Metric Linear Spaces (Mathematics and its Applications)

(Pm(x}= ~(am,nlxni)P < +oo),) n

equipped with the topology determined by the sequence of £-pseudonorms {llxllm} in M(am, n), {Pm(x)} in LP(am,n) for 0

1. Example 1.3.IO Let X be a linear space. Let (x,y) be a two-argumentscaiar-valuedfunction satisfying the following conditions: (il) (x,x) ~ 0, (i2) (x,x) = 0 if and only if x = 0, (i3) (x1 +x2 ,y) = (xl>y)+(x2 ,y), (i4) (ax,y) = a(x,y) for all scalars a, (i5) (x,y) = (y,x), where adenotes the conjugate number to the number a.

Chapter 1

18

The function (x,y) is called an inner product. The space X with the inner product is called a pre-Hilbert space. It is easy to verify that llxll = (x, x) is an F-norm.

y

1.4.

COMPLETE METRIC LINEAR SPACES

Let X be a metric space with a metricp(x,y). A sequence {xn} of elements of X is said to be a Cauchy sequence or to satisfy the Cauchy condition, or to be fundamental if lim

p(xn, Xm)

= 0,

n,m-+co

The space (X,p) is called complete if each Cauchy sequence {Xn} is convergent to an element x 0 EX, i.e., lim p(xn, x 0)

= 0.

It is easy to verify that a subset A of a complete metric space (X,p) is complete if and only if it is closed. A set A contained in a metric space X is called nowhere dense if the closure A of the set A does not contain any open set. A set A is said to be of the first category if it can be represented as a union of a countable family of nowhere dense sets ; otherwise, it is said to be of the second category. From the definition of the set of the category it trivially follows that a closed set of the second category contains an open set. Theorem 1.4.1 (Baire), A complete metric space (X,p) is of the second category. Proof. Suppose that X is of the first category. Then, by definition X= 00

=

U Fn, where the sets Fn are nowhere dense and Fn C Fn+I· Since the

n=l

set F 1 is nowhere dense, there exists an open set K 1 such that K 1 and the diameter of K1 d(K1) =sup p(x, y)

<

fl

F1 =

0

1.

z;uek1

K2

The set F 2 is nowhere dense, hence there exists an open set K2 such that C K~> d(K2) < 1/2 and K 2 fl F 2 = 0.

Basic Facts on Metric Linear Spaces

19

Continuing this process, we can find by induction a sequence of open sets Kn- such that Kn C Kn+l, d(Kn) < 1/n, and

Kn

fl

(1.4.1)

Fn = 0.

Let Xn E Kn. Since d(Kn)--:;..0, {Xn} is a Cauchy sequence. The space X is complete; thus th:=e is an element x 0 E X such that {xn} tends to x 0 • Since Kn C Kn+l, x 0 E Kn, n = I, 2, ... Hence, by (1.4.1) x 0 ¢ Fn, n = 1, 2, ... GO

This implies that x 0 ¢

U Fn =

X and we obtain a contradiction.

D

a=l

1.4.2. Let (X,p) be a complete metric space. Let E C X be a set of the first category. Then the set CE = X"" E is of the second category. Proof. X= E u CE. Suppose that the set CEis of the first category. Then X as a union of two sets of the first category is also of the first category.

CoROLLARY

This contradicts Theorem 1.4.1.

D

A metric linear space X is said to be complete if it is complete as a metric space. A closed linear subspace of a complete metric linear space is also complete. Let (X,p) be a complete metric linear space. Let p'(x,y) be another metric on X equivalent to the metric p(x,y). Then (X,p') is not necessarily complete, as follows from Example 1.4.2 Let X be a real line and letp(x,y)

= Jx-yJ. Obviously (X,p) is complete. Letp'(x,y) = Jarctanx-arctanyJ. It is easy to verifythatthemetricsp and p' are equivalent. The space (X,p') is not complete, since the sequence {xn} = {n} is a Cauchy sequence with respect to the metricp'(x,y) but, evidently it is not convergent to any element x 0 E X. The situation is different if we assume in addition that the metric p'(x,y) is invariant. Namely, the following theorem holds 1.4.4 (Klee, 1952). Let (X,p) be a complete metric linear space. Let p'(x,y) be an invariant metric equivalent to the metric p(x,y). Then the space (X,p') is complete. THEOREM

Chapter 1

20

The proof is based on the following lemmas : LEMMA 1.4.5 (Sierpinski, 1928). Let (E,p) be a complete metric space. Let E be embedded in a metric space (E', p'). Suppose that on the set En E' the metrics p(x,y) andp'(x,y) are equivalent. Then Eisa Go set (i.e. it is an intersection of a countable family of open sets) in E'. Proof Since p and p' are equivalent, for any x E E there exists a positive number r11(x) < 1/n such that y E E, p'(x,y) < rn(x) implies p(x,y) < < 1/n. Let

Un(x) = {y E E': p'(x, y) Gn

=

<

rn(x)},

U Un(x),

xEE

The sets Un(x) are open. Thus the sets Gn are also open. Therefore G0 is a G0 set. Evidently E C G0 • It remains to show that E :::> G0 • Let x 0 E G0 • Then x 0 E Gn, n = I, 2, ... By the definition of the sets Gn, there exist elements Xn E E such thatp'(xn,x0) < r(xn). Since rn(x) < 1/n, the sequence {xn} tends to x 0 in the metric p'(x,y). Let e be an arbitrary positive number. Let n be a positive integer such that 2/n < s, and let k 0 be a positive integer such that

fo < rn(Xn)-p'(xn, x Then, for k

>

0 ).

k 0,

p'(xk, Xn)
< fo +p'(x

0,

Xn)

<

rn(Xn).

Hence, by the definition of the number rn(xn),p(xk,Xn) < 1/n. This implies that the sequence {xn} is a Cauchy sequence with respect to the metric p(x,y). Since (E,p) is complete, there is an element x 0 E E such that lim p(xn,xO) = 0. The metrics p(x,y) and p'(x,y) are equivalent on the set E. Therefore x 0 = xO and x 0 E E.

0

Basic Facts on Metric Linear Spaces

21

LEMMA 1.4.6 (Mazur and Sternbach, 1933). Let(X, p) be a complete metric linear space. Let the metric p(x,y) be invariant. Let X 0 be a dense linear subset of X. If X 0 is a Go set, then X 0 = X. ct)

Proof If X 0 is a G0 set, then by definition X 0 =

U G,, where Gn are open n=l

sets, G, C Gn+ 1 . Since X 0 is dense in X, Gn are also dense in X. This impliesthatthesetsX""Gn arenowheredense. ThusX""X0 is the set of the first category. Then the set X 0 is of the second category. Suppose that the set X"'-X0 is not empty. Let x EX"'- X 0 • Since X 0 is a linear subset,

This leads to a contradiction, since the invariance of the metric p(x,y) implies that x+Xo is of the second category and X"'-X0 is of thefirstcate-

D

~~

LEMMA 1.4.7 Let(X, p)be a metric linear space. Let p(x, y) be an invariant metric. Then there is a complete metric linear space (Y,p') such that X is a dense subset in Y and the metrics p(x,y) and p'(x,y) are equivalent on X. The space Y is called the completion of the space X. Proof We complete X in the classical way, defining Y as the space of all sequences {xn} satysfying the Cauchy condition. We identify two sequences, {xn} and {Yn}, iflim p(xn, Yn) = 0. A metric in Y is defined by the formula p'({xn}, {Yn}) =lim p(Xn,Yn). 11r-+ct;J

It is well known that Y is a complete metric space. X can be embedded in Y as the set of stationary sequences x = {x, x, ... } ; X is dense in Y and on X the two metricsp(x,y) andp'(x,y) coincide. The fact that the metric p(x,y) is invariant implies that Yis a linear space and that the operations of addition and of multiplication by scalars are continuous in the metric p'(x,y). D

Proof of Theorem 1.4.4. Let (Y,p') be the completion of(X,p). By Lemma 1.4.5. Xis a dense G0 set in Y. Hence, by Lemma 1.4.6, X= Y. D

Chapter 1

22

1.4.8. Let {xn} be a Cauchy sequence of elements of a metric space (X,p). If it contains a subsequence {xnk} convergent to x 0 eX, then {Xn} also converges to x 0 • Proof. Let e be an arbitrary positive number. Since Xnk-+x0 , there is an index k 0 such that fork > k 0 , p(xn.,x0) < ef2. On the other hand, thesequence {xn} is a Cauchy sequence. Therefore, there exists a number N such that, for n, m > N, PROPOSITION

p(xn, Xm)

e

< 2·

Putting m = nk, k

> k 0 , we obtain

p(xn, x 0) ~p(xn,Xn.)+p(xn., X0)

e

e

< 2+ T =e.

The arbitrarines of e implies the proposition. PROPOSITION

1.4.9. Let (X,

II II) be an

D

F*-space. Let {ei} be an arbitrary 00

2

fixed sequence ofpositive numbers such that the series

ei is convergent.

i=l

If for each sequence {Xi} of elements of X such that

llxill < C:i the series

00

2

Xi is convergent, then the space (X,

II II) is complete.

i=l

Proof. Let {Yn} be an arbitrary Cauchy sequence of elements of the space X. We can choose a subsequence {Ynk} such that llxkll < ek, where Xk 00

=

Ynk+l-Ynk· The assumption implies that the series

2

Xk is convergent

k=l

to an element xeX. We shall show that {Yn} tends to x+y 111 • In fact Sk k

=

2 X;= Ynk-Yn

1

tends to x. Thus Yn. tends to x+yn 1 • By Proposi-

i=l

tion 1.4.8 {Yn} tends to x+yn1 • D A complete F*-space is called an F-space. A complete pre-Hilbert space (Example 1.3.10) is called a Hilbert space. 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). THEOREM

Basic Facts on Metric Linear Spaces

23

Proof Let {Zn} be an arbitrary sequence of elements of X/Y such that IIZ11II < 1/211• By the definition ofthe norm in the quotient space there are X11 e Z11 such that llx11ll < 1/211 •The space X is complete, hence the series 00

L X11 is convergent to an element x eX. Let Z denote the coset containing n=l

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

k

112z,--zll~li2x,-xll < i=m i=m

22-1·

00

Therefore, the series

2Z

11

is convergent to Z, and by Proposition 1.4.9

n=l

the space X/ Y is complete.

D

1.4.11. The product (X, II II) of n F-spaces is an F-space. Proof Let xm = (xi) be a Cauchy sequence. Then the sequences {xi}, i = 1, ... , n also is a Cauchy sequence. Since (Xt, II liD are complete, there are Xi eXt such that !!xi-xt 11-+0, i = 1, ... , n. Let x = (xt). Then PROPOSITION

n

l!xm-x!!

=

};!!xi-xt lli-+0.

D

i=l

1.5. COMPLETE METRIC LINEAR SPACES. EXAMPLES 1.5.1. The spaces N(L(Q, L:, p,)) are complete. Proof Let {xn} be a Cauchy sequence in N(L(Q, L:, p)). It is easy to verify that the sequence { 11 } is a Cauchy sequence with respect to the measure (this means that, for each a > 0, PROPOSITION

x

lim p({t: !xn(t)-xm(t)! >a})= 0). n,m~co

Therefore, by the Riesz theorem, the sequence {xn} contains a subsequence {Xnk} convergent almost everywhere to a measurable function x(t). Let e be an arbitrary positive number. Since the sequence {x11 } is a Cauchy sequence, there is a positive integer N such that for n, m > N PN(Xn-Xm)

=

JN(!xn(t)-xm(t)!)dp ~B.

a

24

Chapter I

Put m

=

nk and

let k-+oo. By the Fatou lemma, we obtain

PN(Xn-X)

~e.

This implies that Xn-X E N(L(Q, r, p). Since N(L(Q, r, /1)) is linear, X E N(L(Q, r, /1)). The arbitrariness of e implies that Xn-+X. 0 1.5.2. The space M(Q, £, f.l) is complete. Proof Let {Xn} be a Cauchy sequence in M(Q, £, p). Then the sequence {Xn(t)} is convergent for almost all t. Let x(t) denote the limit of the sequence {xn(t)}.lt is easy to verify that x(t) E M(Q, £, p). Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n,m > N PROPOSITION

llxn-Xmll

= esssup lxn(t)-xm(t)l <e. tED

Hence, when m tends to infinity, we obtain llxn-xll = esssup lxn(t)-x(t)l

~e.

tED

The arbitrariness of e implies that Xn-+X.

0

1.5.3. The space C(Q) is complete. Proof Let {xn} be a Cauchy sequence in C(Q). This implies that at each t the sequence of scalars {Xn(t)} is also a Cauchy sequence. Its limit x(t) is continuous as a limit of a uniformly convergent sequence of continuous functions. Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n, m > N PROPOSITION

llxn-Xmll = sup ixn(t)-xm(t)i <e. tED

Hence, when m tends to infinity, we obtain llxn-xll = sup lxn(t)-x(t)i

~e.

tED

The arbitrariness of e implies the proposition. 1.5.4. The space C(QjQ0) is complete. Proof C(QjQ0) is a closed subspace ofthe space C(Q).

0

PROPOSITION

0

Basic !<'acts on Metric Linear Spaces

25

e

1.5.5. The space 0(!J) is complete. Proof Let {Xn} be a Cauchy sequence in eo(D). Then {Xn} is also a Cauchy sequence in each pseudonorm II lit, i.e., PROPOSITION

i = 1, 2, ...

llxn-Xmlli = 0,

lim m,~oo

The sequence {Xn(t)} is convergent for each t.lts limitx(t) is a continuous a)

function on each Qi, hence it is continuous on the set Q =

U Q« and i=l

lim llxn-xllt = 0,

i = 1, 2, ...

n->-a>

0

1.5.6. The space C"'(Q) is complete. Proof Let {xn} be a Cauchy sequence in the space C"'(Q). Then {xn} is a Cauchy sequence in each pseudonorm II Ilk, i.e., PROPOSITION

llxn-Xmilk = 0.

lim m,n-+co

Let k = (0, ... , 0). This implies that the sequence {xn(t)} tends uniformly to a continuous function x(t). In a similar way we can prove that the sequence of derivatives tends uniformly to the corresponding derivative of x(t). Thus C 00 (!J) is complete. 0 PROPOSITION 1.?.7. The space c5(EP) is complete. Proof Let {Xn} be a Cauchy sequence in c5 (EP). Then

lim

llxn-Xn' llm,k = 0

for all m and k.

(1.5.1)

n,n~oo

Putting m = (0, ... , 0), we infer by Proposition 1.5.6 that the sequence {xn(t)} is uniformly convergent to an infinite differentiable function together with all its derivatives. Let s be an arbitrary positive number. Formula (1.5.1) implies that there is a positive integer N such that, for n, n' > N,

26

Chapter 1

Hence, when n' tends to infinity, we obtain Jlxn-xll = sup tEEP

It

k,

a~·,

1 •••

kp

tp

The arbitrariness of e implies that

(xn(t)-x(t))

n ittim'

/

p

~e.

t=l

D

Xn-+X.

1.5.8. The spaces M(am,n) and V'(am,n) are complete. Proof Let {xk} = {{x!}} be a Cauchy sequence. Since supam,n > 0, the

PROPOSITION

m

sequence {x!} converges for any fixed n to a limit x~. Write x0 = x~. The sequence {xk} is a Cauchy sequence, hence it is a Cauchy sequence in each pseudonorm llxllm, i.e. for any e > 0, there is a positive integer K such that for k,k' > K Jlxk-xk'llm = supam,n lx!-.x!'l n

< e,

(for0
Jlxk-xk'llm =};(am, n lx!-x!'I)P

and for 1
<

+=

<

8

n

llx!-x~\lm = (};Cam, nl x!-x!' I)P) 11P <e). n

Hence, when k' -+oo we obtain [[xk-x0 ll

~e.

The arbitrariness of e implies the proposition.

1.6.

[]

SEPARABLE SPACES

A metric space (X, p) is called separable if it contains a countable dense set. PROPOSITION 1.6.1. A metric space (X, p) is separable if and only if for an arbitrary positive e there is a countable set E. such that for an arbitrary x eX, there is an element y e E. such that p(x, y) <e.

27

Basic Facts on Metric Linear Spaces

Proof. Necessity. Let (X,p) be a separable space and let E be a dense countable set. Let E8 = E for all e. Sufficiency. Let E8 be a family of sets satisfying the property described 00

above. Then the set E =

U E 11n is a dense countable set in X.

D

n~I

CoROLLARY 1.6.2. Let (X,p) be a metric space. If there are a non-countable set A and a number {J > 0 such that

p(x, y)

>

{J

for all x, yEA, x #- y,

(1.6.1)

then the space (X,p) is non-separable. Proof. Suppose that (X,p) is separable. Then by Proposition 1.6.1 there is a countable set E812 , such that for each x E A, there is an x E E.12 , such thatp(x,x) < e/2. Thus by (1.6.1) there is a one-to-one correspondence between x and This leads to a contradiction since A is non-countable and E,12 is countable. D

x.

PROPOSITION 1.6.3. A subset X 0 of a separable metric space (X,p) is a separable metric space (X0 ,p'), where p' is the restriction of the metric p to X 0 • Proof. Let e be an arbitrary positive number. By Proposition 1.6.1 there is a countable set E,12 such that for any x E X there is a zx E E,12 such that p(x,zx) < ef2. We associate with each element z E E,12 an element y(z) E X 0 , such thatp(z,y(z)) < e/2, provided that such an y(z) exists. Then

p(x, y(zx))

< p(x,

zx)+p(zx, y(zx))

e

<2

+ 2e

Thus, by Proposition 1.6.1 X 0 is separable.

=e.

D

Let (.Q, L:, p) be a measure space. The measure J.1 is called separable if there is a countable family of sets {An}, An E £, such that, for any set BEL: of finite measure and for an arbitrary positive e, we can find a set An0 such that p(B""'An 0)+p(An.~B)

<e.

PRoPOSITION 1.6.4. A space N(L(Q, L:, p)) is separable measure p. is separable.

if and only if the

28

Chapter 1

Proof. Sufficiency. Let {An} be a countable family of sets with the property described above. Let m: be the set of all simple functions K

x(t) =

L alcXAnk

k=l

where, as usual, XB denotes the characteristic function of the set B, a1c are rationals in the real case and complex rational (i.e., of the form an = bn +en i, where bn and en are rational) in the complex case. It is easy to verify that the set m: is countable and that it is dense in N(L(Q, E, p)). Necessity. If the measure J1 is non-countable, then there are a noncountable family of sets {A"} of finite measure and a positive constant b such that

Let x" = XAa· ThenpN(x",xp) > N(l)b for cx:j=fl. Since the set {x"} is non-countable, by Corollary 1.6.2 the space N(L(Q, E, p)) is not separable. 0 1.6.5. A space M(Q, E, p) is separable if and only if the measure J1 is concentrated on a finite number of atoms PI> ... , PTe· Proof. Sufficiency. Suppose that the measure J1 is concentrated on a finite number of atoms PI, .•. , PTe· Then the space M(Q, E, p) is finite dimensional, and thus separable. Necessity. If the measure J1 is not concentrated of a finite number of atoms, then there is a countable family of disjoint sets {An} (n = I, 2, ... ) of positive measure. Let ex= {ni> n2 , • •• } be a subset of the set of positive integers. Let A = Ani u An 2 u . . . The family {A"} is non-countable and for ex =I= {3, p(A"""Ap)+p(Ap""A")> 0. Let x.,. = XAa· Then llxa-xpll = I for ex =F fl. Therefore by Corollary 1.6.2 the space M(Q, E, p) is not separable. 0 PROPOSITION

1.6.6. A space C(Q) is separable if and only if the topology in the compact set Q is metrizable (i.e., it can be determined by a metric d(t, t')). PROPOSITION

Basic Facts on Metric Linear Spaces

29

Proof. Let (Q,d) be a metric compact space. Then for each n = 1, 2, ... , there is a finite system of sets {An,k}, k = 1, ... , Kn, such that An,k (') An,k'

=

fork ::j::. k',

121

Kn

UAn,k= Q, n=l

sup{d(t, t'): t, t' E An,k}

<

1/n.

The family {An, ~c}, n = I, 2, ... , k = 1, ... , Kn, is of course countable. Let X be the space offunctions x(t) of the form M

x(t) =

L amXAnm'

(1.6.2)

km'

m=l

where am are scalars and, as usual, XY denotes the characteristic function of a set Y. Let X be the completion of X with respect to the norm flx!l = sup [x(t)[. tEO

Let m: be the set of all functions of the form (1.6.2) with coefficients am either rational in the real case or complex rational in the complex case. The set m: is countable and it is dense in X. Thus Xis separable. Kn

Let x(t) E C(Q). Let Xn(t)

=

L an,kXAn,k k=l

in such a way that inf [x(t)-an,k[

<

E

m:, where an,k are chosen

I/n. Since the function x(t) is con-

tEAn,•

tinuous, the sequence {Xn(t)} tends uniformly to x(t). Thus it is fundamental in X. Therefore C(Q) can be considered as a subspace of the space X Thus, by Proposition 1.6.3, C(Q) is separable. Necessity. Suppose that the space C(Q) is separable. Let {xn} be a sequence dense in the unit ball K = {x: llxll < 1}. For t,t'

E

Q, let d(t,t')

1 . = ~ ~ 2 n [xn(t)-xn(t')[. Smce [xn(t)[ < 1, n=l

d(t, t') is always finite. It is easy to verify that d(t, t') is a metric on Q. We

shall show that the topology determined by this metric is equivalent to the original topology on Q.

Chapter 1

30

Let t 0 e Q and let e be an arbitrary positive number. Let m be a positive integer such that 112m < e/4. Since the functions Xn(t) are continuous, there is a neighbourhood V of the point t0 such that, for t e V, lxn(t)-xn(to)

e

I<2

(n

=

1,2, ... ,m)

Then, for t e V, 00

d(t, t 0 ) =

.2,; 2~ lxn(t)-xn(to)l n=l m

~ .2,; 2~

oo

lxn(t)-xn(to)l +

.2,; 2~ lxn(t)-xn(t )1. 0

n=m+l

n=l

Conversely, let V be an arbitrary neighbourhood of t 0 e !J. Since Q is compact, there is a continuous function x(t) such that lx(t)l < 1 x(t0 ) = 0, x(t) = I

forte V, fort¢ V.

The set {xn} is dense in the unit ball of the space C(!J), therefore there is Xn such that llxn-xll = suplxn(t)-x(t)l tEV

This implies that, if lxn(t)l lxn(t)-xn(t0)1

1

< -4 .

< 3/4, then t e Vand lxn(t0)1 < 1/4. Therefore 1

<2

(1.6.3)

implies that t e. V. Let d(t, t0 )

< 1/2n+I. Then (1.6.3) holds and t E

V.

0

PROPOSITION 1.6.7. A space C(!Ji!J0) is separable if and only if the set Q~Q0 is metrizable. Proof The proof follows the same line as the proof of Proposition 1.6.6. 0

e

PROPOSITION 1.6.8. A space 0 (Q) is separable if and only metrizable. Proof Sufficiency. According to the definition of the space

if the set

Q is

e (Q), the set 0

Basic Facts on Metric Linear Spaces

31

Q is the union of an increasing sequence of compact sets Dm. For each

m and n we can find a finite system of sets {A!::k} such that

A:.~c n

A:.1c' =

UAmL k n,,.. =

fork#- k',

0

Qm,

sup{d(t, t'): t, t' E

A:,,J <

1/n.

Let X be the space of functions of type (1.6.2). Let X be the completion of the space X with respect to the metric induced by the F-norm.

where llxllt = sup lx(t)l. Let m: be defined in the same way as in Proposition 1.6.6. The set m: is countable and dense in X. Therefore X is separable. Then by Proposition 1.6.3. e0(Q) is separable. Necessity. Let {xn} be a dense sequence in the space e 0(D). Let for t,t' E Q A

A

00

d(t t')

'

=

~ _1 lxn(t)-xn(t')! L.J 211 l+lxn(t)-xn(t')l'

n=l

It is easy to verify that d(t, t') is a metric. In the same way as in the proof of Proposition 1.6.7 we can show that the metric d(t, t') induces a topology

equivalent to the original one.

D

Let (Xt, II lit) be a sequence of F*-spaces. Let X= (Xt)(s) be the space of all sequences x = {Xt Xt EXt}. The topology in X is given by a sequence of pseudonorms llxll; = llxtllt. Of course the space X is an F*-space. 1.6.9. It the spaces (Xt, II lit) are separable, then the space X= (Xt)(s) is separable. Proof Let denote a dense countable set in Xt. Let m: be the set of

PROPOSITION

m:i

• all sequences of the form

(a, a?, ..• , a., 0. 0 .... }.

Chapter 1

32

mis countable and that it is dense in the

It is easy to verify that the set

o

~~x

Let X be a linear metric space with topology determined by a sequence of F-pseudonorms llxlli· Let XJ = {x: llxlli = 0}. Let Xi be the quotient space X/XJ. The pseudonorm llxlli induces in the space Xi the F-norm

11x11;. PROPOSITION 1.6.10. If all the spaces Xi are separable, then the space X is separable. Proof Let l: = (Xt)(s)· By Proposition 1.6.9 the space l: is separable. The space X can be identified with the subspace X 0 C l: of all sequences

{[x]i}, where [y]i denotes the coset in Xi containing y. Obviously llxlli = ll[x]ill;, i.e. the topology of X 0 inherited from l: and the original topology of X coincide under this identification. Therefore X, as a subspace of a separable space, is, by Proposition 1.6.3, also separable. 0 1.6.11. The spaces C (.Q) are separable. Proof Let x E C"'(.Q). Then PROPOSITION

00

= II a~,'

xllk

.~~k~a~;:

x(t)llo.

Hence, the space Xk can be considered as a linear subset of the space C(.Q). Since .Q is a bounded domain in the n-dimensional Euclidean

space, .Q is compact. Thus by Proposition 1.6.6 the space C(.Q) is sepaD rable. Hence by Proposition 1.6.10 the space c=(.Q) is separable. PROPOSITION

Proof Let x

1.6.12. The space c5 (En) is separable. c5 (En). Then

E

llxllm, k

=II t'/.'' ··· t'::" ·

a~,' ... a~;: a1k1

II

x(t) o,

0

and lim t~' ...

t-+oo

t:"

a1k1

ak,

t1 • • •

akn tn

x(t) = 0.

Therefore Xm,k can be considered as a linear subset of the space C(.QI.Q0), where Q is the one point compactification of En and .Q0 is the added

Basic Facts on Metric Linear Spaces

33

.Q"'-

point. The space .Q0 is metrizable. Hence by Proposition 1.6.6. C(.Qj.Q0) is separable. Therefore, by Proposition 1.6.3 c5 (E11) is also separable. D 1.6.13. The spaces LP(am,n) are separable. Proof Elements of the type {x1,x2 , ••• , Xn,O,O, ... },where Xt are rational

PROPOSITION

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

.

am,n

I1 m - - = n-oo Om',n

0.

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

1.7.

TOPOLOGICAL LINEAR SPACES

This book deals principally with metric linear spaces but, in many cases the notion of topological linear spaces can be a very useful tool. A linear space X is caJied a topological linear space if it is a Hausdorff space and if the operation of addition of elements and the operation of multiplication by scalars are continuous. Since the addition is continuous, the set of neighbourhoods of the form x+ U, where U runs over the set of neighbourhoods of 0, determines a topology in X equivalent to the original topology. Of course, each metric linear space is a topological linear space with a countable basis of neighbourhoods of 0. As follows from the K.akutani construction (see Theorem 1.1.1 ), if there is a countable basis of neighbourhoods of 0, then there is a metric determining a topology equivalent to the original one. Moreover, this metric is invariant. In this case the topological linear space is called metrizable. A point a is called a cluster point of a set A if, for any neighbourhood U of the point a the intersection U n A is not empty. We say that a is a cluster point ofa family ofsets {A .. } if it is a cluster point of each member of the family.

Chapter 1

34

Let X be a linear topological space. A family ~ of non-void subsets of X is called fundamental, if for every two sets M, N e ~. there exists an E e ~ such that E c M n N, and for each neighbourhood of zero U there is a set Me ~ such that M- M C U. Example 1.7.1 Let (X,p) be a metric space. Let {Xn} be a Cauchy sequence in X. The family of sets {An}, where An= {xn,Xn+I• . .. },is fundamental.

1. 7.2. A fundamental family~ has at most one cluster point. Proof Suppose that x and y are two cluster points of the fundamental family ~- Let U be an arbitrary balanced neighbourhood of 0. Since ~ is fundamental, there is an Me~ such that M-M CU. Since x,y are cluster points of ~. they are also cluster points of the set M. This implies that there are x1 ,y1 EM such that x-x1 , y-y1 eM. Hence PROPOSITION

x-y = (x-xi)-(y-yi)+(xi-y1 )e U+U+U.

The arbitrariness of U implies that x = y.

D

A subset A of a topological linear space (in a particular case, the space itself) is said to be a complete set if every fundamental family of subsets of A has a cluster point a eA. 1.7.3. A subset E of a complete topological linear space X is complete if and only if it is closed. Proof Sufficiency. Let ~ be an arbitrary fundamental family of subsets of the set E. Since X is complete the family ~ has a cluster point a e X. The set E is closed, and thus a e E. Necessity. Let a be a point in the closure of E. The family {(a+ U) n E}, where U runs over all neighbourhoods of 0, is a fundamental family of subsets of the set E. Since E is complete, it has a cluster point a' e E. By Proposition 1.7.2 a fundamental family has at most one cluster point. Thus a' = a. Therefore the set E is closed. D THEOREM

1. 7.4. For any topological linear space X, there is a complete topologicallinear space X such that X is a dense linear subset of the space X, THEOREM

•.

