U(p)). Let us write
1X
1
=
f(U(p)) ifaij·
Then (9.4.20) implies
fa(p1Xa) ::F 0.
(9.4.21)
By Lemma 9.4.5 there is a real t such that
l!IXa+t(pa)Jl <
!JIXa!!.
(9.4.22)
It is clear that t ::;C 0. Let us put {3 = IX/t. Then (9.4.22) implies
llp(1X{3)all ~ llf3all.
(9.4.23)
Chapter 9
406
By (9.4.17), (9.4.19) and (9.4.23) we obtain [I,Bcx[[
=lex! [[a[[(cx.B)[[a[[ = [f(U(p))f(U((ap)a))[ ~
[[U(p) U((atJ)a)[[ = [[p(atJ)a[[
<
[[tJa[[
which leads to a contradiction.
0
9.4.7. Let X be afinitedimensional real Banach space. Let U be an isometric embedding of the space X into a Banach space Y such that U(O) = 0. Then there is a continuous linear operator F mapping the linear hull of the set U(X), lin U(X), onto X and such that
LEMMA
F(U(x))
= x.
The operator F is uniquely determined and it is ofnorm one. Proof Let us denote the dimensional of X by n. Let a 1 , ••• , an be linearly independent smooth points. Such points exist, as follows from Proposition 9.4.3. Let fa. be a continuous linear functional of norm one such that i
=
I, 2, ... , n.
Without loss of generality we may assume that at are chosen in such a way that the functionals {fa,}, i = I, 2, ... , n, are linearly independent. Let {.fi} be continuous linear functionals of norm one such that i=l,2, ... ,n
for all real r. The existence of such functionals follows from Lemma 9.4.4. Lemma 9.4.6 implies that .fi(U(x))
= fa,(x),
i=l,2, ... ,n.
(9.4.24)
The mapping G of the space X into Rn given by the formula G(x)
= {fa,( X), ... , la.(x)}
(9.4.25)
is an isomorphism, because the linear functionals {fa,, ... JaJ are linearly independent. Now we define an operator F mapping lin U(X) into X as follows : F(y)
=
G 1(
{ /1
(y), ... ,/k(y)}).
FNorms and Isometries in FSpaces
407
It is obvious that Fis a continuous linear operator. For x EX we have by (9.4.24) and (9.4.25) F(U(x))
= G 1 ( {.h(U(x)), ... ,.fn(U(x))}) = G 1 ( {.fa,(x), ... Jan(x)}) = x.
(9.4.26)
Formula (9.4.26) implies that the operator F is uniquely determined on U(X) and thus on lin U(X). We shall now show that IIFII = 1. Let x E Ybe such a point that F(x) is a smooth point. By Lemma 9.4.4 there is a continuous linear functional f of norm one such that for all real r f(U(rF(x)))
=
riiF(x)ll·
(9.4.27)
By Lemma 9.4.6 (9.4.28)
f(U(y)) =fF(x)(y).
Therefore, the superpositions f(U( · )) and f(U(F( · ))) are continuous linear operator Moreover, we have by (9.4.26) f(U(F(U(y))))
= f(U(y))
for ye X.
Hence, for x = U(y) such that F(x) is a smooth point, llxll
> lf(x)l =
(9.4.29)
lf(U(F(x)))l = IIF(x)ll.
Since F is a continuous linear operator, the set of all x such that F(x) is a smooth point of dense in X (cf. Proposition 9.4.3). Therefore, by (9.4.29), IIFII ~ 1. on the other hand, by (9.4.26), IIFII ;;::, 1. Thus IIFII = 1. D Proof of Theorem 9.4.2. Let {Xn} be a sequence of finitedimensional 00
spaces such that dimXn
=
n, Xn C Xn+l and the set Z
=
U Xn is dense n=l
in X. By Lemma 9.4.7 there are continuous linear operators Fn, IIFnll = 1 mapping lin U(Xn) into X and such that Fn(U(x))
=
x
for
XE Xn,
n = 1,2, ...
In view of the uniqueness ofthe operators, Fn, the operator F(y)
= Fn(Y)
for
y E lin U(Xn)
Chapter 9
408
is a continuous linear operator of norm one, well defined on lin U(Z). The extension of the operator F to the closure lin U(Z) = lin U(X) has the required properties. D
9.5. GROUP OF ISOMETRIES IN FINITEDIMENSIONAL SPACES Let (X, II II) be a finitedimensional real Fspace. Let G(ll II) denote the set of rotations mapping X into itself. By Theorem 9.3.4 all those rotations are linear. It is easy to verify that G(ll II) is a group with the superposition of operators as the group operation. THEOREM 9.5.1 (Auerbach, 19331935). Let (X, II II) be an ndimensional real Fspace. Then there is an inner produc((x,y) defined on X such that G(ll II) C G(ll 11 1), where llxll 1 = (x,x/ 12 is the norm induced by the inner product (x,y). Proof Let K = {x: llxll ~ 1}. Of course, for any isometry, U E G(ll II), U(K) = K. Let E denote an ellipsoid with the smallest volume containing K. We shall show that this ellipsoid is uniquely determined. Indeed, let E 1 be another ellipsoid with the smallest volume containing K. Let
E
{x: (x,x)
=
<
1}
and
E1
=
{x: (x,x) 1 ~ 1},
where the inner products (x,y) and (x,y)1 are determined by Eand E 1 . Let a be an arbitrary real number contained between 0 and 1. Let
Ea
=
{x: (x,x)a
~
1},
where (x,y)a = a(x,y)+(1a)(x,y) 1 • Of course Ea is an ellipsoid. Moreover, K is contained in Ea. Indeed, since K C En E1 if x e K, then (x,x) ~ 1 and (x,x) 1 ~ 1. Therefore
(x,x)a
= a(x,x)+(1a) (x,x)
1
~
1
and
xe Ea.
Since the ratio of the volumes is an invariant of affine transformations, we may assume without loss of generality that E is a ball in the usual Euclidean sense and E 1 is an ellipsoid with the equation b 1 x~+ ... +bnx!
Since vol(E)
=
<;:
1,
i = 1, ... , n.
vol(E1),
b1 ... bn
=
1.
(9.5.1)
FNorms and Isometries in £Spaces
409
The volume of the ellipsoid Ea is expressed by the formula vol (Ea)
=
Cn([a+(la)b1]
•..
