Metric Linear Spaces (Mathematics and its Applications)

in LP(am,n) for p > 1. Example 1.3.10 Let X be a linear space. Let (x, y) be a two-argument scalar-valued function satisfying the following conditions : 0, (i l) (x, x)

(i2) (x, x) = 0 if and only if x = 0, (i3) (x1+x2,Y) = (x1,Y)+(x2,Y),

(i4) (ax,y) = a(x,y) for all scalars a, (i5) (x, y) = (y, x), where a denotes 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 pre-Hilbert space. It is easy to verify that Ilxll =1/(x, x) is an F-norm.

1.4. COMPLETE METRIC LINEAR SPACES

Let X be a metric space with a metric p (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--+00

The space (X,p) is called complete if each Cauchy sequence {xn} is convergent to an element x0 E X, i.e.,

lim p(xn, x0) = 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 = W

= U Fn, where the sets F. are nowhere dense and Fn C Fn+1 Since the ft=1

set F1 is nowhere dense, there exists an open set K1 such that K1 n F1 = o and the diameter of K1

d(K1) = sup p(x, y) < 1. z,vek3

The set F2 is nowhere dense, hence there exists an open set K2 such that K2 C K1i d(K2) < 1/2 and K2 n F2 = 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+1, d(Kn) < 1/n, and

Kn n Fn = o.

(1.4.1)

Let xn e Kn. Since d(Kn)-O, {xn} is a Cauchy sequence. The space X is complete ; thus there is an element x0 E X such that {xn} tends to x0. Since

KnCKn+1,xoEKn,n=1,2,...Hence, by (1.4.1)x00Fn,n=1,2,... This implies that x0 U Fn = X and we obtain a contradiction. a=1

COROLLARY 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 CE is of the first category. Then X as a union of two sets of the first category is also of the first category. This contradicts Theorem 1.4.1. 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 let p(x,y) = Ix-yj. Obviously (X,p) is complete. Let p'(x, y) = jarc tan x-arc tany1. It is easy to verify that the metrics p and p' are equivalent. The space (X, p') is not complete, since the sequence {xn} = {n} is a Cauchy sequence with respect to the metric p'(x, y) but, evidently it is not convergent to any element xo a X.

The situation is different if we assume in addition that the metric p'(x,y) is invariant. Namely, the following theorem holds THEOREM 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.

Chapter 1

20

The proof is based on the following lemmas : LEMMA 1.4.5 (Sierpiriski, 1928). Let (E,p) be a complete metric space. Let E be embedded in a metric space (E', p'). Suppose that on the set E n E' th e metrics p(x, y) and p'(x, y) are equivalent. Then E is a Ga 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 rn(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) < rn(x)}, Go

G. = U Un(x), ZEE

Go = 1 1 Gn n=1

The sets UU(x) are open. Thus the sets G,, are also open. Therefore Go is a GS set.

Evidently E C Go. It remains to show that E J Go. Let xo e Go. Then x° a Gn, n = 1, 2, ... By the definition of the sets Gn, there exist elements x,, e E such that p'(xn,xo) < r(xn). Since rn(x) < 1/n, the sequence {xn} tends to xo in the metric p'(x, y). Let e be an arbitrary positive number. Let n be a positive integer such that 2/n < e, and let ko be a positive integer such that k° < rn(xn)-p'(xn, xo).

Then, for k > ko, p'(xk, xn)
k Hence, by the definition of the number rn(xn), p (xk, xn) < l 1n. 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° e E such that lim p (xn, x°) = 0. The metrics p (x, y) and p'(x, y) are equivalent on the 00

set E.

Therefore xo = x° and xo e E.

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 X0 be a dense linear subset of X. If X0 is a G. set, then X0 = X. Go

Proof. If Xa is a Gb set, then by definition Xo = U Gn, where G. are open

sets, G. C Gn+,. Since X0 is dense in X, Gn are also dense in X. This implies that the sets X \G. are nowhere dense. Thus X\XO is the set of the first category. Then the set X0 is of the second category.

Suppose that the set X\Xo is not empty. Let x e X\XO. Since Xo is a linear subset,

x+X0 e X\Xo This leads to a contradiction, since the invariance of the metric p(x,y) implies that x+Xu is of the second category and X \Xo is of the first category. 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 Yis 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}, if lim p(xn, yn) = 0. A metric in Y is defined by co

the formula

p'({xn}, {Yn}) = lim p(Xn, yn). It is well known that Y is a complete metric space. X can be embedded in Yas the set of stationary sequences x = {x, x, ...} ; X is dense in Yand on X the two metrics p (x, y) and p'(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) Proof of Theorem 1.4.4. Let (Y,p') be the completion of (X,p). By Lemma 1.4.5. X is a dense Ga set in Y. Hence, by Lemma 1.4.6, X = Y.

Chapter 1

22

PROPOSITION 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 x0 a X, then {xn} also converges to x0. Proof. Let e be an arbitrary positive number. Since xnk-)-xo, there is an index ko such that for k > ko, p (xn,t, xo) < e/2. On the other hand, the sequence {x.,,} is a Cauchy sequence. Therefore, there exists a number N

such that, for n, m > N,

p(xn, xm) < 2 Putting m = nk, k > ko, we obtain p(xn, xo)
+ 2 = e.

The arbitrarines of a implies the proposition. PROPOSITION 1.4.9. Let (X, II II) be an F*-space. Let {Ei} be an arbitrary 00

fixed sequence of positive numbers such that the series

ei is convergent. i=1

If for each sequence {xi} of elements of X such that IIxiII < Fi the series w

xi is convergent, then the space (X, II II) is complete. i=1

Proof. Let {yn} be an arbitrary Cauchy sequence of elements of the space X. We can choose a subsequence {Y-k} such that IIxklI < ek, where xk

Ynk+1-Ynk The assumption implies that the series E xk is convergent k=1

to an element x e X. We shall show that {yn} tends to

In fact sk

k

i=1

X. = Ynk-Ynl tends to x. Thus ynk tends to x+yn1. By Proposi-

tion 1.4.8 {yn} tends to x+ynl. A complete F*-space is called an F-space. A complete pre-Hilbert space (Example 1.3.10) is called a Hilbert space. THEOREM 1.4.10. Let (X, II II) be an F-space. Let Y be a subspace of X.

Then the quotient space X/Y (see Section 1.1) is an F-space (i.e. it is complete).

Basic Facts on Metric Linear Spaces

23

Proof. Let {Zn} be an arbitrary sequence of elements of X/Y such that IIZnII < 1/2n. By the definition of the norm in the quotient space there are xn e Zn such that Ilxnli < 1/2n. The space X is complete, hence the series co

Y xn is convergent to an element x e X. Let Z denote the coset containing n=1

x. Then, by the definition of F-norm in the quotient space k

k

Zi--Z

-x

i=m

Therefore, the series

Z is convergent to Z, and by Proposition 1.4.9 n=1

the space X/Y is complete. PROPOSITION 1.4.11. The product (X, II II) of n F-spaces is an F-space.

Proof Let xm = (xi) be a Cauchy sequence. Then the sequences {x;'}, i = 1, ..., n also is a Cauchy sequence. Since (Xi, II Ili) are complete, there are xi e Xi such that IIxm-x{ IHO, i = 1, ..., n. Let x = (xi). Then n

IIxm_xII = I Iix'i -xi II{-*O. i=1

1.5. COMPLETE METRIC LINEAR SPACES. EXAMPLES

PROPOSITION 1.5.1. The spaces N(L(Q, E, u)) are complete. Proof. Let {xn} be a Cauchy sequence in N(L(Q, E, p)). It is easy to verify

that the sequence {xn} is a Cauchy sequence with respect to the measure

(this means that, for each a > 0,

lim p({t: Ixn(t)-xm(t)I > a}) = 0). Therefore, by the Riesz theorem, the sequence {xn} contains a subsequence {xnk} convergent almost everywhere to a measurable function x(t). Let a 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

PN(xn-X.) = f N(I xn(t)-xm(t)I )dl1 < 6. a

Chapter I

24

Put m = nk and let k-*oo. By the Fatou lemma, we obtain

PN(xn-x) < E. This implies that xn-x E N(L(Q, E, p). Since N(L(Q, E, p)) is linear, x c- N(L(Q, E, p)). The arbitrariness of E implies that x,-->x. PROPOSITION 1.5.2. The space M(Q, E, p) is complete.

Proof. Let {xn} be a Cauchy sequence in M(Q, E, p). Then the sequence {xn(t)} is convergent for almost all t. Let x(t) denote the limit of the sequence {xn(t)}. It is easy to verify that x(t) e M(SQ, E, 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

Ixn-x.11 = esssup Ixn(t)-xm(t)I < E. LEO

Hence, when m tends to infinity, we obtain

Ilxn-x11 = esssup Ixn(t)-x(t)I < E. tES2

The arbitrariness of s implies that x.->x.

El

PROPOSITION 1.5.3. The space C(Q) is complete.

Proof. Let {X-n} 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 a 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

Ilxn-xmll = sup I xn(t)-xm(t)I < E. LEO

Hence, when m tends to infinity, we obtain

Ilxn-xII = sup Ixn(t)-x(t)I < The arbitrariness of s implies the proposition. PROPOSITION 1.5.4. The space C(QI Q0) is complete.

Proof. C(QIQ0) is a closed subspace of the space C(Q).

Basic racts on Metric Linear Spaces

25

PROPOSITION 1.5.5. The space eo(Q) is complete.

Proof. Let {xn} be a Cauchy sequence in Co(Q). Then {xn} is also a Cauchy sequence in each pseudonorm II Ili, i.e.,

i = 1, 2, ...

Ilxn-xmlli = 0,

lim m,n-* co

The sequence {xn(t)} is convergent for each t. Its limitx(t) is a continuous co

function on each Di, hence it is continuous on the set Q = U A and

lim Ilxn-xlli = 0,

i = 1, 2, ...

n-aco

PROPOSITION 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., urn m,n-

IIXn-XmIIk = 0. 4D

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°°(SQ) is complete. PROPOSITION 1.5.7. The space cS(EP) is complete.

Proof. Let {xn} be a Cauchy sequence in cS (ED). Then

lim Ilxn-x.' II m,k = 0

for all m and k.

(1.5.1)

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 a be an arbitrary positive number. Formula (1.5.1) implies that there is a positive integer N such that, for n, n' > N, IIXn-x. II a,ei

= sup (t,.... , ta) E En

I

ti k1 ... k,

P

(xn(t)-xn,(t))I JItilm< <e. ti=1

Chapter 1

26

Hence, when n' tends to infinity, we obtain a,ki

11x--x11 = sup teEP

ti

P

.k

(xn(t)-x(t)) 11 Ittlm, < S.

iT,

t=1

The arbitrariness of e implies that xn-*X. PROPOSITION 1.5.8. The spaces M(am, n) and LP(am, n) are complete.

Proof. Let {xk} = {{xn}} be a Cauchy sequence. Since sup a.,, > 0, the m

sequence {xn} converges for any fixed n to a limit xn. Write x° = xn. The sequence {xk} is a Cauchy sequence, hence it is a Cauchy sequence in each pseudonorm IIXIIm, i.e. for any s > 0, there is a positive integer K such

that for k, k' > K Ilxk-Xk'IIm

= supam,n Ixn-xn'l < e, n

(for 0
IIXk-Xk'II m = I (am, n Ixn-xn' I)P < E n

and for 1 < p < +oo 11Xn

I

lm = ( , (am, nI

xn-xn'

I )P)1/P

<e).

n

Hence, when k'-±oo we obtain IIXk-x°II < e.

The arbitrariness of a 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 a there is a countable set E. such that for an arbitrary x e X, there is an element y e Ee such that p(x, y) < e.

Basic Facts on Metric Linear Spaces

27

Proof. Necessity. Let (X,p) be a separable space and let E be a dense countable set. Let EE = E for all e. Sufficiency. Let EE be a family of sets satisfying the property described 00

above. Then the set E = U E1l, is a dense countable set in X. n=1

COROLLARY 1.6.2. Let (X,p) be a metric space. If there are a non-countable

set A and a number S > 0 such that

p(x,y)>6

for allx,yeA,x#y,

(1.6.1)

then the space (X,p) is non-separable. Proof. Suppose that (X,p) is separable. Then by Proposition 1.6.1 there is

a countable set EE/2, such that for each x e A, there is an x e EE/2, such that p(x,z) < e/2. Thus by (1.6.1) there is a one-to-one correspondence between x and x. This leads to a contradiction since A is non-countable and EE12 is countable. PROPOSITION 1.6.3. A subset Xo 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..

Proof Let e be an arbitrary positive number. By Proposition 1.6.1 there is a countable set E812 such that for any x e X there is a zz a EE12 such that

p(x,zz) < E/2. We associate with each element z E EE/2 an element y(z) e X0, such thatp(z,y(z)) < E/2, provided that such an y(z) exists. Then P (X' Y (Z.)) < AX, zz) +P (zz, Y (Z.)) < 2 +

2 =E.

Thus, by Proposition 1.6.1 Xo is separable. Let (S2, E, !e) be a measure space. The measure p is called separable if there is a countable family of sets A. e E, such that, for any set B E E of finite measure and for an arbitrary positive e, we can find a set An0 such that

p(B\A..)+ju(A..\B) < E. PROPOSITION 1.6.4. A space N(L(.Q, E, p)) is separable if and only if the measure u is separable.

Chapter 1

28

Proof. Sufficiency. Let {An} be a countable family of sets with the property described above. Let S?X be the set of all simple functions

x x(t) _ E akXAnk k=1

where, as usual, XB denotes the characteristic function of the set B, ak are

rationals in the real case and complex rational (i.e., of the form an = bn+cn i, where bn and cn are rational) in the complex case.

It is easy to verify that the set U is countable and that it is dense in N(L(Q, E. p)). Necessity. If the measure p is non-countable, then there are a noncountable family of sets {Aa} of finite measure and a positive constant 6

such that

p(AQ \Ap)+p(AI\Aa) > 6 for a # P. Let xa = XAa. Then pr,(xa, xfl) > N(1)6 for a

i9. Since the set {xa} is

non-countable, by Corollary 1.6.2 the space N(L(Q, E, p)) is not separable.

PROPOSITION 1.6.5. A space M(Q, E, p) is separable if and only if the measure p is concentrated on a finite number of atoms p, ..., pk Proof. Sufficiency. Suppose that the measure p is concentrated on a finite number of atoms pl, ..., pk. Then the space M(S2, E, p) is finite dimensional, and thus separable. Necessity. If the measure p is not concentrated of a finite number of atoms, then there is a countable family of disjoint sets {An} (n = 1, 2, ...) of positive measure. Let a = {n1, n2, ...} be a subset of the set of positive integers. Let A

= Ant V Ant v ... The family {Aa} is non-countable and for a p(Aa\A,6)+p(AB\Aa) > 0. Let xa = XAa. Then Ilxa-xxII = 1 for a =;-4 9. Therefore by Corollary 1.6.2 the space M(Q, E, p) is not separable. 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')).

Basic Facts on Metric Linear Spaces

29

Proof. Let (.2, 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 n An,k' = 0

for k

k',

K. n=1

sup{d(t, t'): t, t' c- An,k} < 1/n. The family {Aa,k}, n = 1, 2, ..., k = 1, ..., Kn, is of course countable. Let X be the space of functions x(t) of the form M

X(t) = 11 amXAnm, km ,

(1.6.2)

M=1

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 jjxjj

= sup Mt)I. ten

Let W 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 ¶U is countable and it is dense in X. Thus X is separable. K.

Let x(t) e C(Q). Let xn(t) =

an,kXAn,k E 2Y, where an,k are chosen

in such a way that inf Jx(t)-an,ki < 1/n. Since the function x(t) is conteAn.k

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(SC) is separable. Necessity. Suppose that the space C(Q) is separable. Let {xn} be a sequence dense in the unit ball K = {x: jIxjj < 1}.

For t, t' e Q, let d(t, t') _ Y Zn Ixn(t)-xn(t')I. Since Ixn(t)j < 1, n=1

d(t, t') is always finite. It is easy to verify that d(t, t') is a metric on Si. We shall show that the topology determined by this metric is equivalent to the original topology on Q.

Chapter 1

30

Let to a S2 and let E be an arbitrary positive number. Let m be a positive

integer such that 1/2m < E/4. Since the functions xn(t) are continuous, there is a neighbourhood V of the point to such that, for t e V,

xn(t)-xn(to) I < 2

(n = 1, 2, ..., m)

Then, for t e V, 00

d(t,

to)

1

n=1

Ixn(t)-xn(to)I co

xn(t)-xn(to)I + Y-L I xn(t)-xn(to)I n=n:+1

n=1

Zn

Conversely, let V be an arbitrary neighbourhood of to e D. Since S2 is compact, there is a continuous function x(t) such that Ix(t)I < 1 forte V, x(to) = 0, x(t) = 1 for t 0 V. The set {xn} is dense in the unit ball of the space C(Q), therefore there re iis xn such that Ilxn-X11 =tc-V

xn(t)-x(t)I < 4 . I

This implies that, if I xn(t) I < 3/4, then t e V and I xn(to)I < 1/4. Therefore

I xn(t)-xn(to)I <

(1.6.3)

implies that t e V. Let d(t, to) < 1/2n+1 Then (1.6.3) holds and t e V. PROPOSITION 1.6.7. A space C(QI S2o) is separable if and only if the set S2\920 is metrizable.

Proof. The proof follows the same line as the proof of Proposition 1.6.6.

PROPOSITION 1.6.8. A space e0(12) is separable if and only if the set S2 is metrizable. Proof. Sufficiency. According to the definition of the space eo(Q), the set

Basic Facts on Metric Linear Spaces

31

Sl is the union of an increasing sequence of compact sets Qm. For each m and n we can find a finite system of sets {A'"k} such that

Am n An k, = 0

for k

k',

Um

k

An,k = Slm,

sup{d(t, t'): t, t' c- An k} < 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 F-norm. OD

1

i=1

2i

IIxHIs, 1+IIxIIs,

where IIxik{ = sup lx(t)l. tEa,

Let Q1 be defined in the same way as in Proposition 1.6.6. The set 9C is countable and dense in X. Therefore X is separable. Then by Proposition 1.6.3. eo(Q) is separable. Necessity. Let {x.} be a dense sequence in the space e,(Q). Let for t, t' e Q

I xn(t)-xn(t')I

d(t, t') = n=1

1+Ixn(t)-xn(t')I

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. Let (Xi, II Ili) be a sequence of F*-spaces. Let X = (Xi)(,) be the space of all sequences x = {xj x{ e X{}. The topology in X is given by a sequence of pseudonorms IIxII' = IIxiII{. Of course the space Xis an F*-space.

PROPOSITION 1.6.9. It the spaces (Xi, II Ili) are separable, then the space X = (Xi)(8) is separable.

Proof. Let 9I1 denote a dense countable set in Xi. Let 9I be the set of all sequences of the form

{a,, a,, .... a,, 0. 0....}.

Chapter 1

32

It is easy to verify that the set 91 is countable and that it is dense in the space X. Let X be a linear metric space with topology determined by a sequence of F-pseudonorms IIxjIi. Let Xo = {x: Ilxlls = 0}. Let X{ be the quotient space X/Xo. The pseudonorm IIxlk£ induces in the space Xi the F-norm IIx1Ii'

PROPOSITION 1.6.10. If all the spaces XI are separable, then the space X is separable. Proof. Let X = (Xi)(8). By Proposition 1.6.9 the space X is separable. The

space X can be identified with the subspace X0 C I of all sequences {[x]i}, where [y]i denotes the coset in X{ containing y. Obviously IlxII = II[x]4I , i.e. the topology of X0 inherited from £ 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. PROPOSITION 1.6.11. The spaces CW(S2) are separable.

Proof. Let x e C°°(Q). Then xIIk =

II a k,

X(t)I J0. al kla'..

Hence, the space Xk can be considered as a linear subset of the space C(S2). Since Q is a bounded domain in the n-dimensional Euclidean space, Q is compact. Thus by Proposition 1.6.6 the space C(Q) is separable. Hence by Proposition 1.6.10 the space C°°(Q) is separable. PROPOSITION 1.6.12. The space c5 (En) is separable.

Proof. Let x e c5 (En). Then k

IIxIIm, k

in

M

tl .. akn to

ak=

x(t)110,

0

and k

lim ti ... t".

3k1

al 3k,

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

point. The space Q\Q0 is metrizable. Hence by Proposition 1.6.6. C(Q1Q0) is separable. Therefore, by Proposition 1.6.3 cS (En) is also separable. PROPOSITION 1.6.13. The spaces LP(am,n) are separable.

Proof. Elements of the type {xl, x2, ..., xn, 0, 0, ... }, where xi are rational

in the real case and complex rational in the complex case, constitute a dense countable set in LP(am,n). PROPOSITION 1.6.14. A space M(am,n) is separable provided that, for any m, there is an m' such that

lim

am,n

n-- am,, n

= 0.

Proof We construct a dense countable set in the same way as in the proof of Proposition 1.6.13.

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 called 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 Kakutani 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 of a family of sets {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 91 of non-void subsets of X is called fundamental, if for every two sets M, N e 91, there exists an E e 21 such that E C M n N, and for each neighbourhood of zero Uthere is aset M_ e W 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 {A.n}, where An = {xn, xn+l, ...}, is fundamental. PROPOSITION 1.7.2. A fundamental family 9 has at most one cluster point.

Proof. Suppose that x and y are two cluster points of the fundamental family W. Let U be an arbitrary balanced neighbourhood of 0. Since R1 is fundamental, there is an M e 91 such that M-M C U. Since x, y are cluster points of 91, they are also cluster points of the set M. This implies

that there are xl,yl e M such that x-x1, y-yl a M. Hence

x-y = (x-x1)-(Y-YJ+(x1-Y1) e U-{- U+ U. The arbitrariness of U implies that x = y. 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 e A.

THEOREM 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 21 be an arbitrary fundamental family of subsets of the set E. Since X is complete the family 91 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' a E. By

Proposition 1.7.2 a fundamental family has at most one cluster point. Thus a' = a. Therefore the set Eis closed. THEOREM 1.7.4. For any topological linear space X, there is a complete topological linear space X such that X is a dense linear subset of the space X,

Basic Facts on Metric Linear Spaces

35

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 W and B if 0 is a cluster point of 91-0. Further steps are similar to the proof in the metric case (cf. Lemma 1.4.7).

El

Chapter 2

Linear Operators

2.1. BASIC PROPERTIES OF LINEAR OPERATORS Let (X, 11 IIx) and (Y, II IIY) be two F-spaces. We shall denote the norms

lix and II IIY by the same symbol II 11 whenever no confusion result. A mapping A transforming a linear subset DA e X into Y is called an additive operator if II

A(x+y) = -A(x)+A(y)

for all x, y e DA.

The set DA 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 a continuous linear (additive) operator if it is continuous. If the spaces Xand 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 (x) = A I n)

-F A

(n

n-fold

(x)= 1 A (x). Consequently, for arbitrary rational r, we obtain that A n

n

A(rx) = rA(x). 36

Linear Operators

37

Let a be an arbitrary positive number. Let t 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 11.4((t-r)x)II < 2

and

11(t-r) A(x)II <

2

Hence

IIt A(x)-A(tx)II

C 11(t-r)A(x)II+IIrA(x)-A(rx)II+IIA(rx-tx)II < e. Therefore, the arbitrariness of a 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 x0 belong to the closure DA of the domain D.A. Then, by the definition of closure, there is a sequence {xn}, xn e DA, tending to x0. 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 com-

plete, therefore, there is a limit y e Y of the sequence {A (x.)}. 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 closure D.A. This is the reason why in the theory of continuous linear opera-

tors we shall restrict ourselves to the operators defined on the whole space X. Let X be a metric linear space. A set B C Xis 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 {tn xn} tends to 0. In other words, a set B is called 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 F-norms in X and in Y by the same symbol 11 II. 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 IIaA(x)II > e

ZEE

for all scalars a different from 0. The set E is bounded, therefore, for an arbitrary positive 6 there is a number b such that sup IIbxjl < 6.

ZEE

This implies that, for an arbitrary positive 6, there is an element xb such that and IIA(xs)II > E. IIxnII < 6 Therefore, the operator A is not continuous. Necessity. Suppose that the operator A is not continuous. Then there are a positive number 6 and a sequence {xn} of elements of Xtending to 0 such that

IIA(xn)II>S>0. Let Xn

xn

j,IxnII

By the subadditivity of F-norm we immediately obtain IIxnII

IIxnII < 1

IXnII

+sup 0<- 1,1

II txnll

Hence x.->O implies x;,->0.

Let to = IIxnII. Evidently the sequence {tn} tends to 0. On the other hand, IItnA(xn)II = IIA(tnxn)II = IIA(xn)II > 6 > 0.

Therefore, the bounded set {x} is transformed onto an unbounded set {A (x,)}. This implies that the operator A is not bounded.

Let Bo(X--Y) denote the set of all continuous linear operators mapping X into Y. The set Bo(X-*X) we shall denote briefly by B0(X). It is easy to verify that Bo(X- .Y) is a linear set.

Linear Operators

39

Let a be a family of bounded sets in X. By B,(X->Y) we denote the set B,(X->Y) with the topology determined by neighbourhoods of the following form : U(AO, B, e) = {A a B0(X->Y): sup IIA(x)-Ao(x)II < e, B e a, ZEB

e > 0}. The space B,(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 a is the family of all bounded sets, then the topology generated by a is called the topology of bounded convergence. In this case

the space B,(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 Kis 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. BANACII-STEINHAUS THEOREM FOR F-SPACES Let (X, II IIx) and (Y, II IIY) be two F*-spaces. Let X be complete (i.e. be

an F-space). Since there is no danger of misunderstanding, we shall denote the both norms by II II A family 91 of operators belonging to Bo(X-- Y) is said to be equicontinuous if for each positive a there is a positive S such that sup {IIA(x)II : A e 91, IIxii

< b} < s.

The following theorem is an extension of the Banach-Steinhaus theorem (Banach and Steinhaus, 1927) on F-spaces. THEOREM 2.2.1 (Mazur and Orlicz, 1933). Let X21 be a family of linear operators belonging to the space Bo(X--> Y). For each x e X, let the set {A(x): A e 21} be bounded. Then the family X1 is equicontinuous.

Chapter 2

40

The proof is based on the following lemma : LEMMA 2.2.2. Let (X, II II) be an F-space. Suppose that a closed set V is

absorbing, i.e. for each x e X there is a positive number a such that, for b, 0 < b < a, bx e V. Then the set V contains an open set. Proof. Since the set V is absorbing,

X= U nV. U-1

By the Baire theorem (Theorem 1.4.1) the space X is of the second category. Therefore, there is a positive integer no such that noV is of the second category. Since noV is closed, it contains an open set U. Thus V con-

tains the open set

1

no

U.

F-1

Proof of Theorem 2.2.1. Let e be an arbitrary positive number. Let

U1= n {xe X: IIA(x)II < E}. A E R[

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 ¶U} is bounded. Hence there is a positive number a such that, for b, 0 < b < a, II bA (x)II < E for all A e 21. Thus bx e U1. By Lemma 2.2.2 the set U1 contains an open set U2. Let xo e U2. The

set {A(xo): A e U} is bounded. Hence there is a positive number b, 0 < b < 1 such that II bA (xo)I I < E. Thus bxo a U1. Let U = b (U2-xo)

Of course U is a neighbourhood of 0. Let x e U. Then x = by-bxo. where y e U2. Therefore IIA(x)II < IIbA(xo)II+IIbA(y)II = E+sup {IIbzII: IbI < 1, IIzII<E}.

Hence the continuity of multiplication by scalars implies the theorem. COROLLARY 2.2.3 (Mazur and Orlicz, 1933). Let X, Y be two F-spaces. Let sequence of continuous linear operators {An} be convergent at each point x to an operator A,

lim A.(x) =A (x).