A

A

35

Basic Facts on Metric Linear Spaces A

A

and the topology in X and the topology in X coincide on X. The space X is called the completion of the space X. Proof We define the points of X as fundamental families of subsets of the space X. We shall identify two fundamental families mand m if 0 is a cluster point of m-m. Further steps are similar to the proof in the metric case (cf. Lemma 1.4.7). 0

Chapter 2

Linear Operators

2.1.

BASIC PROPERTIES OF LINEAR OPERATORS

Let (X, II llx) and (Y, II IIY) be two F-spaces. We shall denote the norms II llx and II IIY by the same symbol II II whenever no confusion result. A mapping A transforming a linear subset D.4. e X into Y is called an additive operator if A(x+y)

=

for all x, y

A(x)+A(y)

E

DA.

The set D A is said to be the domain of the operator A. An additive operator A is called a linear operator if A(tx) = tA(x)

for all x

E

DA and all scalars t.

A linear (additive) operator is called q continuous_ linear (additive) operator if it is continuous. If the spaces X and Yare linear spaces over reals, then each continuous additive operator A is a linear operator. Indeed, the additivity of the operator A implies that A(nx) = nA(x)

for every integer n. Since

)+ ... +A (: ),

--

A(x) =A (:

n-fold

we obtain that A (:) =

~

A (x). Consequently, for arbitrary rational r,

A(rx) = rA(x). 36

Linear Operators

37

Let 8 be an arbitrary positive number. Lett be an arbitrary real number. The continuity of the operator A and the continuity of multiplication by scalars imply that there is a rational number r such that //A((t-r)x)//

8

<2

and

//(t-r) A(x)ll

8

<2 .

Hence //t A(x)-A(tx)// ~ //(t-r)A(x)//+//rA(x)-A(rx)//+//A(rx-tx)/1

<

8.

Therefore, the arbitrariness of 8 implies the linearity of the operator A. Let X and Y be complete metric linear spaces. Let A be a continuous linear operator defined on a domain DA C X with the image in the space Y. Let x 0 belong to the closureDA of the domain D A· Then, by the definition of closure, there is a sequence {xn}, Xn EDA• tending to x 0 • The sequence {xn} is obviously a Cauchy sequence. Since the operator A is continuous, the sequence {A (xn)} is also Cauchy. The space Y is complete, therefore, there is a limit y E Y of the sequence {A (xn) }. Write A(x0 ) = y. It is obvious that A(x0 ) is uniquely determined in this way. We have thus extended the operator A from the domain DA onto its closureDA· This is the reason why in the theory of continuous linear operators we shall restrict ourselves to the operators defined on the whole space X. Let X be a metric linear space. A set B C X is said to be bounded if, for any sequence of scalars { tn} tending to 0 and for any sequence {Xn} of elements of B, the sequence {tnxn} tends to 0. In other words, a set Biscalled bounded if, for any neighbourhood of zero U, there is a number b such that B C bU. A sequence {xn} is called bounded if the set formed by its elements is bounded. A linear operator A mapping an F*-space X into an F*-space Y is called bounded if it maps bounded sets upon bounded sets. THEOREM 2.1.1. Let X and Y be two F*-spaces. A linear operator A mapping X into Y is bounded if and only if it is continuous. Proof Sufficiency. Since no confusion will result, we denote the F-norms in X and in Y by the same symbol II 11. Suppose that the operator A is

Chapter 2

38

not bounded. Then there exist a bounded set E and a positive number e such that sup xeE

llaA(x)ll >

s

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

llbxll
EE

This implies that, for an arbitrary positive b, there is an element that IIA(xo)ll > s. llxoll < b aPd

x" such

Therefore, the operator A is not continuous. Necessity. Suppose that the operator A is not continuous. Then there are a positive number(> and a sequence {xn} of elements of X tending to 0 such that

IIA (xn)ll >

(>

> 0.

Let ' X n =

By the subadditivity ofF-norm we immediately obtain

llx~ll :::;;;; llxnll

v'llxnll

+sup

lltxnll.

o,;;;t,;;;l

x:

Hence Xn -+0 implies -+0. Let tn = llxnll. Evidently the sequence {tn} tends to 0. On the other hand, lltnA(x~)ll = IIA(tnx~)ll = IIA(xn)ll > b > 0. Therefore, the bounded set {x~} is transformed onto an unbounded set {A (x~) }. This implies that the operator A is not bounded. 0 Let B 0 (X-+ Y) denote the set of all continuous linear operators mapping X into Y. The set B 0(X -+X) we shall denote briefly by B 0 (X). It is easy to verify that B 0(X-+ Y) is a linear set.

Linear Operators

39

Let u be a family of bounded sets in X. By Ba(X--? Y) we denote the set B 0(X--? Y) with the topology determined by neighbourhoods of the following form : U(A 0 ,B,s) = {AEB0(X-?Y): supiiA(x)-A 0(x)ll <s,BEu, 8

> 0}.

XEB

The space Ba(X--? Y) with this topology is a topological linear space, i.e. the operation of addition and multiplication by scalars are continuous. If the family u is the family of all bounded sets, then the topology generated by u is called the topology of bounded convergence. In this case the space B,y(X--? Y) will be denoted briefly by B(X--? Y). Linear operators mapping X into the field of scalars will be called linear functionals. Continuous linear operators mapping X into the field of scalars will be called continuous linear functionals. Sometimes, when there is no danger of misunderstanding, continuous linear functionals will be called briefly linear functionals or even simply functionals. The space B(X-?K), where K is the field of scalars (i.e. the field of reals in the real case and the field of complex numbers in the complex case), is called the conjugate space to the space X. We shall denote it by X*.It may happen that X* = {0}. In this case we say that X has a trivial dual.

2.2. BANACH-ST_EINHAUS THEOREM FOR F-SPACES Let (X, II llx) and (Y, II lly) 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 11. A family ~ of operators belonging to B0(X--? Y) is said to be equicontinuous iffor each positive e there is a positive (J such that sup {IIA(x)ll: A E ~' llxll ~ b} <e. The following theorem is an extension of the Banach-Steinhaus theorem (Banach and Steinhaus, 1927) on F-spaces. THEOREM 2.2.1 (Mazur and Orlicz, 1933). Let~ be a family of linear operators belonging to the space B0(X--? Y). For each x E X, let the set {A (x): A E ~} be bounded. Then the family ~ is equicontinuous.

Chapter 2

40

The proof is based on the following lemma: LEMMA 2.2.2. Let (X, II ID be an F-space. Suppose that a closed set V is absorbing, i.e. for each x EX there is a positive number a such that, for b, 0 < b
U nV.

X=

ft-1

By the Baire theorem (Theorem 1.4.1) the space X is of the second category. Therefore, there is a positive integer n0 such that n 0 V is of the second category. Since n0 Vis closed, it contains an open set U. Thus V conI

D

tains the open set - U. no Proof of Theorem 2.2.1. Let e be an arbitrary positive number. Let

ul =

n {x EX: IIA(x)ll < e}.

AE'll

Since the operator A is continuous, the set U1 is closed. We shall show that U1 is an absorbing set. Indeed, let x be an arbitrary element of X. By assumption, the set {A (x): A E ~} is bounded. Hence there is a positive number a such that, forb, 0 < b < a, llbA (x)ll < s for all A e ~. Thus bxe u1. By Lemma 2.2.2 the set ul contains an open set u2. Let Xo E u2. The set {A (x0): A E ~} is bounded. Hence there is a positive number b, 0 < b < 1 such that llbA(xo)ll <e. Thus bxo Eul. Let u = b(U2-Xo) Of course U is a neighbourhood of 0. Let x E U. Then x = by-bx0 • where y E u2. Therefore IIA(x)ll

< llbA(x0)il+llbA(y)ll =

s+sup {llbzll: lbl

< 1, llzll<s}.

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

lim A 71 (x) = A(x). n--+~

Linear Operators

41

Then the operator A is a continuous linear operator. Proof The linearity of the operator A follows trivially from the arithme-

tical rules of limits. For each x the sequence {An(x)} is convergent, and thus bounded. By Theorem 2.2.1 the family of operators {An} is equicontinuous, i.e., for an arbitrary positive e, there is a {J > 0 such that llxll < {J implies IIAn(x)JI

e

< 2·

Let JlxJI
JJA(x)-An (x)JJ < 2" 0

Hence e

JIA(x)JI ~ IIA(x)-An.(x) li+IIAn.(x)ll < 2

e

+ 2 =e.

Therefore, the operator A is continuous at 0. The linearity of the operator A implies that it is continuous everywhere. D 2.2.4. If (X, is complete.

CoROLLARY

B(X~Y)

II ID

and (Y,

I ID

are F-spaces, then the space

Proof Let \!£ _be a fundamental family of subsets of the space B(X~ Y). This implies that for any x eX the family \!l(x) of the sets { {A (x): A e M} Me \!£} is a fundamental family of subsets of the space Y. The space Y is complete, and thus the family W(x) has a cluster point A 0(x). Corollary 2.2.3 implies that A 0(x) is a continuous linear operator. Let B be a bounded set in X and Jet U be a neighbourhood of 0 in

Y. Let E be an arbitrary member of \!£. Let M C E be a member of \!£ such that, for A, A1 e M, x e B, A(x)-A 1(x) e U.

Then A(x)-A 0(x) e U.

The arbitrariness of Band U implies that A 0 is a cluster point of M. Thus it is also a cluster point of E, and since E was an arbitrary member of \!£, A 0 is a cluster point of the family\!£. D

Chapter 2

42

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

2.3. CONTINUITY OF THE INVERSE OPERA TOR 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 Ube an arbitrary neighbourhood ofO. We shall first show that the closure A(U) of the set A(U) contains a neighbourhood of 0. Since 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=l

Thus Y

= A(X) =

U nA(M). n=l

By the Baire category theorem (Theorem 1.4.1) there is an n0 such that the set n0 A (M) contains a non-void open set V. Hence A(U) ) A(M)-A(M) ) A(M)-A(M) ) _I (V- V),

no

1

and the set- ( V- V) is of course a neighbourhood of 0. no For any positive s we shall write Xe = {xE X: llxll < s}

and

Y. = {y

E

Y: IIYII < s}.

Let s0 be an arbitrary positive number. Let {Bi} be a sequence of positive numbers such that }; Bi < s0 • As we have already shown, there is C-1

a sequence {?Ji} of positive numbers such that i=0,1,2, ...

(2.3.1)

Linear Operators

43

Let y be an arbitrary element belonging to Y110 We shall show that there is an element X E X2,, such that A(x) = y.

(2.32.)

Formula (2.3. I) implies that there is an element x 0 E X,. such that 1/y-A(xo)ll < 1J1, i.e. y-A(x0 ) E Y11,. Putting i = 1 in formula (2.3.1), we can find that there is an element x 1 E X, 1 such that

lly-A(x0)-A(xl)ll <

1J2·

Repeating this reasoning, we may define a sequence {Xn} of elements of X such that Xn e X, and n

lly-A

"

(2.: Xt) II < 1Jn+1·

(2.3.3)

i~O

Since the space X is complete, the series}; Xn is convergent to an elen=l

ment x and, moreover, llxll < 2s0 • Formula (2.3.3) implies (2.3.2). Hence the image of a neighbourhood of 0 contains a neighbourhood of 0. Let G be an open subset of the space X. Let x E G. Let N be a neighbourhood ofO such that x+N C G. Let Mbe a neighbourhood ofO in Y such that A (N) :) M. Then A(G) :) A(x+N) = A(x)+A(N):) A(x)+M.

Therefore, A(G) contains a neighbourhood of each of its points, i.e. A(G) is open. D THEOREM 2.3.2 (Banach, 1932). Let a continuous linear operator A map an F-space X onto an F-space Y in a one-to-one manner. Then the inverse operator A- 1 is continuous. Proof The operator A = (A-1)-1 maps open sets on open sets. Hence the D operator A-I is continuous. COROLLARY 2.3.3. Suppose we are given in an F*-space X two different F-norms II I and II 11 1 • Suppose that the space X is complete with respect to the both F-norms. Suppose that the F-norm II III is stronger than the F-norm II 11. Then the two F-norms are equivalent. Proof The operator of identity mapping (X, II 11 1) into (X, II I[) is con-

Chapter 2

44

tinuous. Then by Theorem 2.3.2 the inverse operator is also continuous. 0 This implies that the F-norms II II and II 11 1 are equivalent. Let X be a linear space. A family of linear functionals F defined on X is called total if the fact thatf(x) = 0 for allfe F implies that x = 0. 2.3.4. Suppose we are given in an F*-space X two different F-norms II II and I lit· 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 lit are equivalent. Proof We shall introduce a new norm in the space X by the formula

CoROLLARY

llxll2 = llxll+llxllt and we shall show that the space X is complete with respect to the norm

II 11 2. Let {Xn} be a fundamental sequence in the F-norm II 11 2 • Then it is also a fundamental sequence in the F-norms II II and II III> because the F-norm II 11 2 is stronger than the F-norm II II and the F-norm II lit· Since X is complete in II II and I 11 1 , the sequence {xn} tends to xO in the norm II II and to x1 in the norm I lit· Suppose thatx0 =/=- x 1 • Since the family Fis total, there is a functional f E F such that f(x 0 ) =/=- f(xt). The continuity off in the both norms implies that

f(x 0 ) = lim f(xn) = f(x 1 ), and this leads to a contradiction. Therefore, x 0 is a limit of the sequence {xn} in the norms II II and II 111 • Thus it is a limit of the sequence {xn} in the norm II 11 2. Hence the space (X, II 11 2) is complete. By Corollary 2.3.3 the norm II 11 2 is equivalent to the norm II II and to the norm II 11 1 . Hence the norms II II and II 11 1 are equivalent. D

2.4.

LINEAR DIMENSION AND THE EXISTENCE OF A

UNIVERSAL

SPACE

Let (X, II llx) and (Y, II IIY) be two F*-spaces. We say that the space X and Yare isomorphic if there is a linear operator A mapping X onto Y in

Linear Operators

45

a one-to-one manner such that the operator A and the inverse operator A-1 are continuous. We say that an F-space (X, 1/ llx) has a linear dimension not greater than an F-space (Y, II IIY), and we write dimzX:::::;; dimz Y

if the space X is isomorphic to a subspace of the space Y. If dimzX:::::;; dimzY and simultaneously dimzY:::::;; dimzX, we say that the spaces X and Y have the same linear dimension (Banach and Mazur, 1933) and we write it as dimzX = dimzY. If two F*-spaces X and Y are isomorphic, then obviously they have the same linear dimension. On the other hand, we shall see later that there are 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 dimzX

<

dimzY

if dimzX:::::;; dimzYand dimzY =I= dimzX. Let X be a family of F*-spaces. An F*-space X 0 is called a universal space with respect to isomorphism, or briefly a universal space, for the family X if for every X EX there is a subspace of X 0 isomorphic to X (i.e. dimzX:::::;; dimzX0). A universal space is called sometimes universal with respect to linear dimension. For each family X of F*-spaces there is a trivial universal space defined in the following way. We enumerate all the spaces belonging to X. Let A be the set of indices. Let X = {(Xa, II lla), a E A}. Let Xu be the space of generalized sequences x = {xa}, a E A, Xa E Xa, such that

llxll =

sup llxa

a E.AJ

lla <

+oo.

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

Xa=Ofora=:j=b}.

Chapter 2

46

Unfortunately the space Xu obtained in this way does not necessarily belong to the family ~. The problem of universality is a problem of finding a universal space belonging to the family~. Kalton (1977) has shown that there is a universal separable 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 bE A. Let E[c] ={a E E: a~ c}. If, for each c E E, the set E[c] is a finite chain, we say that Eis 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 e1ementof E occurs in the sequence infinitely many times. We shall now define by induction an increasing sequence {mn} of positive integers as follows. As m 1 we take the smallest n such that E[an] = {an}. Suppose that elements am 1 , ... , amk- 1 are chosen. As mk we shall take the smallest integer n such that n > mk-l {am 1, ••• , amk_ 1, an} is a section of E, an¢ {ami> ... , amk_ 1 }. Since E[c] are finite for all c E E, it can be proved that each c E E is in the range of a, where a(k) = amk· D Let ':l(E) be the set of all real valued functions defined onE and vanishing everywhere except a finite number of points (the functions of this type will be called finitely supported). By :l(A) we denote the set of those elements of :l(E) which have a support contained in A. By PA we shall denote the natural projection PA: :l(E)~:l(A), PAx = xxA. By a consistent family ofF-norms we mean a collection {II lie, c E E} ofF-norms such that

II lie is an

F-norm on :l(E[c]),

if a~ c, then for x

E

:l(E[a]), [[x[[a = [[x[[e,

[[tx[[e is non-decreasing for

t

> 0.

(2.4.1.i) (2.4.1.ii) (2.4.1.iii)

Linear Operators

47

For a given consistent family II lie, we define the limit F-norm II II on 7(E) as follows n

llxll

=

inf

{L

n

llutllc1 :

i=l

L

Ut

=

x,

UtE

:=l(E[ct]), n

eN}.

i=l

Observe that II II satisfies the triangle inequality llx+yll ~ llxii+IIYIIIndeed, n1

n1

llx+yll = inf

{L

i=l

n,

n8

+ inf

x+y}

llwtllc,: .}; Wt =

i=l

{L

llvtllct:

i=l

L

Vt =

Y} = llxii+IIYII·

i=l

Since (2.4.l.iii) holds, lltxll is non-decreasing for t > 0. Observe that for x E :=l(E[c]), llxll ~ llxllc. Thus llxll is an F-pseudonorm (i.e. satisfies conditions (n2)-(n6). PROPOSITION2.4.2(Kalton, 1977). Let Ebe tree-like and let {II lie, c E E} be a consistent family ofF-norms. Let II II be the limit norm of this family. Then If A = suppx (the support of x), then llxll = inf

{,l llualla: 1

Ua E

If x

E

(2.4.2.i)

:=l(E[a]),.}; Ua = x}.

aEA

aEA

:=l(E[c]), then (2.4.2.ii)

llxll = llxllc·

llxll is an F-normon ::l(E),i.e.llxll = 0 ifand only ifx = 0. (2.4.2.iii) Proof (2.4.2.i) Lets be an arbitrary positive number. Let (1 ~ i ~ n) be a minimal collection such that u1

+ ... +u, =

x,

Ui E

:f(E[c,])

Chapter 2

48

2:" llutllc, ~

llxll+s.

t=l

Since the family {II lie, c E E} is consistent, we may assume without loss of generality that c, = max(supput). To prove {2.4.2.i) it is sufficient to show that for each i there is an atE A = supp x such that c, ~at. Suppose that the above does not hold. Then there exists a maximal c1 = c {j being some index not greater than n) such that c =I= a for any a E A. n

This implies that c f} supp x. Thus

2 Ut(c) =

0. Since UJ(c) =I= 0 and cis

i=l

maximal, there is an index k =I= j such that Ck = c = c1 • Therefore

L ut+(uJ+uk) = x

i*J, k

and

}; llutllc1+llui +ukllc ~ llxll+s i*j, k

contradicting the minimality of the collection (ul> ... , un). (2.4.2.ii) This follows immediately from {2.4.2.i); if ua 2 Ua = x, a .s;;; c, then

aEA

E

C): (E[a])

and

L llualla =}; lluallc;? 11J; Ua lie= llxllc· aEA aEA

aeA

(2.4.2.iii). If llxll = 0, then for each positive integer n there exist u~ E r:f(E[a]), a E A = supp x such that ~ n LJ Ua = X

aEA and

2..: ll~lla < ! ·

aEA

Then, for each fixed a, lim lim

u~(a)

llu~lla

= 0. Since Sl(E[a]) is finite-dimensional,

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

a E A. Thus x = 0.

An F-norm II II on r:f(E) is called monotone if section A C E.

D

IIPA(x)ll ~ llxll for any

Linear Operators

49

2.4.3 (Kalton, 1977). Suppose that E is tree-/ike and that {II lie. c E E} is a consistent family ofF-norms such that IIPE(aJ (x)llc ~ llxllc for a~ c. Then the limit F-norm II II is monotone. Proof Suppose that A is a section of E. Let x E :!(£). For any B > 0 we can find Ut,Ci, i = I, ... , n such that Ui E :l(E[ct]), x = u1 + ... +u11 and llutllcl+ ·· · +llunllc11 ~ llxll+e. Since A is a section, A n E[ci] = E[at] for some at ~ Ct. Hence PROPOSITION

n

11

IIPA(x)ll

~}.; IIPA(Ut)llc1 ~}.; llutllc1 ~ i=l

llxll+e.

i=l

The arbitrariness of s completes the proof.

0

Let N denote as before the set of all positive integers. By :l(N) we de· note the set of all finitely supported sequences. For each n eN we define en={O,O, ... ,O,I,O, ... }. n-th place

The space :l(E[n]) will be denoted more traditionally by R 11 • Of course R 11 =lin{ e~> ... , e11 }. Let 4J be the family of all F-norms defined on :l(N) and let 4Jn be the family of all F-norms defined on R 11 • 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*(Ji, q) =sup arctan llog p((x))l· XEX q X x,.
It is easy to verify that d*(p,q) is well determined and that it is a metric. Indeed, (I) d*(p,q) = 0 if and only if p(x) = q(x),

(2) d*(p,q) = d*(q,p) since /log p(x) I= 1-log p(x) =!tog p(x) \. q(x) q(x) q(x) \ 1'

1

(3) d*(p,q) =sup arc tan /log p(x) xex q(x)

I

11"0

=sup arc tan Ilog p(x) +log r(x) XEX

x,.
1

r(X)

q(X)

I

Chapter 2

50

~sup arc tan Ilog p(x) XEX

Y(X)

X#O

~

I

\+sup arc tan log r(x) xEX

I

q(X)

X#O

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

Thus we can treat ((/),d) and ((/)n,d)as metric spaces. Let ln denote the restriction mapping ln: (/J-+(/)n· It is easy to see that ln is not expanding, i.e., d*(Jnp, lnq)~d*(p,q). In further considerations we shall not use the metric d*(p,q) but a function

I

p(x)) . d(p, q) =sup log-( xEX

I

q X

X#O

The function d(p, q) is not a metric. It can be equal to +oo. Observe thatforanys >Othereis ab >0 such that d(p,q)
<s. We shall now describe the general construction of universal spaces. We shall suppose that 9 is a set ofF-norms defined on CJ(N) satisfying the following conditions: 111(9) is separable for each n e N

for each p e 9, ln(q)

=

qe 1

11

(2.4.3.i)

(9) there is q e 9 such that

q and d(p, q) =

d(Jn(p), q).

(2.4.3.ii)

Take a dense countable subset EI of JI(9). Then, by induction, we pick a countable subset En of ln(9) as follows: if En-tis chosen, choose for each p e En-t a dense countable subset A(p) of ln(9) n ln(Jn-_:t (p)). Then En=

U

A(p).

pEEn-t

Let E

=

U En. We define a partial order ~on E as follows, p ~ q n~l

if p is a restriction of q, i.e., p e Em, q E En, m ~ n and p(x) = q(x) for xeRm. Next we define a consistent family {II liP, p e E} ofF-norms. If p e Em, E[p] = {PI, ... , Pm}, where PI = p IR'' ... , Pm = PIRm = p. We define m

llxlljl =

P( ,2; x(p,)e,). i~l

51

Linear Operators

Let llxll be the limit F-norm of the family

{II llv,p e

E}.

PROPOSITION 2.4.4 (Kalton, 1977). For each p e 9 and e > 0 there exist a maximal chain C in E and an order preserving one-to-one mapping rp of C onto N such that the linear operator T: ~(N)~~(C) defined by

T(x) Ia = x(tp(a)) satisfies the following inequality :

(1-e) p(x) :::( IIT(x)ll :::( (l+e)p(x).

(2.4.4)

Proof Let {f)n} be a sequence of positive numbers such that

n 00

(l+rJn)

< 1+e.

(2.4.5)

n=l

Let p 0

=

p. We define by induction a sequence Pn e 9 such that

(1 +rJn)-1Pn-I(x) :::( Pn(X) :::( (1 +rJn)Pn-I(x) for n ln-lPn)

=

for n ;;?: 2,

ln-I(Pn-1)

for n? 1.

> 1, (2.4.6.i) (2.4.6.ii) (2.4. 6.iii)

Suppose that {Pk: k < n} are chosen. Then, by (2.4.6.iii) ln- 1(pn- 1 ) E e En-1 (this condition is vacuous if n = 1). Find q E A (Pn-I I Rn- 1) such that (1 +rJn)-1Pn-I(x) :::( q(x) :::( (1 +rJn)Pn- 1(x) By (2.4.3.ii) there is a Pn(x)

E

for X E Rn.

9 such that

(1 +rJn)-1Pn-I(x) :::( plt(x) :::( (1 +?Jn)Pn-I(x) and ln(pn) = q.

Since q E A(Pn-IIRn-1), by (2.4.6.ii), {ln(Pn)} is a chain in E and it is clearly maximal. Let C = {Jn(Pn): n eN} and we define (Jn(Pn)) = n

for n

> 1.

For xe Rn

II T(x)ll = Pn(x) and by (2.4.5) (1-e)p(x) :::( pn(x) :::( (1 +e)p(x).

D

Chapter 2

52

For the construction of a proper consistent family ofF-norms we need the following notions and lemmas. We say that an F*-space (X, II ID is a {JF*-space if X does not contain arbitrarily short lines, i.e. inf sup lltxll > 0. xeX teR

x;CO

The {JF*-spaces are also called spaces without arbitrarily short lines (see Bessaga, Pelczynski and Rolewicz, 1957b). LEMMA 2.4.5 (Kalton, 1977). Let X be a separable F*-space. The product space Xx / 2 contains then a dense {JF*-space. Proof(Pelczynski, see Kalton, 1977). Let {Yn} be a dense sequence in X. Let en = {0, ... , 0,1,0, ... }. Let {xn, m} be the sequence {en} enumerated n·th place

as a double sequence. Let Zm,n

=

(yn,+Xn, m)· Let X 0

=

lin{zn,m,

n = 1, 2, ... , m = 1,2, ... }. Elements Zn,m are linearly independent and each z E X 0 is a finite sum N Z =

.}; tn,m n,m=!

Zn,m.

Therefore, if z o:j::. 0, then the natural projection of z on l 2 ,Pz, is different from zero. This implies, by the definition of the norm in the product of spaces, that inf

suplltzll

~

z;eo teR zeXxN

inf

suplltPzll

=

+oo.

z;e 0 teR zeXxN

Thus X 0 is a {JF*-space. Observe that (ym, 0)

=

lim Zm, n·

Hence (ym,O) E X 0 • Therefore (O,xm,n) is dense in Xx / 2 •

=

n(zm,n-(ym,O)) E X 0 • Thus X 0 D

II ID be a finite-dimensional F-space. Let the norm II I be non-decreasing, i.e., let lltxll be non-decreasing for t > 0 and each x o:j::. 0. Let

LEMMA 2.4.6 (Kalton, 1974). Let (X,

c = inf suplltxll· xex teR

x;eo

53

Linear Operators

Then,Jor all a < c, the set K = {x: llxll ~ a} is compact. Proof Since X is 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-+oo, Xn E K. By the Bolzano-Weierstrass theorem we can Xnk choose a subsequence {xn } such that-1 - 1 -+x0 , where IYI denotes the k Xnk Euclidean norm in X. We shall show that

(2.4.7)

suplltx0 ll
Since Xn k E K and the norm II II is non-decreasing, txn k E K for It I < 1. Let e = (c-a)/2. Let () > 0 be such that lxl < r5 implies l!xll <e. Then, for ally such that ly-tyXn k I < r5 for a certain ly, ltul < 1, we have ltyl ~

(c+a)/2. By the homogeneity of the Euclidean norm

Itx -t 0

;:: 1 1

I I, we obtain

I < r5

if and only if ()

It I < ,....------.-

l x0-~l !xnkli

Since the right-hand side of the inequality tends to infinity as k-'?oo, we obtain (2.4.7) and this leads to a contradiction with the definition of c. 0 THEOREM 2.4.7 (Kalton, 1977). There is a universal separable F-spacefor all separable F*-spaces. Proof We shall first prove the theorem for separable (3F*-spaces. Without loss of generality we may assume that we consider (3F*-spaces with norms p(x) such that p(tx) are non-decreasing and sup p(tx) = 1

for all

X

EX,

X=/=

tER

Indeed, if p does not satisfy (2.4.8), we write p*(x) =min ( 1,

p~) ),

0.

(2.4.8)

Chapter 2

54

where

o


inf sup p(tx). zeX teR z;o
The norm p*(x) is equivalent to the norm p(x) and satisfies (2.4.8). Let '}J be the class of all F-norms on :f(N) satisfying (2.4.8) and such that for each n eN there exists a neighbourhood Un of Rn such that if x e Un and it/ < 1 thenp(tx) = /t/p(x).

(2.4.9)

We shall show that the family '}J satisfies conditions (2.4.3.i) and (2.4.3.ii). Let n eN and let Vm = {x eRn: i~

/xi/~ ~}·Let Q(m,n)

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

supp(tx) = 1 for all x =F 0, x eRn, if xe Vm, /t/ Observe that Jn('JJ) C

< 1, thenp(tx) = /t/p(x).

U Q(m,n).