[a+(la)bn]) 112 ,
where Cn is a constant dependent on n. Let us observe that the function vol (Ea) reaches its minimum at those points at which the function V(a)
=
(a+(la)b 1)
••.
(a+(la)bn)
reaches its maximum. The function V(a) is differentiable, and by a simple calculation we get n ~
d
da V(a) =
V(a)?
1bt a+(la)bt .
•=1
We have assumed that E and £ 1 are ellipsoids with the minimal volume. d Therefore, the derivative da V(a) should be equal to 0 for a= 0 and
a
=
1. The second case implies
b1 + ... +bn
=
n.
(9.5.2)
Equations (9.5.1) and (9.5.2) have a unique solution b1 = b2 = ... = bn = 1 within positive numbers. This follows from the fact that the function W(b) = b1 ..• bn(b1 + ... +bn) has only one extremal point b1 = ... = bn = 1 and it is positive for other positive b. Therefore the ellipsoid E is uniquely determined by K. Thus any isometry U E (}maps E onto itself. D
9.6.
SPACES WITH TRANSITIVE AND ALMOST TRANSITIVE NORMS
Let (X, II IJ) be an Fspace. By G(ll IJ) we denote the group of rotations mapping X onto itself. A norm II II is called maximal if, for any equivalent norm II 11 1 such that G(ll IJ) C G(ll 11 1), we have equality G(JI ID = G(ll 1/J. 9.6.1. Infinitedimensional real Fspaces the maximal norms are induced by inner products. Proof The proposition is an immediate consequence of a theorem of Auerbach (Theorem 9.5.1). D PROPOSITION
Chapter 9
410
Let (X, II ID be an Fspace. The norm llxll is called transitive if, for all positive rand each x EX, llxll = r, {A(x): A
E
G} = {y: IIYII = r},
(9.6.1)
where G(ll ID denotes the group of rotations with respect to the norm llxll. A norm llxll is called almost transitive, if for each positive r and for all x E X of norm r, {A(x): A
E
G(ll ID} = {y: IIYII = r}.
(9.6.2)
If the space X is locally bounded and the norm llxll is phomogeneous, then the conditions of transitivity and almost transitivity are simpler, namely, it is sufficient that (9.6.1) (respectively (9.6.2)) should hold for one fixed r, for example r = 1. Strictly connected with transitive norms is the classical problem of Banach.
Problem 9.6.2 (Banach, 1932). Let (X, I ID be a separable Banach space with a transitive homogeneous norm llxll· Is X a Hilbert space and is llxll a Hilbert norm? As we shall show later, without the assumption of separability the answer is negative. THEOREM 9.6.3 (Pelczynski and Rolewicz, 1962). Let (X, II II) be an Fspace and let llxll be an almost transitive norm. Then the norm llxll is maximal. Proof Let llxll 1 be a norm equivalent to the norm llxll and such that G(il ID C G(ll lit). Let
llxll~ =
Jiitxilt dt. 0
The norm llxll~ is equivalent to the norm llxllt and such that the function litxll~ is strictly increasing for positive t (cf. Theorem 9.1.1). It is easy to verify that a linear isometry U with respect to the norm llxllt is also an isometry with respect to the norm llxll~ Hence, G(ll lit) C G(ll II~). Since
FNorms and lsometries in FSpaces
411
the norm llxll is almost transitive for any fixed x 0 such that llxoll llxoll~ = r1, we have S(r)
=
{x: llxll
C {A (xo): A
E
=
r}
=
rand
=
{A(xo): A E G(ll II)}
G(ll II~)} C {Y: IIYII~
=
r1}
=
S'(r1).
(9.6.3)
Since the function lltxll~ is growing for positive t, (9.6.3) implies that S(r)
=
S'(r1).
(9.6.4)
Hence the isometries belonging to G(ll II) map S(r) onto itself. This implies that G(llll~) C G(ll II), i.e. G(ll II) is a maximal group of isomet
0
~.
THEOREM 9.6.3 (Pelczynski and Rolewicz, 1962). In the spaces LP[O, 1], 1
< p < =,
the standard norm 1
llxll
=
(J lx(t )IPdt )11
P
0
is almost transitive. Proof Letf(t) E LP[O, 1] be a function of norm one such that a= inf f(t)
> 0.
(9.6.5)
0
Let us consider the operator t
T,(x) = x(F(t))f(t),
where
F(t)
=
Jlf(t)IPdt. 0
The operator Tt is an isometry acting in the space LP[O, 1]. Indeed, 1
Jlx(F(t))f(t)l Pdt = Jlx(F(t))IPdF(t)
ll1f(x)IIP =
0
1
=
(9.6.6)
llxiiP
0
because F(t) is a strictly increasing function such that F(O) F(l) = 1. Let us observe that by (9.6.5) the inverse isometry
1
Tj'(y) = y(F1(t)) J(F 1(t))
=
0 and
(9.6.7)
Chapter 9
412
is well defined on the whole LP[O, 1]. Therefore, the isometry Tt maps LP[O, 1] onto itself. Letfand g be two arbitrary elements of norm one. Let 8 be an arbitrary positive number. Obviously, there are two functions f' and g' such that inf /'(t) > 8/4, inf g'(t) > 8/4, 11//'11 < 8/2, [[gg'[[ < 8/2. O
=
O
Let U = Tg,Tj 1• The operator U is an isometry. Since lj.(l) f', Tu,(1) = g', we get Uf' = g'. Hence
IIU(f)g[[ < [[U(/)U(f')II+IIU(f')g')ll+llg'g[[ <
8.
The arbitrariness of 8 implies that the norm xis almost transitive.
D
Another example of a separable Banach space (X, II [I) such that the norm I II is almost transitive are Gurarij spaces. Gurarij (1966) gave an example of a Banach space (X, II [I) with the following extension property. For each finitedimensional subspaces £, F, E C F C X and for an operator T: £+X and each 8 > 0, there is an extension T8 ofT, T8 : F+X such that
(18)[[x[[ <; [[Te(x)[[ <; (l +8) [[x[[. Lusky (1976) showed that all Gurarij spaces are isometric and that the norms in a Gurarij space is almost transitive. Later Lusky (1979) showed that every Banach space (X, II [I) is a subspace of a space (Y, II [I) (its norm being an extension of [[ [I) such that in Y the norm II II is almost transitive and there is a continuous projection of Yonto X. THEOREM 9.6.4 (Pelczynski and Rolewicz, 1962). In the spaces LP[O, 1], 0 < p <; I, the standard norm 1
[[x[[
=
J[x(t)[Pdt 0
is almost transitive.