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 equi-

continuous, i.e., for an arbitrary positive e, there is a 6 > 0 such that IIxJI < 6 implies

IIAn(x)II < 2

Let IIxII < 6. Since A (x) is convergent to A (x), there is an index no such that

IIA (x)-Ana(x)II < 2 Hence

IIA(x)II < IIA(x)-Ano(x)II+IIAn,(x)II <2+ 2 = s. Therefore, the operator A is continuous at 0. The linearity of the operator A implies that it is continuous everywhere. COROLLARY 2.2.4. If (X, II II) and (Y, II II) are F-spaces, then the space

B(X-+Y) is complete.

Proof. Let 2Y be a fundamental family of subsets of the space B(X-Y). This implies that for any x e X the family 21(x) of the sets {{A(x): A G M} ME 2i} is a fundamental family of subsets of the space Y. The space Y is complete, and thus the family 91(x) has a cluster point A0(x). Corollary 2.2.3 implies that A0(x) is a continuous linear operator. Let B be a bounded set in X and let U be a neighbourhood of 0 in Y. Let E be an arbitrary member of W. Let M C E be a member of 2Y such that, for A, Al a M, x e B,

A(x)-AI(x) e U. Then

A(x)-Ao(x) e U. The arbitrariness of B and U implies that Ao is a cluster point of M. Thus it is also a cluster point of E, and since E was an arbitrary member of 2[, Ao is a cluster point of the family W.

Chapter 2

42

In the particular case where X is an F-space the conjugate space X* is complete.

2.3. CONTINUITY OF THE INVERSE OPERATOR IN F-SPACES

,

THEOREM 2.3.1 (see Banach, 1932). Let (X, II II) and (Y, II II) be two

F-spaces. 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 U be an arbitrary neighbourhood of 0. We shall first show that

the closure A(U) of the set A(U) contains a neighbourhood of 0. Since a-b 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=1

Thus

Y=A(X)=UnA(M). n=1

By the Baire category theorem (Theorem 1.4.1) there is an no such that the set no A(M) contains a non-void open set V. Hence

A(U) D A(M)-A(M) ) A(M)-A(M) j no (V- V), and the set

1

n no

(V- V) is of course a neighbourhood of 0.

For any positive E we shall write XE _ {x e X: IIxJI < e}

and

YY = {y E Y: IIYII < e}.

Let so be an arbitrary positive number. Let {ei} be a sequence of pos-

itive numbers such that Z Ei < so. As we have already shown, there is

i-1

a sequence {iii} of positive numbers such that

A(Xet) ) Y,,,

i = 0,1, 2, ...

(2.3.1)

Linear Operators

43

Let y be an arbitrary element belonging to Y,,. We shall show that there is an element x e X2EO such that

A(x) = y.

(2.32.)

Formula (2.3.1) implies that there is an element xo a Xep such that

Ily-A(xo)II < '7i' i.e. y-A(xo) e Y,,1. Putting i = 1 in formula (2.3.1), we can find that there is an element xl a XE such that

Ily-A(xo)-A(xi)II < 272Repeating this reasoning, we may define a sequence X such that xn e Xen and

of elements of

n

(2.3.3)

Ily-A (z xi) II < nn+1. s=o

Since the space X is complete, the series

xn is convergent to an ele-

ment x and, moreover, IIxil < 2so. 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 of 0 such that x+N C G. Let M be a neighbourhood of 0 in Y

such that A (N) ) M. Then

A(G) D A(x+N) = A(x)+A(N) D A(x)+M. Therefore, A(G) contains a neighbourhood of each of its points, i.e. A(G) is open. THEOREM 2.3.2 (Banach, 1932). Let a continuous linear operator A map an

F-space X onto an F-space Y in a one-to-one manner. Then the inverse operator A-' is continuous. Proof. The operator A = (A-')-' maps open sets on open sets. Hence the operator A-' is continuous. COROLLARY 2.3.3. Suppose we are given in an F*-space X two different F-norms II II and II II, Suppose that the space X is complete with respect to the both F-norms. Suppose that the F-norm II I Ii is stronger than the F-norm II II. Then the two F-norms are equivalent. Proof. The operator of identity mapping (X, 11 III) into (X, II II) is con-

Chapter 2

44

tinuous. Then by Theorem 2.3.2 the inverse operator is also continuous. This implies that the F-norms 1111 and 11 ill are equivalent.

Let X be a linear space. A family of linear functionals F defined on X is called total if the fact that f(x) = 0 for all f e F implies that x = 0. COROLLARY 2.3.4. Suppose we are given in an F*-space X two different F-norms II II and II 111. Suppose that X is complete with respect to both F-norms. Suppose that there is a total family F of linear functionals which

are simultaneously continuous with respect to both F-norms. Then the F-norms II II and II III are equivalent.

Proof. We shall introduce a new norm in the space X by the formula 11x112 = IIxII+IIxIIl

and we shall show that the space X is complete with respect to the norm 11

112. Let {xn} be a fundamental sequence in the F-norm II 112. Then it is also

a fundamental sequence in the F-norms II II and II III, because the F-norm 112 is stronger than the F-norm II II and the F-norm II III. Since X is complete in II II and II J. the sequence {xn} tends to x° in the norm II II and to xl in the norm II III. Suppose that x° r x1. Since the family F is total, there 11

is a functional f e F such that f(x°) both norms implies that

f(xl). The continuity of f in the

f(x°) = lim f(xn) =.f(xl). n--.-

and this leads to a contradiction. Therefore, x° is a limit of the sequence {xn} in the norms 11 11 and 11 III. Thus it is a limit of the sequence {x"} in the norm 11 II2 Hence the space (X,II 112) is complete. By Corollary 2.3.3 the

norm II II2 is equivalent to the norm II II and to the norm II IIi Hence the norms II II and II 11, are equivalent. El

2.4. LINEAR DIMENSION AND THE EXISTENCE OF A UNIVERSAL SPACE

Let (X, II IIx) and (Y, 11 IIY) be two F*-spaces. We say that the space X and

Y are isomorphic if there is a linear operator A mapping X onto Y in

Linear Operators

45

a one-to-one manner such that the operator A and the inverse operator A-1 are continuous. We say that an F-space (X, II II x) has a linear dimension not greater than an F-space (Y, II IIr'), and we write

dimzX < dim,Y if the space X is isomorphic to a subspace of the space Y.

If dim,X < dimi Y and simultaneously dim, Y < dim,X, we say that the spaces X and Y have the same linear dimension (Banach and Mazur, 1933) and we write it as

dimiX = dim, Y.

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 non-isomorphic 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

dimiX < dim, Y

if dim,X < dim,Y and dim,Y # dimtX. Let X be a family of F*-spaces. An F*-space X0 is called a universal space with respect to isomorphism, or briefly a universal space, for the family X if for every X e X there is a subspace of X0 isomorphic to X (i.e. dimjX < dimjX0). 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, 11 I1a), a e A}. Let Xu be the space of generalized sequences x = {xa}, a e A, xa a Xa, such that IIxuI = Sup Ilxa IIa < +0o. aEAl

We determine the addition and the multiplication by scalars as follows {xa}+{ya} _ {xa+ya}, t{xa} = {txa}. It is easy to verify that (X.., II II) is an F-space. Space Xb is isomorphic to the following subspace of Xu

Xa= (xeX.: xa=0fora=,Z= b).

Chapter 2

46

Unfortunately the space X. obtained in this way does not necessarily belong to the family X. The problem of universality is a problem of finding a universal space belonging to the family X. Kalton (1977) has shown that there is a universal separable F-space 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 b e A. Let E[c] = {a a E: a < c}. If, for each c e E, the set E[c] is a finite chain, we say that E is tree-like. PROPOSITION 2.4.1 (Kalton, 1977). If E is countable and tree-like, then there is a one-to-one non-decreasing mapping of the set N of positive integers onto E. Proof. We form a sequence {an} such that each element of E occurs in the sequence infinitely many times. We shall now define by induction an increasing sequence {mn} of positive integers as follows. As m1 we take

the smallest n such that E[an] = {an}. Suppose that elements am,, ..., amk_1 are chosen. As Mk we shall take the smallest integer n such that

n > mk_I {amt, ..., amk_1, an} is a section of E, an 0 {amt, ..., 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. Let J(E) be the set of all real valued functions defined on E and vanishing everywhere except a finite number of points (the functions of this type will be called finitely supported). By T(A) we denote the set of those elements of 9(E) which have a support contained in A. By PA we shall denote the natural projection PA: 7(E)-->Y(A), PAx = xXA. By a consistent family of F-norms we mean a collection {11 tic, c e E} of F-norms such that 11

11, is an F-norm on J(E[c]),

(2.4.1.i)

if a < c, then for x e 9'(E[a]), lixHHa. _ IixIic,

(2.4.1.ii)

lltxjJc is non-decreasing for t > 0.

(2.4.1.iii)

Linear Operators

47

For a given consistent family II IIc, we define the limit F-norm II II on Y(E) as follows n

IxII =inf {

n

uj = x, ui c- 9(E[ci]), n EN}.

Iluillcf:

i=1

a=1

Observe that II II satisfies the triangle inequality

Iix+yll < IIxII+IIxII In deed, n,

x+yll = inf

It,

Ilwillci:

wi = x+y}

i=1

i=1

n,

n,

n,


i=1

ti=1

n,

vi = y} i=1

n,

= inf

Iluillct: Y ui = x} i=1

i=1 n,

n,

+ inf =1

Ilvillct: Y vi = y} = IIxII+llyll i=i

Since (2.4.1.iii) holds, IItxii is non-decreasing for t > 0. Observe that for x e ¶(E[c]), IIxII < IIxIIc Thus IIxII is an F-pseudonorm (i.e. satisfies conditions (n2)-(n6). PROPOSITION 2.4.2 (Kalton, 1977). Let Ebe tree-like and let {11 I Ic, c E E} be a consistent family of F-norms. Let I I II be the limit norm of this family. Then

If A = suppx (the support of x), then IxII = inf {Z Ilualla:

ua E J(E[a]), I ua = x}.

(2.4.2.i)

aed

If x e 9(E[c]), then IIxII = IIxIIc.

(2.4.2. ii)

IxII is an F-norm on 9(E), i.e. I lxl l = 0 if and only if x = 0. (2.4.2.iii)

Proof. (2.4.2.i) Let e be an arbitrary positive number. Let ui E 9(E[ci]) (1 < i < n) be a minimal collection such that

ul+...+un=x,

48

Chapter 2 n

t=1

Iludllci < IIxII+E.

Since the family {II IIc, c E E} is consistent, we may assume without loss

of generality that cd = max(supput). To prove (2.4.2.i) it is sufficient to show that for each i there is an at E A = supp x such that cd < at. Suppose that the above does not hold. Then there exists a maximal cj = c

(j being some index not greater than n) such that c r a for any a e A. n

ud(c) = 0. Since uj(c) * 0 and c is

This implies that c 0 supp x. Thus d=1

maximal, there is an index k r j such that ck = c = cj. Therefore

Y, ud+(uj+uk) = x i#2, k

and

IIudllct+Iluj +ukllc < IIxII+E i#j, k

contradicting the minimality of the collection (u1, ..., un). (2.4.2.ii) This follows immediately from (2.4.2.i); if ua E J (E[a]) and

ua=x,a
IIuaIIa = I, IIuaIIc > II f U. IIc = IIXIIc aEA

aEA

aEA

(2.4.2.iii). If IIxII = 0, then for each positive integer n there exist ua e J(E[a]), a c- A = supp x such that

Iua=X aEA

and

Y, Ilualla< aEA

n

Then, for each fixed a, lim IIuaIIa = 0. Since 9(E[a]) is finite-dimensional,

lim ua(a) = 0. Since the support A = suppx is finite, x(a) = 0 for all

>H_

aeA.Thusx=0. An F-norm II II on J(E) is called monotone if IIPA(x)II < IIxII for any section A C E.

49

Linear Operators

PROPOSITION 2.4.3 (Kalton, 1977). Suppose that E is tree-like and that {II

Ile, c e E} is a consistent family of F-norms such that II PE[a] (x)II, < Ilxlle

for a < c. Then the limit F-norm II II is monotone. Proof. Suppose that A is a section of E. Let x e T(E). For any s > 0 we can find ut, ci, i = 1, ..., n such that ut e 9(E[c{]), x = u1+ ... +un and IIutIIc1+ ... +Ilu, Ian < IIxII+s. Since A is a section, A n E[ct] = E[at] for some ai < ct. Hence n

n

IIPA(x)II <

IlutLt < IIxII+s.

IIPA(ut)IIc< < =1

i=1

The arbitrariness of s completes the proof. Let N denote as before the set of all positive integers. By 9(N) we denote the set of all finitely supported sequences. For each n e N we define

en={0,0,...,0,1,0,...}. n-th place

The space 9(E[n]) will be denoted more traditionally by R'n. Of course Rn = lin{e1i ..., en}. Let 0 be the family of all F-norms defined on `9(N) and let 0n be the family of all F-norms defined on R. Now we shall determine the distance between two F-norms defined on an F-space X as follows. Letp(x), q(x) be two F-norms on X. Let

d*(p, q) = sup arc tan tog P(x) q(x) xex x#0

It is easy to verify that d*(p,q) is well determined and that it is a metric. Indeed,

(1) d*(p,q) = 0 if and only if p(x) = q(x), (2) d*(P,q) = d*(q,p) since log

p(x)

-log

(3) d*(p,q) = sup arc tan zEX

a*0

sup arc tan zE%

z#0

log P(x) +log r(x) r(x) q(x)

p(x) =11 q(x)

Chapter 2

50

sup arc tan log P(!)- + sup arc tan log r(x) r(x)

zEX

zoo

ZEX

q(x)

X760

d *(p, r) +d *(r, q).

Thus we can treat (0, d) and (0n, d) as metric spaces. Let Jn denote the It is easy to see that J,, is not expanding, restriction mapping Jn: i.e., d*(Jnp, Jnq)
d(p, q) = sup log P(x) ZEX

x#0

q(x)

The function d(p,q) is not a metric. It can be equal to -boo. Observe that for any e > 0 there is a S > 0 such that d(p, q) < S implies d*(p, q)

<e. We shall now describe the general construction of universal spaces. We

shall suppose that 9) is a set of F-norms defined on 9(N) satisfying the following conditions :

Jn(9) is separable for each n E N

(2.4.3.i)

for each p e 9, 4 E Jn(9) there is q c- 9 such that (2.4.3.ii) Jn(q) = 4 and d(p, q) = d(JJ(P), q). Take a dense countable subset El of J1(9). Then, by induction, we

is chosen, choose pick a countable subset E. of Jn(9) as follows : if for each p E E._1 a dense countable subsetA(p) of J.(9) n Jn(J,,-11(p)).

Then E. = U A(p). PEEn-1

Let E = U E. We define a partial order < on E as follows, p < q n=1

if p is a restriction of q, i.e., p E Em, q E En, m < n and p(x) = q(x) for x e R". Next we define a consistent family { II 11p, p E E} of F-norms. If p c- Em, E[p] = {pl, Pm}, where p, = P IB-1 .. Pm = PI R. = p. We define

,

m

IixII? = P(

'x(p)e). i=)

Linear Operators

51

Let IIxli be the limit F-norm of the family {II IIp, 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 one-to-one mapping (P of C onto N such that the linear operator T: 9 (N)-->9 (C) defined by

T(X)la = x((p(a)) satisfies the following inequality :

(1-E) P(x) < IIT(x)II < (1+e)P(x).

(2.4.4)

Proof Let {fin} be a sequence of positive numbers such that 00

(1+17n) < 1+E.

(2.4.5)

n=1

Letpo = p. We define by induction a sequence p,, e 9 such that

(1+nn)-1Pn-1(x) ( Pn(x) < (1+n.)Pn-1(x) for n > 1, (2.4.6.i)

for n , 2, (2.4.6.ii) Jn-1(pn) = Jn_1(pn_,) for n , 1. (2.4.6.iii) Jn(pn) a E. Suppose that {pk: k < n} are chosen. Then, by (2.4.6.iii) Jn_1(p7_1) E e En_1(this condition is vacuous if n = 1). Find q e A(pn_1 I R1) such that for x e Rn. (1+77n)-1Pn-1(x) < q(x) G (1+i)n)p _1(x) By (2.4.3.ii) there is a pn(x) e J) such that

(1+yln)-1Pn-1(x) CPh(x) < (1+nn)Pn_1(x) and

Ja(Pn) = q. by (2.4.6.ii), is a chain in E and it is Since q e clearly maximal. Let C = {JJ(pn): n e N} and we define

(Jn(pn)) = n

for n > 1.

For x e Rn II T(x)II = Pn(x)

and by (2.4.5)

(1-E)P(x)
Chapter 2

52

For the construction of a proper consistent family of F-norms we need the following notions and lemmas. We say that an F*-space (X, II II) is a #F*-space if X does not contain arbitrarily short lines, i.e. inf sup IItxII > 0. ZEX tGR X00

The flF*-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 X x 12 contains then a dense #F*-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 = (ynJ_xn, m) . Let X0 = lin {zn , m, n

n = 1, 2, ..., m = 1,2, ...}. Elements zn, m are linearly independent and each z e X. is a finite sum N

Z = I tn,m Zn,m n,m=1

Therefore, if z 0, then the natural projection of z on 12, Pz, is different from zero. This implies, by the definition of the norm in the product of spaces, that inf suplltzll < inf suplltPzll = +oo. Z960

ZEXXN

tER

Z960

teR

ZEX X N

Thus X0 is a #F*-space. Observe that

(ym, 0) = lim Z.'.. Hence (y,., 0) E X0. Therefore (0, x.,.) = n(zm,n-(ym,0)) a X0. Thus X0 is dense in Xx 12. LEMMA 2.4.6 (Kalton, 1974). Let (X, II II) be a finite-dimensional F-space. Let the norm II II be non-decreasing, i.e., let IItxII be non-decreasing for

t > 0 and each x # 0. Let c = inf suplltxII. ZEX ZER

Z#0

Linear Operators

53

Then, for all a < c, the set K = {x: IIxII < a} is compact. Proof. Since X is finite-dimensional 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 a K. By the Bolzano-Weierstrass theorem we can choose a subsequence {xnk} such that

xnk x,tkl

-*xo, where Iyl denotes the

Euclidean norm in X. We shall show that suplltxoll < c.

(2.4.7)

Since xnk e K and the norm I I I I is non-decreasing, txnk e K for it l< 1. Let

e = (c-a)/2. Let S > 0 be such that Ixi < b implies IIxII < e. Then, for ally such that I y-tyxnkI < 6 for a certain ty, I tyl < 1, we have I tyl (c+a)/2. By the homogeneity of the Euclidean norm I I, we obtain

txo-t

xnk

<6

Ixnkl

if and only if

Itl <

6

xo-

xnk Ixnkl

Since the right-hand side of the inequality tends to infinity as k-aoo, we obtain (2.4.7) and this leads to a contradiction with the definition of c. E THEOREM 2.4.7 (Kalton, 1977). There is a universal separable F-space for all separable F*-spaces. Proof. We shall first prove the theorem for separable #F*-spaces. Without loss of generality we may assume that we consider I9F*-spaces with norms p (x) such that p (tx) are non-decreasing and

sup p(tx) = 1

for all x e X, x 0 0.

tER

Indeed, if p does not satisfy (2.4.8), we write

p*(x) = min (1, pax)1,

(2.4.8)

Chapter 2

54

where

o < a < c = inf sup p(tx). zE% tER X00

The norm p*(x) is equivalent to the norm p(x) and satisfies (2.4.8). Let 9 be the class of all F-norms on R(N) satisfying (2.4.8) and such

that for each n e N there exists a neighbourhood Un of Rn such that if x e Un and its < 1 then p(tx) _ ItIp(x).

We shall show that the family

satisfies conditions (2.4.3.i) and Ixt

(2.4.3.ii). Let n e N and let V,, = {xERn: +=1

be the set of all F-norms p (x) defined on Rn such that

supp(tx) = 1 for all x

(2.4.9)

0, x e Rn,

if x e Vm, I t I < 1, then p(tx) = It Ip(x).

<1

1(.

Let Q (m, n)

m/ (2.4.8') (2.4.9')

Observe that AM C U Q (m, n). We shall now show that Q (m, n) is m=1

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). Let p e Q (m, n). Let 0 > 0. Let

K= {x: x e Rn, p(x) <

e-

e 2

}'

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

y = inf p (x) .

(2.4.10)

zE Sat V-

Then choose q e Q such that for x e K

Ip(x)-q(x)I < e --f By(l-e T B). Now we consider three cases :

(2.4.11)

Linear Operators

55

(a) x 0 K. Then there is a t < 1 such that p (tx) = e -T' . Since y < e T and e_x is a convex function

q(tx)> a-TB-yeTB(1-e TB)>e TO-e-0 ll-e TB) 1

e T B+e TB-e- e> a B.

Therefore q(x) > e-0 and q(x)

< ear

since p(x) < 1. On the other hand

A) <

e-0/z = e72 0 < ee.

(b) x e K\Int Vm. Then, by (2.4.10), (2.4.11) and the convexity of the function e l,

q(x)<1+ e T B(1--e T B) < 1+e p(x)

p (x)

(1-e B)

<2-e TB<e TB<e , e T 0 (1-e T B) -T1 B-e-To) q(x)<1+ q(x) <1+ 1-e Tle (1-e-10 T)

P(x)

1-e Te+e°

<e TB<e°.

(c) x e Int Vm. Then there is s > 1 such that sx a aVm. Hence sx < log q(S-)

0

and by (2.4.9')

log p (x) < 0. q(x) Thus in all three cases d(p, q) < 0 and Q is dense in Q (m, n). This implies that Jn(7) is separable and (2.4.3.i) holds. We shall now show (2.4.3.ii). Let p e 9) and let q E J(9)) be such that (2.4.12) q) = 0 > 0.

Chapter 2

56

Let

p*(x) = min(eep(x), 1).

(2.4.13)

Of course p* e 9). Let

q(x) = inf{q(y)+p*(x-y): y e Rn}. q(x) is an F-norm. 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* (x-yk)->O. 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 F-norm. Indeed, q (x) can be identified with the quotient norm of the space Xx Rn with the norm II(x,y)II = q(y)+p*(x-y), with subspace (0, Rn).

Observe that p*(x-y) > q(x-y) for x e R. This follows immediately from (2.4.12) and (2.4.13). Hence for x e Rn,

q(x) = inf {q(y)+p*(x-y): y e Rn} inf {q(y) } g(x-y):y a Rn} = q(x). The converse inequality is obvious. Thus q(x) = q(x) for x e Rn, i.e. Jn(q) = q. By (2.4.13) q(x)
q(y)+p*(x-y) > e-Bp(y)+p(x-y) e-B(p(y)+p(x-y)) > e-Bp(x) so that q(x) > e Bp(x). Thus d(p,q) = 0. We have to show that q e 9). Clearly, for x e T (N),

sup q(tx) < supp*(tx) < 1. teE

IER

Suppose that supq(tx) < b < 1. Then for any t e R there are ut and vt tER

such that tx = ut+Vt, ut a Rn, q(ut) < b and p(vt) < b. The set V = {z a Rn: g(z) < b} is compact by Lemma 2.4.6. Let W = {z e lin (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) = 1 for all x 0 0, and (2.4.8) holds.

Linear Operators

57

Let m > n. By condition (2.4.9') there exist em > 0 such that

if y e Rn, q(y) <em, ItI < 1, then q(tx) = Itlq(x),

(2.4.14.i)

if y e Rm, p(y) < em, I t I < 1, then p*(tx) = I tip*(x). (2.4.14.ii)

Now suppose that y e Rm and q(y) < em. Suppose that It I < 1. If y = u+v, u e Rn, q(u)+p*(v) < em, then q(tu)+p*(tv) = I tI (q(u)+P*(v)) and

q(tv) < I tI q(y)

We shall show that the strong inequality does not hold. Indeed, suppos

that ty = u+v and q(u)+p*(v) < t a(y). Then q(u) < emI tI and p*(v) < EmI tI. Let s denote the largest real number such that q(su) < em. Then

q(u) = s'em, and so s > ItI-1. Thus q(t-lu) < em and p(t-lv) < em. Hence q(t-lu)+p*(t-lv) < q(y) and this contradicts with the definition of q.

Thus, by Proposition 2.4.4, there exists a countable dimensional' ID such that, for all p e 9, (9(N),p) is isomorphic to

F*-space (Z, 11

a subspace of Z. Now let p be any F-norm on 9(N) such that( J(N),p) is a ,9F*-space. We shall show that p is equivalent to some q e 95. Without loss of generality we may assume that p(tx) > I tI p(x)

for ItI < 1.

(2.4.15)

Indeed, if p(x) does not hold (2.4.15), we can replace p by a new equivalent norm POX) ) - sup Pl(xt>1 t

Let p*(x) = (p(x))1i2. Of course

p*(tx) > ItI1i2p(x)

for Iti < 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 = tin {a,, } for a sequence {a,4} of linearly independent elements.

Chapter 2

58 n

p(x) = inf {T Okp*(uk): uk a Rk, u1+ ... +un = x, n e N}. k-1

By the triangle inequality we have

p(x) < p(x) < 2p(x). If x e Rn, it is easy to show that n

Okp(uk): u1+ ... +un =

p(x) = inf I

l

x}

k=1

and

p (x) < Onp (x)

Let {M{} be an increasing sequence of positive numbers. We define

q(x) = inf {p(Y)+ 00j,Mil x(i)-Y(i)I : Y e 7 (N)). i=1

It is easy to show that q(x) is an F-pseudonorm.

If P*(Mnlen) < 2n

(2.4.17)

where en = {O, ..., 0,1,0, ...}, then q(x) <j(x) < 2p*(x). Suppose that n-th place

for a sequence {xn}, q(xn)--0. Then, from the definition of q, xn can be represented as a sum xn = un+vn such that p(un)-+0 and E M{lvn(i)I-*0. i=1

If X MMIvn(i)I < 1 then jvn(i)j < M;1 for all i, and hence, for any r, we i-1 have under condition (2.4.17) vn(k)ek I + 2r

p*(vn) < p*( k=1

.

/

Hence lim sup p*(vn) < 1/2r. Thus p*(vn)-->0. Since p(un)-*0, p(xn)-*0.

n--

This shows that under condition (2.4.17) the F-norms p and q are equivalent.

59

Linear Operators

We shall now refine the selection of Mn by the following induction procedure. We define qk(x) on Rk as follows : k

qk(x) = inf {P(y)+f MiIx(i)-y(i)I: y E Rk}. r=1

Let b be chosen in such a way that for each m the set Rn+ n {x: p*(x) bo} is compact. The existence of such a bo follows from the fact that (J(N), p) is an P F*-space. Now we shall choose an increasing sequence of positive numbers {Mn} and a decreasing sequence of positive numbers {bn}, so that

P*(Mn en) < Zn ,

n = 1, 2, ...,

bo, n = 1, 2, ... ,

bn <

if u e Rn-1 and p*(u) <

(2.4.18.a)

(2.4.18.b)

bo, then qn(u) = gn_1(u),

for n > 2,

(2.4.18.c)

if u e Rn, p*(u) < bn, then gn(tu) = It I gn(u)

fortl<1.

(2.4.18d)

To start the induction pick an M1 such that (2.4.18.a) holds and 61 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 M1 < M2 < ... < Mn_1 and b1 > b2 >

... > 6n-1

have been chosen. The set {x a Rn : 6n__1
Ilxlln = fix(=)I, 1=1

we may choose Mn so that (2.4.18.a) holds, and moreover if 4_1 < p*(x) < bo and IIYIIL < Mn 1 6, then p*(x-y) > Bn 1 Bn-1p*(x) and P*(Y) < 4 bo. 1

We shall now show that (2.4.18.c) holds. Let x e Rn-1 and p*(x) < 4 bo.

Chapter 2

60

Suppose that qn(x) < gn_1(x). From the definition of q, we have

qn(x) = inf{gn_1(u)+MnIIvII1n+0np*(w): u+v+w = x, u e Rn-1, v, w c- Rn}.