(2.4.8') (2.4.9')

We shall now show that Q(m,n) is

m=l

separable with respect to the metric d defined previously. Choose in Q(m,n) a countable set Q which is dense with respect to the topology of uniform convergence on compact subsets of Rn (see Proposition 1.6.8). Letp e Q(m,n). LetO > 0. Let K=

{x:

1

x eRn, p(x)

<

e-2 1}

By (2.4.8') and Lemma 2.4.6 the set K is compact. We assume that 0 is chosen so small that Vm C K. Let

r=

inf p(x).

(2.4.10)

zelnt Vm

Then choose q e Q such that for x e K (2.4.11) Now we consider three cases:

Linear Operators

55 1

1

(a) x¢ K. Then there is at< 1 such that p(tx) = e-211 . Since y and e-x is a convex function _!_11

q(tx)~e

-ye

2

_!..11( _!:_11) 2 1-e 2

_!_11+ -~6

....__

:;::::- e

e

2

2

~e

_!_11 2

-e

-11 (

< e-2 6 -

16 )

1-e 2

-e-6....__ :;::::- e-6 .

Therefore q(x) ~ e- 8 and p(x) ~ o q(x) """'e '

since p(x)

~

I. On the other hand

:~? ~ e-~/2 = e-}o <eo. (b) x E K~ Int Vm· Then, by (2.4.1 0), (2.4.11) and the convexity of the function e-x,

1

------,----- < e-2 8 < e 8 • 1-e

(c)

-!..o 2

+ e-o

Int vm· Then there is s

X E

logp(sx) q(sx) and by (2.4.9') l

> 1 such that SX E avm· Hence

I~ fJ

/log~~:~ I~ fJ. Thus in all three cases d(p,q) ~ fJ and Q is dense in Q(m,n). This implies that Jn(1>) is separable and (2.4.3.i) holds. We shall now show (2.4.3.ii). Let E 7> and let E Jn(1>) be such that

p

d(Jn(p),

q)

= fJ

> 0.

q

(2.4.12)

Chapter 2

56

Let p*(x)

Of course p*

E

=

min(eBp(x), 1).

(2.4.13)

']J. Let

q(x) = inf{q(y)+p*(x-y): y ERn}. 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)-+0. Since q is a norm in Rn, Yk-+0. Then, by the continuity of p*(x), p* (x) = 0 and x = 0. It is easy to observe that q(x) satisifes conditions (n2)-(n6) of the F-norm. Indeed, q(x)can be identified with the quotient norm of the space Xx Rn with the norm ll(x,y)ll = q(y)+p*(x-y), with subspace (0, Rn). Observe thatp*(x-y);;:::: q(x-y) for x ERn. This follows immediately from (2.4.12) and (2.4.13). Hence for x ERn, q(x) = inf {q(y)+p*(x-y): y ERn} ~ inf {q(y)+q(x-y):y ERn}= q(x).

The converse inequality is obvious. Thus q(x) Jn(q) = q. By (2.4.13) q(x) ~p*(x) ~ e8p(x). On the other hand, for x E c:;: (N), y E Rn, q(y)+p*(x-y)

teR

q(x) for x eRn, i.e.

> e-0p(y)+p(x-y) ;;:::: e- 0(p(y)+p(x-y));;:::: e-8p(x).

so that q(x) ~ e- 8p(x). Thus d(p,q) = 0. We have to show that q E ']J. Clearly, for x sup q(tx)

=

E

c:;: (N),

~sup p*(tx) ~I. teR

Suppose that supq(tx)

~b

<

I. Then for any t

E

R there are Ut and Vt

teR

such that tx = Ut+Vt, UtE Rn, q(ut) ~ b and p(vt) ~b. The set V ={zeRn: q(z)~b} is compact by Lemma 2.4.6. Let W={z E Iin(Rn,x: p*(z) ~ b}. This set is also compact for the same reason. However, lin{ x} C W + V and we obtain a contradiction. Thus sup q(tx) = I tER

for all x =I= 0, and (2.4.8) holds.

Linear Operators

Let m

~

n. By condition (2.4.9') there exist em

57

> 0 such that

if y ERn, ij(y) ~Em, [t[

~

1, then q(tx) = [t[q(x),

Rm,p(y) ~Em, [t[

~

1, then p*(tx)

if y

E

=

(2.4.14.i)

[tfp*(x). (2.4.14.ii)

Now suppose that y E Rm and q(y) ~Em. Suppose that [t[ = u+v, u ERn, q(u)+p*(v) ~Em, then

<

1. If y

q(tu)+p*(tv) = [t[ (ij(u)+p*(v))

and q(tv)

~

[t[q(y).

We shall show that the strong inequality does not hold. Indeed, suppose that ty = u+v and q(u)+p*(v) < t q(y). Then q(u) < em[t[ and p*(v) < em[t[. Lets denote the largest real number such that q(su) ~Em. Then q(u) = s-1Em, and so s ~ [t[- 1 • Thus q(t- 1u) ~Em and p(t-1 v) ~Em. Hence q(t- 1u)+p*(t- 1 v) < q(y) and this contradicts with the definition of q. Thus, by Proposition 2.4.4, there exists a countable dimensional 1 F*-space (Z, II [[) such that, for all p E ']>, (Y(N),p) is isomorphic to a subspace of Z. Now letp be any F-norm on Y(N) such that( Y(N),p) is a {JF*-space. We shall show that p is equivalent to some q E 'J>. Without loss of generality we may assume that p(tx) ~ [t[p(x)

for

ltl

~

(2.4.15)

1.

Indeed, if p(x) does not hold (2.4.15), we can replace p by a new equivalent norm pl(x) = sup p(tx) . t~l t

Let p*(x) = (p(x)) 112 • Of course p*(tx)

~

[t[ 112p(x)

for [t[

~

1.

(2.4.16)

Now select any strictly increasing sequence On, 1 ~On, On-+2. For any x E (N), let 1 We recall that a linear space X is called countable dimensional if X= lin { a 1 , ... } for a sequence {a,.} oflinearly independent elements.

Chapter 2

58 n

p(x) =

inf{~Okp*(uk): Uk e Rk, u1 + ... +un = x, n eN}. k=l

By the triangle inequality we have

p(x)

~p(x) ~

2p(x).

If x e R 11 , it is easy to show that n

p(x) = inf {~ Okp(uk): u1 + ... +un =

~}

k=l

and

p(x)

~ Onp(x).

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

q(x) = inf{p(y)+ ~Mcix(i)- y(i)l: y e c:! (N)}. i=l

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

p *(M~l en )

~

1 2n,

(2.4.17)

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

for a sequence {Xn}, q(xn)--J>-0. Then, from the definition of q, Xn can be =

represented as a sum Xn = Un +vn such that p(un)--.1>-0 and ,1; Mtlvn(i)i--J>-0. i=l

00

If ,1; MtiVn(i)i

< 1 then

lvn(i)l

< M? for all

i, and hence, for any r, we

i=l

have under condition (2.4.17) r

p*(vn)

~p*(l; Vn(k)ek)+ k=l

;r.

Hence lim sup p*(vn) ~ 1/2r. Thus p*(vn)--J>-0. Since p(un)--J>-0, p(xn)--J>-0. This shows that under condition (2.4.17) the F-norms p and q are equivalent.

Linear Operators

59

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

qk(X)

= inf {iJ(y)+}; Mt/x(i)-y(i)/: y

E

Rk}.

1=1

Let o be chosen in such a way that for each m the set Rm n {x: p*(x) < o0 } is compact. The existence of such a 00 follows from the fact that (c:f(N),p) is anp .F*-space. Now we shall choose an increasing sequence of positive numbers {Mn} and a decreasing sequence of positive numbers {On}, so that *( -1 ) p Mn en 1

1 < 2n'

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

= 1, 2, ... ,

00 , n = 1, 2, ... ,

On<4 for n

n

<

!

(2.4.18.a) (2.4.18.b)

00 , then qn(u) = qn-1(u),

2,

if u eRn, p*(u)
(2.4.18.c) (2.4.18d)

To start the induction pick an M 1 such that (2.4.18.a) holds and 01 such that (2.4.18.b) holds. Condition (2.4.18.c) ought to be satisfied for n ~ 2. Condition (2.4.18.d) follows from (2.4.18.a), and the definition of q and (2.4.16). Now suppose that M 1 < M 2 < ... < Mn- 1 and 01 > 02 > ... > On-1 have been chosen. The set {xe Rn: On-- 1
llxJJ! =

};lx(i)/, i=1

we may choose Mn so that (2.4.18.a) holds, and moreover if On-1
<

!

00 • 1

We shall now show that (2.4.18.c) holds. Let x e Rn-1 andp*(x)< 4 00 •

Chapter 2

60

Suppose that qn(x)

<

qn_ 1(x). From the definition of q, we have

qn(x) = inf{qn-1(u)+Mnllvii~+Onp*(w): u+v+w = x, ueRn-t,v,weRn}.

And by our hypothesis there are u, v, w such that x = u+v+w and (2.4.19) Hence Mnllvll~ +Onp*(w)

<

qn-1(x)-qn-1(u) ::::( qn- 1(v+w).

(2.4.20)

On the other hand, qn-l(x) ::::( On-1P*(x) ::::( t bo.

Hence llvll~ :::(

M;; 1 bo

and, by the choice of Mn,p*(v)::::;;; { b0 • Therefore p*(v+w) :::(p*(v)+p*(w) :::(

First suppose that p*(v+w) and hence

>

fbo+t b0 = b0 •

bn-~> then p*(w) ~ 0;; 1 (;ln_ 1p*(v+w),

Mnllvll~ +Onp*(w) ~ On-1P*(v+w) ~ qn-1(v+w),

which contradicts (2.4.20). Next suppose p*(v+w) < bn_ 1 and choose t ~ I so that p*(t(v+w)) = bn_ 1. Then either lltvll~ > M;; 1lJ 0 or p*(tw) ~ On- 10;; 1lJn_ 1. In both cases

and hence, as v+w E Rn-t, Mnllvll~ +Onp*(w) ~ ltl- 1On-1 bn-1 ~

ltl- 1qn-l(tv+tw) = qn-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 {~Mtlx(i)J: x ERn, p*(x) = b0 }. i~1

Linear Operators

Choose bn < min(bn-I> .Lb~). If p*(x) ~ 2bn and if y E Rn and

61

< bn, then

qn(x) ~p(x) ~ 2p*(x)

n

p(y)+ .rMtix(i)-y(i)l

~ 2<5n, p*(y)~2<5n.

i=l

Hence, for some t ~ I, p*(ty) =<5 0 • Then by (2.4.I6) <5 0 ~

ltl 1/ 2 p(y) and

n

.rMtiY(i)i

~ Liti-1 ~ L<50 2(p*(y)) 2 ~ 2<5nH0 2p*(y) ~p*(y).

i=l

Hence

.r n

.r n

Mtix(i)i

~

i=l

.r n

Mtiy(i)i+

Mtix(i)-y(i)[

i~l

t=l

n

~p*(y)+ l~ Mtix(i)-y(i)l i=l n

~p(y)+

_r Mtix(i)-y(i)l. i=l

n

Then, by the definition of qn(x), we obtain qn(x)

=

2Mtix(i)l, provided i=l

p*(x)

~

bn, and so qn(tx)

=

it Iqn(x) for

(2.4.I8.d) is also satisfied. To conclude, we observe that if x

E

It I ~

I and p*(x) ~ bn. Thus

Rm, then q(x)

=

limqm+n(x). n-+oo

Hence ifp*(x) ~ bm and It I~ I, q(tx) = ltiq(x) by conditions (2.4.18.c) and (2.4.I8.d). Thus q satisifies condition (2.4.I4.ii). To ensure (2.4.14.i), observe that, as q is equivalent top, inf sup q(tx) = d

>0

x;CO teR

and hence q(x) = min(d-1 q(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 {JF*-spaces. Now we shall show that the completion of the space Z is universal for all separable F*spaces. Let X be an arbitrary separable F*-space. By Lor.ma 2.4.5 X X / 2 contains a dense countable dimensional subspace X 0 , being {JF*-space. Thus there is a subspace X0 , 1 of the space Z isomorphic to X 0 • Since X 0 is

Z

62

Chapter 2

dense in Xx 12, the completion X0 , 1 of the space X 0 , 1 contains a subspace isomorphic to the space Xx / 2 • Therefore the completion Z of the space Z contains a subspace isomorphic to the space Xx / 2 and it trivially implies that Z contains a subspace isomorphic to X. D 2.5.

LINEAR CODIMENSION AND THE EXISTENCE OF A 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, codimzX ~ codimz Y,

if Y contains a subspace Y0 such that the space X is isomorphic to the quotient space Y/Y0 • PROPOSITION 2.5.1. Let X andY be two F-spaces. codimzX~ codimzY if and only if there is a linear continuous operator T mapping Y onto X. Proof Necessity. By the definition of linear codimension, there is a subspace Y0 of the space Y such that the space X is isomorphic to the space Y/Y0 • Let T 0 denote this isomorphism. Write T(y) = T0 ([y]), where [y] denotes the coset containing y. Since T0 1 is continuous, Tis also continous. It is easy to observe that T maps Y onto X. Sufficiency. Let Y0 = T-1 (0). The transformation T induces the continuous linear operator T 0 mapping Y/Y0 onto X defined as follows

T 0([y])

= T(y).

T 0 is one-to-one, and hence, by the Banach theorem (Theorem 2.3.2) its inverse is continuous. Therefore X and Y/Y0 are isomorphic. D

If codimzX ~ codimz Y and simultaneously codimz Y ~ codimzX, we say that the spaces X and Y have the same linear codimension and we write itcodimzX = codimzY. If two spaces X and Y are isomorphic, then obviously. codimzX = = codimzY. We say that the linear codimension of an F-space X is less than the linear

Linear Operators

63

codimension of an F-space Y, codimzX < codimz Y, if codimzX ~codimz Y, but codimzX =F codimz Y. Problems connected with linear codimension are in general more difficult than problems connected with linear dimension. For example the quotient space LP([O, 1] x [0, 1])/H, 0 < p < 1, where H denotes the subspace of the space LP([O, 1] x [0, 1]) ofthe functions depending only on the first variab e is not isomorphic to the space LP([O, 1]) (see Kalton, 1981b). Problem 2.5.1. Let (Y, II II) be an F-space. Does there exist a subspace Y0 C Y such that the quotient space is an infinite dimensional separable space ?1

Let X be a family ofF-spaces. We say that an F-space Xu is universal for a family X with respect to the linear codimension (or briefly co-universal for family X) if, for each X eX codimzX ~ codimzXu. In the same way as in the case of linear dimension we can construct a trivial co-universal space for any family X ofF-spaces. Namely we enumerate all members offamily X. We construct the space Xu as in the previous section. Let ..h = {{xa}: Xb = 0}. The space Xu/Xb is isomorphic to the space Xb. The problem of co-universality for a family X consists in finding a space Xu eX co-universal for the family X. THEOREM 2.5.2 (Kalton, 1977). There is a separable F-space which is couniversal/or 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 {N1t(u)} of continuous 'non-decreasing functions (defined for u ~ 0). Suppose that Nn(O) = 0 if and only if u = 0, n = 1,2, ... Suppose that the sequence Nn(u) satisfies a uniform (~ 2) condition, i.e. there is a K > 0 (which does not depend on n) such that Nn(2u) ~ KNn(u). (2.5.1) 1

Added in proof· The answer is negative (Popov, 1984).

Chapter 2

64

It is easy to verify that co

PN(X)

= }; Nn(Xn) n=l

is a metrizing modular on the space of all sequences. Let Nn(/) be the space of all such xthat PN(x) < +oo. By Theorem 1.2.4 the space Nn(/)is linear. Moreover, the modular PN(x) induces a metric on this space such that it is a metric linear space. In examples which wiii be interesting for us the functions N, will satisfy a stronger condition than (2.5.1), namely (2.5.2)

Nn(u+w):::;;;;; Nn(u)+Nn(w).

In this case the modular PN(x) is an F-norm. It is easy to verify that Nn(/) is complete. Indeed, let {xk} = {x7} be a fundamental sequence. Then, for each i, the sequence {x~} is also fundamental. This implies that there is a limit i =I, 2, ...

Observe that for an arbitrary s k,k'>K

>0

there is an index K such that, for

co

PN(xk-xk')

= }; Nn(_x!-x~):::;;;;; s. n=l

Then passing, with k' to infinity, we get PN(xk-x):::;;;;; s,

where x = {xi}. The arbitrariness of s implies the completeness of the space Nn(l). PROPOSITION 2.5.3 (Turpin, 1976). For any separable F-space (X, II II), there are a space Nn(l) and a continuous linear operator T such that T maps Nn(l) onto X. Proof Let {Xn} be a dense sequence in X. Let Nn(u) = lluxnll. LetT({ at}) n

=.I; atXt. By the definition of N

11

(1) it follows that the operator Tis well

i=l

determined and continuous. The proof will be complete when we show

Linear Operators

65

that T maps Nn(l) onto X. This immediately follows from the fact that each x EX can be represented in the form ct)

X=

2:

Xnt,

i=l

1

where llxnill

< lf•

D

Modifying the construction of Turpin we can obtain a stronger result. 2.5.4 (Bessaga, 1982, oral communication). For any separable F-space (X, II ID there is a space Nn(l) such that for any subspace Y C X there is a continuous linear operator T mapping N n(/) onto Y.

PROPOSITION

The proof is based on the following LEMMA 2.5.5. Let (Y, II II) be a separable F-space. Letisup {IIYII: y E Y} > b. Then there is a sequence {Yn} dense in Y such that sup lltYnll > b for tER

n = 1,2, ... Proof Let y

E

Y be an element such that suplltyll

>

2<5. Let {xn} be an

teB

arbitrary dense sequence in Y. We put if suplltxnll

Xn

Yn =

!

xn+

tER

~

y

> b,

otherwise.

It is easy to verify that Yn has the requested property.

D

Proof of Proposition 2.5.5. Without loss of generality we may assume that the norm II II is non-decreasing, i.e. lltxll ::::;;; llxll for It I ::::;;; 1. Basing on Lemma 2.5.5 we can find a number b > 0 and a sequence {Yn} C Y such that IIYnll > b for n = 1,2, ... and 1

UU n

teR

{tyn} = Y.

(2.5.3)

Chapter 2

66

Let the space Nn(l) be constructed as in Proposition 2.5.3. For each Ym we can find Xn 111 such that 1

IIYm-Xnm II~ 2m

(2.5.4)

•

Let T: Nn(l)-+ Y be defined as follows 00

T({tn})

=}; tnmYm· m-1

Since the norm II II is non-decreasing and IIYnll < lJ, inequalities (2.5.4) imply that the operator Tis well defined and continuous. Since (2.5.3), for every y E Ythere is a sequence ofreals {am} such that 00

y=}; amYm

(2.5.5)

m=l

and 1

(2.5.6)

llamYmll < 2 m · By (2.5.6) the sequence t = {tn},

am tn = {O

for n = nm, otherwise

belongs to the space Nn(l) and T(t) = y.

D

Proof of Theorem 2.5.2 (Bessaga, 1982, oral communication). By Theorem 2.4.7 there is a universal separable F-space (X, II II). Using for this space D 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 lltxnll > 0 and inf sup lltYnll = 0. then obn

teR

n

teR

viously the spaces induced by the sequences {Xn} and {Yn} are not isomorphic. On the other hand if {Xn} and {Yn} are dense sequences such that inf sup lltxnll > 0 and inf sup iitYnll > 0, then the spaces induced by sen

teR

"

teR

Linear Operators

67

quences {Xn} and {Yn} are isomorphic. Indeed, there are b > 0 and sequences of scalars {rn} and {s11 } such that llrnXnll > b, llsnYnll > (), n = I, 2, ... (2.5.7) Now we define by induction a permutation of positive integers k (n) in the following way: k(n) is the smallest positive integer different than k(l), ... ... , k(n-I) such that I llrnXn-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 F-SPACES

Let (X, II ID be an F-space. A sequence {en} of elements of the space X is called a Schauder basis (Schauder, I927) (or simply a basis) of the space X if every element x E X can be uniquely represented as the sum of a series (2.6.1)

c

A sequence {g11 }

X is called a basic sequence if it is a basis in the

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

Pn(x) =

.2; ftet. i=l

2.6.1. The operators Pn are equicontinuous. Proof Let X1 be the space of all scalar sequences y = {ni} such that the

THEOREM co

series }; 'l']t et is convergent. The arithmetical rules of the limit trivially i=l

imply that Let

xl is a linear space. n

IIYII* =sup II n

X'l']tetll·

i=l

Chapter 2

68

The space X1 with the norm II II is an F*-space. We shall show that it is complete, i.e. that it is an F-space. Suppose that a sequence {yk} E X 1 is a Cauchy sequence. Let yk = {nn. Since {yk} is a Cauchy sequence, for an arbitrary e > 0 there is a positive integer m0 such that, for m,k > m0 , n

llym-ykll* = sup n

/I}; (n'[' -nf)et

/J

<e.

(2.6.2)

i=l

Hence n-I

ll(n~-n!)enll ~ 1/l\n'l'-n~)ei 1/ i=l

n

+ Jj}; Cn'l'-nf) ei [/ < 2e i=l

Consequently lim (n';'-n~) m,k-+oo

=

o,

n =I, 2, ...

Thus the sequences of scalars {'11~} are convergent for all n. Write 'l']n

= lim

'11~·

~00

Passing with k to infinity in inequality (2.6.2), we obtain that for m>m0 n

sup\\}; n

(n'!'-nt) et II ~ 2e.

(26.3)

i=l

Let and Taking into account inequality (2.6.3) we obtain

llsr-snll ~ lls';'-s~ll+2e for all m > m0 and all n and r. Fix m1 > m0 • The sequence {s;:''} is a Cauchy sequence. Therefore there is a number n0 such that, for n,r > n0 ,

lls::''-s"/''11 <e.

69

Linear Operators

Hence, for n, r > n0 ,

00

Thus the series 2 rJt et is convergent andy = {nt} e X1 • i-1

From inequality (2.6.3) follows n

sup n

II? (rJi -nt)e,IJ ~ 2e \=1

for m > m0 • Hence the space X1 is complete. Let A be an operator mapping X1 into X defined as follows A(y)

=}; rJtet. i=l

By the definition of the space ~1 the operator A is well-defined on the whole space X 1 • The arithmetical rules of the limit imply that the operator A is linear. Since {en} is a basis, the operator A is one-to-one and maps X1 onto X. The operator A is continuous, because · n

oo

IIA(y)ll = [[}; rJtet II~ sup i=l

n

II}; niet I = IIYII* · i=1

By the Banach theorem (Theorem 2.3.2) the inverse operator A-1 is also continuous. Hence n

IIPn(x)ll

=If LrJtetl[ ~ IIYII* =IIA- {y)ll 1

i=1

and the operators Pn are equicontinuous.

D

Let x eX be reperesented in the form (2.6.1). Let fn(x)

= tn.

It is easy to see thatfn are linear functionals. They are called basis junetionals.

Observe that fn(x)en =· Pn(x)-Pn'- 1(x):

Thus from Theorem 2.6.1 immediately follows

Chapter 2

70

2.6.2. The basisfunctionals are continuous. Suppose we are given two F-spaces X and Y. Let {en} be a basis in X and let {fn} be a basis in Y. We say that the bases {en} and {in} are equivalent COROLLARY

00

00

i=I

i=l

if the series 2 ttet is convergent if and 'only if the series 2 tdi is convergent. Two basic sequences are called equivalent if they are equivalent as bases in the spaces generated by themselves. 2.6.3. If the bases e{n} and {fn} are equivalent, then the spaces X and Yare isomorphic. Proof Let Tn: X-+Ybe defined as follows

THEOREM

oo

Tn(x)

n

= r(~ ttet) = ~ ttfi. i=l

i=l

By Corollary 2.6.2 the operators Tn are linear and continuous. The limit T(x) = lim Tn(x) exists for all x. Thus, by Theorem 2.2.3, T(x) is continuous. Since the bases are equivalent, the operator, Tis one-to-one and maps X onto Y. Thus, by the Banach Theorem (Theorem 2.3.2), the 0 inverse operator r-1 is continuous. The following theorem is, in a certain sense, converse to Theorem 2.6.1. THEOREM 2.6.4. Let (X, II /D be an F-space. Let {en} be a sequence of linearly independent elements in X. Let XI be the set of all elements of X which can be represented in the form

Let n

Pn(x) =

.J; ttet. i=l

be a sequence of linear operator defined on X 1 • If the operators Pn are equicontinuous, then the space XI is complete and the sequence {en} constitutes a basis in this space.

Linear Operators

71

Proof. Since the operators P, are equicontinuous, each element x of X 1

can be expanded in a unique manner in the series

Let X2 be the space of all sequences {ti} such that the series

is convergent. Let n

ll{tt}ll*

=sup n

Jj}; t,e,Jj. i=l

In the same way as in the proof of Theorem 2.5.1, we can prove that II II* is an F-norm and that (X2 , II II*) is an F-space. Observe that

llxll ~ ll{tt}ll*.

(2.6.4)

On the other hand, the equicontinuity of the operators P, implies that if X--+0 then II{ ti}ll*-+0. Hence the space x2 is isomorphic to the space X]. Therefore X 1 is. an F-space. By (2.6.4) the sequence {e,} constitutes a basis in X1 • D 2.6.5. Let X be an F-space. A sequence of linearly independent elements {e,} is a basis in X if and only if

CoROLLARY

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

Pn(x)

= }; ttet i=l

are equicontinuous on the set lin{ e,} of all linear combinations of the set

{e,}.

Chapter 2

72

2.6.6. Let X be an F-space with a basis {en}. Let t 1 , t 2 , ••• be an arbitrary sequence of scalars. Let p 1 ,p2, ••• be an arbitrary increasing sequence of positive integers. Let

COROLLARY

n = 1, 2, Let Pn+t

e~=}; t,e,. i=pn+l

Let X1 = {line~} be the space spanned by the elements {e~}. Then the space X 1 is complete and the sequence {e~} constitues a basis in X1 • n

oo

Proof Let x EX, x =

2

ttet. Let Pn(x) =

i=l

2

ttet.

i-1

For ye Xb let n

P~(y)

=}; a,e·,. i=l

Then P~ (y) = PP

n+l

(y) for ally E X1• Therefore, by Theorem 2.6.1, the

operators P~ are equicontinuous and, by Theorem 2.6.4, X1 is complete and {e~} is a basis in X1 • D A basis {e~} of the type described above is called a block basis with respect to the basis {en}· PROPOSITION

2.6.7. Let (X,

II

jl) be an F-spacewith a basis {en}. Let {xn} be

a sequence of elements of X of the form 00

Xk = }; fk,tet

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

i=l

k--->ro

If {en} is an arbitrary sequence of positive numbers, then there exist an increasing sequence of indices {Pn} and a subsequence {xkn} of the sequence { Xk} such that Pn+t

llxkn- :L>kn,tetjj <en. i=p.. +l

Linear Operators

Proof (by induction). Let p 1 = 0,

Xk1

=

73

x1. We denote by p 2 an index

satisfying the inequality

Suppose that the element Xkn-l and the index Pn are already chosen. The hypothesis lim fk, t = 0 implies the existence of an element Xkn such k->-oo

that

Let Pn+l be an index satisfying the inequality Pn+t

llxkn-}; tkn,tetll
Then obviously Pn+t

!ixkn---:};tkn,tetJI

<

D

8n.

'=Pn+l

Observe that in Proposition 2.6.7 we can additionally require that for an arbitrary given sequence {qn} there should be {Pn} satisfying the conclusion of Proposition 2.6. 7 such that qn < Pn· Let X be a complex 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(t) belongs to X and the norm in the space X depends only on lx(t)l. Let X,. be the space of real-valued functions belonging to X. Since X is a C*-space, each x E X may be uniquely represented in the form

x

where

x\ x2 E X,..

PROPOSITION 2.6.8. Let X be a C*-space. Let {en} be. a basis in X,.. Then {en} is a basis in X.

74

Chapter 2

Proof Since {en} in a basis in X,, there are unique expansions of elements x 1 and ,x2 with respect to this basis:

Hence x can be uniquely extended with respect to the basis {e11 }

:

D We shall now give examples of bases. Example 2.6.9 The sequence

-

en= {0, ... , 0, 1, 0, ... },

n

=

1, 2, ...

n -th place

is a basis in the spaces c0 , lP(O < p < +oo), (s). This basis is called a standard basis. Example 2.6.10 (Schauder, 1927) There is a basis in the space C(O, 1]. It is constructed in the following manner. We define a function Uk,,(t) (0 ~ i < 2", k = 0, 1, ...) in the following way:

Uk,t(t) =

0

for 0 ~ t

2k+l ( t- ;k)

for

-2k+l

(t- i~1)

i

i+1

< 2k and ----zk ~ t ~ 1,

i

2i+1 2k+ 1

2k ~ t <

2i+1 for 2k+ 1 ~ t

,

i+1

< -v·

Each continuous function x(t) defined on the interval [0, 1] can be uniquely written in the form

• x(t)

=

00

a0 t+a1 (1-t)+

2~-1

.2; }; ak,IUk,,(t), k=O

i-o

Linear Operators

75

where a0 = x(l), a 1 = x(O) and the coefficients ak,t are defined as follows. We draw the chord l(t)ofthe arcy = x(t)through the points ofthe i i+I abscissae 2k , ~ and we define ak, t

=

x

(2i+l) ( 2i+l) 2k+l -1 2k+l .