The proof follows the same lines as the proof of Theorem 9.6.3 ; only in formula (9.6.6) it is necessary to replace [[Tt(x)[[P by [[T,(x)[[ and [[x[[P by[[x[[.
FNorms and Isometries in £Spaces
413
9.6.5 (Pelczynski and Rolewicz, I962). In the spaces LP[O, I], +oo, the standard norms are maximal.
CoROLLARY
0
<
PROPOSITION 9.6.6 (Pelczynski and Rolewicz, 1962). There is a normed (noncomplete) separable space with a transitive norm which is not a preHilbert space. N
Proof Let LP, 1 ~p ~ +oo, be a linear subset of the space LP[O, I] consisting of all those functions which vanish close to the point I. The space
LP with the standard norm is the required space. Let x and y be two elements of LP of norm one. We shall construct a rotation which maps x ony.
Let hx be a transformation of the closed interval [0, I] which preserves measures, i.e., lhx(E)I = lEI for each measurable set E and, moreover, hx maps the sets Ea = {t: lx(t)l >a} onto the intervals [O,IEal). Let Uxf = f(hx(t)).
Obviously, Ux is an isometry and maps the function x(t) on a function x'(t) such that x'(t) is a nonincreasing function and
>0 lx'(t)l { = 0
for for
0 < t
where ax= ]{t: x(t) # 0}1. Since x E LP[O, 1], ax< I. In the same way we construct an operator Uy mapping y on y' with the same properties as x'. Let us observe that the operator TT 1 defined by formula (9.6.7) is well defined also if we replace condition (9.6.6) by the condition f(t) =!= 0. Indeed, the operator TT 1 is well defined on the set X0 : X0 = {x E LP[O, I]: there is a positive a such thatf(t)
ux'
uy
Chapter 9
414
tions x" and y", where a1fp
for 0 < t
x"(t) = { o"'
a1fp
y"(t) = { 011
Let U be an operator defined by the formula
U(f)
=I( :J't[; t)
for 0
( lay) !((lay)(lt)+I) lax Iax
~ t
foray
It is easy to verify that the operator U is an isometry and that U(x")
= y".
u; 1 U;/UVc,Ux maps the element X on y.
D
Hence the isometry A
=
PROPOSITION 9.6.7 (Pelczynski and Rolewicz, 1962). There is a nonseparable Banach space with a transitive norm which is not a Hilbert space. Proof Let Q be the product of a non~countable set A by the closed interval [0, I], Q = A X [0, 1]. Let I: be the algebra of all such subsets E of Q that for each a E A the set Ea = En ({ax [0, 1]) is measurable in the Lebesgue sense. Let a measure J1 be defined by the formula J1(E) =
_2; I(Ea)l, aEA
where I I is the Lebesgue measure. If x = x(a, t), a E A, t E [0, I] is an element of LP(Q, 1:, Jl), then the support Sx= {aEA: l{t: x(a,t):;FO}I} of the element xis at most countable. We shall show that the standard norm in LP(Q, 1:, Jl) is transitive. Let x = x(a,t), y = y(a,t) be arbitrary elements of LP(Q, 1:, Jl) of norm one. Let {an} be a sequence containing the supports of x andy. Let {A.n} be a sequence disjoint with {an}. Let us take a sequence {/1n} such that /1 2nl =an and /1 2n = An. Let U be an operator defined in the following way: g(p, t)= U(f(a, t)), where g(a, t) = t(a, t) for a =F Pn, · {2 11Pf(p2 n1,2t) g CPn, t) = 21/pl'( R2n, 2t) J fJ
for 0 ~ t
FNorms and Isometries in FSpaces
415
Let us observe that. the operator U is an isometry and that it maps x on x' and y on y', so that x'(a, t), y'(a, t) belong, for any fixed a, to the space LP[O, 1] defined in the preceding proposition. Thus, by Proposition 9.6.6, there is an isometry V mapping x' on y'. Hence, the isometry u 1 VU maps x on y. Hence the standard norm in the space LP(Q, E, tt) is transitive. If p =F 2, the space LP(Q, E, tt) is not isomorphic to a Hilbert space. D
9.7. CONVEX TRANSITIVE NORMS
In this section we shall consider only Banach spaces with homogeneous norms. Let X be a Banach space with a homogeneous norm JJxJJ. The norm 1/x/1 is called convex transitive if, for each element x of norm one, conv {A(x): A
E
G(/1 //)} = {x: Jlxll ~
THEOREM 9.7.1 (Pelczynski and Rolewicz, 1962). Let X be a Banach space. Each convex transitive norm 1/x/1 equivalent to the original one is maximal. Proof Let 1/x/11be a norm equivalent to 1/x/1. Let G(/1 //) C G(/1 1/ 1). Then
1/A(x)/1 = 1/x/1
for all
A
E
G(/1 1/1).
(9.7.2)
Suppose that (9.7.2) does not hold. Then there are A 0 E G(/1 1/ 1) and x EX, 1/x/11= 1 such that I/A 0(x)ll =F JJxjl. Without loss of generality we can assume that IIA 0(x)ll > llxll· Indeed, otherwise we might consider the operator A 0 1• Let r=
Since
iiAo(x)ll. llxll
G(ll II) C G(il 11 1) and the norm llxll is convex transitive, {y: IIYII1 ~ 1} = conv {A(x): A E G(ll ll1)} = conv {A(A 0 (x)): A E G(ll ll1)} :) conv {A(A 0 (x)): A E G(ll II)} = {y: IIYII = rllxll}.
Repeating this argument for the element rx we can prove that and by induction llrnxll 1 ~ 1 for n = 1, 2, ...
llr2xl/1 < 1
416
Chapter 9
Since r > 1, this leads to a contradiction. Therefore (9.7.2) holds and this implies that A E G(ll II), i.e. G(ll II)=G(II III). D 9.7.2 (Cowie, 1981). Let (X, II II) be a Banach space. Suppose that II II is not convex transitive. Then there is a norm II 111 equivalent to the norm II II and such that (X, II II) and (X, II 111) are not isometric and PROPOSITION
G(ll II)
c
G(ll III).