And by our hypothesis there are u, v, w such that x = u+v+w and qn-1(u)+MnIIvIIn+enP*(w) < q.-,(x).

(2.4.19)

Hence

MnIIvIIn+enp*(w) < q.-1(x')-q.-1(u) < qn-1(v+w).

(2.4.20)

On the other hand, q.-1(x)
a

60.

IIv!In<Mn180 and, by the choice of Mn, p*(v) < -4160. Therefore

p*(v+w) < p*(v)+p*(w) < a 80+z 80 = b0. First suppose that p*(v+w) and hence

bn-1, then p*(w) > 0-' 0. _,p*(v+w),

MnIIvIIn+Onp*(w) > 0n-ip*(v+w) > q,-1(y+w),

which contradicts (2.4.20).

Next suppose p*(v+w) < 6n_1 and choose t > I so that p*(t(v+w)) Then either 11tvill > M,-,'60 or p*(tw) > 0n_10n '6,, -v In both cases MnIIvIIn+Bnp*(tw)

en-1 bn-

and hence, as v+w a Rn-1, MnIIvIIn+Onp*(w)

ItI-ten-1 bn-1

> I tI -1gn_t(ty+tw) = gn_1(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 {jM{Ix(i)I : x e Rn, p*(x) = i=1

Linear Operators

61

Choose S < min (bn_1i MD. If p*(x) < bn, then qn(x) < p (x) < 2p*(x) 26n and if y e Rn and

P(Y)+,MsIx(i)-Y(i)I < 2bn, P*(Y)<2bn. i=1

Hence, for some t > 1, p*(ty) = So. Then by (2.4.16) 6, < It 1112 p(y) and n

YMMIY(i)I
Hence n

n

n

Mtlx(i)I <MiIY(i)l+fMix(i)-Y(i)I $=1

i=1

l=1 0



Then, by the definition of qn(x), we obtain qn(x) _ EMtIx(i)I, provided i=1

p*(x) < bn, and so gn(tx) = It I qn(x) for I t I < 1 and p*(x) < bn. Thus (2.4.18.d) is also satisfied.

To conclude, we observe that if x e R", then q(x) = limq,n+n(x). Hence if p*(x)
inf sup q(tx) = d > 0 z#0 tER

and hence q(x) = min(d-lq(x), 1) belongs to 9 and it is equivalent to q. The space (Z, II II) constructed above is countable-dimensional and it is universal for all countable-dimensional /9F*-spaces. Now we shall show

that the completion Z of the space Z is universal for all separable F*spaces. Let X be an arbitrary separable F*-space. By Lemma 2.4.5 X x l2 contains a dense countable dimensional subspace X0, being #F*-space. Thus there is a subspace Xo,1 of the space Z isomorphic to A, Since X0 is

Chapter 2

62

dense in Xx 12, the completion Xo,1 of the space X0,1 contains a subspace isomorphic to the space Xx 12. Therefore the completion Z of the space Z contains a subspace isomorphic to the space Xx 12 and it trivially implies that Z contains a subspace isomorphic to X. 2.5. LINEAR CODIMENSION AND THE EXISTENCE OF A CO-UNIVERSAL SPACE

In this section a notion in a certain sense dual to the notion of linear dimension will be considered. We say that an F-space X has the linear codimension not greater than an F-space Y,

codimlX < codiml Y,

if Y contains a subspace Yo such that the space X is isomorphic to the quotient space Y/Yo. PROPOSITION 2.5.1. Let X and Y be two F-spaces. codimlX < codimiY 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 Yo of the space Y such that the space X is isomorphic to the space Y/Yo. Let To denote this isomorphism. Write T(y) = To ([y]), where [y]

denotes the coset containing y. Since To 1 is continuous, T is also continous. It is easy to observe that T maps Y onto X. Sufficiency. Let Yo = T-1(0). The transformation T induces the continuous linear operator To mapping Y/Yo onto X defined as follows To([y]) = T (y)

To is one-to-one, and hence, by the Banach theorem (Theorem 2.3.2) its inverse is continuous. Therefore X and Y/ Yo are isomorphic.

If codimlX < codiml Y and simultaneously codiml Y < codimlX, we say that the spaces X and Yhave the same linear codimension and we write it codimlX = codiml Y.

If two spaces X and Y are isomorphic, then obviously. codimlX = = codiml Y. We say that the linear codimension of an F-space X is less than the linear

Linear Operators

63

codimension of an F-space Y, codim1X
space of the space LP([O, 1] x [0, 1]) of the functions depending only on the first variab e is not isomorphic to the space LP([0,1]) (see Kalton, 1981b). Problem 2.5.1. Let (Y, II II) be an F-space. Does there exist a subspace Yo C Y such that the quotient space is an infinite dimensional separable space ?1

Let I be a family of F-spaces. We say that an F-space X. is universal for a family X with respect to the linear codimension (or briefly co-universalfor family .) if, for each X e X codimjX < codimiXX. In the same way as in the case of linear dimension we can construct a trivial co-universal space for any family X of F-spaces. Namely we enumerate all members of family X. We construct the space Xu as in the previous section. Let Xb = {{X.}: xb = 0}. The space Xu/Xb is isomorphic to the space Xb. The problem of co-universality for a family X consists in finding a space X. e I co-universal for the family X. THEOREM 2.5.2 (Kalton, 1977). There is a separable F-space which is couniversal for all separable F-spaces.

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 {N,,(u)} of continuous

'non-decreasing functions (defined for u > 0). Suppose that N.(0) = 0 if and only if u = 0, n = 1, 2, ... Suppose that the sequence N (u) satisfies a uniform (L12) 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 PN(x) _

Nn(xn) n=1

is a metrizing modular on the space of all sequences. Let Nn(1) be the space of all such x that pN(x) < +oo. By Theorem 1.2.4 the space Nn(1) is linear. Moreover, the modular pN(x) induces a metric on this space such that it is

a metric linear space. In examples which will be interesting for us the functions Nn will satisfy a stronger condition than (2.5.1), namely

Nn(u+w) < Nn(u)+Nn(w).

(2.5.2)

In this case the modular pN(x) is an F-norm. It is easy to verify that Nn(l) is complete. Indeed, let {xk} _ {xk} be a fundamental sequence. Then, for each i, the sequence {xk} is also fundamental. This implies that there is a limit

i = 1, 2, ...

lim xi = xi, co

Observe that for an arbitrary e > 0 there is an index K such that, for k, k' > K 00

Nn(xn-Xn <

PN(Xk-Xk') _ n=1

Then passing, with k' to infinity, we get PN(Xk-X) < E,

where x = {x:}. The arbitrariness of a implies the completeness of the space Nn(1).

PROPOSITION 2.5.3 (Turpin, 1976). For any separable F-space (X, 11 11), there are a space Nn(1) 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) = 1juxnJI. Let T({a{})

_

n

aix{. By the definition of Nn(1) it follows that the operator T is well r=1

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 e X can be represented in the form 00

x=IXni, i=1 1

where IlxnilI < y j. Modifying the construction of Turpin we can obtain a stronger result. PROPOSITION 2.5.4 (Bessaga, 1982, oral communication). For any separable F-space (X, I II) there is a space Nn(1) such that for any subspace Y C X there is a continuous linear operator T mapping Nn(1) onto Y. I

The proof is based on the following LEMMA 2.5.5. Let (Y,

11

11) be a separable F-space. Let -1 sup {IIYII: Y e Y}

> 6. Then there is a sequence {yn} dense in Y such that sup Iltynll > 6 for £ER

n = 1,2, .. Proof. Let y e Y be an element such that suplltyll > 26. Let {xn} be an WE

arbitrary dense sequence in Y. We put xn

Yn =

xn+

if sup II txnll > 6, tell

1y

otherwise.

It is easy to verify that yn has the requested property. Proof of Proposition 2.5.5. Without loss of generality we may assume that the norm II II is non-decreasing, i.e. IItxII < IIxII for It I < 1. Basing on Lemma 2.5.5 we can find a number 6 > 0 and a sequence {yn} C Y such that IIYnII > 6 for n = 1, 2, ... and

U tt

{tyn} = Y.

(2.5.3)

Chapter 2

66

Let the space Nn(1) be constructed as in Proposition 2.5.3. For each ym we can find xnnti such that (2.5.4)

IIYm-X.. II < 2m . Let T: N.(1) ->. Y be defined as follows 00

T({tn})

= I tnmYm M-1

Since the norm II II is non-decreasing and Ilynll < 6, inequalities (2.5.4)

imply that the operator Tis well defined and continuous. Since (2.5.3), for every y e Y there is a sequence of reals {am} such that

I 00

Y=

am ym

(2.5.5)

m=1

and

IIamymll <2m.

(2.5.6)

By (2.5.6) the sequence t = {t.}, tn =

an {0

for n = nm, otherwise

belongs to the space Nn(1) and T(t) = y. Proof of Theorem 2.5.2 (Bessaga, 1982, oral communication). By Theorem 2.4.7 there is a universal separable F-space (X, II II). Using for this space Proposition 2.5.4 we obtain a co-universal space.

For a given separable F-space (X, II II) the construction of Turpin can give two non-isomorphic spaces. Indeed, if {xn} and {yn} are two dense sequences such that inf sup II txnll > 0 and inf sup Iltynll = 0. then obn

£ER

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 Iltxnll > 0 and inf sup Iltynll > 0, then the spaces induced by sex

tER

n

teR

Linear Operators

67

quences {xn} and {yn} are isomorphic. Indeed, there are 6> 0 and sequences of scalars {rn} and {sn} such that n = 1, 2, ... IlSnynll > 6, lrnXnll > 6,

(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(1), ...

..., k(n-1) such that 1rnX.-Sk(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 FFSPACES Let (X, II I) be an F-space. A sequence {en} of elements of the space X is

called a Schauder basis (Schauder, 1927) (or simply a basis) of the space X if every element x e X can be uniquely represented as the sum of a series Co

x = f tcei.

(2.6.1)

A sequence {gn} C X is called a basic sequence if it is a basis in the space Ygenerated by itself, i.e. it is a basis in Y = lin{gn}. Evidently, if an F-space has a basis, then it is separable. For any x of the form (2.6.1), let n

Pn(x) _ I it e{. i=1

THEOREM 2.6.1. The operators Pn are equicontinuous.

Proof. Let Xl be the space of all scalar sequences y = {77{} such that the W

series

niei is convergent. The arithmetical rules of the limit trivially

imply that X1 is a linear space. Let

IIyII* = supll YnteiII. n

ti=1

Chapter 2

68

The space X1 with the norm II I I is an F*-space. We shall show that it is

complete, i.e. that it is an F-space.

Suppose that a sequence {yk}e X1 is a Cauchy sequence. Let yk _ {4}. Since {yk} is a Cauchy sequence, for an arbitrary s > 0 there is a positive integer mo such that, for m, k > mo, n

Il ym-ykll * = sup n

i=1

<e.

(gym - q/ )et

(2.6.2)

Hence n-i

ifit) et

IAn -nn)enl l < II i=1

nl [ + u (gym-,J ) et

< 2s

i=1

Consequently

Jim ('2m-'))=0,

m,k- oo

n=1, 2, ...

Thus the sequences of scalars {fin } are convergent for all n. Write 27n = lim 71' . n- co

Passing with k to infinity in inequality (2.6.2), we obtain that for m > mo n

(r/i -,qt) et iI < 2e.

sup n

(2 6.3)

i=1

Let n

s"' =

and

i?m et

sn =

?It et.

t=1

t=1

Taking into account inequality (2.6.3) we obtain Il sr-snll < II

s'n-sm II+2s

for all in > mo and all n and r. Fix m1 > mo. The sequence {s"} is a Cauchy sequence. Therefore there is a number no such that, for n, r > no, I

sn'-sn' II < s.

Linear Operators

69

Hence, for n, r > no,

Ilsn-s, If < 3e. Thus the series' nt es is convergent and y = {,71} E X1.

t-1

From inequality (2.6.3) follows n

sup n

(rlm-t7,)es <2e ti=1

for m > mo. Hence the space X1 is complete. Let A be an operator mapping X1 into X defined as follows W

A(Y)=1171e{. =1

By the definition of the space X1 the operator A is well-defined on the whole space Xt. The arithmetical rules of the limit imply that the operator A is linear. Since {en} is a basis, the operator A is one-to-one and maps X, onto X. The operator A is continuous, because co

i7iet I < sup

IA(Y)jj _

n

%=1

niet

= 11Y11*-

4=1

By the Banach theorem (Theorem 2.3.2) the inverse operator A-' is also continuous. Hence n

lipn(x)Il _ 1271 es < ILYII* = 11A1(Y)!I 11

%=1

and the operators P. are equicontinuous. Let x e X be reperesented in the form (2.6.1). Let

fn(x) = tn. It is easy to see that fn are linear functionals. They are called basis functionals.

Observe that

fn(x)en = Pn(x)-Pn-,(x). Thus from Theorem 2.6.1 immediately follows

Chapter 2

70

COROLLARY 2.6.2. The basis functionals are continuous.

Suppose we are given two F-spaces X and Y. Let {en} be a basis in X and let be a basis in Y. We say that the bases {en} and {f.} are equivalent co

if the series

OD

ties is convergent if and only if the series

tifi is con-

vergent. Two basic sequences are called equivalent if they are equivalent as bases in the spaces generated by themselves. THEOREM 2.6.3. If the bases

and {f.} are equivalent, then the spaces

X and Y are isomorphic. Proof. Let T.: X->- Y be defined as follows n

..(x)= Ttiei)ttfs i=1

i=1

By Corollary 2.6.2 the operators T. are linear and continuous. The limit T(x) = lim T,,(x) exists for all x. Thus, by Theorem 2.2.3, T(x) is continuous. Since the bases are equivalent, the operator, T is one-to-one and maps X onto Y. Thus, by the Banach Theorem (Theorem 2.3.2), the inverse operator T-1 is continuous. The following theorem is, in a certain sense, converse to Theorem 2.6.1.

be a sequence of linearly independent elements in X. Let X, be the set of all elements of X THEOREM 2.6.4. Let (X, 11 11) be an F-space. Let

which can be represented in the form 00

x=Itiei. i=1

Let P-(x)

ti ei. i=1

be a sequence of linear operator defined on X. If the operators P. are equicontinuous, then the space X, is complete and the sequence {en} constitutes a basis in this space.

Linear Operators

71

Proof. Since the operators Pn are equicontinuous, each element x of X, can be expanded in a unique manner in the series co

x=Ittec. Let X2 be the space of all sequences {t{} such that the series

is convergent. Let sup n

In the same way as in the proof of Theorem 2.5.1, we can prove that 11* is an F-norm and that (X,,

is an F-space.

Observe that IIxIl < 1I{ti}JI*.

(2.6.4)

On the other hand, the equicontinuity of the operators P. implies that if x-*0 then Hence the space X2 is isomorphic to the space X. Therefore X, is an F-space. By (2.6.4) the sequence {en} constitutes a basis in X. COROLLARY 2.6.5. Let X be an F-space. A sequence of linearly independent elements {en} is a basis in X if and only if

(1) linear combination of elements {en} are dense in X, (2) the operators

Pn(x) _

ttet

are equicontinuous on the set lin{en} of all linear combinations of the set {en}.

Chapter 2

72

COROLLARY 2.6.6. Let X be an F-space with a basis {en}. Let t1i t2, ... be an

arbitrary sequence of scalars. Let p1,p2, ... be an arbitrary increasing sequence of positive integers. Let

n=1,2, ...

Ittl>0,

i=pw+1

Let Pn+i

e;,

ti et. i=Pn+1

Let X2 = {lin e;,} be the space spanned by the elements {e,,}. Then the space X2 is complete and the sequence {en} constitues a basis in X2. n

ao

ttet.

ttet. Let PP(x)

Proof. Let x e X, x

t=1

ti=1

For y e X1, let n

ao

y=

n=1

an e,

Then P (y) = Pp

n +1(y)

P,,(y) _ V at e t t=1

for all y e X1. Therefore, by Theorem 2.6.1, the

operators Pn are equicontinuous and, by Theorem 2.6.4, X2 is complete and {e;,} is a basis in X1. 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 II) be an F-space with a basis {en}. Let {xn} be

a sequence of elements of X of the form 00

xk = I tk,tet

where lim tk, t = 0, i = 1, 2, ...

i=1

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 Pe+1

xkn-s=P,,+1tkn, t et l'l < En.

Linear Operators

73

Proof (by induction). Let p, = 0, xkl = x1. We denote by p$ an index satisfying the inequality Ps Xk,-[

7 fff

<e1.

Suppose that the element xkti_1 and the index pn are already chosen. The hypothesis lim tk, { = 0 implies the existence of an element xkn such kyw

that P. II

Let

=1

tk"'tell < Z En.

be an index satisfying the inequality Xkn

-f tkn, i es s=1

Then obviously Pn+,

Xkn-ftk.,setI

< en.

i=Pn+l

Observe that in Proposition 2.6.7 we can additionally require that for an arbitrary given sequence {q.} there should be {pn} satisfying the conclusion of Proposition 2.6.7 such that qn < p,,. Let X be a complex F-space 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 = x(t) belongs to X and the norm in the space Xdepends only on lx(t)l. Let It be the space of real-valued functions belonging to X. Since X is a C*-space, each x e Xmay be uniquely represented in the form x = x1-f-ix2,

where x', x$ a 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.

Chapter 2

74

Proof. Since {en} in a basis in X,, there are unique expansions of elements xl and x2 with respect to this basis :

x2=I been.

x1=>anen, n=1

n=1

Hence x can be uniquely extended with respect to the basis {en} : W

X = Y cnen, n=1

p

where cn = an+ibn. We shall now give examples of bases. Example 2.6.9 The sequence

0,...,0,1,0,...,

en

n-1, 2, ...

n-th place

is a basis in the spaces c0, 1P(0 < p < +oo), (s). This basis is called a standard basis. Example 2.6.10 (Schauder, 1927) There is a basis in the space C[0,1]. It is constructed in the following manner. We define a function uk,i(t)

(0 < i < 2k, k = 0,1, ...) in the following way:

for 0 < t < Zk and

0

uk,s(t) = 2k+1(t-

i+1 < t < 1,

for 2k < t < 2 }1

?k l

for 2i+1 2k+1

-2k+1 t- i/+l 2k


i+1 2k

Each continuous function x(t) defined on the interval [0, 11 can be uniquely written in the form Go

2s-1

x(t) = a0t+al(1-t)+z, ak,iuk,i(t), k=0 i=0

Linear Operators

75

where ao = x(1), a1 = x(O) and the coefficients ak,{ are defined as follows. We draw the chord 1(t) of the arcy = x(t) through the points of the 22k1

abscissae 2k

and we define

,

2i+ 1

ak { - x

l 2i+ 1

2k+1 )

2k+1

)'

Therefore, the sequence {t, 1-t, u0,0(t), u1,0(t), u1,1(t), u2,0(t),...}

constitutes a basis in the space C[0,1]. Example 2.6.11 Suppose we are given the following system of functions on interval [0,1] : f

hn,,(t) =

for (2n)1/r

(0

1-1 < t < 2j-1 2n+ 1 2n 2j-1 j

for 2n+1 < t < -2n, elsewhere,

n=0,1,2,..., j= 1,2, ...,2n. The system of functions

is called the Haar system. It is a basis in the spaces L"[0,1] (1 < p < +00) If there is a basis in an F-space 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 non-trivial. 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 a1i ..., an the equality

a1x1+ ... +anxn = 0

(2.6.5)

a1=a2=...=an=0.

(2.6.6)

implies

Chapter 2

76

We say that sequence {xn} is topologically linearly independent if for each sequence of scalars {a1, a2, ...} the equality

alx1+ ... +anxn+ ... = 0

(2.6.7)

a1=...=an=...=0.

(2.6.8)

implies

We say that the sequence {xn} is linearly m-independent (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 (quasi-basis, m-quasi-basis) if (1) the linear combinations of {xn} are dense in X, (2) the sequence {xn} is linearly independent (resp. topologically linearly independent, resp. linearly m-independent). Peck (1968) showed that in each separable F-space there is a quasi-basis. (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 {b1ib2, ...} such that the sequence

{blxl, blxl+b2x2i ... , blxl+ ... +bnxn, ...} is a quasi-basis. Labuda and Lipecki (1982) showed that in each separable F*-space there is an m-quasi-basis which is not a quasi-basis. A sequence {en} in an F-space X is called an M-basis sequence, if there are continuous linear functionals e* defined on E = lin{en} such that 1

e* (et)

_ 50

fori=j, for i

j,

and if x e E and e* (x) = 0, n = 1,2, ...,thenx=0. We say that a sequence {en} is equicontinuous if a*(x)en-a0 for all x e E. Kalton (see Kalton, 1979) showed that every M-basis sequence contains an equicontinuous sequence. Let X be an F-space. We say that a sequence {xn} tends weakly to x if,

for each continuous linear functional f, 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 F-space X which, for each basis

{en}, each of its bounded block bases tends weakly to 0 in X, but no basis in Xhas 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 a-finite measure space. Let A be a subset of the space E, p). We say that the set A is solid if u.e A and Ivi < Jul implies v e A. Let (X, II II) be an F*-space contained set-theoretically in E, p). We say that the space X is solid if there is a basis of neighbourhoods of zero consisting of solid sets. PROPOSITION 2.7.1 (Szeptycki, 1980). If (X, II II,,) is a solid F-space, then there is an F-norm II equivalent to 11 IIo such that (i) IIxMI = II Ixi II,

(ii) Hull < llvll if Jul < Ivi. Proof. Observe that if A, B are two solid sets in L°(d2, E, p), then the set. A+B is also solid. Thus the construction of the norm lI II follows from the construction of an invariant metric (Theorem 1.1.1).

We say that X is continuously imbedded in L°(S2, E, p) and we write X Cc L°(S2, E, p) if the identity mapping from X into L°(S2, E, p) is continous. PROPOSITION 2.7.2 (Luxemburg). If (X, II II) is a solid F*-space contained in L°(S2, E, p), then it is continuously imbedded, X Cc L°(S2, E, P).

Proof. (Szeptycki, 1968). Suppose that the proposition does not hold. Then there are a neighbourhood of zero U in L°(S2, E, p) of the form

U = U(E, a) = {u e L°(S2, E, p): p{t e E: I u(t) I > 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 0 U and IIun1I <

.

Zn

Let

En = {t c- E: Iun(t)I > a}. Since un 0 U, by the solidity of the space X we have µ(En) > 0. Let 00

E = U E. m=n

The measure of E is finite. Thus we can find an mn such that

n(En\E) < 2 n-1a, where E = U E.. m=n

Let CO

E'=nE;,,

E"=nE,,. n=1

n=1

Then 0" ,u(E'\E") =.u( n E,\ n=1 n0" E,)
CO

<(En\En) <2 n=1

(2.7.1)

Since E,, is a decreasing sequence of sets and

u(E,)? u(E.)>a, we have u(E') > a. Thus by (2.7.1)

u(E") >

.

(2.7.2)

2

Let X denote the characteristic function of the set E'. Then

aX
(2.7.3)

79

Linear Operators

almost everywhere. The space X is solid. Thus, without loss of generality, we may assume that the norm 11 satisfies the properties (i) and (ii) of Proposition 2.7.1. Therefore by (2.7.3) 11

M. IIaXII

M.

Z I um(t)I

m=n

C Y IIumII <

2-11+1.

m=n

Hence IIaXII = 0, which contradicts (2.7.2).

Let A be a linear space of measurable functions. A set E is called an unfriendly set (Luxemburg and Zaanen, 1971) for A if f 1E = 0 for all f e A. PROPOSITION 2.7.3 (Aronszajn and Szeptycki, 1966). Suppose we are given a

or-finite measure space (Q, E, p). Let A be a solid F*-space in L°(Q, 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

LEMMA 2.7.4. Let (Q, E, p) be a or-finite measure space. Let u,v e L°(S2, E, p). Then, except an at most countable number of reals E, the sets E = {t: I u(t)I +Iv(t)I > 0} and EE = {t: 0} differ by a set of measure 0. Proof. The sets E n (Q'EE) are disjoint for different E. Since p is a-finite, only a countable number of these sets has positive measure. Proof of Proposition 2.7.3. Let (p a L°(Q, E, p) be a positive function such that f (pdp = 1. The existence of such functions follows from the fact that

p is a-finite. Write p (E) = f g;dc. Let E

a = sup{A({t: u(t) :A 0}), u e A}. Let {un} be a sequence of elements of A such that an = %({t: un(t) :5L 0})->.a. By Lemma 2.7.4 there are reals E, ..., $n such that

En = {t: I ui(t)I + ... +I un(t)I > 0} = {t: vn(t)

0},

Chapter 2

80

where

vn(t) = u1(t)+se2u2(t)+ ...

(2.7.4)

Observe that µ (En) increases to a. co

Let Ed = Q\ U En. The set EA is the required unfriendly set. Indeed, n=1

suppose that E., is not an unfriendly set. Then there is a function u(t) e A such that

;u({teEA: u(t) :/= 0}) > b>0. By the definition of {un} we can find an un such that

a-an < b. Using Lemma 2.7.4, we can construct a function of the form

such

that ju({t: u(t)+Eun(t)

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, (EA) = 1-a. Hence Ed is unique up to the sets of µ-measure 0. Thus it is unique up to the sets of 4u-measure 0. PROPOSITION 2.7.5 (Aronszajn and Szeptycki, 1966). Let a measure space

(Q, E, p) be a -finite. Let A C L°(Q, E, p) be a solid F-space. 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 A. be chosen so that A Qt: I un(t)J > Mn} < 2-n,

(2.7.5)

where (E) is defined as in the proof of Proposition 2.7.3, and

2-n for

< An.

Having {Mn} and {An} we shall choose by induction two sequences {fin} and {r)n} in the following way. We define $1 = 1, 0 < 771 < 1. Sup-

Linear Operators

81

pose that 61, ..., n and 771, ..., i]n are chosen. Using Lemma 2.7.4, we can

choose n+1 such that 0 < n+1 < min ()n+1, (2Mn+1)-17n)

(2.7.6)

and

0} _ {t: IVn(t)I+Iun+1(t)I > 0},

{t:

where vn(t) is given by formula (2.7.4). We take

0 < ?ln+' - 91n.

(2.7.7)

By (2.7.5), (2.7.6) and (2.7.7) the sequence {vn} is convergent on S2.

Let W

v(t) = lim Vn(t) = > nun(t) Sr ao

n=1

Observe that 00

Iv(t)I >, IVm(t)I- Z Enlun(t)I >' r]mn=m+1

1f

27n

77m > 0

2n=m+1

on the set

Em = {t; IVm(t)I > n.} n {t: Iuj(t)I < Ms, j = 1, 2, ...}. It is easy to verify that

f[(EA\Em) < a-am+2-+2-m. Hence v(x)

0 almost everywhere on EA.

Let (Q, E, u) be a measure space. We write E ,,

,

if En is a de-

creasing sequence of sets, En c -T, such that

it (E n En) -+ 0 for every set E of finite measure. Let (X, II II) be a solid F*-space contained in L°(Q, E, p). By Xa we denote the set of those u E X for which IIUXE.IH0

for every sequence En\,,o. If Q is of finite measure, (2.7.8) is equivalent to lim uXE = 0. AE) ->o

(2.7.8)

Chapter 2

82

It is easy to verify that X. is a closed solid subspace of X and, if {En} is M

an increasing sequence of sets such that S2 = U En, then xXE* -* x for n=1

allxeX.

If X = N(L(Q, E, p), then Xa = X. If p is non-atomic and X = L°° (S2, E, p), then Xa = {0}. PROPOSITION 2.7.6 (Luxemburg and Zaanen, 1963). Suppose that p is o-finite and that (X, II II) is a solid F-space contained in L°(Q, E, p). Then a set C C X. is compact in X if and only if C is compact in L°(Q, E, p). For any sequence En\a IIuXEaII tends to 0 uniformly

(2.7.9.i)

for u e C.