Therefore, the sequence {t, I-t, u0 , 0(t), u1 , 0(t), u1 , 1 (t),

U 2 , 0(t),

... }

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

r(2n/fp

h.,,(t)

~ 1-(2")~' to

n

j-1 2j-I for~< t < 2n+l 2j-I for 2n+l


,

j 2n '

elsewhere,

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

is called the Haar system. It is a basis in the spaces LP[O, 1] (1 ~ p < +oo) If there is a basis in an 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 a1 , ••• , an the equality (2.6.5) implies a 1 = a2 = ... = an = 0.

(2.6.6)

Chapter 2

76

We say that sequence {xn} is topologically linearly independent if for each sequence of scalars {a 1 , a 2, ••• } the equality (2.6.7) implies al

= ... =

an

= ... =

0.

(2.6.8)

We say that the sequence {xn} is linearly 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 {b~> b2, ••• } such that the sequence {b1 x 1 , b1 x 1 +b 2 x 2 , ... , b1 x 1 + ... +bnxn, ... } is a quasi-basis. Labuda and Lipecki (1982) showed that in each separable F*-space ti:iere 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 onE= lin{ en} such that fori= j, fori =j::.j,

and if x E E and e: (x) = 0, n = 1, 2, ... , then x = 0. We say that a sequence {en} is equicontinuous if e!(x)en-+0 for all x E E. Ka1ton (see Kalton, 1979) showed that every 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 J, f(xn)-+f(x). Since, in general

Linear Operators

77

there is no possibility of extension of continuous linear functionals from a subspace Y of the space X it may happen that a sequence {xn} C Y tends weakly to x in X, but does not tend weakly to x in Y. Drewnowski (1979) constructed an F-space X which, for each basis {e11 }, each of its bounded block bases tends weakly to 0 in X, but no basis in X has the property that its bounded block basis tends weakly to 0 in Y which is a subspace spanned on that block basis.

2.7. SOLID METRIC LINEAR SPACES AND GENERAL INTEGRAL OPERATORS

Let (.Q, E, p,) be a a-finite measure space. Let A be a subset of the space L 0(.Q, E, p). We say that the set A is solid if u.E A and lvl ~lui implies v EA. Let (X, II I[) be an F*-space contained set-theoretically in L0 (.Q, E, p). We say that the space X is solid if there is a basis of neighbourhoods of zero consisting of solid sets. 2.7.1 (Szeptycki, 1980). If (X, II llo) is a solid £_-space, then there is an F-norm II II equivalent to II llo such that

PROPOSITION

llxll =II lxl II, (ii) llull ~ 11~11 if lui (i)

~

Ivi.

Proof. Observe that if A, Bare two solid sets in L0 (.Q, E, p), then the set. A+B is also solid. Thus the construction of the norm II II follows from

the construction of an invariant metric (Theorem 1.1.1).

0

We say that X is continuously imbedded in L0(.Q, E, p) and we write X Cc L 0(.Q, E, p) if the identity mapping from X into L0 (.Q, E, p) is continous. PRoPOSITION 2.7.2 (Luxemburg)./f(X, II I[) is a solid F*-space contained in L 0(.Q, E, p), then it is continuously imbedded, X Cc L 0(.Q, E, p). Proof. (Szeptycki, 1968). Suppose that the proposition does not hold. Then there are a neighbourhoo9 of zero U in L 0(.Q, E, p) of the form

U = U(E, a)= {u

E

L 0(.Q, E, p): p{t E E: lu(t)[ >a}< a},

Chapter 2

78

where E is a set of finite measure and a is a positive number, and a sequence {un} C X such that Un ¢ U and 1

llun// ~ 2n. Let

En= {teE: /un(t)/ >a}. Since Un ¢ U, by the solidity of the space X we have JJ.(En) Let •

> 0.

m=n

The measure of E is finite. Thus we can find an m11 such that ?](E~"'-.E~)

<

2-n- 1 a,

where mn

E==

U E~. m=n

E'=

n E~,

Let

n E~. ct)

ct)

E"

n=l

=

n=l

Then ct)

JJ.(E'"'-.E") =

00

ct)

JJ.(n E~"'-. n E~) ~ JJ.(U E~"'-.E:) n=l n=l n=l 00

~ Lfl(E~"'-.E:) < ~.

(2.7.1)

n=l

Since E~ is a decreasing sequence of sets and fl(E~) ? fl(En)

we have JJ.(E')

> a,

> a. Thus by (2.7.1)

JJ.(E");?: ~.

(2.7.2)

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

ax~

2: /um(t)/ ta=A

(2.7.3)

Linear Operators

79

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

llaxll

mn

~IlL lum(t)l II~ m=n

L llumll ~ 2-n+l.

m=n

Hence llaxll = 0, which contradicts (2.7.2).

D

Let A be a linear space of measurable functions. A set E is called an wifriendlyset(Luxemburg and Zaanen, 1971) for A iff[E = 0 for all/eA. PROPOSITION 2. 7.3 (Aronszajn and Szeptycki, 1966). Suppose we are given a a-finite measure space (.Q, E, p). Let A be a solid F*-space in L 0(!J, E, p). Then there exists a maximal unfriendly set for A, unique up to sets of measure 0. The proof of the proposition is based on the following 2.7.4. Let (.Q, E, p) be a a-finite measure space. Let u, v eL E, p). Then, except an at most countable number of reals ~. the sets E = {t: lu(t)l+lv(t)l > 0} and E~; = {t: u(t)+~v(t) =/= 0} differ by a set of measu~e 0. Proof The sets E n (!J"'-.,E~;) are disjoint for differenU. Since Jl is a-finite, only a countable number of these sets has positive measure. D LEMMA

0 (.Q,

Proof of Proposition 2.7.3. Let qJ e L 0 (.Q, E, p) be a positive function such that qJdJl = 1. The existence of such functions follows from the fact that u Jl is a-finite. Write p(E) = qJdqJ. Let

J

J

E

a= sup{,U({t: u(t) =/= 0}), u e A}.

Let {Un} be a sequence of elements of A such that an= ,U({t: un(t) =/= 0})-+a.

By Lemma 2.7.4 there are reals ~2 ,

... ,

~n

such that

En= {t: lu1(t)l+ ... +lun(t)l > 0} = {t: Vn(t)=/=0},

Chapter 2

80

where Vn(t)

=

u1(t)+;2u2(t)+ ... +;nun(t).

(2.7.4)

Observe that p(En) increases to a. 00

Let EA =

D"'-.U

En. The setEA is the required unfriendly set. Indeed,

n~l

suppose that EA is not an unfriendly set. Then there is a function u(t)E A such that 7-t({t E EA: 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 u+;un such that

f,t({t: u(t)+;un(t) =F 0}) = b+an >a, which contradicts to the definition of a. Now we shall show that EA is a maximal unfriendly set up to the sets of measure 0. Indeed, P, (EA) = 1-a. Hence E A is unique up to the sets of p-measure 0. Thus it is unique up to the sets of .a-measure 0. D PROPOSITION 2.7.5 (Aronszajn and Szeptycki, 1966). Let a measure space (Q, .E, .u) be a-finite. Let A C L 0(Q, .E, .u) 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 An be chosen so that

,U({t: lun(t)l > Mn} < 2-n,

(2.7.5)

where p(E) is defined as in the proof of Proposition 2. 7.3, and

ll;unll < for;,

1~1

2-n

< An.

Having { Mn} and {An} we shall choose by induction two sequences {;n} and {1Jn} in the following way. We define~~ = 1, 0 < 171 ~ 1. Sup-

Linear Operators

81

pose that ~to ••• , ~nand 'f/~o ... , 'f}n are chosen. Using Lemma 2.7.4, we can choose c;n+1 such that 0 < ~n+l < min(A.n+l• (2Mn+I)-~n) (2.7.6) and

{t: Vn(t)+~n+IUn+I(t) ::1= 0}

= {t:

lvn(t)l+lun+l(t)l

> 0},

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

0

<

< i- 'fjn ·

'f/n+l

(2.7.7)

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

v(t) = lim Vn(t) = ,._,. 00

L ~nun(t).

n=l

Observe that

~

lv(t)l

00

L

lvm(t)l~nlun(t)l n=m+l

~ 'fjm- ~

00

}; 'fjn ~ -}'fjm n=m+l

>0

on the set

Em= {t; lvm(t)l

> 'fJm} n {t: luj(t)l < Mj,

j = 1, 2, ... }.

It is easy to verify that

p(EA"\.Em)

< a-am+ 2-m+ 2-m.

Hence v(x) ::1=.0 almost everywhere on EA.

D

Let (D, E, /1) be a measure space. We write En '\,.0, if En 1s a decreasing sequence of sets, En E E, such that J1(E n En)--+- 0

for every set E of finite measure. Let (X, II ID be a solid F*-space contained in L 0 (D, E, J1). By Xa we denote the set ofthose u EX for which (2.7.8)

[luxE.[I--+-0

for every sequence En '\,.0. If Dis of finite measure, (2.7.8) is equivalent to lim

UXE =

p(E}-+0

0.

82

Chapter 2

It is easy to verify that Xa is a closed solid subspace of X and, if {En} is co an increasing sequence of sets such that [J = En, then xx -+ x for

U

n=l

En

all x E Xa If X= N(L(Q, .E, Jl), then Xa =X. If f1 is non-atomic and X = L co (Q, .E, Jl), then Xa = {0}. '

PROPOSITION 2.7.6 (Luxemburg and Zaanen, 1963). Suppose that f1 is a-finite and that (X, II II) is a solid F-space contained in L 0 (Q, 1:, Jl). Then a set C C Xa is compact in X if and only if Cis compact in L 0 (Q, 1:, Jl). For any sequence En~" for u E C.

(2.7.9.i)

lluxE,.II tends to 0 uniformly (2.7.9.ii)

Proof Necessity. The necessity of(2.7.9.i)immediately follows from Proposition 2.7.2. Suppose that (2.7.9.ii) does not hold. Then there ares > 0, a sequence En ~o, and a sequence Un E C such that

llunXE.. II >

(2.7.10) Since Un E C C Xa, we can construct by induction subsequences { Un.i}, { EnJ} such that s.

8

llun,XE,.1 II < 2

fori <j.

(2.7.11)

By (2.7.10) we get fori <j.

Hence the set C is not compact. Sufficiency. Let {vk} be any sequence of elements of C. By the compactness of C in L 0(Q, 1:, Jl) and the Riesz theorem we can find a subsequence {un} = {Vkn} of C tending to u E C almost everywhere. Now we shall show that {un} tends to uin X. Without loss of generality we may assume that X does not have an unfriedly set. Then by Proposition 2.7.5 there is a rp EX positive almost everywhere. Let En,k

=

{r

E

[J:

suplum(t)-u(t)l;:;:::, m~n

+rp(t)}.

Linear Operators

83

Observe that, En, k ~0 for each fixed k andfor any e > 0 we can find by (2.7.9.ii) an index such that llvxn,kll

e

<3

for n

>

nk, v e C,

(2.7.12)

where, for brevity, we write Xn,k = XEn,.· Then llun-ull ~ llxn,k(un-u)ll+ll(l-xn,k)(un-u)ll ~ llxn,kUnll+llxn,kull(1-xn,k)(un-u)ll

~ ; + ; + I ~ ~II <e. provided that k is chosen so that

I ~ ~11 < ; .

0

2.7.7 (monotone convergence theorem, Luxemburg and Zaanen, 1963). Let XC £ 0(!2, 1:, p) be a solid F*-space. Then u e Xa if and only iffor each sequence {vn} such that PROPOSITION

(2.7.13) tending to zero almost everywhere, llvnll-+0. Proof Let u eX, En ~0, Vn = UXEn· By our hypothesis llvnll-+0. Thus ueXa. Conversely, let u e X a and let {Vn} be a sequence satisfying (2. 7.13) and tending to 0 almost everywhere. Let Em,n

=

{t:

Vn(t)

~ ~

u(t) }·

Let Xm,n denote the characteristic function of Em,n· For a fixed m, !2"'-._Em,n ~0 and II(I-xm,n)ull-+0 since u e Xa. We have llvnll

~II~ Xm,nUII +11(1-xm,n)ull ~II;

I

+11(1-xm,n)ull.

(2.7.14)

Chapter 2

84

For each s

> 0 there is an m such that

II; II<~·

(2.7.15)

Formulae (2.7.14) and (2.7.15) imply llvnll < s. This completes the proof.

D

PROPOSITION 2.7.8 (dominated convergence theorem, Luxemburg and Zaanen, 1963). Let (X, II II) be a solid F-space contained in L 0(Q, E, p). Then u E Xa if and only if for each sequence {un} C X such that lun(t)l ~ lu(t)l almost everywhere and un(t)-+u(t) almost everywhere, we have llun-ull-+0. Proof Sufficiency. If En '\.e, then un = (1-XEJ u has the property indicated in the statement. Thus u e X a. Necessity. Suppose that u e Xa and a sequence {un} has the indicated properties. Let Vn(t) = sup{lum(t)-u(t)l, m

>

n}.

The sequence {vn}(t)} is not increasing. It tends to 0 almost everywhere and vn(t) ~ 2lu(t)l. Then by Proposition 2.7.7 llvnll-+0. By the solidity of the space and Proposition 2.7.1 this implies that llun-ull-+0. D Let (Q, E, p) and (QI> E 1 , p 1) be two a-finite measure spaces. Let u e L0 (!21 , E 1 , p 1 ). Let k(t, s) be a function defined on the product space Q X !21 measurable with respect to the product measure fl. Xp1 • Let K(u) =

J k(t, s) u(s)dp (s) 1

D,

be a linear integral operator acting from L 0{Q1 , El> p 1) into L 0 (Q, E, p) By a proper domain DK of the operator Kwe shall mean the set DK = {ue L0 (!21 ,E1 ,p1):

Jlk(t, s)l lu(s)ldp (s) < + oo 1

D,

p-almost everywherej. 00

We recall that (Q, E, p) and (.Q1 , El> p 1) are a-finite, i.e . .Q =

U .Qn, n=l

0,

Linear Operators

85

00

!/1 =

U Dn, n=l

I>

where f1.(!/,n 0)

< + oo and fl.i!Jn, 1) < + oo. Thus the to-

pologies in L 0 (Q, .E, fl.) and in L 0(Q1 , .E1 , f1.1) are described by sequences of pseudonorms, II lin in L 0(Q, E, fl.) and II II~ in L 0 (QI> .El> f1.1) (see Example 1.3.5). We shall introduce F-pseudonorms on the space Dx in the following way

llullm,n =

lluii~+IIIKIIulllm,

where

IKIIul =

J lk(t, s)llu(s)l dfl.l·

(2.7.16)

n,

It is easy to see that the space D xis a solid F*-space with the topology defined by the double sequence of pseudonorms II llm,n· THEOREM 2.7.9 (Aronszajn and Szeptycki, 1966). The space Dx is complete. Proof Let {uk} be a Cauchy sequence in Dx. This means that, for each fixed m, n, lim lluk-Uk•ll = 0. (2.7.17) k,k'-+co

m,n

The sequence {uk} is obviously also a Cauchy sequence in the space L 0 (Q1 , .E1 , Jl.J. Thus there is a u e L 0 (Ql> El> J1.1) such that {un} tends to u in the space L 0(Q1 , E 1 , Jl.J. By (2.7.17), {uk} contains a subsequence {u;} = {uk,} su«h that 00

}; llu~+l-u~llm,n <

+oo.

i=l

Thus the series 00

}; lu~H(t)-ut(t)l i-1

and

00

}; IKI lu~+l-u,l i=l

cqnverge almost everywhere on sets Q,., 1 and !Jn, 0 , respectiveiy. We have CIO

lui~ I~ I+}; lu~+l~u~l. i=l

86

Chapter 2

Thus by the Lebesgue theorem 00

JKJ JuJ ~ JKJ Ju~J+}; JKJ Ju~+~-u~J i=l

almost everywhere on Dn,o. Similarly we can estimate 00

JKIJu-z4J ~}; JKJJu~+l-u~J~O i==r

almost everywhere on Dn,o• as r~=. And, by Proposition 2. 7.8,

II JKJJu;-uJ llm~O. In a similar way we can show that llu;-ulln~O.

Thus

llu;-ullm, n~O. Therefore u; tends to u in Dx· In this way we have shown that each Cauchy sequence contains a convergent sequence. This implies the com0 pleteness of Dx (cf. Proposition 1.4.8). THEOREM 2.7.10 (Aronszajn and Szeptycki, 1966). The operator K is a continuous linear operator from Dx into L 0 (Q, E, p). Proof IIKuJJm ~ II JKJJuiJJm ~ llullm,n for all n. 0 THEOREM 2.7.11 (Aronszajn and Szeptycki, 1966; see also Banach, 1931). Suppose that X C c L0 (D1 , E 1 , p 1) is a solid F-space and suppose that XC Dx. Then the integral operator K maps X into L 0(Q, E, p) is a continuous way. Proof By Theorem 2.7.10 it is enough to show that the identity is a continuous mapping from (X, II ID into (Dx, II llnx), where II llnx is the F-norm induced in the space Dx by-the sequence ofpseudonorms II llm,n (see Section 1.3). Accordingly, we shall introduce a new norm on X, namely

Linear Operators

87

and we shall show that the space X is complete with respect to this new norm. Indeed, let {un} be a Cauchy sequence in (X, II lit). Then it is a Cauchy sequence in (X, 1/ //)and in (Dx, II 1/vK). Since both spaces are complete, the sequence {Un} has limits in both of them. Let u0 be the limit in X and let u1 be the limit in Dx. Since XC c L0 (!:J1 ,I1 ,fit) and Dx C c L 0(Qb 1:1 , fit), the sequence {un} is convergent in the space L 0(Q 1 , It> jt1) simultaneously to u0 and to u1 • Thus u0 = u1• Hence l/un-u0 1/ 1 = l/un-u0 1/+l/un-U0 1/vx-+0.

Therefore (X, 1/ 1/J is complete, and by Corollary 2.3.3 the two norms are 0 equivalent. This implies the continuity of K. THEOREM 2.7.12 (Aronszajn and Szeptycki, 1966). Let (X, 1/ 1/x) CcL0(f:J 1 ,l:t>Jit) and (Y,/1 I/Y)CcL0(Q,l:,Ji) be given F-spaces. If XC Dx and KX C Y, then the operator K: X-+Yis continuous. Proof. We introduce a new norm on X, llxll

= llxl/x+IIK(x)l/y,

and, in the same way as in the proof of Theorem 2.7.11, we show that 1/ //)is complete. We take a fundamental sequence {un} in (X, II //).It is alsofundament~lin(X, II llx). Thus there is a ue X such that llu11 -ul/x-+0. By Theorem 2.7.11 llun-ullvx-+0 and by Theorem 2.7.10 Kun-+Ku in L 0 (Q, 1:, Jl). Suppose that 1/Kun-VIIy-+0. Since Y C c L 0 (Q, 1:, Jl), Ku11 -+v in L 0 (Q, 1:, Jl). This implies that Ku = v. Hence (X, II //)is complete. Thus, by Corollary 2.3.3, the norms llxll and llxl/x are equivalent. 0 This implies the continuity of Ku. (X,

PROPOSITION 2.7.13. (Aronszajn and Szeptycki, 1966). Let K be an integral transformation. Then (Dx)a = Dx. Proof. Letfe Dx. Let Et~0. Thus Jl(Qn,t n E,)--+0,

n = 1,2, ...

Therefore

n = 1,2, ...

88

Chapter 2

By Proposition 2.7.7, IKIIIXE,I tends to 0 almost everywhere and, by the definition of the F-pseudonorm llxllm,n,

II/XEI!Jm,n ~ 0 •

D

Example 2.1.14 Let D = [0,27t] with the Lebesgue measure. Let D 1 = Z be the set of all integers with the discrete measure

Pl{y}) Let k(t, s)

=

1,

y

=

= e'"'. Then Dx =

Example 2. 7.15 Let D = Z and let D 1

0, +I, ±2, ... L(Z)

= fl.

= [0,27t]. Let k(t,s) =

eil8 •

Then Dx

= £1[0,27t].

Example 2.1.16 Let D = D 1 = R with the standard Lebesgue measure. Let k(t,s) = eilB. Then Dx = £1( -oo, +oo).

Chapter 3

Locally Pseudoconvex and Locally Bounded Spaces

3.1.

LOCALLY PSEUDOCONVEX SPACES

Let X be a metric linear space. A set A C X is said to be a starlike (starshaped) set if tA C A for all t, 0 < t ~ I. The modulus of concavity (Rolewicz, I957) of a starlike set A is defined by c(A)

= inf {s > 0:

A+A C sA},

with the convention that the infimum of empty set is equal to +oo. A starlike set A with a finite modulus of concavity, c(A) <+oo, is called pseudoconvex. PROPOSITION

3.1.1. Let A be an open pseudoconvex set. Then (3.1.1)

A+A C c(A)A.

Proof. Suppose that (3.1.1) does not hold. Then there is an x ¢ c(A)A such that x e A:+-A. Since the set A+ A is open, there is an r > I such that rx ¢ A+A. Then A+A ¢ c(A) A r '

because x ¢ c(A)A. Since r >I, we obtain a contradiction of the defiD nition of the modulus of concavity. Observe that we always have c(A) ;::;::: 2. A set A is said to be convex if x, yEA, a, b;::;::: 0, a+b = I imply ax+bye A. Of course, for each convex set A the modulus of concavity c(A) of the set A is equal to 2. The condition c(A) = 2 need not imply convexity, but the following proposition holds : 89

90

Chapter 3

3.1.2. Let A be an open starlike set. If c(A) = 2, then the set A is convex. Proof Let x,y EA. Since A is an open starlike set, there is a t > 1 such that tx,ty EA. Since c(A) = 2, A+A E 2tA. Thus tx+ty E 2tA. PROPOSITION

This implies that (x-;-y) EA. Therefore, for every dyadic number r,

rx+(1-r)y EA.

(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 numbers such that

x 0 = t0 x+(1-t0 )y E A

for

ltol < E

and Applying formula (3.1.2) for x = x 1 andy= x 0 , we find that, for every a such that Ia - rl < s, ax+(l-a)y EA. (3.1.3) Since r could be an arbitrary dyadic number, (3.1.3) holds for an arbitrary real a, 0 ~ a ~ I. Thus the set A is convex. D PROPOSITION 3.1.3. Let A be a starlike closed set. If c(A) = 2, then the set A is convex. Proof Since the set A is closed, 2A = sA. Thus c(A) = 2 implies

n

•>2

x+y that A+A C 2A. Therefore, if x,y E A, then - 2- EA. This implies (3.1.2) for every dyadic number r. Since A is closed, (3.1.3) holds. D A metric linear space X is called locally pseudoconvex if there is a basis of neighbourhoods of zero { Un} which are pseudoconvex. If moreover c(Un) ~ 2 11P, we say that the space X is locally p-convex (see Turpin, 1966; Simmons, 1964; Zelazko, 1965). THEOREM 3.1.4. Let X be a locally pseudoconvex space. Then there is a sequence of Pn-homogeneous F-pseudonorms {II lin}, i.e. such that

lltxlln = ltiPnllxlln,

(3.1.4)

Locally Pseudoconvex and Locally Bounded Spaces

91

determining a topology equivalent to the original one. If the space X is locally p-convex, we can assume Pn = p (n = 1,2, ... ). Proof Let {Un}be a basis ofpseudoconvex neighbourhoods ofO. Without loss of generality we may assume that the sets Un are balanced (cf. Section 1.1). From the definition of pseudoconvexity it follows that there are positive numbers Sn such that Un+Un C SnUn.

Let (q = 0, ±1, ±2, ... ). t

For every dyadic number r > 0,-, = }; at 2i, where at is equal either to i=B

0 or to 1, we put Un(r) = asUn(28)+ ... +atUn(21). In the same way as in the proof of Theorem 1.1.1, we show that Un(r 1 +r2)

Un(r 1 )+ Un(r 2)

)

and U11 (r) are balanced. Moreover, the special form of U11 (r) implies U11 (2qr) = s~ U11 (r). Let //xi/~= inf {r > 0: x E Un(r) }. The properties of the sets U11 (r) imply the following properties of 1/xl/~: . (1) 1/x+y//~ ~ 1/x/I~+IIYII~ (the triangle inequality), (2) 1/ax/1~ = 1/x/1~ for all a, /a/ = 1, (3) 1/s~x/1~ = 2ql/xl/~.

Let log 2 Pn =log sn" Let 1/xl/n = sup 1>0

1/tx/1~ . fPn

By (3) 1/xl/n is well determined and finite since 1/xl/n = sup 1>0

1/tx/1~ fPn

=

sup 1,;;;1,;;;sn

1/tx/1~ • fPn

92

Chapter 3

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

llaxlln =sup lltaxll~ =

sup

t>O

t>O

fP•

lltxll~

= llxlln

fP•

and (n2) holds. Let x,y e X. Then II x + y II 11 = sup llt(x+y)ll~ P• = sup t

t>O

lltxll~

=S;; sup

tPn

I.,;;t.,;;sn

llt(x+y)lln

I.,;;t.,;;s,.

+ sup I.,;;;t.,;;sn

lltyll~ tP•

t

P•

=S;; llxlln+IIYIIn

and (n3) holds. Observe that lltxlln = sup llrtxll~ = sup llrltlxll~ r>O

jtjP•

r>O

lf/P• sup llsxll~

=

sP•

•>O

(r It i)P•

=

lt/P•

/t/P•

llxlln.

Hence llxll is Pn-homogeneous. This implies (n4)--(n6). Now we shall prove that the system of pseudonorms {llxlln} yields a topology equivalent to the original one. Indeed, from the definition of llxlln, llx[[~ =S;; llxlln, in other words, {x: llxlln

<

(3.1.5)

r} C Un(r).

On the other hand, the sets Un(r) are starlike. Therefore, the pseudonorms llxl/~ are non-decreasing, which means that 1/txl/~ are non-decreasing functions of the positive argument t for all n and x e X. Then lltx[i~ [[tx[[~ , , llxlln =sup--= sup - - =S;; llsnXIIn = 211xlln, t>O

tP•

I.,;;t.,;;sn

tP•

in other words Un(r)C {x: llxlln

< 2r}.

(3.1.6)

Formulae (3.1.5) and (3.1.6) imply the first part of the theorem. If the space X is locally p-convex, then by the definition there is a basis of neighbourhoods of 0 {Un} such that c(Un) =S;; 2 11P. Since Un are open it implies that Un+UnC2 1 iPUn.

Locally Pseudoconvex and Locally Bounded Spaces

93

Putting s11 = 21 /P and repeating the construction above, we obtain a system of p-homogeneous pseudonorms determining a topology equivalent to the original one. D 3.1.5. Let X be a locally pseudoconvex space with a topology given by a sequence of Pn·homogeneous F-pseudonorms {II 11 11 }. A set A C X is bounded if and only if PROPOSITION

On=

sup {JJxJJn:

XE

A}

<+oo.

(3.1.7)

Proof If a set A is bounded, then for each n = 1,2, ... there is a tn such that tnA C {x: JJxJJ,

>0

< 1}.

Thus I tn

On~-.

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

< e}

C U.

By (3.1.7) I

e-4c u. On

0

A topological linear space X is called locally convex if there is a basis of convex neighbourhoods of 0. The construction given in Theorem 3.1.4 'leads to the fact that if X is a locally convex metric linear space, then the topology could be determined by a sequence of homogeneous (i.e. !-homogeneous) F-pseudonorms. Homogeneous F-pseudonorms will be called briefly pseudonorms whenever no confusion results. Locally convex metric linear spaces are called B~-spaces. As we shall see later, for locally convex spaces there is also another much simpler construction of a sequence of homogeneous pseudonorms determining the topology. A complete B~-space is called a B0 -space. Let X be an F*-space. If the topology in X can be determined by a sequence of Pn·homogeneous pseudonorms, then the space X is locally

Chapter 3

94

pseudoconvex. In the particular case where the pseudonorms are homo-

geneous, the space X is locally convex. Indeed, the sets - 1- Kn, where m Kn

= {x: //x//n < 1},

constitute a basis of neighbourhoods of 0. The modulus of concavity ofthe set Kn is not greater than 2Pn. Indeed, if x,y E Kn, then llx+y//n ~ 2. Hence

i.e. x+ y E 2P• Kn. Thus Kn +Kn C 2P• Kn. In the particular case where the pseudonorms are homogeneous the sets Kn are convex. It is easy to verify that the following spaces are B0-spaces: LP(Q,E,J1) for 1 ~p ~+oo, C(Q), C(QjQ0 ), the Hilbert space, 0 (D), C 00 (Q), c:S(En). We say that a set A in a metric linear space X is absolutely p-convex, 0

e

ax+bye A. Z-1-----,

z-

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

x+y 211P

E

A and

Locally Pseudoconvex and Locally Bounded Spaces

95

A+A C2 11P A. On the other hand, there are pseudoconvex sets which are not absolutely p-convex for any p. For example, the following open set on the plane

{(x,y): 0<x<2,0
Kr = {x:

llxll < 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 1, there is a p-homogeneous F-norm

space. Then, for a certain p, 0 < p II II equivalent to the original one.

<

Proof Let U be a balanced bounded neighbourhood of 0. The set U is pseudoconvex. Indeed, the set U U is bounded and, on the other hand, the set U is open. Therefore there is a number a such that

+

96

Chapter 3

U+U C aU. Hence U is pseudoconvex. Since U is a bounded set, the system of sets -

.