Proof Take any x 0 such that llxoll = 2. The set
U = conv ({x: llxll ~ 1} u {Ax 0 : A
E
G(ll II)})
is open convex and invariant under the group G(ll II). Let II 111 be the norm induced by the set U. Since U is G(ll II) invariant, G(ll II) C G(ll ll1). On the other hand U is not a ball of any radius in the norm II 11. Hence (X, II II) and (X, II 11 1) are not isometric. D Let X= C(.Q"'.Q0) (see Example 1.3.4) be the space of all continuous function defined on a compact set .Q and vanishing on a compact subset .Qo with the standard norm llxll = sup lx(t)l
(9.7.3)
tED
The setS= .Q"'.Q0 is locally compact. Thus the spaceC(.Q"'.Q0)can be considered as the space of continuous functions defined on a locally compact set Svanishing at infinity with the standard norm (9.7.3). For brevity, we shall denote this space C 0 (S). In the space C 0(S) each rotation U of the space onto itself is of the form U(f) = a(t)f(b(t)), (9.7.4) where a(t) is a continuous function of modulus one and b(t) is a homeomorphism of S onto itself. Indeed, the conjugate space to the space C 0 (S) is the space M(S) of measures defined on Borel sets (see Example 4.3.3). The isometry Uinduces the isometry U* in the space M(S). Of course, U* maps the extreme points of the unit ball on extreme points. The extreme points of the unit ball in M(S) are measures concentrated at one point multiplied by scalars of modulus one.
FNorms and Jsometries in FSpaces
417
This implies that Uis of the form (9.7.4). The continuity of a(t) follows from the fact that U maps continuous functions on continuous functions. The same fact implies the continuity of b(t). Since U maps C 0 (S) onto C 0 (S), b(t) is a homeomorphism. THEOREM 9.7.3 (Wood, 1981). Let C~(S) be the space of complexvalued functions defined on a locally compact set S vanishing at infinity with the standard norm
llxl/
=
sup lx(t)j. tES
The norm II II is convex transitive if and only if the group of homeomorphism
r
of the set
s
is almost transitive on S, i.e., that for each t
E
S, Ft
= {b(t): bE F} J S. Proof Necessity. Suppose that F is not almost transitive on S. Then
there is an s0 E S such that Fs 0 does not contain S. Let U C S be an open set disjoint with Fs0 • Take any f ::j: 0 f E C~(S) such that the support of fis contained in U. Thus f(Fs 0 ) = 0. Hence, by the form of isometry (9.7.4), for each g E {Af: A E G(ll II)}, g(s0 ) = 0 and by definition the norm II II is not convex transitive. Sufficiency. Take any norm II 11 1 such that G(ll ID C G(ll 11 1). We shall show that there is a c > 0 such that llxll = cllxll 1 • Indeed, let II II' denote the norm conjugate to the norm II 11 1 defined on M(S). Let 08 denote the unit measures concentrated at the points. Since a homeomorphism b(t) induces isometries in (C~(S), II 11 1), it also induces isometries in the space (M(S), II II'). Therefore IIOs\l' = IIOb(s)ll' for all s E s and all homeomorphisms b E Fix s E s. Take any t E s. Since the orbit Fs is dense inS, Ot belongs to the closure in the weak*to. pology of {o1 : tE F 8 } Hence
r.
liM'~ sup {[[Ob(•)ll': bE F} = 1\0sll'. (9.7.5) Exchanging the roles oft and s, we find that liM' are all equal. Let c = (IIOtl[') 1 • Then
1Jxll 1
sup {l.u(x)j : .U E M(S), ll.ull' ~ 1} 1 ~ c 1 sup {Jx(t)J: t E S} = 1/xl/. c =
(9.7.6)
Chapter 9
418
Now we shall show the converse inequality. Let Jl be a measure of finite support n
fl
= .}; atJh.,
for
i =I= j.
i=l
Take g(s) E C~(S) such that [[gf[ = 1 and g(!i) = at/[ai[. By (9.7.6) [[g[[ 1 ~ Ijc. Thus n
[[,u[[' ?: c,u(g)
=
c.}; [at[. i=l
On the other hand, n
n
[[,ull' ~.}; [atlllbe,!l' ~c.}; [ail· i=l
i=l
Therefore, for measures of finite support, ll,ull' = cll,ull,
(9.7.7)
where II II denotes the norm in the space M(S) induced by the standard norm. Since the measures of finite support are dense in the weak*topology in the space M(S), (9.7.7) implies ll,ull' ?: cll,ull.
(9.7.8)
llxll
(9.7.9)
Hence ~ cllxiii,
and finally llxll 9.7.2) the norm
=
cllxll 1 • Thus, by the Cowie proposition (Proposition D
I II is convex transitive.
COROLLARY 9.7.4 (Wood, 1981). The standard norm (9.7.3) is maximal in a space C~(S), provided the group of homeomorphisms of S onto itself is almost transitive. Example 9.7.5 (Pelczynski and Rolewicz, 1962) Let Q = [0, 1]. Let Q 0 = {0} u {1}. The standard norm (9.7.3) is convex transitive in the space Cc(Q"'!20 ).