(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 are e > 0, a sequence

En\R, and a sequence un e C such that (2.7.10)

II unXE.II > e.

Since un E C C Xa, we can construct by induction subsequences {En,} such that

for i <j.

II un,XEnj II < 2

(2.7.11)

By (2.7.10) we get II(un,-un,)XE.,I1 >

for i <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°(Q, E, p) 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 u in 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 (p e X positive almost everywhere. Let

En,k = t e S2: supIum(t)-u(t)I min

ap(t)

Linear Operators

83

Observe that, En, k\O for each fixed k and for any e > 0 we can find by (2.7.9.ii) an index such that E

IIvXn,kll <

3

for n > nk, v e C,

(2.7.12)

where, for brevity, we write Xn, k = XE,,,: Then

IIun-uII < IIXn,k(un-u)II +II (1-Xn,k) (un-u)II < IIXn,kunll+IIXn,kull(1-Xn,k)(un-u)II

3+3+ provided that k is chosen so that

' <e. 1

k

E

< 3

PROPOSITION 2.7.7 (monotone convergence theorem, Luxemburg and Zaanen, 1963). Let X C L°(Q, E, p) be a solid F*-space. Then u e X. if and only if for each sequence {vn} such that

I u(t)I > vl(t) > v2(t) > ... > 0,

(2.7.13)

tending to zero almost everywhere, IIvnII-*0 Proof. Let u e X, E \ , vn = uXE,,. By our hypothesis IIvnII->0. Thus

ueXa. Conversely, let u e X. and let {vn} be a sequence satisfying (2.7.13) and tending to 0 almost everywhere. Let

Em,n = {t: vn(t) <

m u(t)

Let Xm,,, denote the characteristic function of Em,n. For a fixed m,

Q\Em,n\, and

II(1-Xm,n)uII--HO since u e Xa. We have 1

IIvnII <

m

Xm,nu +II(1-Xm,n)uII

m I+II(l-Xm,n)uII

(2.7.14)

Chapter 2

84

For each s > 0 there is an m such that u

m

<

II(1-Xm.n)uII < 2

(2.7.15)

Formulae (2.7.14) and (2.7.15) imply IIvnII < E. This completes the proof.

PROPOSITION 2.7.8 (dominated convergence theorem, Luxemburg and Zaanen, 1963). Let (X, II II) be a solid F-space contained in L°(S2, E, P). Then u e X. if and only if for each sequence {un} C X such that I un(t)I Iu(t)I almost everywhere and un(t)-*u(t) almost everywhere, we have Ilun-ulI -A. Proof. Sufficiency. If En\H, then un = (1-XEA) u has the property indicated in the statement. Thus u e Xa. Necessity. Suppose that u e Xa and a sequence {un} has the indicated properties. Let

vn(t) = sup{Ium(t)-u(t)I, m > n}. The sequence {vn}(t)} is not increasing. It tends to 0 almost everywhere and vn(t) < 2I u(t)I. Then by Proposition 2.7.7 IIvnII-30. By the solidity of the space and Proposition 2.7.1 this implies that Ilun-uJI>0.

Let (Q, E, u) and (Q1, E1, p1) be two a-finite measure spaces. Let u e L°(521, E1, y,). Let k(t, s) be a function defined on the product space Q x S21 measurable with respect to the product measure 4u x ii1.

Let

K(u) = f k(t, s) u(s)dp1(s) n,

be a linear integral operator acting from L°(521, E1, pl) into L°(d2, E, It) By a proper domain DR of the operator Kwe shall mean the set

DK = lu a L°(Q1,E1,it,):.1 I k(t, s)I I u(s)I dli1(s) < + oo n,

p-almost everywhere

.

We recall that (92,X, ii) and (Q1, E1, p1) are a-finite, i.e. S2 = U 00 Stn, °, n=1

Linear Operators

85

Sl, = U Q,,,,, where p (Q,n °) < + oo and p1(Sln,1) < + oo. Thus the ton=1

pologies in L°(Q, E, p) and in L°(Q1, E1, p,) are described by sequences of pseudonorms, 11n in L°(D, E, p) and I1n in L°(Q,, E1i p1) (see Example 1.3.5). We shall introduce F-pseudonorms on the space Dg in the following way 11

11

Ilullm,n =11u1In+II IKI Jul Jim,

where IKI Jul =

Ik(t, s)I Iu(s)I dpi.

.l

a,

(2.7.16)

It is easy to see that the space D. is a solid F*-space with the topology defined by the double sequence of pseudonorms 11 II m,n THEOREM 2.7.9 (Aronszajn and Szeptycki, 1966). The space Dg is com-

plete. Proof. Let {uk} be a Cauchy sequence in D. This means that, for each fixed m, n, lim JIuk-ua'lI k,k'-- oo

(2.7.17)

= 0.

m,n

The sequence {uk} is obviously also a Cauchy sequence in the space L°(Q1i Ep1). Thus there is a u e L°(Q1i E1, p1) such that {un} tends to u in the space L°(91i E1, p1). By (2.7.17), {uk} contains a subsequence {u;} = {uk,} such that +00. ti=1

Thus the series

ivl and

1 u;+1(t)-u{(t)1

00

1KI

1u:+1-usI

ti=1

cqnverge almost everywhere on sets Q,,,1 and Qn, °i respectively. We have Go

Jul < Iuti1

+1Iui+1-u'I

i-1

Chapter 2

86

Thus by the Lebesgue theorem

IKI Jul < IKI Iuil+ j IKI i-1

lui+1-uil

almost everywhere on Similarly we can estimate 00 IKI I u-u,l .0

i-r

almost everywhere on Q.,°, as r--goo. And, by Proposition 2.7.8, II IKI Iur-ul Ilm-*0.

In a similar way we can show that II u,-ulln->o. Thus

Ilu.-ullm,n-0. Therefore u; tends to u in DR. In this way we have shown that each Cauchy sequence contains a convergent sequence. This implies the completeness of DR (cf. Proposition 1.4.8).

THEOREM 2.7.10 (Aronszajn and Szeptycki, 1966). The operator K is a continuous linear operator from DR into L°(Q, E, ,u). Proof. IIKuIIm < II IKI Iulllm < llullm,n

for all n.

THEOREM 2.7.11 (Aronszajn and Szeptycki, 1966 ; see also Banach, 1931).

Suppose that X C c L°(Q1, Ei, p1) is a solid F-space and suppose that X C DR. Then the integral operator K maps X into L°(Q, E, lC) 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 II) into (Dg, II IIDH), where II IIDE is the F-norm induced in the space Dg by-the sequence of pseudonorms II Ilm,n (see Section 1.3). Accordingly, we shall introduce a new norm on X, namely Ilxlll = IIkII+IIkIID"

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 II1) Then it is a Cauchy sequence in (X, II 11) and in (Dg, 11 II Dx) Since both spaces are complete,

the sequence {un} has limits in both of them. Let u° be the limit in X and let ul be the limit in D. Since X C c L°(521, El, p1) and DR C c L°(521, El, pl), the sequence {un} is convergent in the space L°(521, E1, pl) simultaneously to u° and to ul. Thus u° = ul. Hence IIun-u°II1 =

Therefore (X, II IIJ is complete, and by Corollary 2.3.3 the two norms are

equivalent. This implies the continuity of K. THEOREM 2.7.12 (Aronszajn and Szeptycki, 1966). Let (X, II IIx) C c L°(521, El, pl) and (Y, II IIY) C, L°(52, E, p) be given F-spaces. If

X C Dg and KX C Y, then the operator K: X-+ Y is continuous. Proof. We introduce a new norm on X, IIxII = IIxIIx+IIK(x)IIY,

and, in the same way as in the proof of Theorem 2.7.11, we show that II) is complete. We take a fundamental sequence {un} in (X,11 II). It is also fundamental in (X,11 IIx) Thus there is a u e X such that IIun-ullx->0. (X,11

By Theorem 2.7.11 IIun-ullDx-+0 and by Theorem 2.7.10 Ku.->Ku in L°(52, E, p). Suppose that IIKu.-vlly->0. Since Y C, L°(52, E, p), Ku.->v in L°(52, E, p). This implies that Ku = v. Hence (X, II II) is complete. Thus, by Corollary 2.3.3, the norms IIxII and IIxIIx are equivalent. This implies the continuity of Ku. PROPOSITION 2.7.13. (Aronszajn and Szeptycki, 1966). Let K be an integral transformation. Then (Dg)0 = D.K.

Proof. Let f e Dg. Let Eg\,e. Thus

p(Dn,l r) Eg)-.0,

n = 1, 2, ...

Therefore IIfxE,Iln->o,

n = 1,2,...

Chapter 2

88

By Proposition 2.7.7, IKI IfX$,I tends to 0 almost everywhere and, by the definition of the F-pseudonorm IIXIIm,n, IfXE

Example 2.7.14

Let Q = [0,2n] with the Lebesgue measure. Let 521= Z be the set of all integers with the discrete measure #1({y}) = 1,

y = 0, +1, ±2,...

Let k(t, s) = e'. Then Dg = L(Z) = 11. Example 2.7.15 Let 52 = Z and let Q1 = [0, 2n]. Let k (t, s) = e"'. Then Dg = L1 [0, 27t]. Example 2.7.16

Let 52 = S21 = R with the standard Lebesgue measure. Let k (t, s) = e't'. Then Dg = L1(-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 to C A for all t, 0 < t < 1. The modulus of concavity (Rolewicz, 1957) 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 -boo. 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

A+A C c(A)A. (3.1.1) Proof. Suppose that (3.1.1) does not hold. Then there is an x 0 c(A)A such that x e A+A. Since the set A+A is open, there is an r > 1 such that rx 0 A+A. Then

A+A ,

c(A) A,

because x 0 c(A)A. Since r > 1, we obtain a contradiction of the defi0 nition of the modulus of concavity. Observe that we always have c(A) > 2.

A set A is said to be convex if x, y e A, a, b > 0, a+b = 1 imply ax+by c- 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

Chapter 3

90

PROPOSITION 3.1.2. Let A be an open starlike set. If c(A) = 2, then the set A is convex.

Proof. Let x,y e A. Since A is an open starlike set, there is a t > 1 such that tx, ty e A. Since c(A) = 2, A+A e 2tA. Thus tx+ty e 2tA. (x+y) e A. Therefore, for every dyadic number r, This implies that 2

rx+(1-r)y e A.

(3.1.2)

The set A is open. Then the intersection of A with the line

L = {tx+(1-t)y: t real} in open in L. Therefore there is a positive number a such that

xo = tox+(1-to)y e A

for It,I < X

x1 = t1x+(1-tl)y a A

for 11-t1I < E.

and

Applying formula (3.1.2) for x = x1 and y = x0, we find that, for every

a such that ja - rI < E,

ax+(1-a)y e A.

(3.1.3)

Since r could be an arbitrary dyadic number, (3.1.3) holds for an arbitrary real a, 0 < a < 1. Thus the set A is convex. 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 = n sA. Thus c(A) = 2 implies s>2

that A+A C 2A. Therefore, if x, y e A, then

X +Y

y a A. This implies

(3.1.2) for every dyadic number r. Since A is closed, (3.1.3) holds. A metric linear space X is called locally pseudoconvex if there is a basis which are pseudoconvex. If moreover of neighbourhoods of zero {

c(U.) < 211p, we say that the space X is locally p-convex (see Turpin, 1966; Simmons, 1964; 2elazko, 1965). THEOREM 3.1.4. Let X be a locally pseudoconvex space. Then there is F-pseudonorms {11 jjn}, i.e. such that a sequence of IItxII" = ItIP"IIxIJn,

(3.1.4)

Locally Pseudoconvex and Locally Bounded Spaces

91

determining a topology equivalent to the original one. If the space X is locally p-convex, we can assume pn = p (n = 1, 2, ...). Proof. Let { Un}be a basis of pseudoconvex neighbourhoods of 0. Without

loss of generality we may assume that the sets U. 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 Un(24) = s°nUn

(q = 0, ±1, ±2, ...). e

For every dyadic number r > 0, -r =

at 2t, where at is equal either to t=s

0 or to 1, we put Un(r) = asUn(2s)+ ... +atUn(2t).

In the same way as in the proof of Theorem 1.1.1, we show that UU(rl+r2) ) Un(r1)+ Un(r2)

and Un(r) are balanced. Moreover, the special form of UU(r) implies Un(24r) = sn UU(r).

Let IIxIIn = inf {r > 0: x e Un(r)}.

The properties of the sets UU(r) imply the following properties of IIxIIn:

(1) Ilx+ylln < IIxjIn+IIyJIn (the triangle inequality), (2) Ilaxlln = IIxIIn for all a, Ial = 1, (3) llsn1xlln = 2°IIxIIn

Let

log 2 pnlogsn. Let IIxIIn = sup t>o

Iltell n t

By (3) IIxIIn is well determined and finite since

114. =Sup t>o

Iltxlln

tn"

= 1\t\d. sup

IItxIIn

tn.

Chapter 3

92

We shall show that IIxIIn are F-pseudonorms. Indeed, if lal = 1, then

= sup

Iltaxlln tn"

llaxlln = sup t>0

9>0

IItxIIn tP..

= Ilxlln

and (n2) holds. Let x,y E X. Then = Sup IIt(x+Y)Iin IIx+Ylln = sup llt(x+Y)Iin tP,. tP.. 1-t_
t>O

sup

IItxIIn tPn

1ACt<_8n

+ sup

IItxIIn

t P,

(IIxIIn+lIYlln

and (n3) holds. Observe that IItxIIn = sup

Ilrtxlln

= sup Ilrltlxlln It1P" r>o (rltl)P"

r>o ItIP"

= I t IP" sup 8>0

I ISXI In

Si"'

= I t I P° IIxIIn.

Hence IIxII is p.-homogeneous. This implies (n4)--(n6).

Now we shall prove that the system of pseudonorms {Ilxlln} yields a topology equivalent to the original one. Indeed, from the definition of IIxIIn+

Ilxlln < IIxIIn+

in other words, {x: kiln < r} C Un(r).

(3.1.5)

On the other hand, the sets Un(r) are starlike. Therefore, the pseudonorms IIxIIn are non-decreasing, which means that Iltxll,, are non-decreas-

ing functions of the positive argument t for all n and x e X. Then

Ilxn = Sup t>0

Itx= tPn

sup 1_< t<_ e,.

Iltxll> tPn

< IlsnXln = 2 llxlln

in other words

UU(r) C {x: kiln < 2 r} .

(3.1.6)

Formulae (3.1.5) and (3.1.6) imply the first part of the theorem.

If the space X is locally p-convex, then by the definition there is a basis of neighbourhoods of 0 {Un} such that c(Un) <21/P. Since Un are open it implies that Un + Un C 21/P Un.

Locally Pseudoconvex and Locally Bounded Spaces

93

Putting s = 2112' and repeating the construction above, we obtain a system of p-homogeneous pseudonorms determining a topology equivalent to the original one. PROPOSITION 3.1.5. Let X be a locally pseudoconvex space with a topology F-pseudonorms {II IIn}. A set given by a sequence of

A C X is bounded if and only if

an = sup {IIxIIn: x e A} <+oo.

(3.1.7)

Proof. If a set A is bounded, then for each n = 1, 2, ... there is a t > 0 such that C {x: IIxII" < 1}. Thus an < -. to

Conversely, let U be an arbitrary neighbourhood of 0. Then there are e > 0 and n such that

{x: IIxII < e} C U. By (3.1.7) e

an

Ac U.

El

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. 1-homogeneous) F-pseudonorms. Homogeneous F-pseudonorms will be called briefly pseudonorms whenever no confusion results. Locally convex metric linear spaces are called Bo 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

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: Jjxjjn < 1},

constitute a basis of neighbourhoods of 0. The modulus of concavity of the set Kn is not greater than 21". Indeed, if x, y e Kn, then IIx+yI In <2. Hence

x+Y 2P"

n

2,

i.e. x+y e 2P"Kn. Thus Kn+K,, C 2P".,,. In the particular case where the pseudonorms are homogeneous the sets K. are convex. It is easy to verify that the following spaces are B,-spaces: LP(Q,2,p) for 1

We say that a set A in a metric linear space X is absolutely p-convex,

0

Z-

L'

Fig. 3.1.1

If a set A is absolutely p-convex, 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 p-convexity it follows that

if x,y e A, then 21 P e A and

Locally Pseudoconvex and Locally Bounded Spaces

95

A+Ac211"A. On the other hand, there are pseudoconvex sets which are not absolutely p-convex for any p. For example, the following open set on the plane

{(x,y):0<x<2,0
Suppose we are given a p-homogeneous pseudonorm II II in a metric linear space X. Then the sets Kr = {x : IIxII < r} are absolutely p-convex.

As a consequence of Theorem 3.1.4 we find that if an F-space X is locally p-convex (locally pseudoconvex), then there is a basis of absolutely p-convex neighbourhoods of 0 (a basis of absolutely p-convex neighbourhoods of 0 with p dependent on a neighbourhood). Albinus (1970) showed that, if a space X is locally p-convex (locally pseudoconvex), then there in an F-norm II II determining a topology equivalent to the original one and such that the sets

Kr = {x: IIxII < r} are absolutely p-convex (absolutely p-convex 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 space. Then, for a certain p, 0 < p < 1, there is a p-homogeneous F-norm equivalent to the original one. II II

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

Chapter 3

96

U+U C aU. Hence U is pseudoconvex. I

Since U is a bounded set, the system of sets - U constitutes a basis m

of neighbourhoods of 0. Basing ourselves on Theorem 3.1.4 for each n we can construct a pseudonorm II II,,. Observe that the F-pseudonorms II IIn differ from one another by a constant coefficient. Therefore II II, is an F-norm with the required properties.

log c(U)

Let us remark that pn is equal to 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 < po, there is a p-homogeneous F-norm II equivalent to the original one. II

The strict inequality p < po in Theorem 3.2.1' cannot be replaced by

the conditional inequality p <po, 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 I) be an F*-space. Let 11 11 bep-homogeneous. Then the space X is locally bounded. Indeed, the unit ball

K = {x: IIxII < 1}

Locally Pseudoconvex and Locally Bounded Spaces

97

is a bounded set, because if x e K and {tn} is a sequence of scalars tending to 0, then Iitnxnll = ItnlP

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(Q,E,p), C(Q), C(QI Sao),

LP(D,E,p) for 1

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 Tand T-1 map open bounded sets onto open bounded sets. Thus

c(Y) = c(T(X)) < c(X) = c(T-1(Y)) < c(Y). Let (X, II II).be a locally bounded space. Let 11 11 be a p-homogeneous

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 lIxll <+00. zed PROPOSITION 3.2.4. Let (X, II II) be a locally bounded space. Let Y be

a subspace of the space X. Then c(Y) < c(X). Proof. Without loss of generality we may assume that the norm II II is p-homogeneous. Let A be an open set in Y. Let

B = U {x a X: IIx-yll < inf {IIy-zII : z e Y\A}} The set B is open and bounded. Moreover,

A= Yt B.

Chapter 3

98

Thus

A+A=(BnY)+(BnY)C(B+B)nYCc(B)BnY=c(B)A. Hence c(A) < c(B) and c(Y) < c(X). PROPOSITION 3.2.5. Let (X, II II) and (Y, II 11y) be two complete locally

bounded spaces. Let T be a continuous linear operator mapping X onto Y. Then

c(Y) < c(X). Proof. Without loss of generality we can assume that II II is p-homogeneous. Let A be an arbitrary open bounded starlike set in Y. Since T induces a one-to-one operator T mapping X/ker onto Y, by Theorems 2.3.2 and 2.1.1 RA = sup inf {IIxII : x e T-1(y)} <+oo. yEA

Let

B = T-1(A) n {x: IIxII < 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). COROLLARY 3.2.6. Let (X, II

II) 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. 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 and Y be two locally bounded spaces.

If dime X < dime Y, then

c(X) < c(Y). If dimjX = dime Y, then c(X) = c(Y).

Locally Pseudoconvex and Locally Bounded Spaces

99

THEOREM 3.2.8. Let X and Y be two complete locally bounded spaces. If codimjX < codim= Y, then c(X) < c(Y). If codimlX = codim: Y,

then c(X) = c(Y). THEOREM 3.2.9 (Kalton, 1977). There is a separable locally bounded complete space (X, 11 11) with an a-homogeneous norm II which is uniII

versal for all separable locally bounded spaces with a-homogeneous norms.

Proof: Let 9 be the set of all a-homogeneous norms defined on 9(N) (for the notation see Section 2.4). The set 9 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(Ja(p),q) we have e-Bp (x) < -q (x) < eep (x)

for x E R. Let

q(x) = inf {9 y)+e°p(x-y): y E Rn}. For x E Rn we have q(x) = q(x) and for all x E 9(N) we have

e Bp(x) < q(x) < eep(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 (9(N),p) for a certain

p. Now we construct the space (9(E), II II) as in Section 2.4. For each a c- E we denote by ea an element of 9(E) such that

for x = a, for x # a. Choose o : N-*E to satisfy Proposition 2.4.1 and write wn = e 1

e. (x) _

{0

Then, by Proposition 2.4.4, for any e > 0 there is an increasing sequence of indices (nk) such that for any m in

(1-e)p( Observe that II the theorem.

=1

II

in

taxi)
I < (I +6)p

(3.2.1)

is a-homogeneous. Since lin {xn} = X (3.2.1) implies

Chapter 3

100

Using a different method Banach and Mazur (see Banach, 1932) showed that the space C[0,1] is universal for all separable Banach spaces, hence also for 1-homogeneous F-spaces.

From Theorem 3.2.9 follows the existence of universal spaces for separable locally p-convex (and locally pseudoconvex) spaces. To show this fact we introduce the following notion. Let (Xi, II Ili) be a sequence of F*-spaces. By (Xi)(,) we denote the space of all sequences x = {xi}, xt e Xi. The topology in (Xi)(,) is defined by F-pseudonorms n

Ilxlln =,' Ilxills In the case where all Xi are identical, Xi = X we denote briefly (Xi)(,) by (X)(8).

It is easy to verify the following facts. (Xi)(,) is an F*-space. If all spaces (Xi, II Ili) are complete, then the space (Xi)(,) is also complete. If all spaces (Xi, II Ili) are locally p-convex (locally pseudoconvex), then (Xi)() is also locally p-convex (locally pseudoconvex).

THEOREM 3.2.10. There is a separable locally p-convex space, which is universal for all separable locally p-convex spaces. Proof. Let (XP, II II) be a separable locally bounded space with a p-ho-

mogeneous norm universal for all separable locally bounded spaces with p-homogeneous norms. Let Xi = XP, i = 1, 2, ... The space (XP)(,) is a separable locally p-convex space. We shall show that it is universal for all separable locally p-convex spaces. Indeed, let X be an arbitrary separable locally p-convex space. By Theorem 3.1.4 there is a sequence of p-homogeneous pseudonorms {II IIn} determining a topology equivalent to the original one. Let

Xn,o = {x: kiln = 0} and let Xn = X/X,,, be the quotient space. The pseudonorm Ilxlln induces a p-homogeneous norm in Xn. Since no misunderstanding can arise, we U

shall denote this induced norm also by II IIn. The space (Xn, II a locally bouded space with a p-homogeneous norm II Let X = (X")(.).

IIn

IIn) is

Locally Pseudoconvex and Locally Bounded Spaces

101

The space X is locally p-convex. The space X is isomorphic to a subspace of the space X consisiting of elements where [x] is the coset in X. induced by x. On the other hand by the universality of XP, X is isomorphic to a subspace of the space (XP)(8).

Basing themselves on the conclusion of Banach and Mazur (1933), that C[0,1] is universal for all separable Banach spaces, Mazur and Orlicz (1948) showed that C(-oo,+oo) is universal for all separable Bo spaces.

THEOREM 3.2.11. There is a separable locally pseudoconvex space which is universal for all separable locally pseudoconvex spaces. Proof. Let (XP, II II) denote a separable locally bounded space with a p-homogeneous norm, being universal for all separable locally bounded spaces with p-homogeneous norms. Let {pi} be a sequence tending to 0,

for example pi = 1/i. The space (X1')(8) 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 pi-homogeneous pseudonorms {II IIi}. In the same way as in the proof of Theorem 3.2.10 we define Xi and X. The space X is locally pseudo-

convex. The rest of the proof is the same. THEOREM 3.3.12 (Shapiro, 1969; Stiles, 1970; for Banach spaces see Banach, 1932). Every separable locally bounded space X with a p-homogeneous 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 N (1) is just the space 1P. THEOREM 3.2.13. Let (X, II IIx) and (Y, II IIi') be two locally bounded spaces. Let II IIx be pg homogeneous and let II IIY be py-homogeneous.

A linear operator A from X into Y is continuous if and only if

IIAII = sup IIA(x)IIY <+00. UZI z,I

(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 {IIxljY: x e E} <+oo. If px = py then the definition of IIAII implies IIA(x)IIY < IIAII IIxIIx.

In this case the number IIAII is called the norm of the operator A. Let (X, II IIx) and (Y, II IIY) be two locally bounded spaces. Let II IIx be px-homogeneous and let IIY be py-homogeneous. Then we can always introduce p-homogeneous norms, p < min (px, py), in X and Y equivalent to the original norms. Indeed, the norms II

IIxIIX = (IIxJIx)P"X,

IIy1IY = (IIYIIY)P'PY

have the required properties. We shall now show that the norm IIAII is a py-homogeneous F-norm in the space B(X--*Y). It is obvious that IIAII = 0 if and only if A = 0. Let A, B e B(X-* Y). Then IIA+BII = sup IIA(x)+B(x)IIY < sup IIA(x)IIY+ sup IIB(x)IIY pxMIX<-1

IIzIIX<-1

IIxIIX<-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 < E} D U(O,B,e) = {A: I A(x)IIY < r for x e B},

where the set B = {x : IIxIIx < 1} is bounded. On the other hand for an arbitrary positive and an arbitrary bounded set B C X, U(O,B,E) D U(O,Kr,E)

{A:

IIAII < rT IPX

where r = sup{IIxIIx: x e B} and Kr = {x: IIxIIx < r}. If the spaces X and Y are complete, then by Corollary 2.2.4 the space

B(X-*Y) is complete, i.e. it is an F-space. In this particular case the conjugate space X is always a Banach space. Let (X, II IIx) and (Y, II IIY) be two locally bounded spaces. Let II IIx be px-homogeneous and let Ily be py-homogeneous. From the definiII

Locally Pseudoconvex and Locally Bounded Spaces

103

tion of the norm IIA II it follows that a family RI of linear operators is equicontinuous if and only if sup {IIAII: A e 91[} <+oo.

Then Theorem 2.6.1 implies THEOREM 3.2.14. Let Z be a complete locally bounded space with a p-homogeneous norm II II and with a basis {en}. Then

K = sup IIPnII <+oo, n

where, as before, w

ttet.

P. (,.Y t{ e{) i=1

L=1

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 II) be a locally bounded space. Let II II be p-homogeneous. 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+m


THEOREM 3.2.16 (Krein, Milman and Rutman, 1940). Let (X, II II) be a complete locally bounded space. Let II II be p-homogeneous. Let {en} be a basis in X of the norm K. Let IIenII = 1, n = 1, 2, ... If, for a sequence {gn} of elements of X, 00

C =YIIgn-enli < 2K' n=1

(3.2.3)

Chapter 3

104

then {gn} is a basis in the space Xo generated by {gn }, Xo = lin { gn }, equiv-

alent to the basis {en}. Proof. The triangle inequality implies n

f ties

-

i=1

n

it

f, tigi

,' ItiIPIlgi-esll i=1

ti=1

'

n

+ I ItilPllgi-eill i=1

i=1

From the definition of the norm of the basis it follows that : n

9-1

J

t{ei+

It1IP=1Itietll< ti=1

t{eijl<2K 11 i=1

i=1

tieil

Therefore n

f IttlPIlgi-eill < max

1
i=1

ItiJPI Ilgi-eill i=1

2KC

Hence (3.2.3) implies

tiei

f

i=1

where S = 2KC < 1. Thus, the elements {gn} consitute a basic sequence equivalent to the basis {en}. PROPOSITION 3.2.17. Let (X, II II) be a complete locally bounded space. Let II II be p-homogeneous. Let {en} be a basis in X. Suppose that X is

infinite-dimensional and that Xo is an infinite-dimensional subspace of X.