1

m

U constitutes a basis

of neighbourhoods of 0. Basing ourselves on Theorem 3.1.4 for each n we can construct a Pn-homogeneous pseudonorm II lin· Observe that the F-pseudonorms II lin differ from one another by a constant coefficient. D Therefore II 11 1 is an F-norm with the required properties. Let us remark that Pn is equal to

log c(U) log 2 . Let c(X) = inf {c(U):

U runs

over the open bounded balanced sets}. The number c(X) is log c(X) called the modulus of concavity of the space X. Let Po = log 2 · Looking at the proof of Theorem 3.2.1 we can see that this theorem may be formulated more precisely. Namely THEOREM 3.2.1' (Rolewicz, 1957). Let X be a locally bounded space. For each p, 0 < p < p 0 , there is a p-homogeneous F-norm II II equivalent to the original one. The strict inequality p < p 0 in Theorem 3.2.1' cannot be replaced by the conditional inequality p ~ p 0 , as follows from Proposition 3.4.8 given later. For locally convex spaces the construction of the norm implies THEOREM 3.2.2 (Kolmogorov, 1934). Let X be an F*-space. If there is in X a bounded convex neighbourhood of 0, then there is a homogeneous norm II II equivalent to the original one. An F*-space in which there is a bounded convex neighbourhood of 0 is said to be a normed space. A complete normed space is called a Banach space. Let (X, II ID be an F*-space. Let II II be p-homogeneous. Then the space X is locally bounded. Indeed, the unit ball K = {x:

llxll ~ 1}

Locally Pseudoconvex and Locally Bounded Spaces

97

is a bounded set, because if Xn e K and {tn} is a sequence of scalars tending to 0, then lltnXnll = ltniP llxnll-+0. In the particular case where the norm II II is homogeneous we find the set K is convex (cf. Section 3.1). Those facts are inverse to Theorems 3.2.1 and 3.2.2. As follows from the definition, the spaces M(D,.E,J1), C(.Q), C(DID0), l:P(.f.J,.E,J1) for 1 < p < +oo and the Hilbert space are normed spaces. Since they are complete (see Section 1.5), they are Banach spaces. Examples of locally bounded spaces which are not Banach spaces will be given in the next section. 3.2.3. Let X, Y be two locally bounded spaces. If the space X is isomorphic to the space Y, then they have equal moduli of concavity, c(X) = c(Y). Proof Let T be a linear operator mapping X into Y. If A+ A C sA, then T(A)+ T(A) C sT(A). Therefore c(T(A)) ~ c(A). If T is an isomorphism, then T and T-1 map open bounded sets onto open bounded PROPOSITION

sets. Thus c(Y)

= c(T(X)) ~ c(X) = c(T-1(Y)) ~ c(Y).

D

Let (X, II ll).be a locally bounded space. Let II II 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 llxll <+oo. :tEA

3.2.4. Let (X, II II) be a locally bounded space. Let Y be a subspacf! of the space X. Then c(Y) ~ c(X). Proof Without loss of generality we may assume that the norm II II is p-homogeneous. Let A be an open set in Y. Let PROPOSITION

B=

U {x eX:

IlEA

llx-yll ~ t inf {lly-zll: z e Y"'A}}

The set B is open and bounded. Moreover, A= Y ()B.

Chapter 3

98

Thus A+A = (Bn Y)+(Bn Y) C (B+B)n YC c(B) Bn Y

Hence c(A)

~

c(B) and c(Y)

~

c(X).

= c(B) A. 0

II II) and (Y, II IIY) be two complete locally bounded spaces. LetT be a continuous linear operator mapping X onto Y. Then c(Y) ~ c(X).

PRorosmoN 3.2.5. Let (X,

Proof Without loss of generality we can assume that II II is 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 T onto Y, by Theorems 2.3.2 and 2.1.1 RA =sup inf {llxll:

X E

T-1(y)} <+oo.

lfEA

Let B = T-1 (A)n{x:

llxll <

2RA}·

It is easy to see that the set B is an open bounded starlike set and T(B) =A. Then c(A) = c(TB)) ~ c(B). Therefore

c(Y) ~ c(X).

0

CoROLLARY 3.2.6. Let (X, II ID be a locally bounded space. Let Y be a subspace of the space X. Then

c(X/Y) ~ c(X). Proof Basing ourselves on Lemma 1.4.7 we can assume without loss of generality that X is complete. Thus we use Proposition 3.2.5. 0

As an immediate consequence of Propositions 3.2.3, 3.2.4 and Corollary 3.2.6 we obtain THEOREM

3.2.7. Let X andY be two locally bounded spaces.

If dimz X~ dimz Y, then c(X) ~ c(Y). IfdimzX = dimzY, then c(X) = c(Y).

99

Locally Pseudoconvex and Locally Bounded Spaces

THEOREM 3.2.8. Let X and Y be two complete locally bounded spaces. If codimzX::::;;; codimzY, then c(X)::::;;; c(Y). If codimzX = codimzY, then c(X)

= c(Y).

THEOREM 3.2.9 (Kalton, 1977). There is a separable locally bounded complete space (X, II ID with an a-homogeneous norm II II which is universal for all separable locally bounded spaces with a-homogeneous norms. Proof: Let ']>be the set of all a-homogeneous norms defined on c:J(N)

(for the notation see Section 2.4). The set ']J satisfies conditions (2.4.3.i) and (2.4.3.ii). Condition (2.4.3.i) is obvious. For the verification of (2.4.3.ii) let us observe that putting 3- = d(J11(p),q) we have e-Bp(x)::::;;; q(x)::::;;; eBp(x)

for x ERn. Let q(x)

= inf {q(y)+eBp(x-y):

y ERn}.

For x ERn we have q(x) = q(x) and for all x

E

c:J(N) we have

e-Bp(x)::::;;; q(x)::::;;; eBp(x).

Hence (2.4.3.ii) holds. Let {Xn} be a sequence of linearly independent elements such that lin{ Xn} is dense in X. Then lin{ Xn} is isomorphic to (c:J(N),p) for a certain p. Now we coU:struct the space (c:J(E), II ID as in Section 2.4. For each a E E we denote by ea an element of c:J(E) such that ea(x)

= {~

for x =a, for x =1= a.

Choose a: N~E to satisfy Proposition 2.4.1 and write Wn = ea(n> Then, by Proposition 2.4.4, for any e > 0 there is an increasing sequence of indices (nk) such that for any m m

m

m

i=l

i=l

i=l

(1-e)p{~ tcxc) ::::;;;!l.r tcwn,//::::;;; (1+e)p ~ tcx,). Observe that the theorem.

II II

(3.2.1)

is a-homogeneous. Since lin {x11 } = X (3.2.1) implies D

Chapter 3

100

Using a different method Banach and Mazur (see Banach, 1932) showed that the space C[O, 1] is universal for all separable Banach spaces, hence also for !-homogeneous 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 (Xt, II lit) be a sequence of F*-spaces. By (Xt)c,> we denote the space of all sequences x = {xt}, Xt eXt. The topology in (Xt)c,> is defined by F-pseudonorms n

llxll~ = };llxtllt. i=l

In the case where all Xt are identical, Xt = X we denote briefly (Xt)c,> by (X)(s)• It is easy to verify the following facts. (Xt)cs> is an F*-space. If all spaces (Xt, II lit) are complete, then the space (Xt)<•> is also complete. If all spaces (Xt, II lit) are locally p-convex (locally pseudoconvex), then (Xt)<•> is also locally p-convex (locally pseudoconvex). 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 ID be a separable locally bounded space with a p-homogeneous norm universal for all separable locally bounded spaces with p-homogeneous norms. Let Xt = 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 lin} determining a topology equivalent to the original one. Let THEOREM

Xn,o = {x: llxlln = 0} and let Xn = X/Xn, 0 be the quotient space. The pseudonorm llxlln induces a p-homogeneous norm in Xn. Since no misunderstanding can arise, we shall denote this induced norm also by II lin· The space (Xn, II lin) is a locally bouded space with a p-homogeneous norm II lin· Let

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 {[x]n}, where [x]n is the coset in Xn induced by x. On the other hand by the universality of XP, X is isoD morphic to a subspace of the space (XP)<•>· Basing themselves on the conclusion of Banach and Mazur 0933), that C[O, I] is universal for all separable Banach spaces, Mazur and Orlicz (1948) showed that C(-oo,+oo) is universal for all separable B0-spaces. THEOREM 3.2.11. There is a separable locally pseudoconvex space which is universal for all separable locally pseudoconvex spaces. Proof Let (XP, I ID denote a separable locally bounded space with ap-homogeneous norm, being universal for all separable locally bounded spaces with p-homogeneous norms. Let {Pt} be a sequence tending to 0, for example Pi = 1/i. The space (XP')<•> is the required universal space. It is easily seen to be a separable locally pseudoconvex space. Let X be an arbitrary pseudoconvex space. Without loss of generality we may assume that the topology in X is determined by a sequence of Pt-homogeneous pseudonorms {II lit}. In the same way as in the proof of Theorem 3.2.10 we define Xt and X. The space X is locally pseudoconvex. The rest of the . proof is the same. D A

A

THEOREM 3.3.12 (Shapiro, 1969; Stiles, 1970; for Banach spaces see Banach, 1932). Every separable locally bounded space X with a 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 Nn(/) is just the space lP. D

THEOREM 3.2.13. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be Px-homogeneous and let II IIY be py-homogeneous. A linear operator A from X into Y is continuous if and only if

IIAII =

sup

IIXIX.;;l

IIA(x)IIY

<+oo.

(3.2.2)

Chapter 3

102

Proof The theorem is a trivial consequence of the fact that a set E is bounded if and only if sup {llxiiY: x E E} < +oo. D

If px = py then the definition of IIAII implies IIA(x)IIY:::::;; IIAII llxllx. In this case the number IIAII is called the norm of the operator A. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be px-homogeneous and let II 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

11x11:r = (llxllx)PIPx,

IIYII~ = (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,BE B(X-+Y). Then IIA+BII = sup IIA(x)+B(x)IIY ::( sup IIA(x)IIY+ sup IIB(x)liY 11x11X:;;;1

11x11X:;;;1

11x11X:;;;1

= IIAII+IIBII· The original topology in B(X-+ Y) is equivalent to the topology induced by the norm IIAII. Indeed, for arbitrary e > 0, {A: IIAII

< s} J

U(O,B,s) = {A: IIA(x)IIY

< e for

x

E

B},

where the set B = {x: llxllx:::::;; 1} is bounded. On the other hand for an arbitrary positive and an arbitrary bounded set B C X, U(O,B,s) J U(O,Kr,e) ={A: IIAII

<_ _!_I_\. rPy Px (

where r = sup{llxllx: x E B} and Kr = {x: llxllx ::( r}. If the spaces X and Yare complete, then by Corollary 2.2.4 the space B(X-+ Y) is complete, i.e. it is an F-space. In this particular case the conjugate space X is always a Banach space. Let (X, II llx) and (Y, II IIY) be two locally bounded spaces. Let II llx be px-homogeneous and let II IIY be py-homogeneous. From the defini-

Locally Pseudoconvex and Locally Bounded Spaces

tion or'the norm IIAII it follows that a family equicontinuous if and only if sup{IIAII: A

Em:}

103

m: of linear operators is

<+oo.

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

liPnil

< +oo,

where, as before, ""

Pn

n

(,2 ttet) =}; ttet. i=l

i=l

The number K is called the norm of the basis {en}. As a consequence of Corollary 2.6.5. we obtain THEOREM 3.2.15. Let (X, II ID be a locally bounded space. Let II II be 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

n+m

[[}; ttetj[

~K[[2ttet[[

i=l

i=l

for all positive integers m and n. THEOREM 3.2.16 (Krein, Milman and Rutman, 1940). Let (X, II II) be a complete locally bounded space. Let II II be p-homogeneous. Let {e11 } be a basis in X of the norm K. Let llenll = 1, n = 1,2, ... If, for a sequence

{gn} of elements of X, <X>

C

=

L llgn-enll <

n=l

2K,

(3.2.3)

Chapter 3

104

then {gn} is a basis in the space X 0 generated by {gn}, X 0 = alent to the basis {en}. Proof. The triangle inequality implies n

n

i=l

i=l

lin{gn}, equiv·

n

112 tcecJJ-}; Jt,JPJJgc-ecJI ~ IJ}; tcgcl[ i=l n

n

~II}; t1etll +}; JtcJPIIgt-ecJI. i=l

i=l

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

ltiiP =

n

j-1

llt1e1ll~ l!l' ttecll + 1\.L ttet// ~2K ll};ttetJI. i=l

i=l

i=l

Therefore n

n

_l

1

Jt,JPIIgt-etll

~ m~x Jt,JP}; llgt-etll I.,;~.;;n

i=l

i=l n

~ 2KC 112: ttetll i=l

Hence (3.2.3) implies n

o-!5)

n

n

112: t,e,ll ~ 112: t,gi II~
where b = 2KC < 1. Thus, the elements basis {en}.

i=l

{gn} consitute a

i=l

basic sequence equivalent to the D

PRQPOSITION 3.2.17. Let (X, I II) be a complete locally bounded space. Let I II be p-homogeneous. Let {en} be a basis in X. Suppose that X is infinite-dimensional and that X 0 is an infinite-dimensional subspace of X. Then there is in X 0 a sequence of elements {xn}, llxnll = I, Xn 00

=

~ ~

t n,1..e.t such that

i=l

lim tn,i = 0

(i

=

1,2, ... ).

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

Locally Pseudoconvex and Locally Bounded Spaces

positive integer k and a positive e such that, for each x J

co

x =

2

E

105

X0,

llxl/ = 1,

t~,e,

i=l

llxll =max lttl >e. lo;;;io;;;k

Hence there exists a linear homeomorphism of X 0 and k-dimensiona space. This contradicts the fact that X 0 is infinite-dimensional. D By Proposition 2.6.7, Theorem 3.2.16, Proposition 3.2.17 we obtain PROPOSITION 3.2.18. Let (X, II II) be an infinite dimensional complete locally bounded space with a basis {en}. Let X 0 be an infinite-dimensional subspace of the space X. Then X 0 contains an infinite-dimensional subspace Y with a basis {in}, which is equivalent to a block basis in X. PROPOSITION 3.2.19 (Stiles, 1971). There is a locally bounded space (X, II II) such that there is no p-homogeneous norm in X determining a topology equivalent to the original one, but every infinite-dimensional subspace Y contains an infinite-dimensional subspace Y 0 in which that topology can be determined by a p-homogeneous norm. Proof Let qn

= ~ ( 1 - log :ogn ) '

n

= 30, 31, ...

Let Nn(u) = uqn. Let X 0 be an infinite-dimensional subspace Qf the space Nn(l). Then by Proposition 3.2.18 there is an infinite-dimensional subspace Y of X 0 such that there is in Y a basis {/11 } equivalent to a block basis 1/e~l/

= 1.

Without loss of generality we may assume that p 11 > ee 2n (see the remark after Proposition 2.6.7). Then, for It I < 1

ltiP ~ 1/te~l/ ~ ltl 11 (1 -

2: ).

Chapter 3

106

Since sup ltP(l- 2';.)-tPI:::;;; 21n,

o.;;;t.;;;I

by Theorem 3.2.16 {e~} and the standard basis {en} in/Pare equivalent. On the other hand the space Nn(/) is not isomorphic to /P. What is more, there is no p-homogeneous norm II II in Nn(l). Indeed, suppose that such a norm II II exists. It is easy to verify that that the sequence en= {0, ... , 0, 1,0, ... }

-

n -th place

is bounded. Hence supllenll

<

n

Let Xn

= (nl: 2n)I/p. Then g

+oo.

.i; lxniP


n=30

00

Therefore the series }; Xnen is convergent in II II· This leads to a contran=so diction since

n~lxniPn

=go 00

~

(n lolg2 1

n) ~ P ( 1 log n

2 loglogn =L.J-ee n=30

and

log

~og n)-

21og logn

__ +oo

n

{xn} ¢ Nn(l).

D

Pelczynski (see Rolewicz, 1957) has shown that in the space Nn(l) given above for each q with p < q :::;;; 1 there is a q-homogeneous norm determining the original topology. This has been the first example showing that the inequality p > p 0 in Theorem 3.2.1' is strict. Let X be a complete locally pseudoconvex space. As follows from Theorem 3.1.4, there is a sequence of Pn-homogeneous pseudonorms {II lin} determining the original topology. Without loss of generality we can assume that [[xllt :::;;; [[xll2 :::;;; llxlls :::;;; ... Suppose that {en} is a basis in X. Let

[[x[[~ =

r

sup II~ r

m=l

tmemllt,

Locally Pseudoconvex and Locally Bounded Spaces

107

where

By Theorem 2.6.1 the system of pseudonorms original topology. Let Xt,o = {x: 1/x/1; = 0}. Let

{II //;} determines the

Xt = X/x,,,

be the quotient space. The pseudonorm norm 1/ /It on Xt in the following way

1/ //~

induces a Pi-homogeneous

1/[x]i/1~' = 1/x/1;,

where {x}t denotes the coset containing x. By the definition of 1/ 11;' we obtain l/t1[e1]i+ ... +tn[en]i //;' = l/t1e1+ ... +tnen/1; ~ l/t1e1+ ... +tn+men+m/1; = l/t1[e1]t+ ... +tn+m[en+m]/1~'

for all positive integers nand m. Therefore, by Theorem 3.2.15, we obtain: 3.2.20. The cosets [en]i consitute a basis of norm one in the space Xt, wfzich is the completion of the space Xi, provided we omit all those [en]i which are equal to 0.

PROPOSITION

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

3.3.

BOUNDED SETS IN SPACES

N(L(fJ, 1:, t-t))

Let a space N(L(fJ,l:,J1)) be given. Let

n(t) = inf {a> 0: N(at) Of course n(t)

~ N~t) }·

> 0 for all t, 0 < t <+oo.

(3.3.1)

Chapter 3

108

THEOREM 3.3.1 (Rolewicz, 1959). inf

r=

If

n(t)>O,

O
then the space N(L(D,E,J1)) is locally bounded. Proof From (3.3.1) it follows that PN(rx) ~ PN(X) for each 0 < r

t

<

r.

Then by induction PN(rnx)

~ ;n PN(x),

n

= 1,2, ...

Therefore, the open set Ka = {x: PN(X)

< d} is bounded.

0

We can weaken the hypothesis on N(t) by assuming additional properties on the measure fl. Namely THEOREM 3.3.2 (Rolewicz, 1959). Let 11 be a finite measure. If liminfn(t) > 0, t-.-oo

then the space N(L(D,E,fl)) is locally bounded. Proof Let for t > 1, · {N(t) N'(t) = tN(1) for 0 ~ t ~ 1.

Since the measure 11 is finite, PN(Xn)--*0 if and only if PN'(xn)-70. Let n'(t)=inf{{a>O:

N'(at)~

Nit)}·

It is easy to verify that infn'(t) > 0. Therefore, Theorem 3.3.1 implies

Theorem 3.3.2.

0

THEOREM 3.3.3 (Rolewicz, 1959). Let the measure 11 be purely atomic and let (3.3.2) r = inf fl(Pl)> 0, where p, runs over all atoms. If

liminfn(t) > 0, t--+0

then the space N(L(D,E,/1)) is locally bounded.

Locally Pseudoconvex and Locally Bounded Spaces

·109

Proof. Let

for 0:::;:;; t:::;:;; 1,

N(t) { N'(t) = tN(1)

for t

> 1.

By (3.3.2), PN(Xn)-+0 if and only if PN• (xn)-+0. The rest of the proof is the same as the proof of Theorem 3.3.2. 0 The following fact is, in a sense, inverse to Theorem 3.3.2. THEOREM

3.3.4.

If the

measure Jl is not purely atomic

and

lim inf n(t) = 0,

(3.3.3)

L•

then the space N(L(Q, fl)) is not locally bounded. Proof By (3.3.3) there is a sequence of numbers {tn}-+oo such that an= n(tn)-+0. Let e be an arbitrary positive number. The measure fl is not purely atomic, hence for sufficiently small e there are sets e A 71 , n = 1,2, ... such that JL(An) = 2N(tn). Let Xn = tnXA.• where, as

usual, XA. denotes the characteristic function of the set An. It is easy to e verify that PN(Xn) = 2. On the other hand, · PN(anxn) =

J

N(n(tn)tn)dJl =

An

1

2

J

N(tn)dJl =

e

4·

An

Hence {anxn} does not tend to 0. Therefore no set K = {x: PN(x)

< e}

is bounded. Since e is an arbitrary positive sufficiently small number, the space N(L(D,E,JL)) is not locally bounded. D 3.3.5. If the measure is not purely atomic and the function N(u) is bounded, then the space N(L(D,E,JL)) is not locally bounded. Proof Since N(u) is bounded, (3.3.3) holds. 0

CoROLLARY

3.3.6. If there is a positive number k such that, for a sufficiently small r, there is a set Ar such that

THEOREM

110

Chapter 3

<

k

rJ-l(Ar)

< 2,

and if

lim inf n(t) = 0

(3.3.5)

t-+-oo

then the space N(L(Q,l:,J-l)) is not locally bounded. Proof By (3.3.4), there is a sequence tn-+0 such that an= n(tn)-+0. Let

be an arbitrary positive number. The assumption on the measure implies that for sufficiently large n there are An such that ks < N(tn)J-l(An) ks. In the same way as in the proof of Theorem 3.3.4 we find that N(L(Q,l:,J-l)) is not locally bounded. D B

The following theorem gives us the connection between local pseudoconvexity and local boundness in the case of spaces N(L(Q,l:,J-l)). 3.3.7. Let the measure J-l be finite and not purely atomic. Then the space N(L(Q,l:,J-l)) is locally bounded provided it is locally pseudoconvex. Proof Suppose that the space N(L(Q,l:,J-l)) is not locally bounded. Then, by Theorem 3.3.2. liminf n(t) = 0, i.e. thereis a sequence {tn}-+oo THEOREM

t-+co

such that an = n(tn)-+0. Let A be an arbitrary open set such that r

= SUPPN(X) <+oo. XEA

We shall show that the set A has infinite modulus of concavity c(A)

= +oo. Let p = inf PN(x). Since the set A is open, p

> 0. Let k be an integer

XEA

k

4r

< -p . The measure

J-l is not purely atomic, hence for a sufficiently,

large n, we can find sets An,i, i = 1,2, ... , k such that

p

J-l(An, t)

= 2N(tn)

and the sets A 11 ,t and An,J are disjoint for i =f=j, provided lim N(t) = +oo. t-+oo

111

Locally Pseudoconvex and Locally Bounded Spaces

Let Xn, ( = lnX.An,l" Then PN(Xn,t) = On the other hand,pN(Xn, 1 + Therefore

p

2 < p,

... + Xn,k) =

whence Xn,(

kp/2

E

A.

> 2r.

+ ...

PN(n(tn) (xn, 1 +xn,k))> r and n(tn (xn, 1 + ... +xn,k) ¢A.

Since n(t?i)-+0, we do not have A+A+ ... +ACKA

for any K> 0.

This implies c(A) = +oo. Now we shall consider the case where R = lim N(t) < +=. t-+co

Let !21 be a subset of Q on which the measure ll is non-atomic. Let A be an arbitrary open set in N(L(Q, E, It)) such that r = suppN(x) ze.A

<

Let p = inf PN(x). Since the measure ll is non-atomic

N(1)~t(!2 1).

a:'A

on the set Ql> then there are disjoint subsets At> ... , Ak of !21 such that ~t(At) < p/R (i = 1,2, ... , k) and A 1 u ... u Ak = !21 • Let n = 1,2, ... , i = 1,2, ... ,k. Xn,i = nx.A., Then PN(xn) < p and xn,i eA. On the other hand, Xn,I+ ... +xn,k = nxa,

and PN(

~

(xn,l+ ... +xn,k))

>

r.

The rest of the proof is the same as in the case where lim N(t) = +oo.

D

t-+co

3.4.

CALCULATIONS OF THE MODULUS OF CONCAVITY OF SPACES

N(L(Q,

~.

p,))

We recall that a function N(u) is called convex if, for arbitrary nonnegative numbers a, b such that a+b = 1 and arbitrary arguments ul> u2 , we have

Chapter 3

112

3.4.1. Let N(u) = N 0(uP), where N 0 is a convex function and 0

THEOREM

PN(ax+by)

JN(!ax(t)+by(t)!)dp ::s;; JN(!a! !x(t)! +!hi [y(t)!)dp u = JN ((!a!!x(t)!+!h!!y(t)!)P)dp n < JN (!a!P!x(t) !P+ [bjP[y(t)fP)dp u ::s;; la!P JN (jx(t)jP)dp+ lb!P JN ([y(t)jP)dp u u =

u

0

0

0

= jajP

J

0

J

N([x(t)!) dp+ jbjP N(!y(t)!) dp u u = ja[PpN(x)+jbjPpN(Y) <e.

The functionN0(u) is convex, whence if N 0(au) ~ -}N0(u), then a~ 1/2. Therefore if N(au) ~ -}N(u), then a~ (-}) 1/P. Thus n(u) ~ (-}) 1/P and the sets K. are bounded (cf. the proof of Theorem 3.3.1). Hence the sets K. are bounded and absolutely 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. 0 Suppose we are given two continuous positive non-decreasing functions M(u) and N(u) defined on the interval (O,+oo). The functions M(u) and N(u) are said to be equivalent if there are two positive constants A, B such that A

::s;;

M(u) N(u)

::s;; B

(3.4.1)

for all u. We say that M(u), N(u) are equivalent at infinity if there are a,A,B > 0 such that (3.4.1) holds for u >a. We say that M(u), N(u) are equivalent at 0 if there are b,A,B > 0 such that (3.4.1) holds for 0 < u
Locally Pseudoconvex and Locally Bounded Spaces

113

By simple calculation we obtain PROPOSITION 3.4.2. Suppose we are given two continuous positive non-decreasing/unctions M(u) and N(u) defined on the interval (O,+oo). If one of the following three conditions holds : the functions M(u) and N(u) are equivalent, (3.4.2.i) the functions M(u) and N(u) are equivalent at (3.4.2.ii) infinity and the measure f.-l is finite, the functions M(u) and N(u) are equivalent at 0 and the measure f.-l is purely atomic and such that (3.4.2.iii) inf{f.-l(A): A being atoms} > 0, then the spaces M(L(Q,E,f.-l)) and N(L(Q,E,f.-l)) are identical as sets of functions and, moreover, PM(xn)--*0 if and only if PN(Xn)--*0, i.e. the topologies in this spaces are equivalent. Proposition 3.4.2 and Theorem 3.4.1 imply THEOREM 3.4.3 (Rolewicz, 1959). Let M(u)

=

0

M 0(uP),


1,

where M 0 is a convex function. Let N(u) be a function such that one of conditions (3.4.2.i), (3.4.2.ii), (3.4.2.iii) holds. Then there is a p-homogeneous norm in N(L(Q,E,f.l,)) equivalent to the original one.

The followitrg two theorems are, in a sense, inverse to Theorem 3.4.3. THEOREM 3.4.4. Suppose that the measure f.-l is not purely atomic. lim inf N(u) uP

=

0,

If (3.4.3)

u...... oo

then the space N(L(Q,E,f.l,)) is not locally p-convex. Proof By Corollary 3.3.5 we can assume that lim N(u)

=

+oo. By

U--->00

Un

(3.4.3) there is a sequence {un}--*oo such that N(un)IJp --*OO. Let kn be the greatest integer such that kn ~ N(un), kn = [N(un)]. Since the measure f.-l is not purely atomic for sufficiently small s > 0 there are disjoint s sets An,i, i = 1, 2, ... , kn, such that J.l(An,t) = 1 +kn ·

Chapter 3

114

Let Xn,t =

UnXAn,t•

Then

<

PN(Xn,i)

8.

On the other hand, 1 N ( kn1fp

4~ Xn,i ) =

N

( Un ) k,,_!IP

8 kn 1+kn -+oo.

t=1

The arbitrariness of 8 implies the theorem.

0

THEOREM 3.4.5. Let there be a positive number rand ani nfinite family o disjoint sets Aa such that

1

< Jl(Aa) <

(3.4.4)

r

If lim sup N(u)

+=

=

(3.4.5)

uP

U->oo

then the space N(L(Q,E,Jl)) is not locally p-convex. Proof Let 8 be an arbitrary positive number. By (3.4.4) there is a b > 0, such that for each u, 0 < u < b, there is an infinite family of disjoin sets {An,u} such that 8

rN(u)

<

(An,u)

<

8

N(u)'

Let Xn,u

=

UXAn,u•

Then PN(Xn,u)

<

8.

On the other hand, PN (

x1,u+ ... +xk,u) piP

>- ~

(_!!_) _ ~ rN(u)

"" rN(u) N k 11P -

N(k- 11Pu) (k 1 /Pu)P

and by (3.4.5) . sup sup I 1m

8k ) N ( -k u ) N( 11 u P . cuP N(k- 1/P u) = hm sup sup N( ) (k_ 11 ) -++oo.

k~oo

k->oo

O
O
u

PuP

The arbitrariness of 8 implies the theorem.

0

Locally Pseudoconvex and Locally Bounded Spaces

115

3.4.6. If the measure f-l satisifes either the assumption of Theorem 3.4.4. or the assumption of Theorem 3.4.5, then in the space LP(Q,£,f-l), 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(Q,£,p)) = 211P and there is a starlike bounded open set A C LP(Q,£,p) such that c(A) = 2 1/P. CoROLLARY

From Corollary 3.4.6 follows THEOREM 3.4.7. There is no locally bounded space universal (co-universal) for all separable locally bounded spaces. Proof Suppose that such a universal (co-universal) space X exists. Then, by definition,

dimz/P:::;;; dimzX, (codimz/P:::;;; codimzX). Hence, by Theorem 3.2.7 (Theorem 3.2.8) 21 /P = c(lP) :::;;; c(X) Thus c(X) bounded.