FNorms and Isometries in FSpaces
419
Example 9.7.6 tPelczynski and Rolewicz, 1962) Let Q be a unit circle and let !!0 be empty. Then the standard norm is convex transitive. It may happen that in certain spaces C(Q""'Q0 ) the standard norm is not convex transitive, and yet it is maximal. Kalton and Wood (1976) gave conditions ensuring that the standard norm is maximal in the space C~(S). There are two such conditions, namely
the set S contains a dense subset such that each point of the subset has a neighbourhood isomorphic to an open set in an Euclidean space, (9.7.IO.i) the setS is infinite and has a dense set of isolated points (9.7.IO.ii) If either (9.7.10.i) or (9.7.IO.ii) holds, then the standard norm in the space C~(S) is maximal. In particular, the interval [0, l] satisfies (9.7.10.i) and by the result of Kalton and Wood (1976) the standard norm is maximal in the space Cc[O, I]. This is an answer to the question formulated by Pelczynski and Rolewicz (1962). At present the only known example of a space C~(S) of continuous complex valued functions vanishing at infinity in which the standard norm is not max{mal are spaces C~(S) where S has a finite number of isolated points t 1 , ..• , tn. In those spaces the norm n
~ )1/2 llxlft = sup lx(t)l + (L.; lx(ti)l 2 ·~ tES
i=l
t¢1•
obviously has a biger group ofisometries than the standard norm. There are also spaces which do not satisfy conditions (9.7.IO.i) and (9.7.10.ii) and yet their standard norm is maximal. Let D be a closed unit circle on a twodimensional Euclidean plane. Let {sn} be a dense sequence in D. Remove form D by induction the interior of an nblade propellor centred at Sn and the missing boundaries of all the previously removed propellors. The remaining set E is a compact, connected and locally connected metric space. It clearly has no nontrivial
Chapter 9
420
homeomorphism since, for each n, the neighbourhoods of snare unique to that point, and so any homeomorphism must map sn on Sn. Obviously, E does not satisfy either (9.7.10.i) or (9.7.10.ii). However, Wood (1981) showed that the standard norm is maximal in Cc(E). Now we shall pass to investigations of spaces of real valued continuous functions. THEOREM 9.7.7 (Wood, 1981). LetS be locally compact. Let C~(S) denote the space of all continuous realvalued functions vanishing at infinity. The standard norm is convex transitive if and only if S is totally disconnected and the group of homeomorphisms of Sis almost transitive. Proof Necessity. Suppose that Sis not totally disconnected. Then there are s1 and s 2 , s 1 s 2 belonging to the same component. By the form of isometry (9.7.4) the function equal to one on that component can only be transformed into a function equal either to + 1 or to  I constant on that component. Thus the standard norm is not convex transitive. The proof of necessity of almost transitivity of the group of homeomorphisms is precisely the same as the proof of the necessity in the proof of Theorem 9.7.4. Sufficiency. Since S is totally disconnected, for each finite system of points {ti> ... , tn}, ti t1 and each system of numbers {ai> ... ,an}, at= = ±I, there is a continuous function g(t) such that Jg(t)/ ~ I and
*
*
g(ti)
=
ai,
i
= 1,2, ... , n.
The rest of the proof follows the same line as the proof of sufficiency in Theorem 9.7.3. D Example 9.7.8 (Pelczynski and Rolewicz, 1962) Let E be the Cantor set. The standard norm (9. 7.3) is convex transitive in the space Cr(E) of real continuous functions defined on E.
Kalton and Wood (1976) proved that, if Sis a connected manifold without boundary of dimension greater than one, then the standard norm is maximal. Of course, by Theorem 9.7.7, the standard norm is not convex transitive. It is not clear what the situation in the case of manifolds
FNorms and lsometries in FSpaces
421
with boundaries and of manifolds of dimension one is like. For example, it is not known whether the standard norm is maximal in the space of real valued continuous functions defined on the interval [0, 1], Cr[O, I]. There are also spaces S such that the standard norm is maximal in the space Cc(S) but it is not maximal in the space Cr(S). Indeed, let E be the compact sp::.ce, de£cribed above, with trivial homeomorphism only. Wood has shown that in Cr(E) the standard norm is maximal. By the form of the rotations in C 0(S), the unique isometrics in Cr(E) are I and I. On the other hand we have PROPOSITION 9.7.9 (Wood, 1981). Let (X, II ID be a real Banach space with dimension greater than I. Then there is a norm II 11 1 in X such that the group G(ll 11 1) contains isometries different from I and I. Proof Take any x 0 E X and a linear continuous func1 ionc.l f such that llxoll = 1 = IIIII and f(x 0) = I. Let T be a symmetry
Tx
=
x2f(x)x0 •
Of course, T 2 = I. T cj::. I,  I and it is easy to verify that Tis an isometry with respect to the norm llxllt = max {llxll, IITxll}.
9.8.
D
THE MAX IMALITY OF SYMMETRIC NORMS
Let X be a real Fspace with the Fnorm llxll and with an unconditional basis {en}. The norm llxll is called symmetric (see Singer, 1961, 1962) if, for any permutation {Pn} and for an arbitrary sequence {en} of numbers equal either to I or to 1, the following equality holds :
llt1e1+ ...
+tnenll =
llcttleP'+ ...
+cntnepn+···ll
As follc,ws from the definition of symmetric norms, the operator U defined by the formula
U(t 1 e1 + ... +tnen+ ... ) = c1 t 1 ep, +... +cntnepn+ ...
(9.8.1)
is an isometry of the space X onto itself. We shall show that if X is not isomorphic to a Hilbert space, then each isometry is oftype (9.8.1).
Chapter 9
422
We say that a subspace Z of the space X of codimension 1 is a plane of symmetry if there is an isometry U =J=. I such that U(x) = x for x E Z. Let Z be a plane of symmetry and let V be an isometry. Then V(Z) is also a plane of symmetry. Indeed, let W = vuv 1 • The operator W is an isometry different from I. Let x = V(y), y E Z. Then W(x) = VUV 1 V(y) = VU(y) = V(y) = x.
If a Banach space X has a symmetric basis {en}, then the planes Ai =
{X : Xi
= 0},
Ai,i+ = {x:
Xi= XJ},
Ai,j = {x:
Xi= XJ},
where x = x1 e1 +x 2 e2 + ... +xnen+ ...
are planes of symmetry. Let us suppose that P is an arbitrary plane of symmetry. Let n be an arbitrary positive integer and let n
Xo=Pn(nAi). i=l
Let us consider the quotient space X1
=
X/X0 •
The space X1 is (n+ I)dimensional. The symmetries which have planes Ai, Ai,i,+, Ai,J, as planes of symmetry imply that there is a basis {e~}.
k = 1, 2, ... , n+ 1), in X1 such that the group Sn of operators of the type
U(t1 e~+ ... +tn+le~+ 1) = (s 1 tle~ 1 +sntne~.+tn+len+l) (9.8.2)
is contained in the group of isometries G. In virtue of Theorem 9.5.1 there is an ellipsoid invariant with respect to G. Since Sn C G, this ellipsoid is described by the equation (9.8.3)
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423
Since the replacing of e~+ 1 by ae~+ 1 leaves Sn as a subgroup of the group of isometries, we can assume without loss of generality that the invariant ellipsoid has the equation
xi+ ... +x!+I =
1.
