I nt. J. Math. Mh. S.
(1980) 42-42
No.
Vol.
423
GENERALIZED KTHE-TOEPLITZ DUALS
I.J. MADDOX
Department of Pure Mathematics
Queen’s University of Belfast
Belfast BT7 INN
Northern Ireland
(Received November 9, 1979)
.ABSTRACT.
The
where O < p <
.
and B-duals spaces of generalized
The question of when the
and
P
spaces are characterized,
B dual spaces coincide is
also
considered.
KEY WORDS AND PHRASES. Genzed Kthe-Toeplitz dual spaces, Sequences of
op Giz spaces.
p
lin
19gO
MATHEMATICS SUBJECT CLASSIFICATION CODES.
I.
INTRODUCTION.
40C05, 40J05.
X and Y denote complex Banach spaces with zero elements O, and
denotes the norm in either X or Y.
The continuous dual of X is written X*.
By s(X) we mean the space of all X-valued sequences x
for k e N
(1,2,3,...}.
II" II
(Xk)
where x e X
k
424
I.J. MADDOX
,
If O < p <
X
we mean by
p
IXkl
(xk) such that Z
<
P
.
(X) the space of all X-valued sequences
Sums are over k
N, unless otherwise
E
indicated.
By
(X)
(xk) such that sup
we denote the space of all x
C, the space of complex numbers, we write
In case X
P
,P (c).
Let A
(Ak)
lXkll
<
.
instead of
denote a sequence of linear, but not necessarily bounded,
If E is any nonempty subset of s(X) then the e-dual
operators on X into Y.
of E is defined to be
E
The
{A
Z
B-dual of E is defined
EB
Since Y is complete
{A
IAkXklI < ,
E}.
for all x
to be
converges, for all x
ZAkXk
we have E
EB
a
The a and
B
E
E}.
duals of E may be regarded
as generalized Kthe-Toeplitz duals, since in case X
be identified with complex numbers
ak,
Y
C, when the Ak may
the duals reduce to the classical
spaces first considered by Kthe and Toeplitz [I].
Using the notation (l/p) + (l/q)
convention that q
i, where I < p <
I, and q
when p
,o B
P
P
i when p
B(X),
P
<
I the
and
I < p
<
,
the a and
it is well-known that
(i.I)
E(X)
P
However, when O < p
the
,q
We shall see that, in general,
strict.
,
, with
B
where the inclusion may be
duals coincide.
B duals coincide provided
Also, when
that Y is finite dimensional.
GENERALIZED KOTHE-TOEPLITZ DUALS
425
CHARACTERIZATION OF THE DUALS.
2.
EB(X)
p
Then A
Let 0 < p < I.
THEOREM i.
N such that A is bounded, for all k > m,
k
m
H
PROOF.
mllAkl
<
suPk_>
and
.
(2.1)
Let (2.i) hold and Z
Sufficiency.
if and only if there exists
IXkl
p
<
.
By a familiar
inequality, see for example Maddox [2], page 22,
P
k=m
XAkXk
Hence
Necessity.
k=m
_<
p
II
kllPllxkll
k--m
<
HP El l kl
p"
is absolutely convergent, and so convergent.
Let A
B(X)
p
and suppose, if possible, that no such m exists.
Then there are natural numbers k(1) < k(2) <
that for i
l
IXkl
that
lAkXk
zi/i
2/6,
2/p for k
< i
such
(2.2)
0 otherwise.
>
Then x e
gp(X)
since
I for infinitely many k, contrary to the fact
converges.
IAk( )ll
X with
suPk>m
l kll
Then there are natural
with k(1) > m such that for i
numbers k(1) < k(2) <
z.x
lzill
2/p.
k(i) and xk
IAkXkll
but
> i
Now suppose, if possible, that
Choose
X
N,
lAk(i) -ill
Define x
k
p <
and z.
lzi It
> 2i
N,
2/p.
< i such that
(2.3)
2] IAk(i)zil
_>
IAk(i)ll
so by (2.3)
I.J. MADDOX
426
we see that (2.2) holds with the new k(i) and z..
I
We may define x
E (X)
p
Hence (2.1) must hold, and the proof
as above and obtain a contradiction.
is complete.