=

+=

for all p, 0 < p:::;;; 1.

and this contradicts to the fact that X is locally

C

Another consequence of the properties of spaces N(L(Q,I,p)) is PROPOSITION 3.4.8. There is a locally bounded space X such that the topology in X can be determined by a p-homogeneous norm for p, 0

21h•. Proof Let h(u) be a positive decreasing continuous convex function defined on the interval [O,+oo) such that h(O) -CO

Let

N(u) = uPo-h(u). Let J1 be a finite atomless measure. By Theorem 3.4.3 for any p, 0

Chapter 3

116

determining a topology equivalent to the original one. Theorem 3.4.4 implies that there is no p 0 -homogeneous norm with this property. D For locally convex spaces Theorems 3.4.4 and 3.4.5 can be formulated in a stronger way. THEOREM 3.4.9 (Mazur and Orlicz, 1958). Let N(L(Q,E,fl)) be a locally convex space. If the measure f1 is not purely atomic, then the function N(u) is equivalent to a convex function at infinity. Proof. Since the space N(L(Q,E,fl)) is locally convex, there is a positive number e such that PN(Xk) < e, k = 1, ... ,n, implies PN (

x 1 + ... +xn) < I· n

Let q, p, n be positive integers. Let p :::( n. Let t be a positive real number. Since the measure f1 is not purely atomic, then, for sufficiently large q, there are disjoint sets El> ... , En such that fl(Et) = 1/nq, i = 1, ... , n. The smallest such q will be denoted by q0 • Let for s E EJ, wherej = k, k+l, ... , k-l+p(mod n), otherwise. Then 1 PN(Xk) = -pN(t) nq

and PN

(

X1

+ ... + Xn) _1 N (p t ). n

Hence I p - - N(t) :::( e q n

implies

q

n

Locally Pseudoconvex and Locally Bounded Spaces

117

The continuity of the function N(u) and the arbitrariness of p and n imply that co - N(t) ::'( e (0
q

Let N(t) :;::: eq0 and let q be the greatest integer less than coN(t)fe. From the definition of q follows sq < coN(t) ::'( e(q+ 1). Then 1 1 q+lcoN(t)"(s, whenceq+IN(cot)::'(l and N(cot)"((q+l)

2

q+l

= --·eq ::'(- N(t). Putting C = 2/s, we obtain eq e N(cot) ::'( CcoN(t)

Recall that, by Theorem 3.4.4, lim infN(t) t

> 0.

1--+<Xl

Choose{) > 0 and T > 0 such that N(t)/t ;;?.: {) for t > T and Tb Then for cot > T, 0 < co < 1, N(t) coN(t). = cot1 -;;?.: Tb

>

>

eq0 •

eq 0 •

Since cot ;;?; Timplies t;;?; T, N(cot) ::'( CcoN(t)forcot Let (Ncot) sup-for t > T, O<wo;;;I CO P(t) = wt>T

>

T, 0 "(co::'( 1.

1

for 0 ::'( t ::'( T.

N(t)

Let 0

< a ::'( 2, at > P(at) =

T. Then

sup

O<wo;;;l

N(coat) N(coat) =a sup co O<wo;;;l coa

wat~T

wat~T

N(st) ::'(a sup - 0<~1

st>T

s

= a P(t).

(3.4.6)

Chapter 3

118

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

t' ) t' ( p -" t" I I PUU) t ::::::::: _t_-:--_ t' """ t'

P(t') t'

Let

l

P(t)

Q(t)

=

for t

>

for 0

< t~T

;(T)

tT2

P(t")

=--ru-··

T,

and u

M(u)

=

JQ(t)dt. 0

Since Q(t) is a 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, N(wt)/w ~ CN(t), hence P(t) ~ CN(t). Since P(t) is equivalent to N(t) at infinity and N(t) satisfies condition (A 2), P(t) also satisfies that condition, i.e. there is a positive constant K such that for sufficiently large t, P(2t) ~ KP(t). The function Q(t) is non-decreasing, whence~ for sufficiently large t, M(t)

< tQ(t) =

P(t)

and M(t);;?o

J'Q(s)ds;;?o ~ Q(~)=P(~). t/2

Hence, for sufficiently large t, P(t)

~ KP ( ~) ~ KM(t)

and the

functionsM(t) andP(t) are equivalent at infiriity. Therefore the functions M(u) and N(u) are equivalent at infinity. 0 THEOREM 3.4.10 (Mazur and Orlicz, 1958). Suppose that there is a constant K > 0 and an infinite family of disjoint sets {Et} such that 1 ~JJ.(E,)

~

K.

Locally Pseudoconvex and Locally Bounded Spaces

119'

If the space N(L(Q,E,p.)) is locally convex, then the function N(u) is equivalent to a convex function at 0. Proof. Since the space N(L(Q,E,p.)) is locally convex, there is a positive e such that PN(Xk) < e, k = 1,2, ... , n, implies

PN ( x 1 + ·~· +xn)

~

1.

Let p be an arbitrary positive integer and let

{~

xk(s) =

We get PN(Xk) and

~

for sEEk+tn (i=0,1, ... ,p-1), otherwise. (k=1,2, ... ,n)

KpN(t)

__.. pn N ( - t ) ~PN n

(x + ...n +Xn) · 1

Hence N(t)~ efKp implies N(t)/n ~ 1/pn. Let 0 < N(t) < efK, and let p be chosen so that efK(p+1) ~ N(t) < efKp. Then N(tfn) ~ 1/pn ~ 2Kfe N(t)fn. Given p > 0. Choose a q ~ 1 such that N(tfq) < e for it I < p. Condition (LlJ implies that there is a constant D~ such that N(qt) ~DqN(t)for Jti
N(!_).

1 )~ 2 KDq_!_N(t). N(q-t )~DqN(qn qn e n

n

Let 0 < w ~ 1 and let n be such that 1/(n+ 1) 1/n < 2w, putting C = 4KDqfe we obtain N(wt)

~

CwN(t)

Let P (t)

N(wt)

= sup - - , O
Q (t) =

P~t)' t

M(t)

W

= JQ(s)ds.

·•·

for 0

<w~

< w ~ 1/n. Since

1, It!
Chapter 3

120

In the same way as in Theorem 3.4.9 one may prove that M(u) is a convex function equivalent to the function N(u) at 0. D 3.5. INTEGRATIONS OF FUNCTIONS WITH VALUES IN 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 a = t~ < t~

that

. 11m

~oo

< ... <

sup

o.,;;;;i.,;;;;k(n)-1

tk(n)

= b,

n n 0. 1ti+ 1 -ti 1 =

We say that a function x(t) is Riemann integrable if, for each normal sequence of subdivisions of the interval [a,b] and arbitrary sk, tJ:_ 1 < < s~ < t~, there exists a limit of the sum k(n)

~ x(s;)(t;-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

Jx(t)dt. a

By repeating the classical considerations this definition can easily be extended to functions of several variables and to line integrals. 1 It is easy to verify that the Riemann integral defined in the way described above has similar arithmetical properties to those it has in the calculus. Namely, if x(t) and y(t) are integrable functions (i.e. such that the Riemann integrals exist), then the sum x(t)+y(t) is integrable and b

b

b

J(x(t)+y(t)) dt = Jx(t) dt+ Jy(t) dt, a 1

a

a

For arbitrary Radon measure it was done by Klein and Rolewicz (1984).

(3.5.1)

Locally Pseudoconvex and Locally Bounded Spaces

121

if a function x(t) is integrable, then for any scalar C the function Cx(t) is integrable and b

JCx(t) dt

b

= C

a

Jx(t) dt;

(3.5.2)

a

if a function x(t) is integrable on an interval [a,b], then for each c, a< c < b, the function x(t) is integrable on intervals [a,c] and [c,b] and c

b

b

Jx(t) dt Jx(t)dt+ Jx(t)dt. =

a

a

c

3.5.1 (Mazur and Orlicz, 1948). An 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 x~, ... , x;n such that THEOREM

llxlli'

< ;n i= 1,2, ... ,kn)

and

llx + ·k~+xk"il >a. 1

Without loss of generality we can assume that kn+l is divisible by 2kn. Suppose that

xn(t) =

(i- ~)

[xi

fort=

~0

for t = :n, i = 0, 1, ... , kn,

I. •.

;n, i = 1,2, ... , kn,

[j-

. Is - kn1 , kn j .- 1 2 2k] lis mear on t h e mterva 2 2 •.J - , , · · ·, n •

Obviously each of the functions Xn(t) is integrable since its values are in the finite dimensional subspace Xn spa~ned by elements x~, ... x;n.

Chapter 3

; 122

Therefore, there is a number ~n > 0 such that, for any division 0 = 1 < 11 < ... ~n (i = 0, 1, ... , k-1) we have k

11.2 [xn(st)-Xn(s;)] (tt-fi-1)/1 < 3~ i=l

for all Stands; such that ft_ 1 < s1,s;
<

~t,

i

=

1,2,

... ,m-1. a>

Let x(t)

= 2; Xnm(t). The function x(t) is continuous as a uniform m=l

limit of a sequence of continuous functions. We shall show that it is not Riemann integrable. Indeed, let It = i/kn.,, i = 0, 1, ... , kn"' and 2i-1 let St = ft, s~ = 2kn., , i = 1,2, ... , knv· Then kn

li k~., ~ [x(st)-x(s;)]ll

=II k~., ~ 6 =II k~., ~ ~ kn'P

knp

cc

[Xnm(St)-Xnm(s;)]ll

p

[Xnm(st)-Xnm(s;)]ll

k ...,

~II k~., ~ [xn.,(s,)-xnv(s;)]ll

-II k~.,6 ~[Xnm(St)-Xnm(s;)]ll ~II x~+ ~~p+x~: 611 p

kn.,

p-1

.. ll-

p-1

~a-

~

2

m=l

a 3m·

a 2 ·

~:?'

1

k ...,

'

kn.,6Xnm(s,)-Xnm(s;)ll

Locally Pseudoconvex and Locally Bounded Spaces

123

Thus the function x(t) is not integrable. 0 If the 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) =}; atXE., i=l

where Gi EX and Et 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 b

len

JXn(t)dt = 2 at /Et/,

a

i=l

where /E/ denotes the Lebesgue measure of the set E. We assume now that, for every homogeneous continuous pseudonorm 1/x/1, the scalar function 1/x(t)/1 is integrable and that the sequence {xn(t)} tends to x(t) almost everywhere and 1/xn(t)/1 ~ 1/x(t)/1. Then we define b

b

Jx(t)dt = lim JXn(t)dt.

a

n~co

a

This definition is correct since one may show that this integral does not depend on the choice of the functions Xn(t). If the space X is not locally convex, we are not able to use this definition, because there are sequences of simple functions {xn(t)} uniformly b

J

tending to 0 such that the sequences of integrals{ Xn(t)dt} do not tend a

to 0. In fact, let Xn(t) =xi

for

i-1

--~

kn

t

i

< - (i= 1,2, ... ,kn), kn

where x~ and kn and a are defined as in the proof of Theorem 3.5.1. It is obvious that the sequence {xn(t)} tends uniformly to 0 and that

i!

j

Xn(t)dtll

=II

xi+

k~+xk,. ~a> 0. fl

a

Since, if a space X is not locally convex, there are continuous functions which are not integrable, it is reasonable to ask what classes of functions are integrabl• in what classes of spaces.

Chapter 3

124

An important class of functions is exemplified by the class of analytic functions. We say that a function x(t) with values in an F-space X is analytic if, for any t 0 , there is a neighbourhood U of the point t0 such that, forte U, the function x(t) can be represented by a power series 00

x(t)

=;}; (t-t )nxn, 0

(3.5.3)

ii,xn EX.

·n=O

If the space X is locally pseudoconvex, then the topology can be determined by a sequence of Pn-homogeneous pseudonorms II lin· This implies that if the series (3.5.3) is convergent at a point t 1 , then it is convergent for all t such that !t-t0 !
hold. For example, if X= S[O, 1], then there is a power series

2 tnxn n=O

convergent only at a finite number of points (see Arnold, 1966; Zelazko, 1972). 3.5.2 (Gramsch, 1965; 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 II Ilk· Let x(t) be an analytic function. Then for each s 0 e [a,b] there is a neighTHEOREM

co

2 (t-s )nxn is n=O convergent. This implies that there is a neighbourhood V! C U such that for t e V! bourhood U80 such that for t E U80 the series x(t) =

0

0

80

0

k = 1,2, ... ,

s1 ,

n = 0,1, ...

Since the interval [a,b] is compact, there is a finite number of points ••• , Sr such that

[a,b] C

,

U V!.

.

Let

.

V! =

.

(ai> bJ)· N,ow. we ~hall consider the Rie-

j=l

mann integral of the func:ion x(t) on the interval [a1,bj]. Let e be an

Locally Pseudoconvex and Locally Bounded Spaces

arbitrary positive number. Let n0 be such an index that Mk/2no Let x(t) = Yn 0 (t)+zn 0 (t), ai ~ t ~ b,, where

...

Yno (t)

125

< e/2.

00

=}; (t-Sj)n Xn,

Zn0 (t)= }; (t-SJ)nxn. n=n0 +1

n=O

The function Yno(t) is integrable, because it has values in the finite-dimensional space spanned by elements x 0 , ••• , Xno· For any subdivision of the interval [aJ>bJ] a1

=

< '11 < ... <

t0

tp

=

b1

and arbitrary u 1, t,_ 1 ~ u 1 ~ t1, the Riemann sums of Zn0(t) can be estimated as follows: p

II}; Zn (uz) (tz-tz-t) //k ~ 0

l=l

p

00

};

[/_l, (uz-SJ)n Xn(tz-tz-t Ilk

n=n0 +1 l=l 00

00

The arbitrariness of e and the integrability of Yn 0 (t) imply that the Riemann sums_ of x(t) converge for any normal sequence of subdivisions of the interval [aJ>bi] with respect to the pseudonorm II Ilk· Since there is only a finite number of the sets V8~, those sums converge on the whole interval [a,b]. This holds foreverypseudonorm II Ilk, whence the completeness of the space X implies that the function x(t) is integrable. 0 Let C be a complex plane. Let D be an open set in C. Let x(z) be a function defined on D with values in a locally pseudoconvex 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 z 0 ED, there is a neighbourhood Vz0 of the point z 0 such that the function x(z) can be represented in Vz0 as a sum of power series 00

x(z)

=}; (z-z )nxn, 0

n=O

XnEX.

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 Tis smooth and the function x(z) is analytic, then the integral x(z)dz exists. r Using the same arguments as in the classical theory of analytic functions, we can obtain

J

THEOREM 3.5.3. Let Then

r

be a smooth closed curve. Let x(z) be analytic.

Jx(z)dz = 0, _

(3.5.4)

J

r

x(z) - dZ

1

( 0) - XZ 21t

r

z-z 0

Or each Zo inside the domain SUrrounded by

x Where Zo

(n) (

) _

Zo -

1

21t

J

x(z)

(z-zo)n+l

d

r,

z,

(3.5.5) (3.5.6)

r and Fare as in (3.5.5).

As a consequence of (3.5.6) we obtain the Liouville theorem: THEOREM 3.5.4. Let x(z) be an analytic function defined on the whole complex plane C, with values in a locally pseudoconvex space X. If x(z) is bounded on the whole plane, then it is constant. Turpin and Waelbreock (1968, 1968b) generalized Theorem 3.5.2 to m dimensions and to a more general class of functions. We shall present their results here without proofs. Let U be an open domain in an m-dimensional Euclidean space Rm. Let Nm denote the set of all vectors k = (k1 , ••• , km), where kt are non-negative integers, (i = 1 ,2, ... , m). We adopt the folowing notation

Jxl = Jkl =

sup

l.,;t.;;m m

!xtl,

.L lktl' i=l

Locally Pseudoconvex and Locally Bounded Spaces

n

127

m

xk =

x~·.

i=l m

k! =

flkd, i=l

where x = (x1, .•. , Xm) E Rm, k = (kl> ... , km) E Nm. We say that a functionf(x) mapping U into an F-space X is of class Cr(U,X), where r is a positive number, if, for any k E Nm such that lkl ~ r, there are continuous mappings Dk f of U into X and Pr,k f of Ux U into X such that D0 f=J, Pr,kf(x,x) = 0 and

Dd(y)=

~

.L.J

Dk+hf(x)•

lhl~r-lkl

(y-x)h 'h! +iy-xir-lklpr,kf(x,y) •

(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(l-p)fp is a boundedfunction, then it is Riemann integrable. We say that a functionf(x) is of class C00 (U, R) if it is of class Cr(U, X) for all positive r .

. THEOREM 3.5.6 (Turpin and Waelbroeck, 1968b). If the space X is locally pseudoconvex, then each bounded function f(x) E Coo (U,X) is Riemann integrable. Indeed, Turpin and Waelbroeck obtained an even stronger result, namely that the integration is a continuous operator from the space C,( U, X) (Coo ( U, X)) with topology of uniform convergence on compact sets all Dkf, lkl ~r, andpr,kf(all Dkfand allpr,kf) into the space X.

3.6. VECTOR VALUED MEASURES Let (X, II II) be an 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

128

Chapter 3

values in X such that, for any sequence of disjoint sets E1o ... , En, ... belonging to E M(E1 u

... uEnu ... ) = M(EJ+ ... +M(En)+ ...

The function M (E) is called a vector valued measure (briefly a measure). By the variation of the measure Mona set A C Q we shall mean the number M(A) =sup {liM(H)Ii: He A, HeE}.

THEOREM 3.6.1 (Drewnowski, 1972). The variation M of a vector valued measure has the following properties : M(e) = 0,

(3.6.1.i)

if A C B, then M(A) M(Q) <+oo,

~

(3.6.1.ii)

M(B),

(3.6.1.iii)

.M(UHn).;;;:;}; M(Hn) for any sequence{Hn} of sets m=l n=l belonging to E, (3.6.1.iv) M(H) = lim M(Hn),

if His either a union of an increasing

n--+-ro

sequence {Hn} C E of sets or sing sequence {Hn} of sets.

if it is an intersection of a decrea(3.6.1.v)

Proof Formulae (3.6.1.i) and (3.6.l.ii) trivially follows from the definition of the variation. (3.6.l.iii). Suppose that (3.6.1.iii) does not hold. Then there is a sequence of sets Hn E E such that IIM(Hn)li-+oo. Without loss of generality we may assume that IIM(Hn)li

> 1+IIM(Hl)ll+ ··· + IIM(Hn-t)li.

The sets Hn = Hn"'-(H1 u

... UHn-t)

A

are disjoint and IIM(Hn)li

ro

> 1. Thus the series };

n=l

M(Hn) is not con-

vergent, and this leads to a contradiction. ro

(3.6.1.iv). Let A be an arbitrary set contained in

U Hn.

n=l

Let An

Locally Pseudoconvex and Locally Bounded Spaces

129 00

= (A" (H1 u

... u Hn-J) () Hn The sets

An are disjoint and A =

U An. n-1

Then 00

jjM(A)IJ

00

00

II}; M(An) II~}; IJM(A,.)IJ ~}; M(Hn).

=

n=l

n=l

n=l

00

The set A is an arbitrary subset of the set

U Hn.

n=l

Thus 00

M{H1 UH2u ... ) = sup{IJM(A)IJ: AC

U Hn} n=l

00

~}; M(Hn). n=l

(3.6.1.v). Let Hn E l: be an increasing sequence of sets. Thus Hn C H and by (3.6.l.ii) M(Hn) ~ M(H), n = 1,2, ... This implies that lim M(Hn) ~ M(H). Let e be an arbitrary positive number. Let A C H, A e 1: be such that M(H)-e

<

IJM(A)IJ ~ M(H).

As in the classical measure theory we can prove lim M(A()Hn) = M(A). Hence for sufficiently large n M(H)-2 e

<

.

IJM(A()Hn)IJ ~ M(Hn).

The arbitrariness of e implies lim M(Hn) ~ M(H). Now we shall consider the case where Hn is a decreasing sequence

n H,.. Since Hn c H, n = 1,2, ... , we have 00

of sets, Hn

E

1:. Let H

=

n-1

by {3.6.1.ii) M(H) ~ M(Hn)· Thus lim M(Hn) ~ M(H) n--+-00

Chapter 3

130

Observe that, if An C Hn ""-Hn-1 , then the series

2"' M(An) converges tl=l

uniformly with respect to the choice of An. Then for an arbitrary positive· e there is an index n0 such that

II

i

n=n,+l

M(~n)ll <e.

Let A C Hn 0 , A E E, be such a set that [[M(A)/1 C()

A

Let A =A""-(

U An (Hn ""-Hn-J). n=n +1

> M(Hn 0 )-e. A

A

Observe that A C H and [[M(A)

0

-M(A)fl <e. Hence M(A) pletes proof.

~

M(H)-2e. The arbitrariness of e comD

A measure M is called bounded if the set { M(E): E E E} is bounded. As a trivial consequence of (3.6.l.iii), we obtain 3.6.2. Let X be a locally bounded space. Then each measure with values in X is bounded. Proof Without loss of generality we may assume that the norm in the space X is p-homogeneous. Thus, by (3.6.l.iii), the set CoROLLARY

{M(E): EEE}

is bounded.

D

CoROLLARY 3.6.3. Let X be a locally pseudoconvex space. Then every measure M with values in X is bounded. Proof The topology in X can be determined by a sequence II lin of Pn·homogeneous pseudonorms. Let X~= {x: ffxlln = 0}. Let Xn be the quotient space Xn =X/X~. The pseudonorm II lin induces a Pn·homogeneous norm in Xn, and the measure M induces the measure Mn with values in the completion of Xn,Xn. Thus by (3.6.l.iii)

sup{f[M(E)[[n: EEE} <+oo, Therefore the measure M is bounded.

n = 1,2, ... D

Locally Pseudoconvex and Locally Bounded Spaces

131•

Recently Talagrand has proved that each measure with values in the space L 0(Q,E,p,) is bounded. However, there are an F-space Hand an unbounded measure M with values in H, as follows from Example 3.6.4 (Turpin, 1975c) Let Q = [0, 1], let E be the algebra of Lebesgue measurable sets, and let .ll denote the Lebesgue measure. For fe L 0 [0, 1] we write

a(f) = .ll({t ED: f (t) # 0}, and Var {f)= inf {Var (/1): / 1 =/almost everywhere}. Let H be the space of all measurable functions f such that for each >0 there are g,he£0 [0,1] such that f=g+h and Var(g) <+oo and a(h) < 8. Let 11/11 = inf {lfgffco+Var(g)+a(h): f= g+h}, where [[g[[co = ess sup [g(t)[. 8

It is easy to verify that for fi,f2 E H we have

flii+/211::::;;; 11/111+11/211, lis/ II ::::;;;·11/11 for [sf ::::;;; 1, [[sfll-+0 if s-+0 11/11 = 0 if and only if/is equal to 0 almost everywhere. Thus II/II is an 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, ~ince in the opposite case we can simply take the completion of H. Let SEE. The characteristic function Xs E H. Indeed, for each 8 > 0 there is a finite union of intervals D such that a(xs-xn) < 8. Since Xn is of finite variation, Xs E H. . Let {Sn}, Sn E .E, be a sequence of disjoint sets. Thus ./l(Sn)-+0. From the definition of the norm II I~ we infer that IIXs.ll::::;;; .ll(Sn) for sufficiently large n. This trivially implies that the measure M(S) = XsE H,

SEE,

Chapter3

132

is a vector valued measure. We shall show that the measure M is not bounded. Indeed, let n, h be non-negative integers. Let Inn=[.!!__, h+ 1]. ' n n

Let fn

= _!_

2

n 2h
I211,n

= _!_ M( n

U I2h, n)·

2h
We shall show that {fn} does not tend to 0 and this will complete the proof. Let fn = gn +hn. Let A be the union of those intervals Jn,n = I 211 ,n u I211 +1,n for which hn is equal to 0 on subsets of positive measures on both intervals I 211 ,n and I2h+I,n· This implies that the variation of gn on J 211 ,n is not smaller than 1/n = ~ A.(Jn,n) and 2 Var (gn) ;::?: A.(A). On the other hand by the definition of A 1-A.(A) = x(D"'A)::::;; 2a(hn and finally 2Var (gn)+2a(hn);::?: 1. Therefore

/Ifni/

;::?: 1/2 and this completes the example.

A measure M is called compact if the set {M (E): E e L'} is compact. By Corollary 3.6.2 we trivially obtain CoROLLARY

3.6.5. If X is 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 D = [0, 1] and let E be the algebra of Lebesgue measurable sets. Let X= 12• We define a vector measure Mas follows M(E) = {

Jsin2nntd~}.

E

Locally Pseudoconvex and Locally Bounded Spaces

133

It is easy to verify, that M is a measure. By Corollary 3.6.3 the measure M is bounded. We shall show that it is not compact. Indeed, let

En= {t: sin2nnt

> 0}.

Observe that the n-th coordinate of M(En) = 1/TC. This implies that the sequence M(En) does not contain any convergent subsequence. Thus the measure M(E) is not compact. Let f(t) be a scalar valued function defined on Q. We say that the· functionf(t) is measurable if, for any open set U contained in the field of scalars, the setf-1(U) belongs to .E. A sequence of measurable functions {fn(t)} is said to tend to a measurable function f(t) almost everywhere if lim fn(t) = f(t) for all t except a set E such that M(E) = 0. n->oo

PROPOSITION 3.6.7 (Drewnowski, 1972).

If fn

tends almost everywhere to

J, then for each e > 0 00

lim k->oo

M(U

En (e)) = 0,

n=k

where En(e) = {x: //fn(x)-f(x)// ~ e}. Proof The sequence

is a decreasing sequence of measurable sets. Then by (3.6.1.v) lim M(An) =

M (lim

An) = M (lim supEn(e)). n->oo

Observe that if x

E

limsupEn(e), then the sequence fn(x) is not convern

gent to f(x). Thus lim supEn(e) C A= {x: fn(x) is not convergent tof(x)}. By our hypothesis M(A) = 0. Thus, by (3.6.l.ii),

M(lim sup En(e))~ M(A) =

0.

D

11->CO

As a consequence of Proposition 3.6.7, we obtain extentions of the classical theorems of Lebesgue and Egorov

Chapter 3

134

3.6.8 (Lebesgue). If {fn(x)} tends to f(x) almost everywhere, then, for each e > 0,

THEOREM

lim M({x: 1/fn(x)-f(x)// ~ e})

=

0

n---+ro

3.6.9 (Egorov). If {/n(x)} tends to f(x) almost everywhere, then for each e > 0 there is an FE .E such that {fn(x)} tends to f(x) uniformly on F. Proof By Proposition 3.6.7, for each k there is an index Nk such that THEOREM

(3.6.2)

(3.6.3)

M(D"'F) =

M((J

0 En(-k ))~t M( 0 En(-k 1

k-l n-N•

k=l

n-N•

1 ))

OC>

~2

;k =e.

k=l

We shall show that {fn(x)} tends uniformly to f on th set F.· Indeed, let 'YJ be an arbitrary positive number, let k be such that 1/k < 'YJ· If x E F, then

and //fn(x)- f(x)//

or n

~

1

< -k <

'YJ

0

Nk.

For simple measurable functions there is a natural definition of the integral with respect to the measure M. Namely, if g(t)

=

l

N 1

n=l

hnXE.,

En

E

.E, bn-scalars.

Locally Pseudoconvex and Locally Bounded Spaces

135

then we define N

j g(t)dM(t) = !1

}; bnM(En). n~J.

We say that a measur:; M is L"" -bounded if ll1.:: set

{J gdM:

O
is boundecl (Turpin, 1975). Of course, every L ""-bounded measure is bounded. The converse is not true, as will be shown later. However, for a large class of spaces (containing locally pseudoconvex spaces) every bounded measure is L"" -bounded. Let X be an F-space. Let {An} be a sequence of non-negative numbers. We say that the space X has property P({A.n)} if, for any neighbourhood of zero U, there is a neighbourhood of zero V such that n = 1,2, ...

PROPOSITION 3.6.10. If a space X is locally pseudoconvex, then it has property P ( { ;n }) . Proof Let topology in X be determined by a sequence mogeneous pseudonorms. Let

llxnll < e,

n

=

{II Ilk}

of Pk-ho-

1,2, ...

Then D

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

g(t)

=}; hiXE•• i=l

Chapter 3

136

0 ~ bt ~ 1, be a simple measurable function. Let U be an arbitrary neighbourhood of zero. Since X has property P({1/2n}), there is a balanced neighbourhood of zero V such that 1 1 TV+ ... + 2n VC U,

n = 1,2, ...

By our hypothesis the measure M is bounded, i.e. there is an s such that M(E) c sVfor all Ee 2:. We write each bt in the dyadic form

>0

co

~bt,lc

bt = ~ 2/c ' k=l

where bt,lc is equal either to 0 or to 1. Thus N

J

g(t)dM =

.2

btM(Et) =

i=l

D

co

N

k=l

n=l

N

co

i=l

k=l

22

b~~n

M(Et)

=}; 2 b~: M(Et N

co

=

.2

;lc M(

iy bt,lcEt) C sU.

k=l

The arbitrariness of U implies that the set A= {

Jg(t)dM: 0 ~ g(t)~ 1, g-simple mesurable}

D

is bounded.

0

3.6.12 (Maurey and Pisier, 1973, 1976). Each bounded measure M with values in a space L 0(Q,E,p) is L 00 -bounded.

THEOREM

The proof is based on several notions and lemmas. To begin with we shall recall some classical results from probability theory. Let (Q 0 ,E0 ,P) be a probability space (i.e. a measure space such that P(Q0) = 1). A real valued E0-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)

j X(w)dP

=

and V(X) =

JIX(w)-E(X)j dP. 2

o.