(9.8.4)
Now we shall prove LEMMA
9.8.1. Let X 1 be an (n+ I)dimensional real Banach space with norm
Jlxll. Let the group ofisometries G contain the group Sn. If the group G is infinite, then the intersection of the sphere S = {x: llxll = 1} with the subspace X' spanned by elements e~, ... , e~ is a sphere in the Euclidean sense. Proof. Since the space X1 is finitedimensional, the group G is a compact Lee group. Thus G contains a one parameter group g(t). Obviously, there is an element x 0 , llxoll = 1 such that g(t)x0 defines a homeomorphism between an open interval ( c;, e), e > 0, and a subset of points of S. Now we have two possibilities : (1) g(t)x0 x0 ¢X', (2) g(t)x0 x0 EX'. Since Sn C G, we can find in the first case n locally linearly independent trajectories (in the second case (n1)). This implies that there is a neighbourhood U of the point x 0 such that for each x E U (in the second case for x E U nX') there is an isometry A such that A(x0 ) = x. This implies that the group G (resp. the group G' ofisometries of X' is) transitive. This D implies the lemma (cf. Section 5).
Lemma 9.8.1 implies that the group G of isometries of the space X1 is n
finite or that the quotient x;n At are Hilbert spaces for n = 1, 2, ... The i=
second case tri_vially implies that the space X is a Hilbert space. Let us now consider the first case, i.e. the case where the group of isometries G of the space X1 is finite. By (9.8.4) we can assume without loss of generality that the group G is contained in the group Gn+l of orthogonal transformation of the space xl. LEMMA 9.8.2. Let xl be an (n+ I)dimensional real Banach space with the norm llxll. Let Sn C G C Gn+1· Let P be a plane of symmetry determined by
Chapter 9
424
an isometry U £G. 1hen Pis either of type Ai or of type Ai.;± (i,j = 1, 2, ... , n+ 1) provided n is greater than 7.1 Proof To begin with, let us assume that the plane P does not contain the element en+J· Let P 0 be the plane of symmetry determined by an isometry belonging toG and such that en +I¢ P 0 and P 0 is nearest to en+ 1 (nearest in the classical Euclidean sense). Let P0
=
{x: a1 x1 + ... +an+ 1 Xn+
1=
0},
where (9.8.5) Since en+I ¢ P 0 , an+l =F 0. The planes Ai (i = 1, 2, ... , n) contain the element en+I· Therefore, the angle between P 0 and Ai ought to be of type nfn, because otherwise, composing symmetries with respect to P0 and Ai, we could obtain a plane of symmetry P 1 nearer to en+I than P0 • This implies that n i = 1, 2, ... ai =cosn Hence either ai = 0 or [a;[ ~ 1/2. Therefore, by (9.8.5) we have the following possibilities : (1) /an+I[
=
1,
h
(2) /an+Il = }13/2, there is an i 1 such that [%[ = (3) /an+ 1 [ = there is an i 1 such that [ail/ = j/2"3,
h
( 4) /an+ I[ = 1/ (5) /an+ 1 [ = (6) /an+ I[ =
v2.
there are il and i2 such that [ail/ = [a;z/ = 1
h there are ic and i such that [a; vz, [a;,[ = f, there are i i i such that [ai.[ = [a;,[ = [a;,[ 1 [,
2
1,
2,
3
h f, =
h
(7) /an+ 1 [ = Jl~' there is an i 1 such that [a;.[ = jl~' and a;= 0 otherwise. We shall show that only cases (1) and (7) are possible. Let us take indices j1oj2,j3 such that Jjk[ n, A =F im for all k and m. This is possible since n;:;, 7. Let a{= (a~, ... , an+ 1), where a~+ 1 = an+I• a;.= a;. and = 0 otherwise. The plane
<
a;
P" = {x: a' x 1 +
... +a~+ 1 Xn+ 1 =
0}
1 Indeed, Lemma 9.8.2 holds also for n = 2,4,5,6,7. It does not hold for n but for our purpose it is sufficient to show this for sufficiently large n.
= 3,
FNorms and Isometries in FSpaces
425
is also a plane of symmetry. It does not contain en+ I• but its distance from that point is exactly the same as P 0 • Therefore, the angle between P 0 and P' ought be of the type 21t/n, because otherwise, composing symmetries with respect to these two planes, we could find a plane P 1 nearer to en+I• such that en+I¢ P 1 • The cosinus of the angle between P 0 and P' is equal to 3/4 in case (2), to 1/4 in cases (3), (5), (6), and to 1/2 in case (4); this eliminates cases (2), (3), (5), (6). Let us take A = i 1 and j 2 =/= i 2 ; then the co sinus of the angle between the respective plane and P 0 is equal either to 3/4 or to 1/4. This eliminates case (4). Finally, only cases (1) and (7) are possible. So far we have assumed that the planePdoes not contain en+l' Suppose now that en+I E P. Let P 0 = P n X', where X' is the space spanned by the elements e1 , ••• , en. P 0 is a plane of symmetry in the space X', and restricting all considerations to the space X' we are able to prove our lemma. D 9.8.3. Let X be a real infinitedimensional Fspace with a basis {en} and with a symmetric norm llxll· Then either X is a Hilbert space or each isometry is of type (9.8.1). Proof As it follows from the previous considerations, if X is not a Hilbert space, then the planes Ai, Ai,H, Ai,j are all possible planes of symmetry. Let us denote the isometrics corresponding to Ai, Ai,i+, Ai,i by Si, Si.H, si.i, respectively. Let U be an arbitrary isometry. Then U(Ai), U(Ai·H), U(Ai,i) are planes of symmetry corresponding to the isometrics UStu1, USi,H u1, ust,i u1, respectively. Therefore, those isometrics are of type St, si,j+, Si.i. Let us denote the class of all such isometrics by .Let A, BE be such commutative isometrics that there is one and only one isometry c such that AC = CB. Then A = Si, B = Si, C = st,i. This implies that each isometry USi u 1 is of the type Si. Thus U is of the type (9.8.1). D THEOREM
m
m
Em
We shall now consider the spaces over complexes. Let X be a complex Fspace with basis {en} and norm llxll. The norm llxll is called symmetric if, for any permutation of positive integers Pn and for any sequence of complex numbers {en}, lsnl = I, the following holds:
llt1ed ...