If we examine the proof of Theorem I we see that in the sufficiency we
had l
llXkll ,
e(X).
so that A
<
p(X)
constructions involved x
Also, in the necessity, the
P
such that l
IAkXkll
was divergent.
Hence we
have:
THEOREM 2.
If 0 < p < i then
S(x).
P
=(x)
P
_
Next we consider the case i < p < =.
THEOREM 3.
Let I < p < =.
is bounded for all k
m e N such that
M
PROOF.
Then A e
Ea(X)
p
if and only if there exists
m, and
q
[IAkll
k=m
(2.4)
Let (2.4) hold and x e
Sufficiency.
P
(X).
By Hlder’s
inequality,
[lXkl[
k=m
Necessity.
Since
a(X)p
-< M I/q
1/p
(El IXkl
E l(X) when p > I, the existence of the m in the
theorem follows from Theorems I and 2.
Now for k > m we may choose z k
For all %
p we have
__(%kZk
lZkl
X with
p(X),
so
<
such that
GENERALIZED KOTHE-TOEPLITZ DUALS
427
k=m
for all
By (i.I) it follows that
P
q
,
llAkZkll
k=m
H
2qH,
whence M _<
<
so (2.4) holds, and the proof is complete.
Let I < p <
THEOREM 4.
.
8(X)
p
Then A
if and only if there exists
m e N such that A is bounded for all k > m, and
k
sup
k=m
where the
lfl
q
<
supremum is over all f
PROOF.
,
(2.5)
Y* with
lfl]
-<
.
With the restriction that all the A are bounded, and with
k
different notation, this result was proved by Thorp [3].
Only the existence
of m in the necessity needs attention, and this follows from Theorems 1 and 2,
8(X)
P
and the fact that
c
18(X).
Finally, we examine the case p
.
The proofs are left to the reader.
We remark that with the restriction that all the A are bounded, the result
k
concerning
8(X)
THEOREM 5.
was given by Maddox [4].
A
(X)
if and only. if there exists m e N such that A
k
is bounded for all k > m, and
(2.6)
k=m
THEOREM 6.
A
e
8(X)
if and only if there exists m
is bounded for all k -> m, and
N such that A
k
I. J. MADDOX
428
m+n
Z
supl
AkXk[
k=m
< o,
(2.7)
m+n
sup
where the
3.
I AkXkl
ll k--m
suprema
/
O (m + ),
are over all n > O and all x
k
(2.8)
X with
lXkl
_<
1.
COINCIDENCE OF DUALS.
It was shown in Theorem 2 that, when O < p
i,
8(X)
P
a(X)
P
for any
Banach spaces X and Y.
We next shown that, when I < p <
,
the inclusion
a(X)
P
c
B(X)
P
may be
strict.
THEOREM 7.
(X)
P
B(X)
P
PROOF.
If 1 < p <
then there are Banach
space.s X
with strict inclusion.
Take X
Y
and write
P
e
(O,O,...,i,O,O,...)
k
where I is in the k-place and there are zeros elsewhere.
operators A on
k
P
(xk)
Define bounded linear
into itself by
AkX
for each x
and Y such that
P
xkek
Then
I[Akll
1 for all k
N
so A is not in
aCX)
p
by Theorem 3.
Let us now show that (2.5) holds.
for x
P
Take any f
we have
f(x)
7.f.x.
1
1
E* with
P
lfll
<
I.
Then
429
GENERALIZED KOTHE-TOEPLITZ DUALS
for some
(fi)
llfi lq
such that
f(AkX)
(Af) (x)
and so
llAfll--Ifkl.
ZllAfl
A,
fkXk
Hence
q
Zlfk lq <
I,
8(X).P
so by Theorem 4 we have A e
Still with the case 1 < p <
THEOREM 8.
Hence, by definition of
< i.
we have:
and Y is finite dimensional then for any X
If 1 < p <
we have
a(x)
P
PROOF.
A e
8(X)
P
t
8(x).
P
We have to show that A
(X)
P
then by Theorem 4 there exists m
all k e m.
implies A e
Now if
N such that b is bounded for
K
Suppose Y has finite dimension n and that
Hamel base for Y.
a(X).
P
(bl,b 2
,b n) is a
Then y e Y implies
n
Z Xi(Y)bi
i=l
y
Take z e X and k > M.
where each %. e Y*.