Then there is in X0 a sequence of elements {xn}, IIxnII = 1, xn 00 _ E to i ei such that i=1

lim t, "-W

=0

(i = 1,2, ...).

Proof. Suppose that the conclusion does not hold. Then there are a

Locally Pseudoconvex and Locally Bounded Spaces

105

positive integer k and a positive e such that, for each x E Xo, IIxII = 1,

xtsei i=1

IIxII=max Itil>e. 1
Hence there exists a linear homeomorphism of X0 and k-dimensiona space. This contradicts the fact that Xo is infinite-dimensional. 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 X0 be an infinite-dimensional subspace of the space X. Then Xa contains an infinite-dimensional subspace Y with a basis {fn}, which is equivalent to a block basis in X.

1971). There is a locally bounded space II) such that there is no p-homogeneous norm in X determining a topology equivalent to the original one, but every infinite-dimensional subspace Y contains an infinite-dimensional subspace Yo in which that topology can be determined by a p-homogeneous norm. PROPOSITION 3.2.19 (Stiles, (X, II

Proof. Let

qn = P 1-

1

log logn )'

n=30,31,...

Let NN(u) = u4". Let X0 be an infinite-dimensional subspace of the space Nn(l). Then by Proposition 3.2.18 there is an infinite-dimensional subspace Y of X0 such that there is in Y a basis {fn} equivalent to a block basis P"+i

e"

tte1,

Ile"II = 1.

ti=y"+1

Without loss of generality we may assume that pn > ee2" (see the remark after Proposition 2.6.7).

Then, for Itl < 1 Iti' < Ilte"II < ItIP(1

Chapter 3

106

Since

1

sup ItP(1o
201

by Theorem 3.2.16 {e;,} and the standard basis {en} in 1 P are equivalent.

On the other hand the space Nn(1) is not isomorphic to 1P. What is more, there is no p-homogeneous norm 11 11 in Nn(1). 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 < +00. n 00

1/P

Let xn = (nlog2n) . Then I 1X-1P <0030

Therefore the series i xne, is convergent in II II This leads to a contradiction since 00

1

00

fIxnIP*=

n=30

1

uo (nlog2n) 00

-e

2

Pp(1

log n

e

log log n

n=son

_

1

loglogn

-2 log logs

_

= + 00

and {xn} 0 Nn(1).

Pelczyfiski (see Rolewicz, 1957) has shown that in the space Nn(1) given above for each q with p < q < 1 there is a q-homogeneous norm

determining the original topology. This has been the first example showing that the inequality p > po 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 pn-homogeneous pseudonorms {II II } determining the original topology. Without loss of generality we can assume that lIXI12 < Ilxlls < ...

IIxiii <

Suppose that {en} is a basis in X. Let

IIxIi = sup r

m=1

tmemi, ll

Locally Pseudoconvex and Locally Bounded Spaces

where

I

107

00

X=

tmem.

IX-1

By Theorem 2.6.1 the system of pseudonorms {ll Ili} determines the original topology. Let X{. o = {x: IIxIIi = 01. Let Xi = X/8"o be the quotient space. The pseudonorm II IIi induces a pi-homogeneous norm II Ili" on Xi in the following way II[x]illti = IIxIIti,

where {x}s denotes the coset containing x. By the definition of II IIi we obtain II t1[e1]i+ ... +tn[en]i IIi = Iltlei+ ... +tnenll i lltle1+ ... +tn+men+mlli = Ilt1[e1]t+ ... +tn+m[en+m]Il$ for all positive integers n and m. Therefore, by Theorem 3.2.15, we obtain :

PROPOSITION 3.2.20. The cosets [en]i consitute a basis of norm one in the space Xi, which is the completion of the space Xi, provided we omit all those [en]i which are equal to 0.

For locally convex spaces, Proposition 3.2.20 was proved by Bessaga and Pelczyliski (1957).

3.3. BOUNDED SETS IN SPACES N(L(Q, E, 'u))

Let a space N(L(Q,(E,p)) be given. Let

n(t) = inf {a > 0: N(at) > N(t) }. l

2

Of course n(t) > 0 for all t, 0 < t <+oo.

11111

(3.3.1)

Chapter 3

108

THEOREM 3.3.1 (Rolewicz, 1959). If

r=

inf

n(t) > 0,

o«<+co then the space N(L(SZ,E,µ)) is locally bounded.

Proof. From (3.3.1) it follows that pN(rx) < pN(x) for each 0 < < r. $ Then by induction /

pN(r"x) <

2n

pN(x),

n = 1,2....

Therefore, the open set Kd = {x: pN(x) < d} is bounded. We can weaken the hypothesis on N(t) by assuming additional properties on the measure a. Namely THEOREM 3.3.2 (Rolewicz, 1959). Let p be a finite measure. If

lim inf n(t) > 0, then the space N(L(SQ,E,µ)) is locally bounded. Proof. Let

N'(t)

N(t) {tN(l)

for t > 1,

for 0t( 1.

Since the measure µ is finite,

O if and only if pN.(xn)- O. Let

n'(t) = inf {{a > 0: N'(at) <

N t)1 2

It is easy to verify that infn'(t) > 0. Therefore, Theorem 3.3.1 implies Theorem 3.3.2. THEOREM 3.3.3 (Rolewicz, 1959). Let the measure p be purely atomic and let

r = inf p(ps)> 0, where pi runs over all atoms. If

liminfn(t) > 0, tyo

then the space N(L(Q,E,p)) is locally bounded.

(3.3.2)

Locally Pseudoconvex and Locally Bounded Spaces

109

Proof. Let

N(t)

N'(t) _ {tN(l)

for 0 < t < 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. The following fact is, in a sense, inverse to Theorem 3.3.2. THEOREM 3.3.4. If the measure ju is not purely atomic

and

lim inf n(t) = 0,

(3.3.3)

then the space N(L(Q, Z, µ)) 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 ,u is not purely atomic, hence for sufficiently small e there are sets e

A, n = 1,2, ... such that u(A,,) = 2N(tn) . Let xn = t,,XA., where, as usual, XA. denotes the characteristic function of the set A. It is easy to

verify that pN(xn) =

e 2

On the other hand, pN(anxn) =

J A.

N(n(tn)tn)d/1 = 2

J A.

N(tn)dp = 4

.

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(Q,E,µ)) is not locally bounded. COROLLARY 3.3.5. If the measure is not purely atomic and the function N(u) is bounded, then the space N(L(D,E,µ)) is not locally bounded. Proof. Since N(u) is bounded, (3.3.3) holds. THEOREM 3.3.6. If there is a positive number k such that, for a sufficiently

small r, there is a set A, such that

Chapter 3

110

k < rp(A,) < 2, and if

lim inf n(t) = 0

(3.3.5)

then the space N(L(Q,E,p)) is not locally bounded. Proof. By (3.3.4), there is a sequence t.,,->0 such that a = n(tn)-0. Let e be an arbitrary positive number. The assumption on the measure implies

that for sufficiently large n there are A. such that ke < N(t,)p(A,) < e. Let x,, = Then pN(xn) < e and pN(anxn) > ke. In the same way as in the proof of Theorem 3.3.4 we find that N(L(Q,E,p)) is not locally bounded. The following theorem gives us the connection between local pseudoconvexity and local boundness in the case of spaces N(L(Q,E,p)). THEOREM 3.3.7. Let the measure p be finite and not purely atomic. Then the space N(L(Q,E,p)) is locally bounded provided it is locally pseudoconvex.

Proof. Suppose that the space N(L(Q,E,p)) is not locally bounded. Then, by Theorem 3.3.2. liminf n(t) = 0, i.e. there is a sequence {tn}-*oo

such that a = n(t,,)-0. Let A be an arbitrary open set such that

r = suppN(x) <+oo. 2EA

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 zEA

4r k < . The measure it is not purely atomic, hence for a sufficiently, P large n, we can find sets A,,, {, i = 1, 2, ..., k such that

p p(A.,i) = 2N(tn)

and the sets A,,, and An, p are disjoint for i lim N(t) = +oo. tco

j, provided

Locally Pseudoconvex and Locally Bounded Spaces

Let xn, i = tnXA,,,,. Then pN(xn,i) =

111

< p, whence xn,{ e A.

2 = kp/2 > 2r. On the other hand, pN(xn,1+ ...+ xn,k) Therefore

PN(n(tn) (xn,1+ ... +xn,k))> r and n(tn (xn,l+ ... +xn,k) 0 A. Since n(tri)-0, we do not have

A+A+... +A C KA

for any K > 0.

This implies c(A) = +oo. Now we shall consider the case where R = lim N(t) <+oo. Let S2, be a subset of Sl on which the measure p is non-atomic. Let A

be an arbitrary open set in N(L(S2, E, p)) such that r = suppN(x) 0Ea

< N(1)µ(52,). Let p = inf pN(x). Since the measure p is non-atomic x*d

on the set 521, then there are disjoint subsets A,, ..., Ak of 52, such that p(At)

xn,l+ ... +xn, k = nXn, and

PN(n (xn,l+ ... +xn, k)) > r. The rest of the proof is the same as in the case where

lim N(t) = +oo.

3.4. CALCULATIONS OF THE MODULUS OF CONCAVITY OF SPACES N(L (52, _Y, ,u))

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 u1i u2, we have N(aul+bu2) < aN(u1)+bN(u2).

Chapter 3

112

THEOREM 3.4.1. Let N(u) = N0(uP), where No is a convex function and

0

Proof. Let a be an arbitrary positive number. Let K8 = {x: pN(x) < e}.

We shall show that the set K is absolutely p-convex. Indeed, let I aI P-+- I bI P = 1 and let x, y e K. Then

PN(ax+by) = f N(Iax(t)+by(t)Ddu

f N(Ial Ix(t)I+Ibj Iy(t)l)dp

n

= f No((IaI Ix(t)I+Ibl Iy(t)I)P)dp a

< f No(IaIPIx(t)IP+IbjPIy(t)IP)dp D

$ No(u), then a> 1/2.

Therefore if N(au) > z N(u), then a > (2 )1h i. Thus n (u) > (z and the sets K8 are bounded (cf. the proof of Theorem 3.3.1). Hence the sets K. are bounded and absolutely p-convex. This implies (see Theorem 3.2.1' and Theorem 3.1.4) that there is a p-homogeneous norm equivalent to the original one.

Suppose we are given two continuous positive non-decreasing func-

tions M(u) and N(u) defined on the interval (0,+0o). The functions M(u) and N(u) are said to be equivalent if there are two positive constants A, B such that

A < M(u) < B (3.4.1) N(u) 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
Locally Pseudoconvex and Locally Bounded Spaces

113

By simple calculation we obtain PROPOSITION 3.4.2. Suppose we are given two continuous positive non-de-

creasing functions M(u) and N(u) defined on the interval (0, + 00). If one of the following three conditions holds : the functions M(u) and N(u) are equivalent, (3.4.2.1) the functions M(u) and N(u) are equivalent at infinity and the measure p is finite, (3.4.2.ii) the functions M(u) and N(u) are equivalent at 0 and the measure p is purely atomic and such that inf{p(A): A being atoms} > 0, (3.4.2.iii)

then the spaces M(L(Q,E,p)) and N(L(Q,2,p)) are identical as sets of functions and, moreover, pM(xn)-0 if and only if pN(xn)- .O, 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) = MO(up),

0
where Mo is a convex function. Let N(u) be a function such that one of conditions (3.4.2.1), (3.4.2.ii), (3.4.2.iii) holds. Then there is a p-homogeneous norm in N(L(Q,E,p)) equivalent to the original one. The following two theorems are, in a sense, inverse to Theorem 3.4.3. THEOREM 3.4.4. Suppose that the measure p is not purely atomic. If

lim inf N(u) = 0,

U- 00

(3.4.3)

UP

then the space N(L(Q,E,p)) is not locally p-convex.

Proof. By Corollary 3.3.5 we can assume that lim N(u) = +oo. By u- 0

(3.4.3) there is a sequence {un}-oo such that N(un)l/p -*oo. Let k,, be

the greatest integer such that k < N(un), k = [N(un)]. Since the measure It is not purely atomic for sufficiently small e > 0 there are disjoint

sets An,r, i = 1,2, ..., k, such that p(A,,,t) =

e

1 +kn'

Chapter 3

114

Let Xn,t = unXAn,t

Then PN(xn,t) < E.

NlN

On the other hand,

kn

1

(-k-

un

E kn

(knl/P) l+kn moo.

4=1

The arbitrariness of E implies the theorem. THEOREM 3.4.5. Let there be a positive number r and ani nfinite family o disjoint sets Aa such that

1 < p(Aa) < r

(3.4.4)

If

lim sup

u-). J

N(u) UP

= +00

(3.4.5)

then the space N(L(Q,E,p)) is not locally p-convex. Proof. Let e be an arbitrary positive number. By (3.4.4) there is a S > 0,

such that for each u, 0 < u < 6, there is an infinite family of disjoin sets {A.,,,.} such that N(u)

< (An n) < N(u).

Let Xn,u = uXAn,,

Then PN(xn,u) < E.

On the other hand, PN

X1, u+ ... +xk, u > ek k11P

)

and by (3.4.5)

lira sup sup o
(u) N

u

EuP

N(k-1/P u)

rN(u) N(ki/P) = rN(u) (k 1IPu)P

(u )

EuP N(k-1/P u) x nosup o
+°°'

Locally Pseudoconvex and Locally Bounded Spaces

115

COROLLARY 3.4.6. If the measure y satisifes either the assumption of Theo-

rem 3.4.4. or the assumption of Theorem 3.4.5, then in the space LP(Q,.,µ), 0 < p < 1, there is a p-homogeneous norm determining a topology equivalent to the original one, and there is no q-homogeneous (q > p) norm with this property. In other words, c(LP(SQ,E,p)) = 21/p and there is a starlike bounded open set A C LP(S2,E,p) such that c(A) = 21/p. From Corollary 3.4.6 follows THEOREM 3.4.7. There is no locally bounded space universal (co-universal) for all separable locally bounded spaces.

Proof. Suppose that such a universal (co-universal) space X exists. Then, by definition, dim 1/P < dim,X,

(codimilP < codimiX). Hence, by Theorem 3.2.7 (Theorem 3.2.8) 21/P = c(lP) < c(X)

for all p, 0 < p < 1.

Thus c(X) = +oo and this contradicts to the fact that X is locally bounded. E Another consequence of the properties of spaces N(L(Q,E,p)) is PROPOSITION 3.4.8. There is a locally bounded space X such that the topology in X can be determined by a p-homogeneous norm for p, 0

but cannot be determined by a p0-homogeneous norm. In other words, c(X) = 21/p0, but for any open bounded starlike set A, c(A) > 211-'° Proof. Let h(u) be a positive decreasing continuous convex function defined on the interval [0,+0o) such that h(0) < po and lim h(u) = 0. U- M

Let N(u) = uP°-h(u).

Let p 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 p0-homogeneous norm with this property. 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,p)) be a locally convex space. If the measure p 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,p)) is locally convex, there is a positive

number a such that pN(xk) < e, k = 1, ...,n, implies PN

(X'+ ...

-{--X )

n

< 1.

Let q, p, n be positive integers. Let p < n. Let t be a positive real number. Since the measure p is not purely atomic, then, for sufficiently

large q, there are disjoint sets El, ..., En such that p(E{) = 1/nq, i = 1, ..., n. The smallest such q will be denoted by q0. Let forseEj, t

where j = k, k+ 1, ..., k-l+p(mod n).

Xk(S) =

otherwise.

0

Then PN(Xk) = nq pN(t) and

PN

xJ+...-f-xn) _ n

q

Hence

qn

N(t)

<E

implies /

N( I

1

t)<1.

N(

t) .n

Locally Pseudoconvex and Locally Bounded Spaces

117

The continuity of the function N(u) and the arbitrariness of p and n imply that

(0<w<1)

-N(t) < s implies

1q N(wt) < 1. Let N(t) > eqo and let q be the greatest integer less than wN(t)/e.

From the definition of q

+

eq
co N(t) <e,

follows

= qE 1 eq < e N(t). Putting C = 2/e, we obtain 9

N(wt) < CwN(t)

for co >

Eqo

N(t)

Recall that, by Theorem 3.4.4,

lim inf N(t) > 0. t co Choose 6 > 0 and T > 0 such that N(t)/t > 6 for t > T and TS > eqo Then for cot > T, 0 < co < 1, t-

wN(t) = wt

N= t)

> TS > eqo.

Since cot > T implies t > T, N(o) t) < CcoN(t) for cot > T, 0 <w < 1. Let sup

(Ncot)

o<w<1

P(t) = wt>T 1

for t > T,

CO

for 0 < t < T.

N(t)

Let 0 < a< 2, at > T. Then N(coat) _

P(at) = o<w<1

CO

wat'>T

- a sup i <

a sup N(st) = a P(t). o<# -
at>T

S

N(coat) ) coa (3.4.6)

Chapter 3

118

Formula (3.4.6) implies that the function P(t)lt is non-decreasing for t > T. Indeed, let T < t' < t"

(Pt")
P(t')

t,

=

t"

Let

P(t)

for t > T,

It

Q(t) = t

p(T) T2

for 0 < t < T

and U

M(u) = f Q(t)dt. 0

Since Q(t) is a non-decreasing 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, CN(t), hence P(t) < CN(t).

N(cot)1(0

Since P(t) is equivalent to N(t) at infinity and N(t) satisfies condition (Q, 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 non-decreasing, whence, for sufficiently large t,

M(t) < tQ(t) = P(t) and

M(t)>

f

Q(s)ds> 2

Q(2)=P(2).

qz

Hence, for sufficiently large t, P(t) < KP (+) < KM(t) and the functions M(t) and P(t) are equivalent at infinity. Therefore the functions M(u) and N(u) are equivalent at infinity. THEOREM 3.4.10 (Mazur and Orllcz, 1958). Suppose that there is a constant K > 0 and an infinite family of. disjoint sets {Ef} such that

I < p(E4) < 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(S?,E,u)) is locally convex, there is a positive e such that pN(xk) < e, k = 1, 2, ..., n, implies pN

(xi+ ... n

Let p be an arbitrary positive integer and let for seEk+tn (i=0, 1,...,p-1), t xk(s) 0 otherwise. We get (k = 1,2, ... , n) pN(xk) < KpN(t) and

pn N(

t) < pN (xl+ ... n

n

+xn) Hence N(t) < elKK implies N(t)ln < 1lpn. Let 0 < N(t) < e/K,

and let p be chosen so that a/K(p+1) < N(t) < e/Kp. Then N(t/n) 1/pn < 2K/e N(t)/n.

Given p > 0. Choose a q > 1 such that N(t/q) < e for I t I < p. Condition (Q implies that there is a constant Do such that N(qt)
=N

9

q

Let 0 < w < 1 and let n be such that 1/(n+1) < w < 1/n. Since 1/n < 2w, putting C = 4KDq/e we obtain

N(wt) < CwN(t) Let

P (t) = sup

O<w
N(wt CO

Q (t) = P(t) t e

M(t) = f Q(s)ds.

for 0 <w < 1, ItI < p.

Chapter 3

120

In the same way as in Theorem 3.4.9 one may prove that M(u) is p a convex function equivalent to the function N(u) at 0. 3.5. INTEGRATIONS OF FUNCTIONS WITH VALUES IN F-SPACES

Let X be an F-space. 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 that

lim

sup

n-*00 O<-i-
It+1-t;I=0.

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, tk , < < sx < tx, there exists a limit of the sum

J

k(n)

x (Sk) (tk -tk-1)

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

f x(t) dt. a

By repeating the classical considerations this definition can easily be extended to functions of several variables and to line integrals.'

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

f (x(t)+y(t)) dt = f x(t) dt+ f y(t) dt, a

a

a

1 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

b

f Cx(t) dt = C f x(t) dt; a

(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 b

f x(t)dt+ f x(t)dt. b

c

f x(t) dt = a

a

c

THEOREM 3.5.1 (Mazur and Orlicz, 1948). An F-space 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 non-decreasing (i.e. lltxll is a non-decreasing function for positive-t). Since X is not locally convex, there is a positive number a such that for arbitrary n there are elements xi, ..., xka such that Ilxli% <

2n

i = 1,2, ..., k,)

II Xi+ .k +xkn II > a.

and

n

Without loss of generality we can assume that kn+l is divisible by 2kn. Suppose that

(x;

xn(t) _ 1 0

for t = (i- 2) kn, i = 1,2, ..., kn, for t =

kin

i = 0, 1, ..., kn,

lis linear on the intervals

J-1

,

>

[ 2kn 2kn

, j = 1,2, ..., 2kn

Obviously each of the functions xn(t) is integrable since its values are

in the finite dimensional subspace X. spanned by elements xi,

. . . xkw.

Chapter 3

122

Therefore, there is a number Sn > 0 such that, for any division 0 = t < t1 < ... < tk = 1 such that lti+1-t1! >Sn (i = 0, 1, ..., k-1) we have k 1

i=1

[xn(Si)-xn(Si)] (ti-ti-01 1 <

3n

for all si and s; such that ti_, < si, s, < ti. We choose by induction a sequence nm such that 1/knm < 8i, i = 1,2,

..., m-1. Let x(t) _

xnm(t). The function x(t) is continuous as a uniform m=1

limit of a sequence of continuous functions. We shall show that it is not Riemann integrable. Indeed, let ti = i/kn,, i = 0, 1, ..., k.,, and

2i-1

let si = ti, si = 2k, , i = 1, 2, ..., knn.

[x(si)- x(si)]

1

kna

Xnv(Si)-Xnv(Si)] i=1

.(Si) - X-,, (Si)]

P-1

x'+ ... +Xk:,

xn,A(Si) - xn

knn

a- P-1 3m > m=1

m=1

2

.

i=1

Locally Pseudoconvex and Locally Bounded Spaces

123

Thus the function x(t) is not integrable. p If the F-space X is locally convex, we can also define by analogy to the Lebesgue integral, an integral called the Bochner-Lebesgue integral. Accordingly we aproximate a function x(t) by simple functions k.

Xn(t) _ i=1

aiXE

where at e X and Ei are Lebesgue measurable sets of the interval [a, b]. The Bochner-Lebesgue integral of the simple function xn(t) is defined in a natural way as ko

b

f xn(t)dt = f at IEi 1, i=1

a

where IEl denotes the Lebesgue measure of the set E. We assume now that, for every homogeneous continuous pseudonorm llxll, the scalar function llx(t)ll is integrable and that the sequence {xn(t)} tends to x(t) almost everywhere and ll xn(t)ll < llx(t)ll. Then we define

x(t)dt = lim Jxn(t)dt. 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

tending to 0 such that the sequences of integrals{ f xn(t)dt} do not tend a

to 0. In fact, let

xn(t) = xi

for

i-1 < t < kn 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 b

iI

a

r xn(t)dtll

xi+ kn+xk

=

> 0.

II

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 integrable 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 F-space X is analytic if, for any to, there is a neighbourhood U of the point to such that, for t E U, the function x(t) can be represented by a power series OD

x(t) _

1(t-to)nxn,

Fkxn a X.

(3.5.3)

n=0

If the space X is locally pseudoconvex, then the topology can be determined by a sequence of pn-homogeneous pseudonorms 11 jJn. This

implies that if the series (3.5.3) is convergent at a point t1i then it is convergent for all t such that It-tol
hold. For example, if X = S[0,1], then there is a power series

tnxn n=o

convergent only at a finite number of points (see Arnold, 1966; Zelazko, 1972).

THEOREM 3.5.2 (Gramsch, 1965; Przeworska-Rolewicz and Rolewicz 1966). Let X be a locally pseudoconvex F-space. 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 pk-homogeneous pseudonorms 11 Ilk. Let x(t) be an analytic function. Then for each so e [a, b] there is a neighCO

bourhood U80 such that for t E UQo the series x(t)

(t-so)nxn is n=0

convergent. This implies that there is a neighbourhood V8o C Uo such that for t e V.0

ll(t-so)nxnIlk< 2n ,

k= 1,2,...,

n=0,1,...

Since the interval [a, b] is compact, there is a finite number of points

s, ..., s, such that

[a, b] C U V. Let Ve = (a1, b,). Now we shall consider the Riemann integral of the func'ion x(t) on the interval [a1,b1].' Let e be an

Locally Pseudoconvex and Locally Bounded Spaces

125

arbitrary positive number. Let no be such an index that Mk/2no < e/2. Let x(t) = yno(t)+zno(t), a1 < t < b1, where 00

Yno (t)

_.E (t-S1)nXn,

(t-S1)nxn.

Zno (t) _

n=o

n=n°+1

The function yno(t) is integrable, because it has values in the finite-dimensional space spanned by elements xo, ..., xno. For any subdivision of the interval [a1, b1]

a1 = to < 71 < ... < tp = b1 and arbitrary al, t,_1
i

mated as follows :

n

.0

P

Zno(aj) (tl-t{-1) Ilk <

`fit-s1)nxn(tl-tl-1

n=n,+1 l=1

1=1

m 71

Sup

k

ll(a-S1)nxnllk

n=n°+1 af5Q5bj m

Y

n =no+1

Mk

Mk

e

2n

The arbitrariness of a and the integrability of yno(t) imply that the Riemann sums of x(t) converge for any normal sequence of subdivisions of the interval [a1, b1] with respect to the pseudonorm II J. Since there is only a finite number of the sets V8 , those sums converge on the whole interval [a, b]. This holds for every pseudonorm II Ilk, whence the complete-

ness of the space X implies that the function x(t) is integrable.

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 F-space 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 zo e D, there is a neighbourhood Vzo of the point zo such that the function x(z) can be represented in Vzo as a sum of power series Go

X(Z) = I (Z-z0)'Xn, n=o

Xn E X.

Chapter 3

126

As in the real-argument 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 T is smooth and the function x(z) is analytic, then the integral f x(z)dz exists. Using the same arguments as in the classical theory of analytic functions, we can obtain THEOREM 3.5.3. Let I' be a smooth closed curve. Let x(z) be analytic. Then f x(z)dz = 0, r

(3.5.4)

r z - z dz

x(zo)

r or each zo inside the domain surrounded by I', x(n) (zo)

=

2n

r.J

(z

x(z+1 dz,

(3.5.5)

(3.5.6)

where zo and T are 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 m-dimensional Euclidean space R. Let Nm denote the set of all vectors k = (k1, ..., k,,,), where k{ are non-negative integers, (i = 1, 2, ..., m). We adopt the folowing notation

IxI = sup Ixsl 1<_t_<m

Iki = f Iktl, i=1

Locally Pseudoconvex and Locally Bounded Spaces

127

m

xk,

xk = i=1 M

k!

k1 !, ti=1

where x = (x1i .... a Rm, k = (k1, .... km) a Nm. We say that a function f(x) mapping U into an F-space X is of class

C,(U,X), where r is a positive number, if, for any k e Nm such that 1k! < r, there are continuous mappings Dk f of U into X and Pr,k f of Ux U into X such that Do f =f, pr,k f(x,x) = 0 and +IY-xlr-jkj pr,kf(x,Y)

Dkf(Y) _ Ihi _
(see Turpin and Waelbroeck, 1968).

THEOREM 3.5.5 (Turpin and Waelbroeck, 1968b). Let X be a locally

p-convex space. Let U be an open set in Rm. If f(x) e C,(U,X) for r > m(1-p)/p is a bounded function, then it is Riemann integrable. We say that a function f(x) is of class C,,,,(U, R) if it is of class C,(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 C.,, (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) (C. (U, X)) with topology of uniform convergence on compact sets all Dk f, JkJ < r, and p,,k f (all Dk f and all p,,k f) into the space X.