+tnen+ ... 11 = llsltlep.+ ... +entnepn
+ ... 11.
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426
Obviously, if the norm [[x[[ is symmetric, then each operator of the type (9.8.1) (where en are complex numbers of moduli 1) is an isometry. In the same way as in the real case we define planes of symmetry. LEMMA
9.8.4. Let
xl be an (n+ I)dimensional complex Fspace with basis
••• , en+ 1 } and norm fix[[. If the group of isometrics G contains all operators of type (9.8.2) (where et are complex numbers of moduli I), then either G consists of operators of type (9.8.I) or G contains all orthogonal transformations which map the space generated by e1 , ••• , en onto itself. Proof Suppose that an isometry V maps an element et, I ~ i ~ n, on an
{e1 ,
element x 1 which is not of the type e1. Without loss of generality we may assume that the first n coordinates of x 1 are reals. Let us now consider the real space spanned by the elements ei> ... , en, x 1 • Applying Lemma 9.8.2, we find that the intersection of the set {x: [[x[[ = R} with the space spanD ned by e1 , •.• , en is a sphere. This implies the theorem. Lemma 9.8.4 implies in the same manner as in the real case the following: 9.8.5. Let X be an infinitedimensional complex Fspace with basis {en} and the symmetric norm lfx[[. If X is not a Hilbert space, then each isometry is of type (9.8.1).
THEOREM
CoROLLARY
9.9
9.8.6. The symmetric norms are maximal.
UNIVERSALITY WITH RESPECT TO ISOMETRY
We shall say that an Fspace Xu with the Fnorm [JxJJ is universal with respect to isometry for a class m: ofFspaces if, for any Fspace X Em:, there is a subspace Y of the space Xu and a linear isometry U mapping X onto Y. 9.9.1. There is no Fspace Xu universal with respect to isometry for all onedimensional Fspaces.
PROPOSITION
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427
Proof Let Xn be the real line with the following Fnorm :
l
it[ [[t[[n= 2[t[ Ijn
for for for
[t[
~I,
1 ~t< 2I/n,
[t[;;;: 2Ijn.
It is easy to verify that [[t[[n is an Fnorm. Suppose that a space Xu universal with respect to isometry exists. Then there are elements en E Xu such that [[te[[n = fit [[n, where [[x[[ denotes the norm in Xu. Hence [[2en[[ = [[2[[n I
=  > 0. On the other hand, [[en[[ = [[l[[n = 1 does not tend to 0. Theren fore, the multiplication by scalars is not continuous, and this leads to a contradiction since Xu is an Fspace. D
All onedimensional Banach spaces are isometric; hence, the real line (the complex plane in the case of complex one dimensional Banach spaces) with the usual norm [t[ is universal for all onedimensional Banach spaces with respect to isometry. Banach and Mazur (1933) proved that the space C[O, 1] is universal with respect to isometry for all separable Banach spaces. As a particular case we find that C[O, I] is universal with respect to isometry for all twodimensional Banach spaces. The space C[O, 1], however, is infinitedimensional. Hence, the following problem arises. Does there exist an ndimensional Banach space universal with respect to isometry for all twodimensional Banach spaces? The answer is negative. It was given for n = 3 by Griinbaum (1958) and for all positive integers n by Bessaga (1958). Of course, it is enough to restrict ourselves to real twodimensional Banach spaces and in the rest of this section only real Banach spaces will be considered. LEMMA 9.9.2 (Bessaga, 1958). Let Z be a bounded set in thendimensional real Euclidean space. Letj;, ... ,Jm map Z into the (n+I)dimensional real Euclidean space. If the set f1(Z) u /2(Z) u ... u fm(Z)
contains an open set, then at least one of the functions j;_, ... ,fm is not Lipschitzian.
Chapter 9
428
Proof Let
M(A, s) =max {p: there are Xt, i =I, 2, .. . ,p, Xi E A, llxtXJII > 8}
(compare Section 6.1). If the set A is ndimensional and bounded, then by a simple calculation we find that
M(A,s)~Kur
(9.9.1)
On the other hand, if A contains an open set of dimension (n+ 1), we find that there is a positive K1 such that M(A,s) ?': K 1 (
e1 )n+l.
(9.9.2)
Let f(z) be a Lipschitzian function defined on A and let L denote the Lipschitz coefficient of the function f Then M(A,s)
~ K( ~
r
implies M(f(A),s)
~ K Ln
(!)
(9.9.3)
n
Suppose that the functions j;_, ... ,Jm are Lipschitzian. Then by (9.9.3) there is a positive constant K 2 such that M(f1 (Z) u ... u fm(Z), s)
~ K2(!
r
This leads to a contradiction of formula (9.9.2).
0
THEOREM 9.9.3 (Bessaga, 1958). There exists no finitedimensional Banach space X 0 universal with respect to isometry for all twodimensional real Banach spaces. Proof Suppose that such a universal space (X0 , II II) exists. Let dim X 0 = n
and let Q={xeX0
:
llxll~l}
be a unit ball in the space X 0 • Q is a convex centrally symmetric body in thendimensional Euclidean space. Let labl denote the Euclidean dis
FNorms and lsometries in FSpaces
429
tance between the points a and b. Let Ia! = laO!. Let r = inflal and aeQ
R = suplal. aeQ
Let E denote the class of all convex centrally symmetric plane figures A such that r:::::;; a:::::;; R for all a EA. Let A, A' E E. Let
p(L1, A')= inf sup !aa'!,
u
where U is an isometry (in the Euclidean sense) a E A, a' E U(A') and a, a' are collinear. The quantity p(A, Ll') is called the distance between A and A'. We shall now introduce the distance between two subspaces in the following way: d(M,L) = max(sup inf
!xy!,
llx!l=lyeM xeL
!xy!).
sup inf !IYII=l xeL
yeM
The set of all twodimensional subspaces with this metric is a compact (; )dimensional set. By a simple calculation we can prove that there is a positive constant K such that (9.9.4)
p(A(L),A(M):::::;; Kd(M,L),
where A(L) = L n {xE X0 : llxll = 1} and LI(M) = M n {xE X0 : llxll
Let p = (;)
=
1}.