LEMMA 3.6.13 (Tchebyscheff inequality). P({w: !X(w)l;;?; e})

~4E(X'). e

Proof E(X2)

=

JIXI dP 2

o.

j

j

X 2 dP+

{co: IX(co)l~}

X 2 dP

{co: IX(o)l<s}

;;?; s2P({w: IX(w)i;;?; e}). LEMMA 3.6.14. Let X(w)

E

~

j

X 2 dP

{co: IX(co)l~s}

0

L 2(Q0) and let 0
Then . zE2(X) P({w. X(w);;?; A.E(X)});;?; (1-..1.) E(X 2)"

Proof Let X'(w)

= { :(w)

ifX(w);;?; A.E(X), ifX(w)

<

A.E(X).

By the Schwarz inequality we have

E 2(X') ~ E(X' 2)P({w: X'(w) =/= 0}) ~ E(X 2 )P({w: X(w);;?; A.E(X)}). Since E(X) ~ E(X')+A.E(X), we have E(X') ;;?; (1-..I.)E(X). Thus

0

Chapter 3

138

Let (fJ,l:,P) be a probality space. Random variables X 1(co), ... , X 11(co) are called independent, if P({co: X 1(co) e Bt> ... , Xn(co) e Bn}) = P({co: X 1(co) e B1})· ••• ·P({co: Xn(co) e Bn})

(3.6.4)

for all sets BI> ... , Bn E: 1:. A sequence {Xn} of random variables is called a sequence ofindpendent random variables if, for each finite set of indices i 1 , ••• , i11 , the random variables Xfl, ... , Xtn are independent. By (3.6.4) we conclude that, if {Xn(co)} is a sequence of independent random variables, then E(X11 · ... · Xtn) = E(XtJ · ... · E(Xtn)

for i 1 < i 2 < ... in. By a Rademacher sequence we shall mean a sequence of independent random variables r11(co) taking the values +I or -1 with the same probability 1/2. 3.6.15 (Paley-Zygmund inequality; see Paley and Zygmund, 1932). Let {r11(co)} be a Rademacher sequence. For any sequence of numbers {a11 } LEMMA

N

P ({co:

N

I~ rn(co)anl ~A~ lanl 2}) 2

n=l

n=l

Proof We shall write N

X(co) = j ~ rn(co)an 12 n=l

and we shall calculate N

E(X) =

JI~ rn(co)anr dP J ~ rt(co) r1(co) atCii dP u U

n=l N

·=

i,j=l

=

N

N

i-1

i=l

J~ latlldP = ~ latl

n

1•

;?:

+

(1-.1)2 •

Locally Pseudoconvex and Locally Bounded Spaces

139

Using the formula

and observing that

Jrn (w)rn (w)rn (w)rn (w)dP = 0 1

3

2

4

n

if one index is defferent from the others, we obtain N

E(X 2)

=

JI}; rn(w)an 1dP 4

a

n=l

N

=

,2; [an[ +6

};

4

[an[ 2 [am[ 2

l~n~m~N

n=l

2: [an[ r. N

~ 3(

2

n=l

Using Lemma 3.6.14 we obtain the required inequality.

0

Let Q = {-1, 1}N be a countable product of two point sets. The elements of Q are sequences a= {an}, where an take the values either +1 or -1. Let rn(a) be the following random variables rn(a) =On. Let A= {An} be a sequence ofreals such that [An[~ 1. On the two point set { -1, + 1} we define measures P1n as follows

1+A

P;.n({ +1}) = - 2 -, P;.n({ -1 }) =

-

1-A 2 -.

Let P1 be the product measure, P1 = P1 , X ... X P1n X .•. It is easy to verify that the random variables {rn(t)} constitute a sequence of independent random variables and

Jfn(a)dP;. =An.

a

Let A= {0,0, ... }. The measure P1 will be denoted by P0 •

(3.6.5)

Chapter 3

140

3.6.16 {Musial, Ryll-Nardzewski and Woyczynski, 1974). Let P;. and P 0 be as described above. Let LEMMA

t

p =

(Po+P;.).

Then, for each sequence of numbers {x 1 , x 2 , integer n, n

••• }

and for each positive

n

P({a: I~ rt(a)xt I~+ I~ AtXt I})~+· i=l i=l

m =

Proof Without loss of generality we may assume that

n

If L;ixtl 2

>

i,we shall use the Paley-Zygmund inequality

i=l

1 1 ( 1 )2 1 ~23 l-16 ~g· n

If 2

lxtl 2 < t

we shall use the Tchebyscheff inequality (Lemma

i=l

3.6.13). By simple calculation we obtain n

JI~ (Ai-ft (a))xtl dP.t 2

u

i=l

n

=

~ i=l

Ji(-1t-ft(a))xtl dP.t 2

u

n

= ~ A.f i=l

n

lxtl 2 -2

~

n

-1tlxtl 2

i=l

n

=

l\t-A.n!xtl 2 <{. i=l

Jft(a)dP.t+ ~ lxtl

u

i=l

2

Locally Pseudoconvex and Locally Bounded Sraces

141

In the calcution we have used the fact that ;, are independent random variables and formula (3.6.5). Now

PA

({a: It fc(a)xt I~+}) •=1

= PA

({a: It (lt-ft(a))xtl ~ +}) •=1

=

1-PA({a: It (lt-rt(a))xc I~ f}). t=1

n

82 ~ 82 1 I- 72 L.J (I-Ai)lxtl 2 ~ I--p 4· i-1

Hence m

>f

also in this case.

0

Proof of Theorem 3.6.I2 (modified proof given by Ryll-Nardzewski and Woyczynski (I975). Let M be a bounded measure. Thus, by definition, the set n

K = {x: x

=}; etM(At), Et ~qual either - I or+ I, 1=1

=4t e J:, Ac disjoint sets} is bounded. We have to show that the set n

K1 = {x: x

=}; ltM(At), JkJ ~I,

At el:, At disjoint}

i=1

is also bounded. M(At) e L 0(D,J:,p). We shall write M(At) = fi(t). Let [) = {-I, + 1}N, let P be a measure with the property described in Lemma 3.6.16, and let rt be a sequence of independent random variables described in that lemma. Thus by Lemma 3.6.I6 (3.6;6)

Chapter 3

142

for all t. By the arguments given in: Example 1.3.5 we may assume that

fl.(Q) <+oo. Let

n

{t: I~ At/i{t) I> sc}. that fJ.{Tc) > 0. Let fl.c(A) = fl.(A n Tc)/fJ.(Tc).

Tc =

i=I

Suppose fl.c(Tc)

Of course

= 1. For the product measure fJ.cXP we have by (3.6.6) fl.c

xP({(t,a): /4 ft(a)fi(t)l? +14 n

n

t=l

•=11

lctfi(t)

I})?+·

Then, by the Fubini theorem,

and, by the definition of Tc, n

.~a~\ fl.c ({t ETc: I~ Bt/i{t)

I> c})? +·

Hence, by the definition of fl.c, max fl. ••= ±l

({t

E

T:

" I~ Bt/i(t) t=l

?

.~~\fl.({tETc:

=

+fl.({t e T:

I> c})

n

16stfi(t)l >c})?+fJ.(Tc)

It I? sc}). ?ctfi(t)

(3.6.7)

•=I

Let Uc

= {x:

fl.({t:

lx(t)i

> c}) < c}

be a basis of neighbourhoods of zero in L 0(Q, E, fl.). Since the set K is bounded, then for each c > 0 there is an s > 0 such that sK C Uc. Thus, by (3.6.7), sK1 C U 8c. The arbitrariness of c implies that the set 0 K1 is bounded.

Locally Pseudoconvex and Locally Bounded Spaces

143

Now we shall define integration with respect to an L 00 -bounded measure of scalar valued functions. Let (X, II II) be an F-space. Let M be an L 00 bounded measure taking its values in X. Let f be a scalar valued function. Let

M·(f) =sup{ II

J

gdMII: g being simple measurable u functions, lgl ~ 111}.

3.6.17 (Turpin, 1975). M ·(f) has the following properties. If l.hl ~.hi, then M ·(/1) ~ M ·(/2) ~+oo, (3.6.8.i) Iff is bounded, then limM·(if) = 0, (3.6.8.ii)

PROPOSITION

t--+0

Iff is measurable and M ·(f) = 0,

then f = 0 almost everywhere (3.6.8.iii) (i.e. except a set A such that M(A) = 0). If {In} is a sequence of measurable functions tending almost everywhere to J, then M ·(f)~ lim inf M · (fn). (3.6~8.iv) n--+oo

(3.6.8.v) M·(.h+.h) ~ M·(ft)+M·(_h), providedft,.h are measurable. Proof (3.6.8.i), (3.6.8.ii), (3.6.8.iii) are trivial. (3.6.8.iv). Let {fn} be a sequence of measurable functions tending to f almost everywhere. Let g be a simple measurable function such that lg(t)l ~ lf(t)l. Let 8 be an arbitrary positive number. Lets be an arbitrary number such that 0 < s < 1. Let I fn(t)l if g(t) =I= 0, gn(t) = olg(t)l if g(t) = 0.

I

Observe that {gn(t)} converges almost everywhere. Thus, by the Egorov theorem (Theorem 3.6.9) for each 'YJ > 0 there is a set A such that M(D"'A) < 'YJ and {gn} converges uniformly on A. Since g is a simple function, there is an 'YJ > 0 such that for each setH such that M(H) < 'YJ we have I\

JsgdMII <

H

8.

144

Chapter 3

Taking 1J and A as described above, we have II

JsgdMII ~II AJsgdMII +II JsgdM[I ~lim inf M · (/n)+e,

n

n~A

n~~

because, for sufficiently large n, lfn(t)l > lsg(t)l for tEA. The arbitrariness of e and simply (3.6.8.iv). (3.6.8.v). To begin with, we shall assume that ft and / 2 are simple. Let g(t) be an arbitrary simple measurable function such that lg(t)l ~ l.h(t)+};(t)l. Let gt(t)

=

J

l

g,(t) lft(t)j

~ft(t)l+l/2(t)l

if lft(t)l+lf2(t)l =fo 0, i = 1,2. if lfl(t)l+lf2(t)l = 0.

The functions g 1(t) and g 2(t) are simple and g(t) = gl(t)+g2(t).

Hence II

JgdMII~II nJgldMII+II nJg2dMII~M·(fl)+M·(.t;).

n

The arbitrariness of g implies

Suppose now that.ft and,/; are not simple. Then there are two sequences of simple mesurable functions {.h,n} and {h,n} tending almost everywhere to / 1 and}; and such that lfi.nl ~I AI, i = 1,2. Therefore, by (3.6.8.iv),

Let B(f) = (

~

limsup M· (ft.n+An)

~

supM ·(ft,n)+supM ·(h,n)

~

M ·(h)+M·(JJ.

n

n

JgdM: g simple measurable functions, n

0 igl

~Ill}

Locally Pseudoconvex and Locally Bounded Spaces PROPOSITION

145

3.6.18. The set B(f) is bounded if and only if

1imM ·(if) = 0.

(3.6.9)

~0

Proof If the set B(f) is bounded, then limM· (if)= lim (sup{llxll: x t--+0

E

B(tf)})

t-->-0

= lim(sup{llxll: x

E

tB(f)}) = 0.

(3.6.10)

t-->-0

Conversely, if (3.6.9) holds, then by (3.6.10) the set B(f) is bounded. 0 Let X denote the set of those /for which B(f) is bounded. By (3.6.8.v) and (3.6.9), M· (f) is an F-pseudonorm on X. We shall now use the standard procedure. We take the quotient space X/{f: M·(/)=0}. In this quotient space M ·(f) induces an F-norm. We shall take completion of the set induced by simple functions. The space obtained in this way will be denoted by D(D,E, M). Since, for each simple function g,

II

JgdM[J ~ M·(g).

{l

We can extend the integration of simple functions to a linear continuous operator mapping D(D,E,M) into X.

3.7.

INTEGRATION WITH RESPECT TO AN INDEPENDENT RANDOM MEASURE

Let (D0,E0,P) be a probability space. Let Q be another set and let E be a a-algebra of subsets of D. We shall consider a vector valued measure M(A), A E E, whose values are real random variables, i.e. belong to X= L 0(D0 1:0 P). We say that M(A) is an independent random measure if, for any disjoint system of sets {AI> ... , An}, the random variables M(At) EX, i = 1,2, ... , n are independent. We recall that a vector measure M is called non-atomic if, for each A E D such that M(A) =1=- 0, there is a subset A0 C A such that M(A 0) =1=- M(A).

Chapter 3

146

Let X(w) be a random variable, i.e. X(w) we denote Fx(t) = P({w: X(w) ~ t}). The function Fx(t) is non-decreasing, lim

E

L 0(£2 0 ,.E0 ,P). By Fx(t)

Fx(t)

= 0, limFx(t) = I.

t-+- ro

I->-+ ro

It is called the distribution of the random variable X(w).

Let Q = [0, 1], and let .E be the algebra of Borel sets. We say that a random measure M is homogeneous if for any congruent sets A 1 A 2 C C [0, 1] (i.e. such that there is an a such that a+A 1 = A2 (mod 1)) the random variables M(A 1) and M(A 2) have identical distribution. It is not difficult to show that, if M is a homogeneous independent random measure, then for each n we can represent M(A) as a sum M(A) = M(A 1)+ ... +M(An), where the random variables M(Ai), i = 1,2, ... , n, are independent and have the same distribution (Prekopa, 1956). Let X(w) be a random variable. The function +ro

fx(t) =

J

eits dF(s).

(3.7.3)

-ro

is called the characteristic function of the random variable X(w ). If the random variables X1 , ••• , Xn are independent, then fxi+ ... +Xn(t)

= fxit)· ··· -fxn(t).

(3.7.4)

Directly from the definition of the characteristic function !ax(t) =fx(at).

(3.7.5)

We say that a random variable X(w) is infinitely divisible if for each positive integer n there is a random variable xn such that X= Xn+ ... +Xn,

(3.7.6)

in other words, by (3.7.4). fx = (/xn)n.

(3.7.7)

If X is an infirute-divisible random variable, then its characteristic function/x can be represented in the form

fx(t) =

exp(iyt+

]"'(eitx_1- 1 ~xx2 ) 1 ~2x2 dG(x)),

-00

(3.7.8)

Locally Pseudoconvex and Locally Bounded Spaces

147

where y is a real constant, G(x) is a non-decreasing bounded function and we assume that at x = 0 the function under integration is equal to - t 2/2. 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

J

l+u2 (costu-1)----zt2 dG(u).

(3.7.9)

0

Of course, without loss of generality we may assume that G(O) = 0. Suppose that M(A) is a non-atomic independent random measure with symmetric values. Then by (3.7.9) co

fM(t)

= exp

J

1+u2 (cos tu-1)----uz-dGA(u).

(3.7.10)

0

For an arbitrary real number a we obtain by (3.7.5) and (3.7.10) co

/aM(A)

=

exp

J

1+u2 (cos tu-1) ----zt2 dGA(u).

(3.7.11)

0

Let A 1 ,A 2 E I be two disjoint sets. By (3.7.4) (3.7.12) Formulae (3.7.11) and (3.7.12) imply that for a simple real-valued function h(s) fjh(s)dM(t) u

= exp(-

JTM(th(s))ds),

(3.7.13)

0

where (3.7.14) In the sequel we assume Q = [0, 1]. Now se shall prove some technical lemmas.

Chapter 3

148

Let co

UM(x) =

J

min ( xll,

-~ll) {1 +ull)dGM(u)

(3.7.15)

0

and for x

> 0,

(3.7.16)

for x =0. It is easy to see that the two functions Uy(x) and 'Py (x) are equal

to 0 at 0, continuous and increasing. LEMMA 3.7.1. For all x;;;: 0, a;;;: 0, there are positive numbers c1(a) and ell such that max TM(v)::::;;;c 1(a)UM(x) (3.7.17) O~v~ax

and 1

f TM(xt)dt ;;;: c UM(x).

(3.7.18)

2

0

Proof. Observe that

1-cosuv = 2sinll

z;;:::::;;;-} u v

2 2

and 1-cosuv:::::;;; 2. Hence, for c1{a)

=

max(2,-} all), we have

max (1-cosuv):::::;;; c1 (a)min(1,x2u2). O~v~ax

Therefore ( ) . ( ll 1 ) max 1-cosuv :::::;;; c1 a mm x , - 2 • 2

o,;;;v,;;;ax

U

(3.7.19)

U

Integrating (3.7.19) with respect to 1+ull dGy we obtain {3.7.17). ull Observe that sinz . (z, 2 1) 1---;;;: cllmm ,

z

Locally Pseudoconvex and Locally Bounded Spaces

149

where · ( mm . ( 1·- sinz) . z- 2( 1- sinz)) c2 =mm - , mm - . Z

Z>1

0
Z

Integrating TM(tx) with respect to t, we obtain 1 Joo( 1- sinxu) ~ 1+u2 dGM(u) J TM(tx)dt =

-xu

0

0

0 Observe that

Integrating by parts the second integral on the right-hand side, we obtain

J

1+u2 dGM(u)

1/z

J oc

00

~

= ~

GM(u) 2+u21oo +2 ~

1IX

1/X

GM(u) du ~

2'l'M(X).

Thus (3.7.20) On the other hand, integrating by parts both terms, we obtain

(3.7.21) Formulae (3.7.20) and (3.7.21) imply that the functions 'l'.lll(x) are equivalent at infinity.

u.~~~(x)

and

150

Chapter 3

3.7.2 (Urbanik and Woyczyiiski, 1967). Let G{u) be a continuous non-decreasing function such that G(O) = 0 and lim G(u) = 1. Let LEMMA

u-o-+oo

J 00

([>(x)

=

(3.7.22)

G;;) du.

1/X

Then the function ([>(yx) is concave. Proof We apply a change of variable v = l/u 2 in formula (3.7.22), and obtain ([> (yx) =

f G~~) jG( ylv) du =

1/y;

dv.

0

Thus the derivate of ([>(yx) is equal to G(I/yx). It is non-increasing, since G is non-decreasing. Therefore the function ([>(yx) is concave. D 3.7.3 (Urbanik and Woyczyiiski, 1967). Let {In} be a sequence of simple measurable functions belonging to L 0 [0, 1]. The sequence

LEMMA

1

Jfn(s)dM(s) tends to 0 in L [0, 1] 0

0

if and only if 1

J'l'M(/fn(s)l)ds = 0.

lim

n-+oo 0

Proof We shall use the classical statement that a sequence {hn} of functions belonigng to L 0 [0, I] tends to zero in L 0[0, I] if and only if the logarithms of its characteristic functions tend uniformly to 0 on 1

each compact interval. Thus by (3.7.13) {

Jfn(s)dM(s)} tends to zero in 0

1

L 0 [0, 1] if and only if { jTM(fn(s))ds} tends to Ouniformly on each como

pact interval [0, T]. 1

If~

JTM(tfn(s))ds} tends uniformly on each compact interval, then 0

1

lim

1

J JTM(ifn(s))dtds =

n-oo 0

0

1

lim ~oo

1

J JTM(ifn(s))dsdt = 0

0

0.

Locally Pseudoconvex and Locally Bounded Spaces

151

Hence by (3.7.18) 1

lim

JUM(Ifn(s)l)ds = 0.

(3.7.23)

n-+oo 0

Since UM(x) and IJfM(x) are equivalent at infinity, 1

lim

J'PM(Ifn(s)l)ds = 0.

(3.7.24)

x--+co 0

Conversely, if (3.7.24) holds, then (3.7.23) holds. By (3.7.17) we find 1

that { JrM(tfn(s))ds} tends to Ouniformly on each compact interval. 0 0

THEOREM 3.7.4 (Urbanik and Woyczyiiski, 1967). Le M be a symmetric homogeneous independent random measure with values in the space L 0[0, 1]. Then there exists a continuous function N(x) defined on [O,+oo) such that N(O) = 0, N(J./i) is concave and such that the set of all real 1

functions for which the integral J fdM exists is the space N(L). 0

Conversely, if N(yx) is a concave continuous function such that N(O) = 0, then there is a symmetric independent homogeneous random measure M with values in the space L 0 [0, 1], such that the set of real val1

ued functions f for which the integral J fdM exists is precisely the space 0

N(L). Proof The first part of the theorem follows immediately from Lemmas 3.7.2 and 3.7.3. Conversely, if N(yx) is concave, then it can be represented in the integral form
N(yx) = J q(u)du, 0

where q(u) is a non-increasing function. We put G(O) = 0 and G(x) = min(l,q(1/x2)) for x > 0. Let M be a symmetric independent homogeneous random measure such that

152

Chapter 3

(3.7.10) holds. To complete the proof it is enough to show that the functions N(x) and :~:,

00

NG(x)

J G~~)

=

du

=

1/Z

J

min (I ,q(u))du

0

are equivalent at infinity. Since q(u) is non-increasing, q(u) ~ q(1) for u > 1. Thus z

1

<1:

N(y'X)

Jq(u)du = Jq(u)du+ Jq(u)du = Jq(u)du+q(1) J min(I,q(u))du

=

0

0

1

z

1

0

1

~ K+CNG(yx). 1

where K =

Jq(u)du and C =

q(1).

0

On the other hand, trivially NG({x) ~ N(y'x).

0

Woyczynski (1970) extended this result to the case where M(A) take values in the product space L 0 (D 0 ,E0 ,P) X ... XL0 (!J0 ,E0 ,P) and f(s) is n..times

a matrix function.

3.8. UNCONDITIONAL CONVERGENCE OF SERIES

Let Q = N be the set of all positive integers. Let ~ be the algebra of all subsets of N. Let M be a vector measure defined on A E E taking its values in an F-space (X, II II). Let Xn = M({n}). Since M is a vector 00

measure the series

2

Xn

has the property that for each sequence

en

n=1 00

taking values ·either 0 or 1 the series

2 en x~ is convergent. The series n=1

with this property are called unconditionally convergent series. Thus

Locally Pseudoconvex and Locally Bounded Spaces

153

we have shown that each vector measure on N induces unconditionally convergent series. But the converse is also true. Namely, each unconditionally convergent series }; Xn induces a vector measure by the formula n=l

M(E)=};

Xn.

neE

PROPOSITION

3.8.1. The measure M induces by an unconditionally concc

vergent series };

Xn

is compact (and thus bounded).

n=l 00

Proof Since the series };

Xn

is unconditionally convergent, for each

n=l

e > 0 there is an index N such that for each sequence either 0 or 1 00

l

2;

BnXn

n=.N+l

I<

{en}

taking values

(3.8.1)

e.

.N

Thus the finite set };

e11 Xn,

where {e1 ,

••• ,

en}

runs over all finite systems

n=l

taking values either 0 or 1, constitutes an e-net in the range of the meas-

D

~M 00

THEOREM

3.8.2 (Orlicz, 1933). A series};

Xn

is unconditionally convergent

n=l

jf and only if for each permutation p(n) of positive integers the series 00

2 Xp(n) is convergent. ft=l 00

Proof Suppose that the series .for each e

>

2

Xn

is unconditionally convergent. Then

n=l

0 there is an index N such that

k+m

IIJ:

enXn

.n=k

fork> N, m

>

I<

e

0 and Bn taking the value either 0 or I.

Chapter 3

154

Let K be such a positive integer that, for n arbitrary r > K and s ;;::;: 0 r+s

II~ Xp(n) I! = n=r

>

K, p(n)

>

N. Then for

q

114 I < 8 tXt

(3.8.2)

8•

•=p

where p = inf {p(n): r ~ n ~ r+s}, q =sup {p(n): r ~ n ~ r+s}, t = {1 if i = p(n) (r ~ n ~ r+s), 8 0

otherwise.

The arbitrariness of 8 implies that the series

2"'

Xp(n)

is convergent.

n=l

Suppose now that the series 2"'

Xn

is not unconditionally convergent.

n=l

This means that there are a positive number tJ and a sequence {8n}, taking the value either 1 or 0 and a sequence of indices {rk} such that r•+l

il ~

8 nXn n=r•+l

8

I > J.

Now we shall define a permutation p(n). Let m be the number of those 8n, n = rk+ 1 , ... , rk+I• which are equal to 1. Let p(rk+v) = n(v), where n(v) is such an index that 8n(v) is a v-th 8i equal to 1, rk < i ~ rk+I• Q < v ~ m. The remaining indices rk < n ~ rk+1 we order arbitrarily. Then r.+m

r•+l

n=r•

n=r.+l

II~ Xp(n)ll =II~

This implies that the series

8

nXnll >

J.

2"'

is not convergent.

xp(n)

D

n=l

A measure M induced by a series "'

2

Xn

is L"' -bounded if and only if

n=l

for each bounded sequence of scalars {an} the series 2"'

anXn

is con-

n=l

-vergent. The series with this property will be called bounded multiplier convergent.

Locally Pseudoconvex and Locally Bounded Spaces

155

THEOREM 3.8.3 (Rolewicz and Ryll-Nardzewski, 1967). There exist an 00

F-space (X,

II II) and an uncoditionally convergent series}; Xn

of elements

n=1

of X which is not bounded multiplier convergent.

The proof is based on the following lemmas. LEMMA 3.8.4. Let X beak-dimensional real space. There exists an open symmetric starlike set A in X which contains all points PI> ... , p 3k of the type (e~> ... , ek), where ei equals 1 or 0 or -1, such that the set Ak-1 =A+ ... +A (k-1)-fold

does not contain the unit cube C=

{(a1,

... ,

ak):

lail <

1, i = 1,2, ... , k}.

Proof Let A 0 be the union of all line intervals connecting the point 0 with the points PI> ... , Ps•· Obviously the set A~- 1 is (k-1)-dimensional. Therefore there is a positive number e such that the set (A 0 +A.)k-\ where A. denotes the ball of radius e (in the Euclidean sense), has a volume less than 1. Thus the set A= A 0 +A. has the required property. D

LEMMA 3.8.5. There is a k-dimensional F-space (X, II II) such that IIPill ~I, i = 1,2, ... , 3k and there is a point p of the cube C such that IIPII;;:, k-1. Proof We construct a norm II II in X in the way described in the proof of Theorem 1.1.1, putting U(l) =A. Since PtE A = U(I), IIPill ~ 1. Furthermore, since Ak- 1 = U(k-I) does not contain the cube C, there is a point p E C such that p E U(k-1). This implies that II PII ;;:, k-1. D Proof of Theorem 3.8.3. We denote by (Xk, space constructed in Lemma 3.8.4. Let 1

II

II~) the 2k dimensional

'

llxllk = 2k llxllk· Let X be the space of all sequences a

= {an} such that

00

lllalll

=}; ll(a2•-•+t• ... , a2~)/lk-1+la1l <+oo. k=l

Chapter 3

156

It is easy to verify that X is an F-space with the norm

Ill Ill. Let

(0,0, ... , 0, 1,0, ... )

Xn =

n-th place

and let {en}' be an arbitrary sequence of numbers equal either to 0 or to 1.

,2; e

The series

11

x 11 is convergent. Indeed,

n=l r

IIIL

Ill~ Ill Yo 111+111 Yk+llll+ ... +Ill Yk'III+IIIY~III,

1

BnXn n=m

where k is the smallest integer non-greater than log 2 m, k' is the largest less than log2 r, and 2•

,2;

Yo =

BnXn,

n=m 2i

YJ =

,2;

j=k+1, ... ,k',

BnXn, n=2J-'+l r

Yo= . : : ,_; I

\'

BnXn.

n=2•'+1

In virtue of Lemma 3.8.5 1

IIIYolll ~ 2k-l ' 1

IIIYJIII ~ 2i-l ' '

1

II!Yolll ~lk'.

j = k+1, ... , k'

.