3.6. VECTOR VALUED MEASURES Let (X, 11

11) be an F-space. Let Q be a set, and let E be a a-algebra of

subsets of Q. We consider a function M(E) defined for E E E with

Chapter 3

128

values in X such that, for any sequence of disjoint sets E1, ..., En, ... belonging to E M(E1 L ... V En U ...) = M(E1)-}- ...

+M(En)+...

The function M(E) is called a vector valued measure (briefly a measure). By the variation of the measure M on a set A C Q we shall mean the number

M(A) = sup {IIM(H)II: HC A, He E}. THEOREM 3.6.1 (Drewnowski, 1972). The variation M of a vector valued measure has the following properties :

M(O) = 0,

(3.6.1.i)

if A C B, then k(A) < M(B), M(SS) <+oo,

(3.6.1.ii)

(3.6.l.iii)

M(U Hn) < Y M(Hn) for any sequence {H,} of sets m=1

n=1

belonging to E,

(3.6.1.iv)

M(H) = lim M(HH), if H is either a union of an increasing n-ao

sequence {Hn} C E of sets or if it is an intersection of a decreasing sequence {Hn} of sets. (3.6.1 .v)

Proof. Formulae (3.6.1.i) and (3.6.1.ii) trivially follows from the definition of the variation. (3.6.1.iii). Suppose that (3.6.1.iii) does not hold. Then there is a seIIM(Hn)IH-*oo. Without loss of generquence of sets Hn e E such that ality we may assume that IIM(Hn)II > 1+IIM(Hl)II+ ... + II

M(Hn-1)II

The sets

On = Hn\(H1 U ... V Ha-1) are disjoint and IIM(HH)II > 1. Thus the series00 I M(Hn) is not conn=1

vergent, and this leads to a contradiction. 00

(3.6.1.iv). Let A be an arbitrary set contained in U H. Let An IL=1

Locally Pseudoconvex and Locally Bounded Spaces

129

_ (A\(H, v ... V Hn_1)) 0 H. The sets An are disjoint and A = U A,. n-1

Then 00

OD

M(An) <

IIM(A)II = n=1

CO

JIM(An)II < f M(HH). n=1

n=1

W

The set A is an arbitrary subset of the set U H. n=1

Thus

M(H1V H2V ...) = sup{IIM(A)II : A C 00U Hn} n=1

n=1

(3.6.1.v). Let Hn e E be an increasing sequence of sets. Thus H. C H and by (3.6.1.ii) M(Hn) < M(H), n = 1,2,... This implies that

lim M(HH) < M(H). Let e be an arbitrary positive number. Let A C H, A e .E be such that

M(H)-e < IIM(A)II < M(H) As in the classical measure theory we can prove lira M(A n Hn) = M(A). n- 00

Hence for sufficiently large n

M(H)-2 e < IIM(A n Hn)II < M(Hn). The arbitrariness of a implies

lim M(HH) > M(H). nyOD

Now we shall consider the case where Hn is a decreasing sequence

of sets, Hn a E. Let H = n H,,. Since H C H, n = 1, 2, ..., we have OD

by (3.6.1.ii) M(H) < M(HH). Thus

lim M(HH) > M(H) n~GD

Chapter 3

130

Observe that, if An C Hn\Hn_1, then the series

=1

M(An) converges

uniformly with respect to the choice of A. Then for an arbitrary positive, e there is an index no such that 00

I M(An) I < e. n=n,+1

Let A C Ho, A e E, be such a set that IIM(A)II > M(Hno)-e.

Let A= A\(

U00

n=n,+1

A n (Hn\Hn_1)). Observe that A C H and JIM(i)

-M(A)II < e. Hence M(A) > M(H)-2e. The arbitrariness of a completes proof.

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 COROLLARY 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 p-homogeneous. Thus, by (3.6. Liii), the set { M(E) : E E E}

is bounded. 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 IIn of

p,; homogeneous pseudonorms. Let Xn = {x: IIxIIn = 0}. Let X. be the quotient space X. = X/X°. The pseudonorm II IIn 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 (3.6.1.iii)

sup {IIM(E)IIn: Ee E} <+co, Therefore the measure M is bounded.

n = 1,2, ...

Locally Pseudoconvex and Locally Bounded Spaces

131,

Recently Talagrand has proved that each measure with values in the space L°(Q,L',p) is bounded. However, there are an F-space H and 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 .. denote the Lebesgue measure. For f E L°[0,1] we write a(f) = 1({t E SQ: f (t) =/Z 0}, and

Var (f) = inf {Var (f): fl = f almost everywhere}. Let H be the space of all measurable functions f such that for each

s > 0 there are g,h e L°[0,1] such that f = g+h and Var(g) <+oo ando(h) <E. Let

IIIII = inf {IISIIW+Var(g)+a(h): f = g+h}, where

IIgII. = ess sup Ig(t)I.

It is easy to verify that for fl, f2 E H we have IIAII+IIf2II,

Isfji <-IIfII for IsI < 1, lsf II-*0 if s-0 If II = 0 if and only if f is equal to 0 almost everywhere. Thus IIf II is an F-norm 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, since in the opposite case we can simply take the completion of H. Let S E E. The characteristic function Xs E H. Indeed, for each e > 0

there is a finite union of intervals D such that Q(Xs-XD) < E Since XD is of finite variation, Xs E H. Let {S,a}, S. E Z, be a sequence of disjoint sets. Thus ) (Sn)-+0. From the definition of the norm 11 1[ we infer that I I A (S.) for. sufficiently large n. This trivially implies that the measure

M(S) = Xs E H,

SEE,

Chapter 3

132

is a vector valued measure. We shall show that the measure M is not bounded. Indeed, let n, h be non-negative integers. Let Ih,n

h+lJ = rh L-, nJ n

.

Let

fn =

1

1

n 2h
Izh, n =n M (U n

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 J*,. =12h,n U I2h+l,n for which hn is equal to 0 on subsets of positive measures on both intervals I2)1, n and I2h+1.n This implies that the variation of gn on

J2h,n is not smaller than 1/n

2(Jg,n) and

2 Var (gn) > I(A). On the other hand by the definition of A

1-1(A) = X (Q\A) < 20 (hn and finally

2Var (gn)+2a(hn) > 1. Therefore 1Ifn1J > 1/2 and this completes the example.

A measure M is called compact if the set {M(E): E cE} is compact. By Corollary 3.6.2 we trivially obtain COROLLARY 3.6.5. If X is finite-dimensional, then each measure is compact.

If the space X is infinite-dimensional Corollary 3.6.5 does not hold, as follows from Example 3.6.6

Let Q _ [0,1] and let E be the algebra of Lebesgue measurable sets. Let X = 12. We define a vector measure M as follows

M(E) = { f sin2 rnt dt}. R

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

E. = {t: sin2nnt > 0}. Observe that the n-th coordinate of M(EE) = 11it. This implies that the sequence M(EE) 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 function f(t) is measurable if, for any open set U contained in the field of scalars, the set f -1(U) belongs to X. 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- 00

PROPOSITION 3.6.7 (Drewnowski, 1972). If fn tends almost everywhere to

f, then for each e > 0 00

lim M(U En(e))=0, n=k

k-.oo

where

En(e) = {x: II fn(x) f(x)II > E}. Proof. The sequence OD

A. = U Ek(E) b= n

is a decreasing sequence of measurable sets. Then by (3.6.1.v) lim M(An) = M (lim An) = M(lim supEE(e)).

Observe that if x a limsupEn(e), then the sequence fn(x) is not convern

gent to f(x). Thus lim supEE(e) C A = {x: fn(x) is not convergent tof(x)}. n- OD

By our hypothesis M(A) = 0. Thus, by (3.6.1.ii), M(lim supEn(s)) < M(A) = 0.

11

n- co

As a consequence of Proposition 3.6.7, we obtain extentions of the classical theorems of Lebesgue and Egorov

Chapter 3

134

THEOREM 3.6.8 (Lebesgue). If {,,(x)} tends to f(x) almost everywhere, then, for each e > 0, lim M({x: 11f-(X)-f(X)11 >' e}) = 0

THEOREM 3.6.9 (Egorov). If {fn(x)} tends to f(x) almost everywhere, then for each e > 0 there is an F E 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 A

( U En (J-)) < Z .

(3.6.2)

Let ao

as

k) 1

F=Q\Uk=1Un=Nx En

(3.6.3)

By (3.6.1.iv), (3.6.2), (3.6.3)

M(S2\F)=M(U n'kEf\kC k=1

k=1

M(nvEn

()) k

k=1

We shall show that {,,(x)} tends uniformly to f on th set F. Indeed,

let q be an arbitrary positive number, let k be such that 1/k < iq. If x E F, then

xen and

k\ \Q\En 1\ k ))

Ilfn(x)-f(x)II < -k < or n > Nk. For simple measurable functions there is a natural definition of the integral with respect to the measure M. Namely, if

= S(t)

N

I b.XE., n=1

E. c- E, b.-scalars.

Locally Pseudoconvex and Locally Bounded Spaces

135

then we define N

f g(t)dM(t) = f b, M(E,,). n=J

D

We say that a measure M is L°° -bounded if iIic set

if gdM: 0 < g(t) < 1, simple measurable function} is bounded (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 F-space. Let {An} be a sequence of non-negative numbers. We say that the space X has property P({An)} if, for any neighbourhood of zero U, there is a neighbourhood of zero V such that

n= 1,2,...

A1U1+... +AnUncV,

PROPOSITION 3.6.10. If a space X is locally pseudoconvex, then it has

property P \

2n })

.

Proof. Let topology in X be determined by a sequence {I1 Ilk} of pk-homogeneous pseudonorms. Let IIxaII < e,

n = 1,2, ...

Then k < n=1

n1/k IlxnhIk

n=1

<

k

2A -1

E.

THEOREM 3.6.11 (Rolewicz and Ryll-Nardzewski, 1967; Turpin, 1975). Let an F-space X have property P({1/2n}). Then each bounded measure with values in X is L°° -bounded. Proof. Let

N i=1

Chapter 3

136

0 < bi < 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

2 V+ ... -{-

n = 1,2,...

VC U,

By our hypothesis the measure M is bounded, i.e. there is such that M(E) C sV for all E e 2:. We write each bE in the dyadic form

ans>0

CO

bi,k

bi = k=1

2k

where bi,k is equal either to 0 or to 1. Thus N

f J a

g(t)dM i=1 =f

f N

biM(EE)

m

N

k=1

n=1

= i=1

bi, 2 M(EE) k=1

bik M(EE 2

00

1 N _ 0 2k M( U

bi,k EE) C S U .

k=1

The arbitrariness of U implies that the set

A = { f g(t)dM: 0 < g(t) < 1, g- simple mesurable} a is bounded. -THEOREM 3.6.12 (Maurey and Pisier, 1973, 1976). Each bounded measure M with values in a space is L°°-bounded.

The proof is based on several notions and lemmas. To begin with we shall recall some classical results from probability theory. Let (Q°,E°iP) be a probability space (i.e. a measure space such that P(DO) = 1). A real valued E°-measurable function is called a random variable.

Locally Pseudoconvex and Locally Bounded Spaces

137

Let X(w) be a random variable. We write

E(X) = f X(w)dP and

V(X) = f IX((o)-E(X)j2dP. n,

LEMMA 3.6.13 (Tchebyscheff inequality).

P({w: JX(w)j > E}) < I E(X ). Proof.

E(X2) = f IX12dP

f

f

X2dP-{-

{w: I8(w}I>E}

f

X2dP

{w: I%(e}I<E}

X2dP

{w: I%(w}I jE}

E2P({w: IX(w)j > E}). LEMMA 3.6.14. Let X(w) a L2(Qo) and let 0 < A < 1. Then E2(X) P({w: X(w) > AE(X)}) > (1-A)2 E(X2)*

Proof. Let X(w)

{0

ifX(w) > AE(X), ifX(w) < AE(X).

By the Schwarz inequality we have

E2(X') < E(X'2)P({w: X'(w) 0}) E(X2)P({w: X(w) > AE(X)}). Since

E(X) < E(X')-}-AE(X), we have

E(X') > (1-A)E(X). Thus

(1-A)2E2(X) < E(XZ)P({w: X(w) > AE(X)}).

0

Chapter 3

133

Let (SQ,E,P) be a probality space. Random variables X1(w), ..., X,((o) are called independent, if P({co: X1(co) E B1i

..., Xn(co) E Bn})

= P({w: X1(w) E B1}) ... P({w: Xs(w) E Bn})

(3.6.4)

for all sets B1, ..., B, E E. A sequence {Xn} of random variables is called a sequence of indpendent random variables if, for each finite set of indices i1, ..., in, the random variables Xt1i ..., Xi, are independent. By (3.6.4) we conclude that, if {Xn(w)} is a sequence of independent random variables, then E(Xt1 ... Xtn) = E(XXJ ... E(XXn)

for i1 < i2 < ... in. By a Rademacher sequence we shall mean a sequence of independent

random variables rn(co) taking the values +1 or -1 with the same probability 1/2. LEMMA 3.6.15 (Paley-Zygmund inequality ; see Paley and Zygmund, 1932). Let {rn(w)} be a Rademacher sequence. For any sequence of numbers {an} N N 2

P ({W:

>

rn (w)an

n=1

11 lan12))

n=1

Proof. We shall write N

X(w) = I rn(w)an

2

n=1

and we shall calculate N

E(X)

f

= aJ

_

f

I rn(w)an I2 dP n=1

N

n i,j=1 N

rt(w) rj(co) at a1 dP `N

=D f I i-1 pail jdP = Ei=1Iail2.

> 3 (1-2)2.

Locally Pseudoconvex and Locally Bounded Spaces

139

Using the formula t

k,+...+k. =k

k1 ! ... kn !

n

and observing that f rn1(w)rn2(w)rn3(co)rn4(w)dP

=0

D

if one index is defferent from the others, we obtain N

E(X2)

4

= SJf

n=1

rn(w)an dP

1

N

_ : Ian14+6 n=1

lan12Iam12

1_
N

3(Ilan12)2. n=1

Using Lemma 3.6.14 we obtain the required inequality.

Let S2 = {-1,1 }N be a countable product of two point sets. The elements of 0 are sequences a = {an}, where an take the values either

+1 or -1. Let in(a) be the following random variables rn(a) = an. Let ) = {2n} be a sequence of reals such that I'nI < I. On the two point set {-1, +I} we define measures PzA as follows +1 }) =

Pa.({ -I}) =

1

A

2 1 2.1

Let Px be the product measure, Px = P2 x ... X Pz x ... It is easy to verify that the random variables {&(t)} constitute a sequence of independent random variables and f rn (a)dP2 = An. D

Let .% = {0, 0,

... }. The measure P2 will be denoted by P0.

(3.6.5)

Chapter 3

140

LEMMA 3.6.16 (Musial, Ryll-Nardzewski and Woyczytiski, 1974). Let PA and PO be as described above. Let

P = 2 (Po+P2). Then, for each sequence of numbers {x1,x2, ...} and for each positive integer n, n

r"i(a)xi > 8

M = P (ta: i=1

Atxi11)> g

Proof. Without loss of generality we may assume that

= 1. n

If

Ixtl2 > ',we shall use the Paley-Zygmund inequality t=1

m

>2Pa({a:

ri (a)xi

8})

i=1

>

2 Po({a:

ii(a)xt i= 1 \

a (I

Ixt12)112}

t=1

1

2 3 (1-1612> If

IxtI2 < a we shall use the Tchebyscheff inequality (Lemma ti=1

3.6.13). By simple calculation we obtain n

f f (,.i-r"i (a))xi 2dP2 A

i=1 n

E i=1 n

f

.1

I() -r"t(a))xtl2dPA n

n

_

.1; IxxI2-2 i=1

_

,' )1tIxid2 t=1

.y (1-,I;)Ixt12< 4 t=1

n ,1

O

ft(a)dPA+ f Ixtj2 i=1

Locally Pseudoconvex and Locally Bounded Spaces

141

In the calcution we have used the fact that rt are independent random

I_

variables and formula (3.6.5).

Now P,t

({a:

g

)

(At-, (a))x$

= Px ({a:

I_<

8})

t=1 n ( = 1-PA (Sa: l

'(At-iit(a))xi t=1

8 }),

n 82 f2_

(1 -A:) JXjJ2 >' 1-

82

1

-4

tit

Hence m > e also in this case.

Proof of Theorem 3.6.12 (modified proof given by Ryll-Nardzewski and Woyczynski (1975). Let M be a bounded measure. Thus, by definition, the set

K= {x: x=

E{M(Ag), et equal either -1 or+ 1, t=1

At e E, At disjoint sets}

is bounded. We have to show that the set n

K1 = {x: x = JAM(At), 122 < 1, At c- E, At disjoint} t=1

is also bounded. M(Ag) E L°(Q,E,µ). We shall write M(At) = f (t). Let S2 = {-1, +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 3.6.16

lt(a)f(t) 8 tit

t=1

I)

8

(3.6.6)

Chapter 3

142

for all t. By the arguments given in : Example 1.3.5 we may assume that

it(Q) <+oo. Let

Tc{t: f{(t)

> 8c}.

t=1

Suppose that p(TT) > 0. Let p ,(A) = p(Ar Tc)/u(TT). Of course pe(TT) = 1. For the product measure pe X P we have by (3.6.6) n

n

:I(a)J(t)

ll(t,a):

8

.YJ At.r(t) i=1,

i=1

Then, by the Fubini theorem, n

max

uc

E,=±1

n

8 1

Si Ytt(t )

({ t e Tc: l

.if{(t)

{=1

and, by the definition of TT, max

E/=±1

tft(t)

pe ({ t e TT :

l

Hence, by the definition of pc, n

max p ({t e T:

E,=f1

max It

e1=f1

8

Ill

c({te l

e o{(t)

> c})

t=1

Ye{ft(t)

t e Tc:

> c }) > 111

l=1 n

1 P(Tc) 8

l

T: ti=1

Atft(t) > 8c}).

(3.6.7)

1

Let

U, = {x: u({t:

Ix(t)J

> c}) < c}

be a basis of neighbourhoods of zero in L°(Q, E, p). Since the set K is

bounded, then for each c > 0 there is an s > 0 such that sKC Uc. Thus, by (3.6.7), sK1 C U. The arbitrariness of c implies that the set K1 is bounded.

Locally Pseudoconvex and Locally Bounded Spaces

143

Now we shall define integration with respect to an L'-bounded measure of scalar valued functions. Let (X, II II) be an F-space. Let M be an L00-

bounded measure taking its values in X. Let f be a scalar valued function. Let M.(f) = sup { I f gdM 11: g being simple measurable n

functions, IgI < IfI}. PROPOSITION 3.6.17 (Turpin, 1975). M (f) has the following properties. (3.6.8.i) If I.f1I
If f is measurable and

0,

then f = 0 almost everywhere (3.6.8.iii)

(i.e. except a set A such that M(A) = 0). If {f,,} is a sequence of measurable functions tending almost everywhere to f, then (3.6.8.iv)

n- 00

M' (f1-I-f2) < M' (fi)+M. (f2),

(3.6.8.v)

provided fl, f2 are measurable. Proof. (3.6.8.i), (3.6.8.ii), (3.6.8.iii) are trivial.

(3.6.8.iv). Let (f.} be a sequence of measurable functions tending to f almost everywhere. Let g be a simple measurable function such that Ig (t)I < If(t)j. Let E be an arbitrary positive number. Lets be an arbitrary

number such that 0 < s < 1. Let I fn(t)I

gn(t) =

Ig(t)I

if g(t)

0,

if g(t) = 0. Observe that {gn(t)} converges almost everywhere. Thus, by the 0

Egorov theorem (Theorem 3.6.9) for each 17 > 0 there is a set A such

that M(Q\A) <,q and {gn} converges uniformly on A. Since g is a simple function, there is an r7 > 0 such that for each set H such that M(H) < 17 we have

11 f sgdM < a. H

Chapter 3

144

Taking 17 and A as described above, we have

f sgdM < f sgdM I + 11 f sgdM) < lim inf M (fn)+e,

n-.ao A a D \A because, for sufficiently large n, I fn(t)I > Isg(t)I for t E A. The arbitra-

riness of a and s imply (3.6.8.iv).

(3.6.8.v). To begin with, we shall assume that f1 and f2 are simple. Let g(t) be an arbitrary simple measurable function such that Ig(t)I < If1(t)+f2(t)I Let g{(t)If+(t)l

(

gi(t) = J Ifi(t)I+If2(t)I

if 1.f1(t)I+If2(t)I

0, i = 1,2.

if 1f1(t)I+If2(t)I = 0.

10

The functions g1(t) and g2(t) are simple and

g(t) = 91(t)+92(t)Hence

II f gdM < f g1dM n

n

f g2 dM

n

M' (fi)+M' (f2)

The arbitrariness of g implies

M' (f1+f2) < M' (A)+M' (f2) Suppose now that f1 andf2 are not simple. Then there are two sequences of simple mesurable functions { f1,n) and { f2 ) tending almost every-

where to j1 and f2 and such that Ift,nl < Ifi I, i = 1,2. Therefore, by (3.6.8.iv), M ( f 1 + f 2 ) < liminf M (f1, n+f2, n) 00

li m sup M' (f1, n+f2, n) n-ico

< supM' (fi,n)+supM . (f2, n) n

< M ' (fi)+M' (f2) Let

B(f) = I f gdM: g simple measurable functions, a

ISI < 1f1}

Locally Pseudoconvex and Locally Bounded Spaces

145

PROPOSITION 3.6.18. The set B(f) is bounded if and only if

limM (tf) = 0.

(3.6.9)

Proof. If the set B(f) is bounded, then

limM (tf) = lim (sup{IIxII: x e B(tf)}) t- o

t- o

= lim(sup{IIxII: xe tB(f)}) = 0.

(3.6.10)

t->o

Conversely, if (3.6.9) holds, then by (3.6.10) the set B(f) is bounded. Let X denote the set of those f for which B(f) is bounded. By (3.6.8.v)

and (3.6.9), M (f) is an F-pseudonorm on X. We shall now use the standard procedure. We take the quotient space X/(f: M (f) = 0). In this quotient space M- (f) induces an F-norm. We shall take completion

of the set induced by simple functions. The space obtained in this way will be denoted by Ll(Q,E, M). Since, for each simple function g,

f gdM

n

M49).

We can extend the integration of simple functions to a linear continuous operator mapping L1(Q,E, M) into X.

3.7. INTEGRATION WITH RESPECT TO AN INDEPENDENT RANDOM MEASURE

Let (Q0,E0,P) be a probability space. Let Q be another set and let E be a a-algebra of subsets of 0. We shall consider a vector valued measure M(A), A E E, whose values are real random variables, i.e. belong to X = L°(Q0 E0 P). We say that M(A) is an independent random measure if, for any disjoint system of sets {Al, ..., A..}, the random variables M(At) e X, i = 1,2, ..., n are independent.

We recall that a vector measure M is called non-atomic if, for each A E Q such that M(A) # 0, there is a subset A0 C A such that M(A°)

: M(A).

Chapter 3

146

Let X(w) be a random variable, i.e. X(w) E L°(Q°iE°, P). By Fx(t) we denote

Fx(t) = P({w: X(w) < t}). The function Fx(t) is non-decreasing, lim

t-

Fx(t) = 0, limFx(t) = 1. CO

It is called the distribution of the random variable X(w).

Let Q = [0, 1], and let X be the algebra of Borel sets. We say that a random measure M is homogeneous if for any congruent sets Al A2 C

C [0,1] (i.e. such that there is an a such that a+A1 = A2 (mod 1)) the random variables M(A1) and M(A2) 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(A1)+ ... +M(An), where the random variables M(At), i = 1, 2, ..., n, are independent and have the same distribution (Prekopa, 1956). Let X(w) be a random variable. The function +00

fx(t) = f eiesdF(s).

(3.7.3)

is called the characteristic function of the random variable X(w). If the random variables X1, ..., Xn are independent, then

fxl+ ... +X. (t) = fxl(t) .... fxn(t).

(3.7.4)

Directly from the definition of the characteristic function

fax(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). .1x = (fx'.)n

(3.7.7)

If X is an infinite-divisible random variable, then its characteristic function fx can be represented in the form 00

fx(t) = exp (it+

itX I+X2 ](eitx_i_j) dG(x),, X2

(3.7.8)

Locally Pseudoconvex and Locally Bounded Spaces

147

where y is a real constant, G(x) is a non-decreasing bounded function and we assume that at x = 0 the function under integration is equal to -t2/2. This is called the Levy-Kchintchin 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

f (costu-1)

1 u u2

dG(u).

(3.7.9)

0

Of course, without loss of generality we may assume that G(0) = 0. Suppose that M(A) is a non-atomic independent random measure with symmetric values. Then by (3.7.9) 00

fM(A)(t) = exp

f (cos to-1)

1

u'o2 dGA(u).

(3.7.10)

0

For an arbitrary real number a we obtain by (3.7.5) and (3.7.10) AM(A) = exp

f(cos to-1)

z

1 zu dGA(u).

0

u

(3.7.11)

Let A,A, e X be two disjoint sets. By (3.7.4)

f

00

fM(Al u A2)(t) = exp

J+2

(cos to-1)

2

dGA1+A2(u).

(3.7.12)

0

Formulae (3.7.11) and (3.7.12) imply that for a simple real-valued function h (s) f fh(s)dM(t) = exp(- f TM(th(s))ds), n D

(3.7.13)

where W

TM(x)

= I(1 -cosxu) i+u2 u

dGM(u).

0

In the sequel we assume 92 = [0, 1]. Now se shall prove some technical lemmas.

(3.7.14)

Chapter 3

148

Let UM(x) =

min I x2,

J

Z I (1+u2)dGM(u)

(3.7.15)

0

and 00

G(3)

J

Y'M(X) =

u

du

for x > 0,

for x =0

0

(3.7.16)

.

It is easy to see that the two functions UM(x) and VIM (x) are equal to 0 at 0, continuous and increasing. LEMMA 3.7.1. For all x > 0, a > 0, there are positive numbers cl(a) and c2 such that

max TM(v) < cl(a) UM(x)

(3.7.17)

0_
and

i f TM(xt)dt > c, Um (x).

(3.7.18)

0

Proof. Observe that

1-cosuv = 2sin2 2 < 2 u2v2 and

1-cosuv<2. Hence, for cl(a) = max(2, z a2), we have

max (1-cosuv) < cl(a)min(1,x2u2). 0<_v<_az

Therefore

1-cosuv

max 0_
u2

Integrating (3.7.19) with respect to 1 Observe that smz

1-

z

1

< cl(a)min x2,uz-

> c2min(z2,1),

(3.7.19)

uz

U2

dGM we obtain (3.7.17).

Locally Pseudoconvex and Locally Bounded Spaces

149

where siz z

c2 = min I min 11- slz z)

in Z-2 1 o Integrating TM(tx) with respect to t, we obtain 1

f

00

l+uz 1- sinxu) u2 XU

('

TM(tx)dt = J

00

c2

dGM(u)

)

I +U2

f min(x2u2, 1)

u2

dGM(u)

0

= C2UM(x)

Observe that

r

1/z

UM(x) =

f x2(1+u2)dGM(u)+ J

Co

z

1 u2 u- dGM(u).

1/z

0

Integrating by parts the second integral on the right-hand side, we obtain 00

00

f

J +U2 uz

dGM(u) = GM(u)

00

2 u uU2

1/z

du +2 f GM3(u) u 1/z

1/z

2!1' (x). Thus

2Y'M(x) < UM(x).

(3.7.20)

On the other hand, integrating by parts both terms, we obtain 1/z

UM(X)

= x2(1+u2)GM(u) 0z- f 2ux2GM(u)du 0 Go

+GM(u)

I+u u2.

z

+2 f GM(u) du 1/z

< limGM(u)+21VM(x).

wm

1/z

u3

(3.7.21)

Formulae (3.7.20) and (3.7.21) imply that the functions UM(x) and Y'M (x) are equivalent at infinity.