+ 1 and let Q be the set of all 2(p+ 1)gons contained in
E. Let Ll E Q. Let us number all vertices of Ll in the positive orientation
a1,
•.. ,
a2n+ 2 • Let J1i
=
area of the triangle Oaia;+l total area of A
The numbers f11> h(a1 ,
••• , f1p
••• ,
aP+I)
i= 1,2, ... ,p.
are affine invariant. The function
= Cf11> ... , f1p)
Chapter 9
430
is a Lipschitzian function of vertices a 1, . .. , aP+l· In this way we can assign to each A functions h1(A), ... , hp(A) dcpen dent on the point from which we begin the numbering. If A0 E !J, then we can find such positive c:LJ, and a positive M.1, such that if, for A, A' and then we can establish a correspondence between the vertices of /j and A' at~a; so that
[ata;[ os;: M.1,p(A,A'). Let U be an affine transformation of the plane such that, for A, A' E E, we have U(A), U(A') E E. Then p( U(A), U(A')) os;: ~o(A,A'). r·
This implies that all functions ht(A (L)) satisfy the Lipschitz condition, i.e. that there is a positive constant K such that [ht(A(L)hi(A(M))[ os;: Kd(M,L),
i =I, 2, ... ,p.
It is easy to verify that e2.ch of the functions hi maps Q into a set contain
ing the interior. Thus the set h1 (A) u ... u hp(A)
contains an open pdimensional set. Since p
=
(
~)
+I, this leads to
a contradiction of Lemma 9.9.2.
[]
The following problem are strictly connected with Theorem 9.9.3. Problem 9.9.4. Let X be a separable Banach space universal with respect to isometry for all finitedimensional Banach spaces. Is X universal for all
separable Banach spaces with respect to isometry? Problem 9.9.5. Let X be a separable Banach spaces universal with respect to isometry for all twodimensional Banach spaces. Is X universal with respect to isometry for all finitedimensional spaces?
FNorms and Isometries in FSpaces
431
Problem 9.9.6. How can we show a minimal number of twodimensional
real Banach spaces XI> ••• , Xk(n) such that there exists no ndimensional real Banach space universal with respect to isometry ? The existence of such numbers follows from Theorem 9.9.3. There is a conjecture that k(n) = ( ; )
+ 1 (in a particular case k(3) =
4).
Klee (1960) has investigated the following related problem: What is the minimal dimension of a body Kuniversal in the affine sense for all polyhedra of dimension n with r faces (r vertices)? More precisely : We say that a finitedimensional convex body K is auniversal for a class of convex bodies C)( provided each member of C)( is affinely equivalent to some proper section of K; and K is centrally auniversal for C)( provided K is centrally symmetric and every centrally symt;netric member of C)( is affinely equivalent to a central section of K. Replacing affine equivalence by similarity leads to the notion of suniversality and central suniversality. ~x,v (n, r) or ~.! (n, r) is the smallest integer k such that there is a kdimensional convex body K xuniversal for all ndimensional convex polyhedra having r+ 1 vertices, orr+ 1 maximal faces, respectively. r;a,v (n, r), or r;a.! (n, r) is the smallest integer k such that, there is a kdimensional convex centrally symmetric body, centrally auniversal for all centrally symmetric ndimensional polyhedra having 2r vertices or 2r maximal faces, respectively. Klee (1960) has shown the following estimation: += >
~a,v(n,r);;?
n 1 (r+1)
n+
~ ~a.f(n,r)
<
~ ~s.f(n,r)
< +=.
r,
+= > r;a,v(n, r);;? r;a.!(n, r) = r, += >
~s,v(n,r);;?
11
1 (r+2)
n+
The following proposition is strictly connected with Problem 9.9.6. PROPOSITION 9.9. 7 (Rolewicz, 1966). There are three plane, centrally symmetric convex figures P 1 , P 2 , P 3 such that there exists no centrally symmetric convex threedimensional body K admitting a line L through its
Chapter 9
432
centre and sections P~, P~, P; through L such that Pi is affinely equivalent to P;,J = 1, 2, 3. Proof Let PI be a square, P 2 circle and P 3 a square with side 2 with corners rounded off by circular arcs of radius E (see Figure 9.9.1). We shall show that for a sufficiently small E the required body K could not exist.
·£
E
''>· Fig. 9.9.1
Suppose that such a body K exists for all e. Let t be a point at which L intersects the boundary K. Since t E P~ n P~, there is only one supporting hyperplane of Kat t. Thus t cannot be an extremal point (corner) of P~. The point t must belong to one of the curved arcs of P~. Otherwise t would be a flat point of the body K, and this is impossible, became P~ is strictly convex. Since we are only interested in affine properties, we may assume without loss of generality that P~ is a unit disc in the plane z = 0, P; is similar to P3 and situated in the plane y = 0, and P~ is a parallelogram situated in a certain plane z = ay. (Here we put the xaxis as L.) Now we shall investigate the relationship bet.ween a and E. Let Pi denote the boundary of i = 1, 2, 3. The three curves PI, p 2 , p 3 intersect at point t. Since P; is a normal section of K, the normal curvature K 3 of the boundary in the direction of p 3 is equal to the totaJ curvature of p 3 at t, i.e. it is equal to 1/e. The total curvature K 2 of p 2 is equal to 1 ; hence, the normal curvature of the boundary of Kin this direction is at most I. The set K is convex and PI is a straight line the neighbourhood oft, hence, the minimal curvature of the boundary of K is equal to 0. The maximal curvature is in the direction perpendicular to PI· Let us denote
P;,
FNorms and Isometries in FSpaces
433
the maximal curvature by k. Using Euler's formula, we can express the normal curvature of the boundary of Kin the directions of p 2 and p 3 as
where {J is the angle between p 2 and p 1 at point t. Therefore
On the other hand
Hence {J tends to 0 as s+0; this implies that a+0, as s+0. Since for all s the points (1,0,0) and (1,0,0) belong to K, it follows that for s+0 the set P~ tends toP~. This leads to a contradiction, because P~ is a circle and P~ is a parallelogram. D
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