Therefore r

~~~~ n=m

k'

BnXnlll

~2

i=k-1

;i < 2L2 ~ ! ' w

00

and the series

,2;

BnXn

n=l

ditionally convergent.

is convergent. Thus the series

,2; n=l

Xn

is uncon-

Locally Pseudoconvex and Locally Bounded Spaces

151

On the other hand, as follows from Lemma 3.8.5, for each k there is a point ak E Xk, ak = (a2k+1• ... , a2k+1), SUCh that latl < 1, i = 2k+ 1, •.. , 2k+1 and

Let bn =an, n

= 1,2, ... The sequence {bn} is bounded and

2t'+l

Illn=2k+1 }; bnxnlll =

llakllk

>

+·

00

Therefore the series }; bnxn is not convergent. This implies that the n=l 00

series };

Xn

is not bounded multiplier convergent.

D

n=l

Observe that the measure M induced by an absolutely convergent 00

series };

Xn

which is not bounded multiplier convergent is bounded, but

n=l

is not L 00 -bounded. Turpin (I 972) constructed a non-atomic measure with this property.

3.9.

INVARIANT A(X)

We shall prove the results of Turpin (1975), however restricting ourselves to the metric case. Let (X, II II) be an F-space. By A(X) we denote the set of such sequences {An} that the space X has property P({A.n}), i.e. for any neighbourhood of zero U there is a neighbourhood of zero V such that (3.9.1)

Of course, by definition, A (X) always contains all sequences with finite support, i.e. sequences of the form A= {A.1 , A2 , ... , A.n, 0, 0, ... }. If A (X) contains a sequence with an infinite support, in other words : if there is a sequence A= {An}, An> 0 belonging to A(X), we say that the space X is strictly galbed.

Chapter 3

158

3.9.1. Let (X, II II) be an F*-space. Then the set A(X) is equal to the set of all such sequences {A.n} that, for any bounded sequence {xn} C X, the sequence

PROPOSITION

N

{YN}

= {}; AnXn}

(3.9.2)

i=l

is bounded. Proof Necessity. Let {A.n} e A(X). Let U be an arbitrary neighbourhood of zero. By the definition of A(X) there is a neighbourhood zero V such that (3.9.1) holds.

Let {xn} be an arbitrary bounded sequence. Then there is a b > 0 such that Xn e bV, n = 1,2, ... Thus, by (3.9.1) YN E bU, N = 1,2, ... Since U is arbitrary, the sequence YN is bounded. Sufficiency. Suppose that {A.n} ¢ A(X). This means that there is a neighbourhood of zero U such that (3.9.1) does not hold for any neighbourhood of zero V. This implies that for an arbitrary neighbourhood of zero V and an arbitrary positive integer n (3.9.3)

A.nV+A.n+ 1 V+ ... ¢ U.

Now we shall choose a sequence of elements {xn} and an increasing sequence of indices {nm} in such a way that i

=

(3.9.4)

nm+l, ... ,nm+l,

(3.9.5)

Such a choice is possible since (3.9.3) holds. The sequence {xn} tends to 0 by (3.9.4), thus it is bounded. On the other hand by (3.9.5) the sequence

{zm}

=

{A.nm+!Xnm+l+ ... +A.nm+IXnm+I}

is not bounded. This implies that the sequence {YN} is not bounded either. 0 PROPOSITION

3.9.2. lf{A.n}eA(X)

and0~p. 11 ~An,

n= 1,2, ... , then

{P.n} E A(X). Proof It follows immediately from the definition of A(X).

0

Locally Pseudoconvex and Locally Bounded Spaces

159

PROPOSITION 3.9.3. Let {A.i} e A(X). Let!., be a sequence of non-empty finite disjoint subsets of the set of positive integers. Let

Then {.Un} e A(X). Proof. Let {x.,} be an arbitrary bounded sequence. Let

, {x., Xi =

for i e /.,, elsewhere.

0

The sequence {rn} is of course also bounded. Then the sequence N

{YN}

= {};

N•J

,Unxn} =

n=l

{}; }; n=l

Atx;}

iEln

is bounded.

D

3.9.4. Let {-1.,}, {.Un} e A(X) and let (nt,m;) be a double sequence with terms not equal to one another. Then the sequence {vi} = {An,·.U,..} E A(X). Proof. Let {xi} be an arbitrary bounded sequence. Then PROPOSITION

N

YN

N

= }; VtXi = }; An,,llm, X~ t=l

t=l

p

~};An (2; ,llmXin,m)' n=l

where p

=

(3.9.6)

m

max nt and the summation with respect to m is taken over IEO;i~N

all m for which there is an index i = i.,, m ~ N such that (n;, mt) = (n, m). = }; ,UmXtn,m, is bounded. Therefore the sequence The set {y:},

y:

m

{YN} is also bounded.

D

COROLLARY 3.9.5. If an F*-space (X, II II) has property P({qn}) for a certain q, 0 < q < 1, then in the space X bounded multiplier convergence and unconditional convergence are equivalent.

+t}

Proof. We shall show that if {qn} e A(X), then {(

e A(X). Indeed

let (nt,mt) be such a double sequence that (nt+mt) is a non-decreasing

Chapter 3

160

sequence and the sequence (nt,mt) contains all pairs (n,m). Let An

=

qn

=

fln·

It is easy to verify that An.flm, = qn•+m• has the same order of growth as

qv'n. Then {qv'n} E A(X). Since lim (_!_)n q-vr. = 0, by Proposition 3.9.2 '11--->-00

{(~f}eA(X).

2

Theorem 3.6.11 implies the corollary.

D

PROPOSITION 3.9.6. A(X) is an invariant of isomorphisms. Proof An isomorphism maps bounded sets onto bounded sets, and the inverse also maps bounded sets onto bounded sets. Thus by Proposition 3.9.1 A(X) is an invariant of an isomorphism. D PROPOSITION

3.9.7. Let Y be a subspace of an F*-space X. Then

A(Y):) A(X).

Proof Let {.An} E A(X). Let {xn} be an arbitrary bounded sequence in Y. Then it is also a bounded sequence in X. Therefore the sequence N

{YN} = {

.J.: AnXn} n=l

is bounded. This implies that {.An}

E

A(Y).

D

3.9.8. A(X) c [1. Proof. Each space X contains a one dimensional subspace X 0 • Since A(X0 ) = [1, by Proposition 3.9.7 the corollary holds. D CoROLLARY

3.9.9. Let a continuous linear operator T map an F-space X onto an F-space Y. Then PROPOSITION

A(Y):) A(X).

Proof Let {.An} E A(X). Let U be an arbitrary neighbourhood ofzero in Y. The set T- 1( U) is open. Then there is a .neighbourhood of zero V in the space X such that

A1V+A.2 V+ ... C T- 1(U). then

Locally Pseudoconvex and Locally Bounded Spaces

161

By the Banach theorem (Theorem 2.3.1) the sets T(V) are open. The arbitrariness of Uimplies that{A.n} e A(Y). D As an immediate consequence of Propositions 3.9.6, 3.9.7 and 3.9.9 we obtain THEOREM

3.9.10. If

dimz

X~

dimz Y

(codimz X~ codimz Y),

then A(X)~A(Y).

PROPOSITION 3.9.11. If X is a locally p-convex space, then

A(X) ~ lP. Proof Let {llxllm} be a sequence of p-homogeneous pseudonorms determining the topology. Let {xn} be a bounded sequence in X. Let

llxnllm·

Mn =sup n

Let co

C=

};

IA.niP ·

n=l

Let N

YN

=}; AtXt. i=l

Then co

IIYNiim ~}; IA.tiP llxtllm ~ CMm.

(3.9.7)

D

i=l

CoROLLARY 3.9.12 (compare Mazur and Orlicz, 1948). If an F*-space X is locally convex, then A(X) = [1, PROPOSITION 3.9.13. Let X be a locally pseudoconvex space. Then A(X)~ /P.

n

p>O

Chapter 3

162

Proof Let {A.J

n

E

[P.

Let {llxJim} be a sequence of Pm-homogeneous

p>O

pseudonorms determining the topology in X. Let w

Cm = } ; IA.tJPm • i=l

Taking Xn and YN as in the proof of Proposition 3.9.11 we obtain an estimation similar to estimation (3.9.7)

IIYNIJm ::=;:;: CmMm. 0

This implies the proposition.

Now we shall prove propositions converse to Propositions 3.9.11, 3.9.13. PROPOSITION 3.9.14. (cf. Metzler, 1967). If A(X)) [P, then the space X is locally p-convex. Proof Suppose that an F*-space (X, II JJ) is not locally p-convex. Then there is a positive number e such that for any m there are points {Xm, 1 , ••• ... , Xm, km} such that i

= 1,2, ... , km,

(3.9.8)

and (3.9.9) Let for n, k 1 + ... +km-1

1 < n ::=;:;: k 1 + ... +km, let An--=----=-=--.---2mkl/Pm m

2mxm,t, 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 Xn =

N

{YN}

= {

.2; Anxn} n=l

is not bounded. Therefore {..1.n} does not belong to A(X). Since {..1.n} E /11, A(X) does not contain /P. 0

Locally Pseudoconvex and Locally Bounded Spaces

n

PROPOSITION 3.9.15. If A(X) :::>

163

/P, then the space X is locally pseudo-

P>O

convex. Proof The proof follows the same line as the proof of Proposition 3.9.14 if we replace, in the definition of A.n and in formula (3.9.9), p by

Pm tending to 0.

D

As follows from Proposition 3.9.15, locally pseudoconvex spaces are stricly galbed. There are also non-locally pseudoconvex spaces which are strictly galbed, as follows from PROPOSITION 3.9.16 (Turpin, 1975). For any sequence {..1n} belonging to all spaces lP (in a particular case for any geometric sequence qn, 0 < q < 1), there is a non-locally pseudoconvex space X such that {A.n} e A(X). co

Proof Let Cn

= }; IA.tl 1/n. Let X be the space of all sequences x = {xn}. i=l

such that Xn e L 1fn[o, 1] and for all positive integers k c~

lim

llxnllvm[o,l] = 0.

We introduce a topology in X by a sequence of F-pseudonorms llxllk = s~pc!llxnllvrn 10, 11 • It is easy to verify that X is an F-space. The space X is not locally pseudoconvex. Indeed, let Xn = {x eX: Xi= 0 for i =F n }. The space Xn is isomorphic with L lfn[o, 1]. Take any neighbourhood of zero Uin X such that U::PXn, n = 1,2, ... By the definition of topology in X, this is possible. Then the modulus of concavity of U can be estimated as follows c(U)

~

c(UnXn)

~

c(Xn) = 2n,

n

= 1,2, ...

Thus c( U) = +oo. This implies that X is not locally pseudoconvex. On the other hand, let U be an arbitrary neighbourhood of zero in X. U contains a neighbourhood of zero U0 of the form U0

= {x: llxllk < e}.

Let V = {x: llxiiHI

< e}. co

Let JI, Y2• .•. E V, Yi = Yi, n• Let X = •

1: A.,y,. i=l

164

Chapter 3

Then 00

1\x\\ =

supc~ //}; AtYt, n//v,• n

i~1

~ supc~}; \A.t\ 11n 1\Yi, n\\L•I•[O,J] n

i~1

~ supc~+ 1 sup\IY«,n\\L•I•[o,1J n

Therefore x

E

i

U and {An}

E

<e.

A(X).

There are non strictly galbed spaces, as follows from 3.9.17. Let (D,E,p) be a measure space. Suppose that the measure f1 is not purely atomic (i.e. there is a set D 0 such that the measure f1 restricted to the set D 0 is a non-atomic measure which does not vanish identically). Then the space L 0(D,E,p) is not strictly galbed. Proof According to our hypothesis about the measure, we can find D 0 , O
+

11-+00

and 00

D0 C

U En

(3.9.11)

forK= 1,2, ...

n~K

Let {An} be an arbitrary sequence of positive numbers. Let n Xn

= An XEn•

By (3.9.10), {xn} tends to 0, and thus it is bounded. On the other hand, • the seque6ce N

YN(t) = };..tnxn(t) "~1

tends to infinity for

t E

D 0 • Thus it is not bounded.

3.9.18. Let X be the space l{P•}, 0 x = {xn} such that PROPOSITION

00

1\x\1 = }; \xn\P• n~1

< +oo,

D

< Pn ~ 1, of all sequences

Locally Pseudoconvex and Locally Bounded Spaces

165

with a topology given by the F-norm //xll. If p,-+0, then the space X is not strictly galbed. Proof Let {A.n} be an arbitrary sequence of positive numbers. Since p 11 ~0, for each n we can find an index k(n) such that A.~Jc > i-· We shall choose a k(n) such that k(n) < k(n+ 1). Let {~n} be a sequence of positive numbers such that {~~"} tends to 0, but the series }; ~~k is

n=l

divergent. It is obvious that the sequence {x11 } = {~11 e11 }, where {e11 } denotes the standard basis in l{Pn}, tends to 0. Thus it is bounded. On the other hand, N

1/YN/1

N

=112: Anxn// =}; 1/A.nXn/1 n=l n=l

+,}; N

~

n=l

~ 11Pk(n) ~00. N

Therefore the sequence YN =}; A.11 x 11 is not bounded. This implies

n=l

that {A.n} ¢ A(X).

D

3.10. 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 {t11 } tending to 0 the series 00

}; t 11 x 11

is convergent. Of course, a subsequence of a C-sequence is a

n=l

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 {x11 } of elements of an F-space (X, is a C-sequence if and only if the set 00

A= {}; anxn: /an/~ 1, n = 1, 2, ... } n=l

is bounded.

•

1/ //)

-Chapter 3

166

Proof. Sufficiency. Suppose that the set A is bounded. Let {en} be an arbitrary sequence of numbers tending to 0. Let Cn =sup JctJ. Of n~i

course, {C11 } also tends to 0. Let e be an arbitrary positive number. Since the set A is bounded, there is an index N 0 such that for n > N 0

JJCnxJJ <

(3.10.1)

B

for all x EA. Thus for n, m, N 0 < n < m,

since

Necessity. Suppose that the set A is not bounded. Then there are a sequence of numbers {tn} tending to 0, a sequence of numbers {an}, lanl ~ 1, and a sequence of indices {nk} such that the sequence n•+l { fk

};

GnXn}

does not tend to 0.

n=n•+l Let Cn

=

{~an

for nk < n ~ nk+I• elsewhere. 00

The sequence {en} tends to 0, but the series Therefore, the sequence

{xn}

2

CnXn

is not convergent.

n=l

is not a C-sequence.

D

An F-space X is called a C-space if for any C-sequence {Xn} the series 00

2

Xn n=l

is convergent. As an example of an F-space which is not a C-space

we can consider the space c0 (see Example 1.3.4.a). Indeed, the standard 00

basis {e11 } in c0 is a C-sequence, but the series

2 en is not convergent. n=l

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.I0.2 (cf. Bessaga and Pelczynski, I958). 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 c0 • 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 00

that the series

2

Xn is not convergent. Then there are an increasing

n~l

sequence of indices nk and a positive number /J such that IIYkll

> IJ, where

nk+t

Yk

=

.2.)

Xn.

n~n•+l

The sequence {Yk} is also a C-sequence, and it does not tend to 0. Proof of Theorem 3.10.2. If (X,

II II) is not a C-space, then there is a

0 C-se-

CD

quence {xn} such that the series

2 Xn

is not convergent. Basing our-

n~I

selves on Proposition 3.10.3, we can assume without loss of generality that the sequence {xn} does not tend to 0. Let {fk} be basis functionals corresponding to the basis {ek}. Since {xn} is a C-sequence, {fk(Xn)} are C-sequences for k = I ,2, ... Hence lim fk(Xn) = 0,

k =I, 2, ...

(3.I0.2)

.,_..00

By Theorem 1.2.2 we may assume that the norm II II is non-decreasing. Since {xn} does not tend to 0 and (3.10.2) holds~ we can find a subsequence {xflp} = {zp}, a positive number/Jand an increasing sequenceofindices {np} such that · '

1

llzp-Zpll < 2p IJ.

(3.I0.3)

Chapter 3

168

where n,+,

2,;

z~ =

fn(zp)en.

n=np+l

Corollary 2.6.6 implies that {z~} is a basic sequence. We sp.all show that it is equivalent to the standard basis of c0 • Let {tp} be a sequence of coefficients of x belonging to X 0 =lin {z~} with respect to the basis {z~}. Since //z~ll > 6/2, the sequence {tn} tends to 0. On the other hand, (zp} is a C-sequence. Therefore, by (3.10.3), {z~} is also a C-sequence. co

This means that if tn-+0, then the series}; tpz~ is convergent. Therefore, p=l

the basis z~ is equivalent to the standard basis of c0 • Thus, by Theorem 2.6.3, the space X 0 is isomorphic to the space c0 • D PROPOSITION 3.10.4. Let (X, II II) be a locally bounded space. with a basis {en}. Let Y be a subspace of the space X. Suppose Y is not a C-space. Then the space Y contains a subspace Y0 isomorphic to the space c0 • Proof Without loss of generality we may assume that I II is p-homogeneous. Using the same construction as in the proof of Theorem 3.10.2, we find elements {zp} and {z~}. By formula (3.10.3) and Theorem 3.2.16 the sequence {zp} is a basic sequence. The rest of the proof is the same as in Theorem 3.10.2. D Let X be a non-separable F-space. If Xis not a C-space, then it contains an infinite dimensional separable subspace X which is not a c~space. Indeed, if X is not a C-space, then there is a C-sequence {xn} such that 00

the series };

Xn

is not convergent. Let

X= lin{xn}. The space X has the

n=l

required properties. By the classical theorem of Banach and Mazur (1933), space C[O, I] is universal for all separable Banach spaces. The space C[O, 1] has a basis (Example 2.6.10). Thus we have PROPOSITION 3.10.5 (Bessaga and Pelczynski, 1958). If X is a Banach .space which is not a C-space, then X contains a subspace isomorphic to c0 • Now we shall prove

Locally Pseudoconvex and Locally Bounded Spaces

THEOREM 3.10.6 (Schwartz, 1969). The spaces LP(D,E,p), 0 ~ p are C-spaces.

169

< +oo,

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 0(D,E,p,) a C-sequence {x,.(t)} is given. Then on each set D 0 of finite measure the series 00

}; lx,.(t)l 2 convergent almost everywhere. n=l

Proof(Kwapien, 1968). Let {r,.(s)} = {sign(sin2n211s)} be a Rademacher system on the interval [0, 1]. Since under our hypothesis {x11 (t)} is a C-sequence, the set of elements

"

{ Yn,s(t)

=}; rt(s) Xt(t):

n = 1, 2, ... , 0

~ s ~ 1}

i=l

is, by Proposition 3.10.1, bounded in the space L 0 (D,E,p). This implies that for any positive B there is a constant C such that, for all n = 1,2, ... and all s, 0 ~ s ~ 1, ~t({te

D 0 : IYn,s(t)i ~ C}) ~ ~t(D0)-B.

(3.10.4)

Let n be fixed al!d let

At= {s: IYn,s(t)i ~ C}.

(3.10.5)

By the Fubini theorem and (3.10.4) we find that the set E of those t for which the Lebesgue measure IAel of the set At is greater than l-ye has a measure greater than ~t(D0)-ye, i.e. if E

= {t e D0 : jAel

~

2-y'e},

then ~t(E) ~ ~t(D 0)-ye.

Lett e E. Formula (3.10.5) implies n

JI}; rt(s) Xt(t)r ds ~ C

Ao

i=l

2•

Chapter 3

170

Thus n

J _2/xt(t)/ 2 -2 J ( _2 A, i=l

A,

Re(xt(t)x,(t))rt(s) r;(s)) ds

~ C2 •

(3.10.6)

l~i<j~n

Applying the Schwarz inequality to (3.10.6), we obtain n

C ;;?: (1-y'B)}; /xt(t)/ 2 2

i=l

-2 (

_2

/xt(t)x1(t)J2) 112 (

l~i<j~n

_2

( Jrt(s)r1(s)dsrf' 2.(3.10.7)

l~i<j~n

A,

Since rt is a Rademacher system, J rt(s) r;(s)ds = A•

(3.10.8)

J ri(s) r,(s)ds, 1'\..A•

where I denotes the interval [0, 1]. The system {ri(s)rJ(s)}, 1 ~ i <j, is an orthonormal system in the space £2[0, 1]. The Fourier coefficients of the characteristic function X['\..A• of the set /~At are equal to fri(s)r,(s)ds, thus equalto- Jri(s)rJ(s) 1'\..A•

A,

ds, by (3.10.8). Therefore, by the Parseval inequality. _2

(

l~i<j~n

f rt(s) r,(s)dsr ~/!"'At/ <{e.

(3.10.9)

A•

Observe that

( l,

/xt(t)x,(t)/ 2

l~i<j~n

n

= ( };Jxt(t)/ 2 i-1

tt =

f'

2

~

(

_2 l~i,j~n

/xt(t)x1(t)J 2

f'

2

n

(3.10.10)

};Jxt(t)/ 2 • i~l

Formulae (3.10.7), (3.10.9), (3.10.10) imply n

(1-y'e -2tJe) _2/xt(t)/ 2 ~ C 2 •

(3.10.11)

i=l 00

Formula (3.10.11) implies that the series 2;/xi(t)/ 2 is convergent on the i=l

171

Locally Pseudoconvex and Locally Bounded Spaces

set E. Since p(E) > p(.Q0)-yc, the arbitrariness of 'e implies that the D series is convergent almost everywhere on .Q0 • 3.10.8. Let Then {xn} tends to 0.

CoROLLARY

{x11 }

be a C-sequence in the space LO(.Q,E,p).

LEMMA 3.10.9 (Schwartz, 1969). Let {x11 } be a C-sequence in a space LP(.Q,E,p), 0
represent the set .Q in the form .Q

=

U

Dt u N, where, on the set N,

i~I

all functions x 11 (t) are equal to 0 almost everywhere, .Q~, C .Qi+I• and for each i the sequence of functions {x 11(t)} tends uniformly to 0 on Dt. Suppose that the sequence {x 11 (t)} does not tend to 0 in LP(.Q,E,p). Then there is a positive number b' such that, for infinitely many n, J/xn(t)/Pdp

> b'.

Q

Let 0 < b < b'. We shall construct by induction a sequence of disjoint sets {Ak}, k = 1,2, ... , and increasing sequences of indices {nk}, {mk} such that Ak C Dm. and

J/x .(t)/P dp > b, 11

A•

(3.10.12)

forj=/=k.

Let n0 be an arbitrary index such that

J /x D

is an index m 0 and a set A 0 C Dm0 such that

110

(t)/ Pdp > b. Then there

J /x

110

(t)/Pdp >b. Suppose

A,

that Ak, nk, mk are defined for k

=

0,1, ... , r-1. Then there is an index

m~ and a set Br C Dm'T such that Ak C Br for k

k

= 0, 1, ... , r-1 and

= 0, 1, ... , r-1.

Since Br C !Jm'T and the sequence {x11(t)} tends uniformly to 0 on !Jm',T

172

Chapter 3

there is an index n,.

> nr-1 such that

Jlxnlt)l dp ~ 2~,

Br

and

J lxnr(t)IP dp >b.

(3.10.13)

t:J"-Br

Formula (3.10.13) implies that there is an index m, and a set A, C Qmr disjoint with the sets A 0 , ••• , Ar-1 such that

Jlxnr(t)IPdp >b.

(3.10.14)

Ar

In this way we have constructed by induction the required sequence. Let {en} be a sequence of positive numbers tending to 0 such that co

,2;c: = +oo.

(3.10.15)

n=1

Then r

r

r

f\2: CkXn.(t)\P dp ~ 2 J12: CkXn.(t)\p dp (J

k=O

j=O A1

r

k=O

r

J(cJixnlt)IP- L cflxn.(t)IP)dp

~};

j=OA1

k=O k#j

r

~b

2

j=O

r

cJ- C

2 2;~k

j,k=O

r

~ b}; cJ-4C,

(3.10.16)

;~o

where C =sup c!. n co

Then, by (3.10.15), the series}; Ck xn. is not convergent in LP(Q,E,p). k=1

Therefore, {xn} is not a C-sequence and we obtain a contradiction.

0

LEMMA 3.10.10 (Schwartz, 1969). If {xn} is a C-sequence in a space LP(Q,E,p), 1 ~p <+oo, then Xn-+0.

Locally Pseudoconvex and Locally Bounded Spaces

173

Proof The proof is the same as the proof ofLemma3.10.9, but in (3.10.16) the Minkowski inequality is used. D Proof of Theorem 3.10.6. The proof is a trivial consequence of Proposition 3.10.2 and Lemmas 3.10.9, 3.10.10. 0 3.10.11. The space LD[O, 1] is not universal for separable F-spaces. Proof The space L 0 [0, 1] is a C-space. Hence it does not contain a subspace isomorphic to c0 • D CoROLLARY

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

A0 = {

.2;

Xpn:

Pn runs over all finite increasing systems of

n=l

positive integers} is bounded. An F-space X satisifies 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·: xxx~x, called multiplication, which is associative (x·y)·z

=

x·{y·z),

x,y, zeX

(3.11.1)

174

Chapter 3

and bilinear (x1+x2)·y = X1·y+x2·y,

(3.11.2)

X·(JI+Y~

= X·Y1+X·Y2

(3.11.3)

= t(x·y) = x·(ty)

(3.11.4)

(tx)·y

for x,y E X, t being a scalar. For brevity we shall write x·y = xy. We say that an algebra X is commutative if xy

=

yx.

(3.11.5)

We say that an algebra Xhas a unit ifthere is anelemente EX such that ex= xe

= x.

(3.11.6)

It is easy to verify that the unit is unique. If an algebra X has a unit, then (3.11.4) automatically holds. In the sequel only commutative algebras with a unit will be considered. For brevity we shall call them simply algebras. We say that an algebra X is an F*-algebra if it is an F*-space and the operation of multiplication is continuous. We say that an F*-algebra X is locally bounded if it is a locally bounded space. The theory of locally bounded algebras was developed by Zelazko (1960, 1962, 1962b, 1963, 1965). It is a generalization of the classical theory of Banach algebras. THEOREM 3.11.1 (:Zelazko, 1960). Let X be a locally bounded algebra. Then the topology in X can be determined by a p-homogeneous norm which is submultiplicative, i.e.

llxyll ~ llxiiiiYII·

llxll

(3.11.7)

If X is complete, then it is also complete with respect to the norm II 11. Proof By Theorem 3.2.1 the topology in X can be determined by a p-ho-

mogeneous norm

llxll

=

Observe that

llxll'. Let llxyll'

~1o TYIT''

llxll is finite, since the multiplication is a continuous oper-

Locally Pseudoconvex and Locally Bounded Spaces

ator (cf. Theorem 3.2.13). It is easy to verify that submultiplicative and [fell = 1. Since

175

llxll is p-homogeneous,

Jlxll ?: Jjxjj' lle!l' the norm llxll is stronger than the norm llxJ['.

(3.11.8)

Now we shall identify x e X with a continuous linear operator L.e mapping X into itself by formula Lxy = xy. Observe that the norm operator IILxll of the operator Lx is equal to llxll. Thus we can consider (X, II JJ) as a subspace of the space (B(X), II JJ). If { Xn} tends to x in the space (X, II II'), then, by the continuity of multiplication, Lx,. tends to Lx in (X, II II). Thus the norms II II and II II' are equivalent. D Let X be an algebra. An element x e X is called invertible if there is an element x- 1 e X, called inverse to x, such that (3.11.8) It is easy to verify that x- 1 is uniquely determined (provided it exists) and that (x- 1)- 1 = x. PROPOSITION 3.11.2. Let (X, II II) be a complete locally bounded algebra. Let x be an invertible element in X. Then there is a neighbourhood of zero U such that, for y e U, the element x-y is invertible. Proof Without loss of generality we may assume that II II is p-homogeneous and submultiplicative (see Theorem 3.11.1). We can write co

(x-y)-1

= x-1(e-.x-1y)-1 = x-1), (x-Iy)n, ..:....; n=O

where the series is convergent provided

IIYII < - 1-1 . llx- 11

D

Let X be an algebra. A set M C X, {0} =I= M =I= X is called ideal if it is a linear subset of X and for all x e X, xM C M. Let X be an F-algebra, by the continuity of multiplication we conclude that for any ideal M its closure M is also an ideal provided that M =I= X. We say that an ideal M C X is maximal, if for any ideal M1 such that M C M 1 we have M = M 1 •

Chapter 3

176

PROPOSITION 3.11.3. Let X be a complete locally bounded algebra. Then every maximal ideal is closed. Proof Let M be a maximal ideal in X. By Proposition 3.11.2, e ¢ M. Thus M is an ideal. Since M is maximal and M C M, M = M. D PROPOSITION 3.11.4. Let X be a complete locally bounded algebra over complexes. The function (x-.A.e)- 1 is analytic on its domain. Proof Let A.0 belong to the domain of the function (x-.A.e)- 1 • This means that the element x-A.0e is invertible. Let y = (A.-A. 0 )e. Then (x-A. 0 e)-y = x-A.e. Therefore (x-A.e)- 1

=

((x-A. 0 e)-y)-1 CXl

= (x-A. 0 e)-1 } ; (x-A. 0 e)-n(A.-A.0)n, n=1

where the series on the right is convergent provided

JA.-A.oJ ~ JJ(x-A.oe)-1 11-1 • Thus (x-A.e)- 1 is analytic.

D

Proposition 3.11.4 implies an extension of the theorem of Mazur and Gelfand (Mazur, 1938; Gelfand, 1941). THEOREM 3.11.5 (Zelazko, 1960). Let X be a complete locally bounded field over complex numbers. Then X= {A.e: A. being a complex number}. Proof Suppose that there is an x E X which is not of the form .A.e, i.e. x -=1= A.e for all A.. Since X is a field (x-A.e)- 1 is well defined on the whole complex plane. By Proposition 3.11.4 it is an analytic function. Observe that lim (x-A.e)-1 = 0 . .t.-..oo

Thus, by the Liouville theorem (Theorem 3.5.4), (x-A.e)- 1 = 0 and we obtain a contradiction. D Let X be a complete locally bounded algebra over the complex numbers. Let f be a linear functional defined on X (not necessarily continuous). We say that the functional/is a 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 M1 = {x: f(x) = 0}

is an ideal. It is a maximal ideal since it has codimension 1. Since X is locally bounded, it is closed by Proposition 3.11.3. Thus the functional f is continuous. Thus we have proved PROPOSITION 3.11.6 (Zelazko, 1960). In complete locally bounded algebras all 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)

and there is a C N(uv)

> 0 such that for sufficiently small u,

~

(3.11.10)

N(u)+N(v)

CN(u)N(v)

v

(3.11.11)

Chapter 3

178

> 0 such that

and there is a p

N(u) = N 0(uP),

(3.11.12)

where N 0 is a convex function in a neighbourhood of zero. Let X= N(l) be the space of all sequences x = {x0 ,x1 , that

••• }

such

co

llxll = PN(X) =

J: N(lxni) < +oo.

n=O

The space (X, II II) is an 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 = {Yn} then we define n

X·y

=

{

.2: XkYn-k}·

k=O

By (3.11.10) and (3.11.11) we conclude that for sufficiently small xandy (3.11.13) llxyll ~ C llxiiiiYII. Formula (3.11.13) implies that the multiplication is continuous. Thus X is a complete locally bounded algebra. Now we shall give examples of functions satisfying conditions (3.11.10)-(3.11.12). The simplest are functions N(u) =uP, 0