Chapter 3

150

LEMMA 3.7.2 (Urbanik and Woyczynski, 1967). Let G(u) be a continuous

f)

non-decreasing function such that G(0) = 0 and limG(u) = 1. Let u--++oo

00

O(x) =

G(u du.

(3.7.22)

1/x

Then the function O(j/x) is concave. Proof. We apply a change of variable v = 1/u2 in formula (3.7.22), and obtain ao

(1/ z) =

fG(u)d u = fG(_)-)dv.

Thus the derivate of O(j/x) is equal to G(1/j/5c). It is non-increasing, since G is non-decreasing. Therefore the function t(j/x) is concave. LEMMA 3.7.3 (Urbanik and Woyczynski, 1967). Let {f,} be a sequence of simple measurable functions belonging to L°[0,1]. The sequence

Jfa(s)dM(s) tends to 0 in L°[0,1] 0

if and only if lim f WM(J fn(s)I)ds = 0. n-#oo °

Proof. We shall use the classical statement that a sequence {hn} of functions belonigng to L°[0,1] tends to zero in L°[0,1] if and only if the logarithms of its characteristic functions tend uniformly to 0 on 1

each compact interval. Thus by (3.7.13) { f fn(s)dM(s)} tends to zero in 0 1

L°[0,1] if and only if { f TM(fn(s))ds} tends to O uniformly on each com0

pact interval [0, T].

If I f TM(tfn(s))ds} tends uniformly on each compact interval, then lim f f TM(tfn(s))dtds "-'Co 0

0

= lim f f TM(tfn(s))dsdt = 0. "-i°o o

0

Locally Pseudoconvex and Locally Bounded Spaces

151

Hence by (3.7.18)

lim f UM(Ifn(s)I)ds = 0.

(3.7.23)

n- CO 0

Since UM(x) and FM(x) are equivalent at infinity,

lim f 'PM(Ifn(s)I)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 that { f TM(tfn(s))ds} tends to 0 uniformly on each compact interval. 0

THEOREM 3.7.4 (Urbanik and Woyczynski, 1967). Le M be a symmetric

homogeneous independent random measure with values in the space L°[0,1]. Then there exists a continuous function N(x) defined on [0,+00)

such that N(0) = 0, N(1/) is concave and such that the set of all real functions for which the integral f fdM exists is the space N(L). 0

Conversely, if N(1/z) is a concave continuous function such that N(0) = 0, then there is a symmetric independent homogeneous random measure M with values in the space L°[0,1], such that the set of real valued functions f for which the integral f 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(1/ x) is concave, then it can be represented in the integral form

N(1I) = f q(u)du, 0

q(u) is a non-increasing function. We put G(0) = 0 and G(x) = min(1,q(1/x2)) for x > 0. Let M be a symmetric independent homogeneous random measure such that where

Chapter 3

152

(3.7.10) holds. To complete the proof it is enough to show that the functions N(x) and z,

ao

NG(x) =

f

G(u)

du = f min(l,q(u))du 0

1/z

are equivalent at infinity. Since q(u) is non-increasing, &)
foru>1. Thus z

z

1

N( OX = f q(u)du = f q(u)du+ f q(u)du 0

0

1

2

= f q(u)du+q(1) f min(l,q(u))du 0

K+CNG(V X).

where K = f q(u)du and C = q(1). 0

On the other hand, trivially

NG(l') < N( X) Woyczynski (1970) extended this result to the case where M(A) take values in the product space L°(Q0,E0,P) x ... X L°(D°iE0,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 E be the algebra of all subsets of N. Let M be a vector measure defined on A e E taking its values in an F-space (X, Let x = M({n}). Since M is a vector

measure the series 00E x has the property that for each sequence ei n=1 co

taking values ' either 0 or 1 the series Z e xn 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 unconc ditionally convergent series I xn induces a vector measure by the formula n=1

M(E) = I xn. n=E

PROPOSITION 3.8.1. The measure M induces by an unconditionally con-

x is compact (and thus bounded).

vergent series n=1

Proof. Since the series

n=1

xn is unconditionally convergent, for each

e > 0 there is an index N such that for each sequence {en} taking values either 0 or 1 Co

Gl

En Xn

< C.

(3.8.1)

I

n=N+1 N

Thus the finite set

enxn, where {e1, ..., sn} runs over all finite systems n=1

taking values either 0 or 1, constitutes an a-net in the range of the measure M. THEOREM 3.8.2 (Orlicz, 1933). A series 00I xn is unconditionally convergent n=1

if and only if for each permutation p(n) of positive integers the series GO

E xv(n) is convergent. n=1

Proof. Suppose that the series

0'

Xn is unconditionally convergent. Then

n=1

for each e > 0 there is an index N such that k+m

LI EaXn

< e

n=k

for k > N, m > 0 and en taking the value either 0 or 1.

Chapter 3

154

Let K be such a positive integer that, for n > K, p(n) > N. Then for

arbitrary r > K and s > 0 r+a

4

Eixi < E,

xy(n) II = I n=r

(3.8.2)

1=p

where

p = inf {p(n): r < n < r+s}, g = sup {p(n): r < n < r+s}, if i = p(n) (r < n < r+s), 11 Et

= to

otherwise.

The arbitrariness of E implies that the series E xv(n) is convergent. n=1 00

Suppose now that the series

xn is not unconditionally convergent. n=1

This means that there are a positive number S and a sequence {En}, E taking the value either 1 or 0 and a sequence of indices {rk} such that rk+1

LI Enxn > a.

n=rk+1

Now we shall define a permutation p(n). Let m be the number of

those s., n = rk+l ,..., rk+1, which are equal to 1. Let p(rk+v) = n(v), where n (v) is such an index that En(v) is a v-th Ej equal to 1, rk < i < rk+1r

0 < v < m. The remaining indices rk < n < rk+1 we order arbitrarily. Then rk+m

I xp(n) n=rk

rk+1 Enxn>b.

n=rk+1

This implies that the series

xP(n) is not convergent. n=1 00

A measure M induced by a series E xn is L'-bounded if and only if n=1 00

for each bounded sequence of scalars {an} the series

anxn is con-

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 Ryll-Nardzewski, 1967). There exist an 00

F-space (X, II I I) 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 be a k-dimensional real space. There exists an open symmetric starlike set A in X which contains all points pl, ..., pak of the type (e1, ..., sk), where E{ equals I or 0 or -1, such that the set

Ak-1 = A+ ... +A (k-1)-fold

does not contain the unit cube

C = {(a1i ..., ak): Ia{I < 1, i = 1,2, ..., k}.

Proof. Let Ao be the union of all line intervals connecting the point 0 with the points p, ...,per. Obviously the set Ak-1 is (k-1)-dimensional. Therefore there is a positive number a such that the set (Ao+A8)k-1, where A. denotes the ball of radius E (in the Euclidean sense), has a volume less than 1. Thus the set A = Ao+Ae has the required property. LEMMA 3.8.5. There is a k-dimensional F-space (X, II

II)

such that

i = 1,2,..., 3k and there is a point p of the cube C such that IlplI > k-1. IIp{II < 1,

Proof. We construct a norm II II in X in the way described in the proof

of Theorem 1.1.1, putting U(1) = A. Since pi e A = U(1), IIptJI < 1. Furthermore, since Ak-1 = U(k- 1) does not contain the cube C, there is a point p e C such that p e U(k-1). This implies that IIpII > k-1. Proof of Theorem 3.8.3. We denote by (Xk, II IIk) the 2k dimensional space constructed in Lemma 3.8.4. Let IIxIIk = 2k IIxIIk

Let X be the space of all sequences a = {a...} such that 00

IIIaIII = I II(as--+1, ..., k=1

<+oo.

Chapter 3

156

It is easy to verify that X is an F-space with the norm III III Let

X. = (0, 0, ..., 0, 1, 0, ...) n-th place

and let {en} -be an arbitrary sequence of numbers equal either to 0 or to 1. 00

The series I En xn is convergent. Indeed, n=1

< III YO III+III Yk+l III+ ...+III Yk'III+IIIYOI I I,

EnXn

n=m

where k is the smallest integer non-greater than log2m, k' is the largest less than log2r, and 2k

Yo

=1Enxn, n=m

j = k+1, ..., k',

Enxn,

Y1 = V1

Yo = L, Enxn n=2k'+1

In virtue of Lemma 3.8.5 1

2k_1

j=k+1,...,k'

1

2j_1 1

2k' Therefore k'

1

i=k-1

n=m

2t

1

8

< 2k_2< m,

m

and the series

D

xn is uncon-

EnXn is convergent. Thus the series n=1

ditionally convergent.

Locally Pseudoconvex and Locally Bounded Spaces

157

On the other hand, as follows from Lemma 3.8.5, for each k there is a point ak a Xk, ak = (a2a+i, ..., a2k+1), such that Ia{I < 1, i = 2k+1, ...

..., 2k+1 and

II ak Ilk >

2k

(2k-1) >

2

.

Let b = an, n = 1, 2, ... The sequence {bn} is bounded and 2k+1

I bn xn n=1

=IIakIIk>

Therefore the series

2.

bnxn is not convergent. This implies that the n=1

series E x is not bounded multiplier convergent. n=1

Observe that the measure M induced by an absolutely convergent oo

series n=1

x which is not bounded multiplier convergent is bounded, but

is not L°°-bounded. Turpin (1972) constructed a non-atomic measure with this property.

3.9. INVARIANTA(X)

We shall prove the results of Turpin (1975), however restricting ourselves to the metric case. Let (X, II II) be an F-space. By A (X) we denote the set of such sequences {An} that the space X has property P({.i;,}), 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 _ {2, 22, ... 9 An, 0, 0, ...}. If A A (X) contains a sequence with an infinite support, in other words : if

there is a sequence 2 = {An}, A. > 0 belonging to A(X), we say that the space X is strictly galbed.

Chapter 3

158

PROPOSITION 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 {An} that, for any bounded sequence {xn} C X, the sequence (3.9.2)

An xn }

{YN}

i=1

is bounded.

Proof. Necessity. Let {An} 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 b V, n = 1, 2, ... Thus, by (3.9.1) yN e b U, N = 1, 2,... Since U is arbitrary, the sequence yN is bounded.

Sufficiency. Suppose that {An} 0 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 ,ZnV+An+1V+ ...'# U. (3.9.3) Now we shall choose a sequence of elements {xn} and an increasing sequence of indices {nm} in such a way that I1x{II < 2m ,

M

i = nm+1, ..., nm+1 ,

(Anm+iXnm+1+ ... +Anm+ixnm+i) 0 U.

(3.9.4)

(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} = {)nm+lXnm+1+

+Anm+lxnm+l}

is not bounded. This implies that the sequence {yN} is not bounded either.

PROPOSITION 3.9.2. If {A.} e A (X) and 0 < /en < An, n = 1,2,..., then {µn} e A(X).

Proof. It follows immediately from the definition of A(X).

Locally Pseudoconvex and Locally Bounded Spaces

159

PROPOSITION 3.9.3. Let {AJ e A(X). Let In be a sequence of non-empty finite disjoint subsets of the set of positive integers. Let

/In= f

Ai

iEI Then {µn} e 11(X). Proof. Let {xn} be an arbitrary bounded sequence. Let x,

(xn

for i e In,

0

elsewhere.

The sequence {xtn} is of course also bounded. Then the sequence N ) N {YN} n=1

/In Xn

=

X{}

n=1 iEI,

is bounded. PROPOSITION 3.9.4. Let {An}, {/ln} a 11(X) and let (n{, m{) be a double

sequence with terms not equal to one another. Then the sequence {vt}

= {A.,-p.}e11(X). Proof. Let {xi} be an arbitrary bounded sequence. Then N

N

Vi X{ _

YN = t=1 P

_ n=1

2nullm'Xt t=1

n

(3.9.6) m

where p = max n{ and the summation with respect to m is taken over 1
all m for which there is an index i = in, m < N such that (ni, mi) = (n, m). llmxt.,m, is bounded. Therefore the sequence The set {yN}, yn = {yN} is also bounded.

El

nPro

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. We shall show that if {qn} e 11(X), then

I

(_2

e 11(X). Indeed

let (n{, m{) be such a double sequence that (n{+ mi) is a non-decreasing

Chapter 3

160

sequence and the sequence (ni, ms) contains all pairs (n, m). Let An

=qn=Pn It is easy to verify that 2n,11m, = q"+" has the same order of growth as n

qI I. Then (q /-n} e A(X). Since Jim

1

q- I/-- = 0, by Proposition 3.9.2

e A(X). Theorem 3.6.11 implies the corollary. 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. PROPOSITION 3.9.7. Let Y be a subspace of an F*-space X. Then

A(Y) A(X). Proof. Let {A,n} E A(X). Let {xn} be an arbitrary bounded sequence in Y. Then it is also a bounded sequence in X. Therefore the sequence {y1V}

= { "=1 Anxn}

is bounded. This implies that {An} e A(Y). COROLLARY 3.9.8. A(X) C P.

Proof. Each space X contains a one dimensional subspace X0. Since A(X0) = 11, by Proposition 3.9.7 the corollary holds. PROPOSITION 3.9.9. Let a continuous linear operator T map an F-space X onto an F-space Y. Then

A(Y) A(X). Proof. Let IA,,} e A(X). Let U be an arbitrary . neighbourhood of zero in Y. The set T-1(U) is open. Then there is a ,neighbourbood of zero V in the space X such that A,V+A,V+ ... C T-1(U). Then

A1T(V)+A2T(V)+ C U.

Locally Pseudoconvex and Locally Bounded Spaces

161

By the Banach theorem (Theorem 2.3.1) the sets T(V) are open. The arbitrariness of U implies that { t } e A(Y). As an immediate consequence of Propositions 3.9.6, 3.9.7 and 3.9.9 we obtain THEOREM 3.9.10. If

dime X < diml Y

(codimi X < codimi Y),

then

A(X) A(Y). PROPOSITION 3.9.11. If X is a locally p-convex space, then

A(X) ) IP. Proof. Let {IIxIIm} be a sequence of p-homogeneous pseudonorms determining the topology. Let {xn} be a bounded sequence in X. Let Mn = Sup IIxnIIm. n

Let 00

C=

IP.

I1%

n=1

Let N

YN=IAix{. s=1

Then 00

IIYNIIm <

c=

IIP IIxtIIm < CMm.

(3.9.7)

COROLLARY 3.9.12 (compare Mazur and Orlicz, 1948). If an F*-space X is locally convex, then A(X) = 11. PROPOSITION 3.9.13. Let X be a locally pseudoconvex space. Then

A(X)Dnip. P>0

Chapter 3

162

Proof. Let {2i} e n I P. Let {IIxIIm} be a sequence of p.-homogeneous P>0

pseudonorms determining the topology in X. Let Go

i". ti=1

Taking xn and yN as in the proof of Proposition 3.9.11 we obtain an estimation similar to estimation (3.9.7) IIYNIIm < CmMm.

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) DIP, then the space X is locally p-convex. Proof. Suppose that an F*-space (X, II II) is not locally p-convex. Then there is a positive number a such that for any m there are points {xmi1, xm, gm} such that lIxm, ill < 4m ,

i = 1, 2, ..., km,

(3.9.8)

and xm, {

Let for n k 14-

> E.

(3.9.9)

+km -1 < n < k -}1

+k m , let 2n =

1

2mkl/Pm

xn = 2-x.,{, where i = n-(k1+ ... +km-1). 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 TN

{YN}

j

= l Lam,2nxn} n=1

is not bounded. Therefore {A.} does not belong to A(X). Since {2,,} 61P, A(X) does not contain IP.

Locally Pseudoconvex and Locally Bounded Spaces

163

PROPOSITION 3.9.15. If AM j n IP, then the space X is locally pseudoP>0

convex.

Proof. The proof follows the same line as the proof of Proposition 3.9.14 if we replace, in the definition of A. and in formula (3.9.9), p by pm tending to 0. Li As follows from Proposition 3.9.15, locally pseudoconvex spaces are stricly galbed. There are also non-locally pseudoconvex spaces which are strictly galbed, as follows from PROPOSITION 3.9.16 (Turpin, 1975). For any sequence {A } belonging to all spaces IP (in a particular case for any geometric sequence qn, 0 < q < 1), there is a non-locally pseudoconvex space X such that {A,,} e A (X).

Proof. Let c _

00

IA=Ill'. Let X be the space of all sequences x = {xn}.

i=1

such that x. e L11'[0, 1] and for all positive integers k Jim ck, IIxnIIL=,^[o 1]

= 0.

n-OD

We introduce a topology in X by a sequence of F-pseudonorms IIxIIk = supcnl1xnllL,,,,[0,1]. It is easy to verify that X is an F-space. The

space X is not locally pseudoconvex. Indeed, let X. = {x e X: x{ = 0 for i n}. The space Xn is isomorphic with L11"[0, 1]. Take any neighbourhood of zero U in X such that U X,,, 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: IIxIIk < E}. Let V = {x: IIxIIk+1 < E}. Let y1, YB, ... e V, ys = ys, n. Let x = ti=1

Ys-

Chapter 3

164

Then 00

114 =supcn n

{=1

As Ys, nj L'M"

C SUpCn

lAsll/n IIYt. -11V1-[0,11

n

i=1 < SUpcn+' supllYi,

njL',"[0,1] < E.

i

n

Therefore x e U and (An} e A(X).

There are non strictly galbed spaces, as follows from PROPOSITION 3.9.17. Let (.Q,E,p) be a measure space. Suppose that the measure p is not purely atomic (i.e. there is a set D° such that the measure p restricted to the set S2° is a non-atomic measure which does not vanish identically). Then the space L°(Q,Z',p) is not strictly galbed.

Proof. According to our hypothesis about the measure, we can find Do, 0<,u (D°) <+oo, and a sequence of sets {En} such that (3.9.10) lim p(En) = 0 n- 00

and co

D°C U E.

for K = 1, 2, ...

(3.9.11)

n=B

Let { t } be an arbitrary sequence of positive numbers. Let

xn =

n .Z'E"

An

By (3.9.10), {xn} tends to 0, and thus it is bounded. On the other hand, the sequence N

YN(t)_IAnxn(t) n=1

tends to infinity for t e Sl°. Thus it is not bounded. PROPOSITION 3.9.18. Let X be the space 1{p"}, 0 < pn < 1, of all sequences

x = {xn} such that co

IIxII =

IxnlP. <+00, n=1

Locally Pseudoconvex and Locally Bounded Spaces

165

with a topology given by the F-norm IIxlI. If p,,-->O, then the space X is not strictly galbed.

Proof. Let {A,,} be an arbitrary sequence of positive numbers. Since p.-->O, for each n we can find an index k(n) such that 2nk(n) > z. We shall choose a k(n) such that k(n)
divergent. It is obvious that the sequence {xn} = {$nen}, where {en} denotes the standard basis in fir"l, tends to 0. Thus it is bounded. On the other hand, N

N

11f ).nxn

IYNI1

n=1

IlAnxn11

n=1

N

>

G$nnk (n) -00. n=1

N

Therefore the sequence yN =

A. xn is not bounded. This implies n=1

that {a.n,} 0 A(X).

3.10. C-SEQUENCES AND C-SPACES

The following notion of C-sequences is related to bounded multiplier convergence. We say that a sequence {xn} of elements of an F-space X is a C-sequence if for any sequence of numbers {tn} tending to 0 the series 00

to xn is convergent. Of course, a subsequence of a C-sequence is a n=1

C-sequence, and the image of a C-sequence by a continuous linear operator is also a C-sequence. PROPOSITION 3.10.1. A sequence {xn} of elements of an F-space (X, II is a C-sequence if and only if the set OD

A = l' anxn: Ianl < 1, n = 1, 2, ...} n=1

is bounded.

II)

-Chapter 3

166

Proof. Sufficiency. Suppose that the set A is bounded. Let {cn} be an

arbitrary sequence of numbers tending to 0. Let C. = sup Icil. Of n_
also tends to 0. Let E be an arbitrary positive number. Since course, the set A is bounded, there is an index No such that for n > No (3.10.1)

1JCnxlj < E

for all x e A. Thus for n, m, No < n < m, Ci Xi

= 1CnxjI < E,

Cn

since m

x=

Ci

C xieA. i=n+1

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}, lanI < 1, and a sequence of indices {nk} such that the sequence n,,+1

anxn}

{ tk

does not tend to 0.

n=n,+1

Let Cn I

tkan 0

for nk < n < nk+1, elsewhere. W

The sequence {cn} tends to 0, but the series E CnXn is not convergent. n=1

Therefore, the sequence {xn} is not a C-sequence. An F-space X is called a C-space if for any C-sequence {xn} the series X. is convergent. As an example of an F-space which is not a C-space n=1

we can consider the space co (see Example 1.3.4.a). Indeed, the standard co

basis {en} in co is a C-sequence, but the series I en is not convergent. n=1

A subspace of a C-space is also a C-space. Therefore, if an F-space X contains a subspace isomorphic to c0, then X is not a C-space.

Locally Pseudoconvex and Locally Bounded Spaces

167

Now we shall prove the following THEOREM 3.10.2 (cf. Bessaga and Pelczynski, 1958). Let X be an F-space with a basis {en}. If X is not a C-space, then X contains a subspace isomorphic to the space co.

The proof is based on the following PROPOSITION 3.10.3 (Schwartz, 1969). Let (X, II II) be an F-space such that each C-sequence tends to 0. Then X is a C-space. Proof. Suppose that X is not a C-space. Let {xn} be a C-sequence such xn is not convergent. Then there are an increasing

that the series

n=1

sequence of indices nk and a positive number 6 such that IIykII > 6, where nk+,

Xn.

Yk =

n-nk+l

The sequence {yk} is also a C-sequence, and it does not tend to 0. Proof of Theorem 3.10.2. If (X, 11 11) is not a C-space, then there is a C-seW

quence {xn} suc11 that the series Exn is not convergent. Basing ourn=1

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 C-sequence, {fk(xn)} are C-sequences for k = 1,2, ... Hence

lim fk(xn) = 0,

k = 1, 2,...

(3.10.2)

By Theorem 1.2.2 we may assume that the norm II II is non-decreasing. Since {xn} does not tend to 0 and (3.10.2) holds, we can find a subsequence {zv}, a positive number 6 and an increasing sequence of indices {np} such that

Ilzv-z 11 < 2v 6.

(3.10.3)

Chapter 3

168

where nv+1

ZP =

fn(zp)en.

n=ni+1

Corollary 2.6.6 implies that {z,} is a basic sequence. We shall show that it is equivalent to the standard basis of co. Let {tp} be a sequence of

coefficients of x belonging to Xo = lin {z,} with respect to the basis {zp}. Since IIz,II > 6/2, the sequence {tn} tends to 0. On the other hand, (Zp} is a C-sequence. Therefore, by (3.10.3), {z,} is also a C-sequence. This means that if tn->0, then the series E tpz1, is convergent. Therefore, P=1

the basis zP is equivalent to the standard basis of co. Thus, by Theorem 2.6.3, the space X0 is isomorphic to the space co. 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 C-space. Then the space Y contains a subspace Yo isomorphic to the space co. Proof. Without loss of generality we may assume that 11 11 is p-homoge-

neous. 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. Let X be a non-separable F-space. If X is not a C-space, then it contain s

an infinite dimensional separable subspace X which is not a Gspace. Indeed, if X is not a C-space, then there is a C-sequence {xn} such that xn is not convergent. Let X = tin{xn}. The space X has the

the series n=1

required properties. By the classical theorem of Banach and Mazur (1933), space C[0,1] 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 C-space, then X contains a subspace isomorphic to co. Now we shall prove

Locally Pseudoconvex and Locally Bounded Spaces

169

THEOREM 3.10.6 (Schwartz, 1969). The spaces LP(Q,E,p), 0

The proof is based on the following propositions. PROPOSITION 3.10.7 (Kolmogorov-Kchintchin inequality ; see also Orlicz,

1933b, 1951, 1955). Suppose that in the space L°(Q,E,µ) a C-sequence

{xn(t)} is given. Then on each set D° of finite measure the series CO

Ixn(t)I2 convergent almost everywhere. n=1

Proof(Kwapien, 1968). Let {rn(s)} _ {sign (sin 27t2ns)} be a Rademacher

system on the interval [0, 1]. Since under our hypothesis {xn(t)} is a C-sequence, the set of elements n

Y., 40

ri(s) xi(t): n = 1, 2, ..., 0 < s <

1}

i=1

is, by Proposition 3.10.1, bounded in the space L°(Q,E,p). This implies that for any positive e there is a constant C such that, for all n = 1, 2, .. .

and all s,0<s<1, ,u({t a Do: IYn,B(t)I < C}) > /"(Q°)-E.

(3.10.4)

Let n be fixed and 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 IAtI of the set At is greater than 1- rE has a measure greater than p(Qo)-j/E, i.e. if

E = {t a D°: JAtd > 2-1/E }, then

p(E) > p(Do)-r Let t e E. Formula (3.10.5) implies IS

f I ri(s) xt(t)I$ds < C2. At

i=1

Chapter 3

170

Thus

f I Ixi(t)j2 -2 f ( I Re(xi(t)xj(t))ri(s) rj(s)) ds < C2. At i=1

At

(3.10.6)

1<_i<j,
Applying the Schwarz inequality to (3.10.6), we obtain n C2 >

Ixi(t)I2 i=1

-2 (

Ixi(t)xj(t)I2)1/2 (

(f

LJ

ri(s)rt(s)ds)2)112.(3.10.7)

1<_i<j<_n At

Since ri is a Rademacher system, f ri(s) rj(s) ds

=-

f ri(s) r j(s) ds,

(3.10.8)

rya,

At

where I denotes the interval [0, 1].

The system {ri(s)rj(s)}, 1 < i <j, is an orthonormal system in the space L2[0, 1]. The Fourier coefficients of the characteristic function XI\A, of the set I\At are equal to f ri(s) r j(s) ds, thus equal to - f ri(s) rj(s) I\At ds, by (3.10.8). Therefore, by the Parseval inequality.

At

f ( f ri(s) rj(s)ds)2 < II\Atl < /T.

(3.10.9)

1_
Observe that Ixi(t)xj(t)IZ)1/2

<(

Ixi(t)xj(t)IZ)1/2

n

n

R

Ixi(t)I2 )2

,.y

_ i=1

(3.10.10)

Ixt(t)I2.

Formulae (3.10.7), (3.10.9), (3.10.10) imply n

(1-)/e -2j/e)

Ixi(t)I2 < C2.

(3.10.11)

i=1 00

Formula (3.10.11) implies that the series

Ixi(t)I2 is convergent on the ti=1

Locally Pseudoconvex and Locally Bounded Spaces

171

set E. Since p(E) > p(Q°)-j, the arbitrariness of a implies that the series is convergent almost everywhere on D°. COROLLARY 3.10.8. Let {xn} be a C-sequence in the space L°(Q,E,p). Then {xn} tends to 0.

LEMMA 3.10.9 (Schwartz, 1969). Let {xn} be a C-sequence in a space LP(Q, E, p), 0 < p < 1. Then xn -->O.

Proof. If {xn(t)} is a C-sequence in the space LP(D,E,p), then it is also a C-sequence in the space L°(Q,E,p). Using the Egorov theorem, we can 00

represent the set Q in the form Q = U A u N, where, on the set N, i=1

all functions xn(t) are equal to 0 almost everywhere, Q C S2i+ and for each i the sequence of functions {xn(t)} tends uniformly to 0 on Di. Suppose that the sequence {xn(t)} does not tend to 0 in LP(Q,E,p). Then there is a positive number 6' such that, for infinitely many n, f Ixn(t)IPdp > 6'.

n

Let 0 < S < S'. 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 D.,, and

f Ixnk(t)IPdu >6,

f Ixnk(t)IP dI[ < 2t1 k

Ak

forj

Al

(3.10.12)

k.