Then
n
r.
Akz-and h
i
Ak
X*.
i=l
Z
Since
convergent for all x
AkXk
converges for all x
k=m
k---m
P
(2.9)
i(AkZ)bi
Ak)Xk
(X) and each i.
P
(X) we have
I.J. MADDOX
430
Choose z
If t
k
then
P
e
lZkl
X,
_(tkZ k)
gp(X)
so that
k(x i
Ak)Zk
V. t
k--
-< I such that
g
converges for all t
21(I i
Ak)Zkl
>
II
whence for each i,
P
kll q
ll i
k--m
(2.10)
<’-
By (2.9) and HDlder’s inequality,
n
n
i=l
i=l
q/p.
(2.11)
Denoting the final term in (2.11) by H,
llll
k=m
n
q _<
v.
i=l
[li
k=m
ll q
(2.12)
It follows from (2.10) and (2.12) that (2.4) holds, so by Theorem 3 we have
a(x).
P
A
For certain values of p, and any X, the next result is the converse of
Theorem 8.
THEOREM 9.
If 2 < p <
a(X)
p
and
B(X)
P
then Y must be finite
dimensional.
PROOF.
q <
Suppose, if possible, that Y is infinite dimensional.
-2/q then Ec < =.
2, if c k -k
k
By the Dvoretzky-Rogers theorem [5],
there exists an unconditionally convergent series Ey
lykll
2
c
k
for k
N.
Then
X* with
lAkll
lykl
fll
k
in Y such that
Hence
r[l Ykll q
Take f
Since
x
(2.13)
diverges.
and define rank one operators A
k
so by (2.13) and Theorem 3
A is not in
Yk )
a(X).
p
f"
431
GENERALIZED KOTHE-TOEPLITZ DUALS
Now if x
P
(X) then
ZAkXk If(xk)Yk.
But
(f(xk))
and ly is unconditionally convergent, so that
k
pB(X),
converges, whence A
Zf(xk)Yk
which gives a contradiction.
We remark that it would appear that the argument of Theorem 9 cannot be
2, since in a general Hilbert space Y the unconditional
2
convergence of Zyk implies that l
used in the case p
lykl
2 of Theorem 9 when Y is a Hilbert
However, we can deal with the case p
space:
Let Y be a Hilbert
THEOREM IO.
space
and suppose 4
(x).
2(X)
Then Y must be finite dimensional.
Suppose, if possible, that Y is infinite dimensional.
PROOF.
orthonormal sequence (e
X*,llgll
Take g
IAkll
I.
k)
in Y and denote the inner product in Y by
I ad define rank one operators Ak
Now let f
lfll
Y* with
f(z)
for all z
Y, with
IlYll
I[Afl[
-<
l(ek,Y) I,
I for all k.
k
@
(z,y)
Then for x
g(x)
X,
(ek,Y).
so by Bessel’s inequality,
.
2, and so A
B (X).
But A
2 (X)
This contradiction implies our result.
(yl,Y2).
g, so that
Then there exists y
.][fll 2_< ]lyll2_<
Thus (2.5) holds with q
[IAkl]
< I.
(g(x)ek,Y)
(Af) (x)
Hence
llfll
I.
_<
e
Choose an
since
Y such that
I.J. MADDOX
432
The case p
(X) (X)
is due essentially to Thorp [3], who shows that
if and only if Y is finite dimensional.
REFERENCES
I.
Kthe, G. and Toeplitz, O. Lineare Rume mit unendlich vielen
Koordinaten und Ringe unendlicher Matrizen, J. reine angew.
Math. 171 (1934), 193-226.
2.
Maddox, l.J.
Elements of Functional
Analsis, Cambridge University
Press, 1970.
3.
Thorp, B.L.D. Sequential-evaluation convergence, J. London Math.
Soc. 44 (1969), 201-209.
4.
Maddox, l.J. Matrix maps of bounded sequences in a Banach space,
Proc. American Math. Soc. 63 (1977), 82-86.
5.
Dvoretzky, A. and Rogers, C.A. Absolute and unconditional convergence
in normed linear spaces, Proc. Nat. Acad. Sci. (U.S.A.). 36
(1950) 192-197.