Let no be an arbitrary index such that

f I xn0(t)I Pdp > 6. Then there

a

is an index mo and a set A° C Slm° such that f I xn0(t)I Pdp > 6. Suppose A.

that Ak, nk, Mk are defined for k = 0, 1,

..., r-1. Then there is an index

m, and a set B , C Qm'' such that Ak C B r for k = 0,1, ..., r-1 and f Ixnk(t)IPdp< B,

1

2k

'

k=0,1,...,r-1.

Since Br C Dm, and the sequence {xn(t)} tends uniformly to 0 on Slmr,

Chapter 3

172

there is an index n, > n,_1 such that

f lxn.(t)J dfu <' and

f

(3.10.13)

I xx,(t)I P dic > 6.

Formula (3.10.13) implies that there is an index m, and a set A, C Slm, disjoint with the sets A0, ..., A,_1 such that

f Ix,(t)lPdp > 6.

(3.10.14)

A.

In this way we have constructed by induction the required sequence. Let {c.} be a sequence of positive numbers tending to 0 such that Go

CR= +00.

(3.10.15)

n=1

Then

.

. Ckxn.(t)IPdu

If nf k=0

>,j=02Alf fk=0ckx, .

(t)IPdp

.

//

I f (Cj Jx.,(t)JP- Ycklxne(t)lP)dp k=0

j=OAS

.

.

k#j

1cj-C'Y' 2j+k j=0

j,k=0

6' c -4C,

(3.10.16)

J-0

where C = sup cn. n

Then, by (3.10.15), the series E ck xn. is not convergent in LP(Q,E,p). k=1

Therefore,

is not a C-sequence and we obtain a contradiction.

LEMMA 3.10.10 (Schwartz, 1969). If {x,,} is a C-sequence in a space LP(Q,E,.u), 1

Locally Pseudoconvex and Locally Bounded Spaces

173

Proof. The proof is the same as the proof of Lemma 3.10.9, but in (3.10.16) the Minkowski inequality is used.

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. COROLLARY 3.10.11. The space L°[0,1] is not universal for separable F-spaces.

Proof. The space L°[0,1] is a C-space. Hence it does not contain a subspace isomorphic to c°.

The notions of C-sequences and C-spaces introduced by Schwartz (1969) are similar to what Matuszewska and Orlicz (1968) called condition (0). A sequence {xn} of elements of an F-space X is called perfectly bounded

if the set N

A°

I xi,,: pn runs over all finite increasing systems of n=1

positive integers) is bounded. An F-space X satisfies condition (0) if each perfectly bounded sequence

is unconditionally convergent. It is easy to verify that for F-spaces satisfying condition (0) Theorem 3.10.2 holds. Therefore, if an F-space X

can be imbedded into an F-space with a basis, then X is a C-space 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 : X x X-->X, called multiplication, which is associative x, y, z E X

(3.11.1)

Chapter 3

174

and bilinear (x1+x2) Y = x1 Y+x2 Y,

(3.11.2) (3.11.3)

for x,Y, x1, x2,Y1,Y2 e X,

(tx) y = t (x y) = x (ty)

(3.11.4)

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 X has a unit if there is an element e e X 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 p-homogeneous norm IIxII which is submultiplicative, i.e. IIxYII < 11411A.

(3.11.7)

If X is complete, then it is also complete with respect to the norm 11 11. Proof. By Theorem 3.2.1 the topology in X can be determined by a p-homogeneous norm IIxII'. Let

114 = sup

11xY11'

IIYII

Observe that IIxII is finite, since the multiplication is a continuous oper-

Locally Pseudoconvex and Locally Bounded Spaces

175

ator (cf. Theorem 3.2.13). It is easy to verify that IIxII is p-homogeneous, submultiplicative and IIeII = 1. Since

(3.11.8)

IIxII > IIxII' IIeII'

the norm IIxII is stronger than the norm IIxII

Now we shall identify x e X with a continuous linear operator Lz mapping X into itself by formula Lzy = xy. Observe that the norm operator IIL=II of the operator Lz is equal to IIxII Thus we can consider (X, II II) as a subspace of the space (B(X), II II). If {xn} tends to x in the space (X, II II'), then, by the continuity of multiplication, Lz. tends to Lz in (X, 11 11). Thus the norms II II and II II' are equivalent. 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

xx-1=x 1x=e.

(3.11.8)

It is easy to verify that x-1 is uniquely determined (provided it exists) = x. and that (x-1)-1

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 x-y is invertible. Proof. Without loss of generality we may assume that II is p-homogeneous and submultiplicative (see Theorem 3.11.1). We can write II

OD

(x-

y)-1

(x-1y)n,

= x1(e-x1y)_1 = x-1 n=o

where the series is convergent provided IIeII < x-1 1 II

II

Let X be an algebra. A set MC X, {0} M = X is called ideal if it is a linear subset of X and for all x e X, xMC M. Let X be an F-algebra, by the continuity of multiplication we conclude that for any ideal M its closure M is also an ideal provided that M -,/= X. We say that an ideal M C X is maximal, if for any ideal M1 such that M C M, we have M = M1.

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 0 M. Thus M is an ideal. Since M is maximal and M C M, M = M. PROPOSITION 3.11.4. Let X be a complete locally bounded algebra over complexes. The function (x-Ae)-1 is analytic on its domain.

Proof. Let Ao belong to the domain of the function (x-Ae)-1. This means that the element x-Aoe is invertible. Let y = (A-Ao)e. Then (x-Aoe)-y = x-Ae. Therefore (x-Ae)-1

=

((x-Aoe)-y)-1

_ (x-Aoe)-1' (x-Aoe)-'a(A-A0)'' , n=1

where the series on the right is convergent provided IA-Ao' < JK(x-Aoe)-11h-1

Thus (x-Ae)-1 is analytic. 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 Ae, i.e. x Ae for all A. Since X is a field (x-Ae)-1 is well defined on the whole complex plane. By Proposition 3.11.4 it is an analytic function. Observe that

lim (x-Ae)-1 = 0. A OD

Thus, by the Liouville theorem (Theorem 3.5.4), (x-Ae)-1 = 0 and we obtain a contradiction. 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 f is a multiplicative-linear functional if

f(xy) = f(x)f(y).

(3.11.9)

Locally Pseudoconvex and Locally Bounded Spaces

177

There is a one-to-one correspondence between multiplicative-linear functionals and maximal ideals. Namely, for a given multiplicative-linear functional f the set

M,={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 (2elazko, 1960). In complete locally bounded algebras all multiplicative-linear 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 multiplicative-linear 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 multiplicative-linear 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 non-decreasing 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) < N(u)+N(v)

(3.11.10)

and there is a C > 0 such that for sufficiently small u, v

N(uv) < CN(u)N(v)

(3.11.11)

Chapter 3

178

and there is a p > 0 such that N(u) = N0(up), (3.11.12) where No is a convex function in a neighbourhood of zero. Let X = N(1) be the space of all sequences x = {xo, x,, ... } such that OD

N(IxnI) < +oo.

IIXII = PN(X)

n=o

The space (X, II II) is an F-space (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 = {y.} then we define n

x.y = { k=0 xkyn-k J. ((

By (3.11.10) and (3.11.11) we conclude that for sufficiently small x and y (3.11.13)

IIxYII < C IIxIIIIYII

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 < p < 1. There are also other more complicated functions. For example 0 for u = 0, N(u) _ -up logu for 0 < u < e-2/p 2p le-2 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 lim N(u) = 0,

and at point

a-2/p since

N(e-21)

= 2p-1e-2. In the interval (0, a-2/p)

it is continuous as an elementary function. The function N(u) is non-decreasing. Indeed, on the interval (0, a-2/p)

dN

_

du

because logu < -2/p.

-pup-, logu-up-'

= -up-1(P logu+1) > 0,

Locally Pseudoconvex and Locally Bounded Spaces

179

Now we shall calculate the second derivative d2N

=

(p-1)uP-2(Plogu+1)-puP-2

du2

= -uP-2(p(p-l)logu+2p-1) = -Up-2((p-1)(plogu+l)+1) < 0 for 0 < u < e-21P. Thus N(u) is concave on the interval (0, a-2/P). Now formula (3.11.11) will be shown. Suppose we are given u, v 0 < u, v < e-2/P. Then (uv)Pjloguvj = (uv)Pllogu+logvl < uPjlogujvPjlogvJ,

because Ilogu, logvl > 2/p > 2. Let N0(x) = -x2logx. It is easy to see that if x = u'12, then (3.11.14)

NO(uP/2) = 2 N(u).

We shall show that No is convex in a neighbourhod of zero. Indeed, dNo

dx

=

dx2 -

-2xlogx-x,

-21ogx-2-1 = -2logx-3

and the second derivative is greater than 0 on the interval (0, a-312). Therefore the function No is convex in a neighbourhood of 0 and (3.11.12) holds. Example 3.11.8 Let N be as in Example 3.11.7. By N±(1) we shall denote the space of all

sequences of complex numbers x = {xn}, n = ..., -2,-1,0,1,2, ... such that co

1lxii = PN(x)

N(I xnl) < +oo.

(3.11.15)

n=-ao

In a similar way as in Example 3.11.7 we can show that (N±(1), 11

11) is

Chapter 3

180

a complete locally bounded space. We introduce multiplication a by the convolution +W

xy =

{k=-aoI

xn-kYk

In a similar way as in Example 3.11.7 we can show that N±(1) 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

x(t)

n=-

xneint

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 N. is of the form

F(x) = x(to) (3.11.16) Indeed, let z = eit. The element z is invertible in N. We shall show that IF(z)J = 1. Suppose that IF(z)l > 1. Then there is a, 0 < a < 1 such that IF(az)l = 1. Since Fis multiplicative, it follows that IF(anzn)I = 1. On the other hand, a"zn tends to 0 and this is a contradiction since each multiplicative-linear functional in NF is continuous. If IF(z)I < 1, then JF(z 1)I > 1 and we can repeat the preceding considerations. Thus IF(z)J = 1 and F(z) = eti' for a certain to, 0 < to < 27r. Since F is a multiplicative linear functional, we obtain for every polynomial n

P(z)

i=k

a{zt

(here n, k are integers not necessarily positive),

F(p(z)) = p(F(z)) = P(e'°).

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

x(t) = L xnein6 ri=-ao

form a sequence belonging to the space N±(1). If x(t) :i-1: 0 for all t, then the function 11x(t) can also be expanded in a Fourier series 1

x(t)

yneint n=-00

such that {yn} e N±(1).

For N(u) = u this is the classical result of Wiener. For N(u) = uP, 0 < p < 1 it was proved by 2elazko (1960). Let X be a locally bounded complete algebra over complex numbers. Let x e X. By the spectrum a(x) of x we mean the set of such complex numbers A, that (x-)e) is not invertible. By Proposition 3.11.2, the set of such complex numbers 2 that (x-2e) is invertible is open. Moreover, if A > IIxII, the element (x-1e) is also invertible. This implies that the set a(x) is bounded and closed. Hence it is compact. Let 20 e u(x). Then by the definition of a spectrum, the element (x-toe) is not invertible. Then there is a multiplicative linear functional F such that F(x-2oe) = 0, i.e. F(x) = 20. Conversely, if 20 0 a(x), then, for each multiplicative linear functional F, F(x) 2o. Thus o(x) = {F(x): F runs over all multiplicative functionals}. Let O(.1) be an analytic function defined on a domain U containing

the spectrum a(x). Let Tc U be an oriented closed smooth curve containing a (x) inside the domain surrounded by T. We shall define O(x)

2ni

f t(2)(x-)e)-ld2.

r

Chapter 3

182

The integral on the right exists since r o a(x) = 0. It is easy to verify that, for any multiplicative linear functional F,

F(O(x)) = O(F(x)).

(3.11.17)

Applying (3.11.17) to the algebra NF, we obtain THEOREM 3.11.10. Let x (t) e NF . Let 0 (A) be an analytic functions de-

fined on an open set U containing

a(x) _ {z: z = x(t), 0 < t < 27c}. Then the function l(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 2elazko (1960). Let N satisfy all the conditions described in Example 3.11.7. By NH we denote the space of all analytic functions x(z) defined on the open unit disc D such that the coefficients {xn}, n = 0, 1, ... of the power expansion co

x(z) = f xnzn n=o

form a sequence {xn} belonging to N(1). There is a one-to-one corespond-

ence between pointwise multiplication in NH and the convolution in N(1). Thus we can identify NH with N(l). Now we shall show that every multiplicative linear functional F de-

fined on NH is of the form F(x) = x(zo), Izol < 1. To begin with we shall show this for x(z) = z. Suppose that F(z) = a, Ial > 1. Then F(z/a) = 1. Therefore F(a-nzn) = 1. On the other hand, a-nzn-* 0, and this leads to a contradiction with the continuity of F.

Observe that N(l) C 1. Therefore each function x(z) a NH can be extended to a continuous function defined on the closed unit disc D. Thus we have THEOREM 3.11.11. Let x(z) a NH. Let 0 be an analytic function defined on an open set U containing x(D). Then O(x(z)) a NH.

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; Przeworska-Rolewicz and Rolewicz,

1966). Let x1, ..., xn e N. (or NH). Let 0(z1, ..., zn) be an analytic function of n variables defined on an open set U containing the set

a(x) _ {(xl(t), ...,xn(t)): 0 < t < 2n} a(x) _ {xl(z), ..., xn(z)): I z I < 1}). Then the function 1(x1, ..., x,) belongs to N. (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, X, P) be a probability space. Let (X, 11 11) be a locally bounded space. Let the norm 11 11 be p-homogeneous. 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 Xl(t ), ..., Xn(t) are called independent if, for arbitrary open sets Al, ..., An

P({t: X{(t) a At, i = 1, ..., n}) _

P({t: XX(t) e At}).

A sequence of random variables {XX(t)} is called a sequence of independent random variables if, for each system of indices n1, ..., nx, the random variables Xnl(t), ..., Xnt(t) are independent.

Chapter 3

184

THEOREM 3.12.1 (Sundaresan and Woyczynski, 1980). Let X be a locally bounded space. Let II II be a p-homogeneous norm determining the topology in X. Let {XX(t)} be a sequence of independent, symmetric, identically distributed random variables. Then

E(IIXIII) = f IIX1(t)IIdP < +oo

(3.12.1)

a

if and only if

X1(t)+ ... + Xn(t) n1IP

0

(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 X. are identically distributed, all X. admit values x1, ... For each positive integer m we shall define new random variables X k(t)

Xk(t)

if Xk(t) = x1i ..., xm,

0

elsewhere.

By Rk(t) we shall denote Xk(t)-Xk(t). For each fixed m, {IIXk II} constitutes a sequence of independent identically distributed symmetric random variables taking real values. Moreover, E(IIXi II) < E(IIXIII) < +oo.

(3.12.3)

The random variables {Xk } takes values in a finite-dimensional space. Thus we can use the strong law of large numbers for the one-dimensional case (see for example Petrov, 1975, Theorem IX.3.17). Let an = n11P. Then

ak-2 = nL=Jk

n-2IP nL=Jk

= 0(n-2P+I) = O(n an 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 Xx takes values in a finite dimensional space).

Locally Pseudoconvex and Locally Bounded Spaces

185

Thus

n-11P(Xi + ...

(3.12.5)

-->O

almost everywhere. At the same time, IIRn II is a sequence of independent identically

distributed real random variables with finite expectation, so that, by the classical strong law of large numbers, IIRi II+ ... +II Rn II -*E (IIR1 II) n

(3.12.6)

almost everywhere. Since IIRI II tends pointwise to 0 as m tends to infinity

and IIRmJI < IIX1II, by the Lebesgue dominated convergence theorem E(IIRi II) tends to 0.

The set Slo of those t for which (3.12.5) and (3.12.6) converge at t is of full measure, i.e. P(Q0) = P(Q) = 1 Let e > 0. Choose an m > 0 such that E(IIR II) <

4

For any t e Do, we can find an N = N(e, t, m) such that IIn-IIP(X1(t)+ ... X'(t))II <

(3.12.6)

for n > N. Thus, for t e £2 and n > N, IIn-1/P(X1(t)+ ... +X2(t))II

IIn-11P(Xi (t)+ ...

+Xn(t))II+IIn-1IP(R(t)+

... +R(t))

2 +E(IIRi II) < e. This completes the proof under the condition that X. are countable valued. To complete the proof of necessity we shall use the standard approxi-

mation procedure. For each e > 0, there is a symmetric Borel function T. taking values in a countable set in X such that II TT(x)-xII < e.

(3.12.7)

Chapter 3

186

Hence

Iin-IlP(Xl+... +Xn)II <


(IIXi-T+(X1)II+...+IIXn-Te(Xn)ID

+IIn-1IP(Te(XI)+ ... +Te(Xn))II

Since the first term is less than a and the second one tends to 0 we obtain (3.12.2). Sufficiency. Observe that (3.12.2) implies that

n-1IPXn = n-llP(XI+... +Xn)-

(n-1 n

lIP

(n-1)-1/P (XI

)

+ ... +Xn-1) tends to 0 almost everywhere. This means that II X-11 ->0 almost everywhere. The random variables {IIXnII} are independent. Then, using the classical result from probability theory (see for example Petrov, 1975, Theorem IX.3.18), we obtain that 00

Y P({t: IIX1(t)II > n}) < n=o

Hence 00

E(IIXIID <

. , n=1

nP({t: n-1 < IIXi(t)II < n}) P({t: IIX,(t)II > n}) < +oo.

n=0

Other results concerning convergence of random variables in nonlocally convex spaces the reader can find in Woyczynski (1969, 1974), Ryll-Nardzewski and Woyczynski (1974) ; and Marcus and Woyczynski (1977, 1978, 1979).

Chapter 4

Existence and Non-Existence of Continuous Linear Functionals and Continuous Linear Operators

4.1. CONTINUOUS LINEAR FUNCTIONALS 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 non-trivial iff:0. If there is a nontrivial linear continuous functional defined on X, we say that X has a non-trivial dual space X* (briefly X has a non-trivial dual). If each linear continuous functional defined on X is equal to 0, we say that X has a trivial dual. Let

U={x: If(x)I <1}. The set U is open as an inverse image of the open set {z: IzI < 1} under a continuous transformation.

Moreover, the set U is convex, since if x,y e U, a,b > 0, a+b = 1, then

I f(ax+by)I < alf(x)I +blf(y)I < 1. This implies that if an F*-space X has a non-trivial dual, then there is an open convex set UC 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 non-trivial 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 = 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

IxIIu=inf{t>0: X EU}. Evidently, (4.1.1)

IIxIIu > 0

and U = {x: IIxIIu < 1}. Moreover IltxlIu = tllxllu

for t > 0

(positive homogeneity)

(4.1.2)

and

(4.1.3)

x+YIIU < IIxIIu+IlYllu

Formula (4.1.2) is trivial. We shall prove formula (4.1.3). Let a be an arbitrary positive number. The definition of I xl I a implies

(1-e)

eU

and

(1-E)

IIxIIu

e U. IlYllu

The set U is convex ; therefore

(1-E)

IIxIIu_

X

IIxIIu+IlYllu IIxIIu

= (1-e)

x+y

IlYllu

} (1-E)

_

Y

IIxIIu+IlYllu IlYllu

E U.

IIxIIu+IIYIIU

Since e is an arbitrary positive number, we obtain IIx+Yllu_ IIxIIu+IIYIIU

and this implies (4.1.3).

A functional satisfying conditions (4.1.1)-(4.1.3) is called a Minkowskifunctional. If we replace (4.1.2) by IxIIu = Itl IIxIIu

for all scalars t,

(4.1.2')

then a Minkowski functional becomes a homogeneous pseudonorm (see Section 3.1).

Let us remark that Ixlla 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 I f(x)1 < I IXIIU,

then the functional f(x) is continuous. On the other hand, iff(x) is a continuous linear functional and

U= {x: If(x)I < 1}. then

If(x)1 < IkXllu.

THEOREM 4.1.1 (Hahn 1927, Banach 1929). Let X be a real linear space. Let p (x) be a real-valued functional (generally non-linear) such that: (1) p(x+y) < p(x)+p(y) (subadditivity), (2) p(tx) = tp(x) for t > 0 (positive homogeneity). Let Xo be a subspace of the space X. Let fo(x) be a linear functional defined on X0 such that

fo(X) < p(X) Then there is a linear functional f(x) defined on the whole space X such that

f(x)=fo(x)

for xeXo

f(x)
for x e X.

and (4.1.4)

Proof. We say that a linear 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 9T be the set of all extensions of the functional fo such that

g(x) < p(x)

for x belonging to the domain of g.

We can introduce a relation of partial order in 91 in the following way. We say that h --S g if g is an extension of h. By the Kuratowski-Zorn lemma there is a maximal element of the family 91. 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 yo e Y and let us denote by Yo the space spanned by Y and yo.

Obviously each element x e Yo can be written uniquely in the form x = y+ayo, where y e Y and a is a scalar. Let x', x" a Y. Then

f(x')+f(x") =f(X +x")
Hence

f(x")-P(x"-Yo) < f(x')+P(x'+Yo). Since x' and x" are arbitrary elements of Y, we obtain

A = sup (f(x)-p(x-yo)) < B = inf ( f(x)+p(x+yo)). ZEY

ZEY

Let c be an arbitrary number such that A < c < B. We define a functional f1 in Yo in the following way: f1(y+ayo) = f(Y)+ca. Obviously, f1 is an extension of the functional f. We shall show that fi e U. Let a be a positive number. Then r

1

fi(Y+aYo) = ca+f(Y) < aB+f(Y) < aI f (a ) +P(Yo+ a +f(y) = -f(Y)+P(ayo+Y) \+f(Y)

))

l

= 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.

COROLLARY 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 non-trivial continuous linear functional 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 xo such that

Existence of Continuous Linear Functionals and Operators IIxoIIu

191

0. Let X0 be the space spanned by x0. Let fo be a functional

defined on the space X, in the following way : lo(x) = fo(axo) = allxoIJu. Obviously, fo(x) < IIxIIv for x e X0. By Theorem 4.1.1 we can extend the functional fo to a linear functional

f(x) defined on the whole space X and satisfying the condition f(x) < IIxIIu for x e X. Therefore, the functional f(x) is continuous. 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 non-trivial 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 non-trivial functional g(x), continuous and linear (with respect to reals).

Let f(x) = g(x)-ig(ix). Obviously, f(x) is a non-trivial functional, continuous and linear (with respect to reals). We shall show that f(x) is also linear with respect to complex numbers. Indeed, let z = 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). 0 there is a continuous linear functional f such that f(xo) is different from 0. Proof. Since the space X is locally convex, there is a convex neighbourhood of zero U such that x0 0 U. Then, of course, Ilx0 # 0. The rest of the proof follows the same line as the proof of Corollaries 4.1.2 and THEOREM 4.1.4. Let X be a B,-space. Then for any xo

4.1.3.

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 fo(x) be a continuous

linear functional defined on Y. Then there is an extension f(x) of the functional fo(x) to the whole space X such that 11f 11 = IIf 1.

Proof. In the real case the theorem is a trivial consequence of Theorem

Chapter 4

192

4.1.1. Indeed, if we put p(x) = IlfolIlIxll, then there is an extension f such that f(x) Ilfoll

Let us now consider the complex case. We shall define two linear real valued functionals11 and f2 on Y by the equality

fo(x) =fi(x)+if2(x) Of course, IIf ll < Ilfoll and IIf2II < Ilfoll. Moreover, for x e Y,

f1(iX)+if2(iX) =fo(iX) = ifo(X) = ifi(x)-f2(x). ,(ix). Therefore fo(x) = fi(x)-if,(ix). Hence f2(x) = Let Fi(x) be a real-valued norm preserving extension fl(x) to the whole space X. Of course, by definition, IIFIII = Ilfill Letf(x) = Fi(x)-iF,(ix).

The functional f(x) is a continuous functional linear with respect to complex numbers (compare Corollary 4.1.3). Since F, is an extension off,, f is an extension of the functional fo. To complete the proof it is enough to show that I f II < Ilfoll

Let x be an arbitrary element of X. Let O = argf(x). Then If(x)I = If(e-iex)I = IF1(e-i1x)I < IIFiiflleiexII < Ilfi II IIxII < IIfthI IIxII Hence IIxII < Ilfoll.

COROLLARY 4.1.6. Let X be a normed space. Then

IIxII = sup

If(X) I

111111

where the supremum is taken over all continuous linear functionals f e X*. Proof. Let x0 be an arbitrary element of X. Let Xo be the space spanned

by x0, i.e. the space of all elements of the type axo. Let us put fo(x) = a I IxoII for x e Xo. The functional fo is of norm one. Basing ourselves on Theorem 4.1.5, we can extend it to a continuous linear function Fo of norm one. Then II xoll = fo(xo) = Fo(xo). Hence

sup I f(xo) I < I Ixol I = Fo(xo) 111111

< IUII41 sup f(xo)

Existence of Continuous Linear Functionals and Operators

193

Duren, Romberg and Shields (1969) gave an example of an F-space

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 12', 0 < p < 1, also contain such subspaces N. Kalton (1978) showed that every separable non-locally convex F-space has this property. 4.2. EXISTENCE AND NON-EXISTENCE OF CONTINUOUS LINEAR FUNCTIONALS

THEOREM 4.2.1 (Rolewicz, 1959). Let

lim inf N(t) > 0. t t->ao Then in the space N(L(Q,E,u)) there are non-trivial 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(Q,E,p)) there is an open convex set U different from

the whole space. Let E e E, where µ(E) = a, 0 < a <+oo. Since lim inf N(t) > 0, there are a positive constant a and a positive number t-aoo

t

T such that, for t > T, N(t) > at. Let U be a convex hull of the set {x: pN(x) < 11. The set U is an open convex set. We shall show that it is different from the whole space N(L(S2,E,µ)). Let x1, ... , x.,, be arbitrary elements such that pN(xt) < 1, i = 1, 2, ..., n

Let Bs = {s: Ixs(s)l > T}. Let xi(s)

for s e BB,

elsewhere

I0

Let X0, =

X1'+ ... +x;,

o

n

and

and

x;' = xs-x;.

... xo" = xi + n

Since Ixa'(s)I < T(i = 1,2,..., n), Ix"(s)I
a,f I x:(s) I dic <.1 N(Ix E

E

Id/u
(i = 1, 2, ..., n).

Chapter 4

194

Hence f Jxo(s) Jdp
not belong to the set U.

If the measure p has an atom E0 of finite measure, then there is a non-trivial continuous linear functional in the space N(L(Q,E,p). Indeed, by the definition of measurable function, x(t) is constant on E0 p-almost everywhere, x(t) = c. Let us put f(x) = c. It is obvious that f(x) is a continuous linear functional.

If the measure p is atomless, the following theorem, converse to Theorem 4.2.1, holds : THEOREM 4.2.2 (Rolewicz, 1959). Let p be an atomless measure. If

lim inf N(t) = 0,

(4.2.1)

t

t- oo

then there are no non-trivial continuous linear functionals in the space

N(L(Q,E,p)) Proof. Let e be an arbitrary positive number. Suppose that (4.2.1) holds.

Then there is a sequence {tm} tending to infinity such that

Ntmm) -->O.

Let E be an

Let km be the smallest integer greater that

arbitrary set belonging to E of the finite measure. Since the measure p is atomless, there are measurable disjoint sets El, ..., Eke, such that km

E= U EE

and

,u(Ei) = (E)

(i = 1, 2, ..., km).

Let xm(s) =

{

tm

for s e EE (i = 1, 2, ..., km),

0

elsewhere.

Obviously, for sufficiently large m, pN(x;,) < e. On the other hand, for s e E km

ym(s) = km

x`m(s) = km Lam, ti=1

Existence of Continuous Linear Functionals and Operators

195

Since km moo, every function of the type aXE belongs to the convex hull U of the set {x: pr,(x) < e}. The set E is an arbitrary set of finite measure. 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,p)).

COROLLARY 4.2.3 (Day, 1940). In the spaces LP(Q,E,u), 0