A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
S TEPHEN V ÁGI
A remark on Plancherel’s theorem for Banach space valued functions
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 3
esérie, tome 23, n
o2 (1969), p. 305-315
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FOR BANACH SPACE VALUED FUNCTIONS
STEPHEN VÁGI
1. Introduction.
Convolution
operators acting
on Banach space valued functions have turned out to be a useful instrument inproving
Lpinequalities,
andhave found a nainber of
applications
in classicalanalyisis ([1], [9], [10]).
Omitting
what is not relevant to thepresent
purpose one can sum-marize this
theory,
which runsparallel
to that of the scalar case, as fol- lows :A)
One has to prove that one’soperator
is bounded on Lpo for sdmeparticular
B)
This information is then used to show that theoperator
is ofweak
type (1,1 ).
C) Finally,
theinterpolation
theorem of Marcinkiewicz allows to conclude that it is bounded on allLp’s,
1~ po .
This note is concer-ned with
Step
A.In the classical case, i. e., when
dealing
with scalar functions thisstep
is carried outby taking
po = 2 andusing
Plancherel’s theorem. This isusually quite
easy and seems to be theonly
effectiveprocedure
to handlethis
step.
The same
approach
succeeds also in theapplications
which the vector method has found so far. This is so because in every case the Banach spaces which contain the values of the L2 functions one isworking
withturn out to be Hillbert spaces, and Plancherel’s theorem does hold for Hil- bert space valued functions. Now it is remarkable that
Steps
B andC,
which
embody
the difficult real variableaspects
of thetheory
and whichaccount for its power, in no way
depend
on the fact that these spacesPervenuto alla Redazione il 12 Novembre 1968.
are Hilbert spaces. To sum np :
Extending
the method from scalar to vector functions does not effect B andC, however, Step
A remains easyonly
ifPlancherells theorem is available for vector valued .L2 functions. The pre-
ceding
observations are intended to motivate the interest inasking
whetherthat theorem - in a sense to be made
precise
below - does hold forquadratically integrable
functions with values in ageneral
Banach space, i. e., one which is notlinearly homeomorphic
to a Hilbert space.The purpose of this note is to
provide
apartial
answer to the abovequestion.
Italways
turns out to benegative
if the Fourier transform isrequi-
red to be an
isometry
ofL2 ;
without this restriction it isnegative
for aclass of Banach spaces which includes many of those
currently
used inclassical
analysis.
Whether the answer isalways negative,
is not known.2.
I)efituitious, Notations,
and Statement of Results.J57 will be a
complex
Banach space withnorm || ||
and dual E’. Forwill denote the Banach space of
E-valued, strongly
mesurable functions defined on the real line B for which
Integration
in thepreceding
formula andeverywhere
else in this paper is withrespect
toLebesgue
measure. General references onanalysis
inBanach spaces are the treatises
[3]
and[4].
For the Fouriertransform
of f
is definedby
the usualintegral
formulaNote
that Ff
is a bounded continuous function which vanishes atinfinity.
Also,
note that Ll(R, E) n
L2(R, E)
is dense in L2(R, E)
Plancherel’s theorem in this context is the statement thatfor f E
L1jR, E) (R, E) 9f belongs
toL2 (R, E)
andthat J
can be extended to a bounded linear map of into itself. If this is thecase 9
will be said to the extendable(for E).
It will followeasily
that if9
isextendable,
it is alinear
homeomorphism
ofL2 (R, E)
onto itself.The results to be established are the
following:
THEORElBI 1.
If
can be extended to anisometry of L2 (R, E)
ontoitself,
then E - itsoriginal
norm - is a Hilbert space.THEOREM 2.
If 7
can be extended to a bounded linearoperator of L2 (R, E)
into
itself
and oneaf’
the conditionsi, ii,
iii listed below issatisfied,
then Eis
linearly homeomorphic
to a Hilbet"t space.i)
E has an unconditional Schauder basis.ii)
Thedual,
or an interated dualof E,
has an unconditional Schauder basis. °iii)
E is thedual,
or an iterated dualof
a Banach space, wlcich hasan unconditional Schauder basis.
Theorem 2 can be used to prove the non
extendability
ofJ
forconcrete spaces E. A number of such results are obtained in Section 4.
3.
Proofs.
Before
proving
the theorems a fewsimple properties
of the Fourier transform have to be established. Denoteby cS
the E-valuedinfinitely
often differentiable functions on R which decreaserapidly
atinfinity ;
i. e., such that(1-~- ~ I x Ik) (x) 11-+
0 forIxl --~
oo, and for all nonnegative intergers k,
l.Exactly
as in the scalar case, one shows thatc5
is dense inLP(R, E); 1
~S p
oo, andthat 97
mapscS bijectively
onto itself. DefineJf
forf E
E Li
(R, E) by (~’f ) (~)
=(-~). Again
it followsexactly
as in the scalar casethat
Trestrieted
tocS
is the inverse ofJ
restricted toc3.
If97
can be extended to a bounded linear map ofL2 (R, E)
intoitself,
then so can7,
and thetwo
extensions,
also denotedby ~’
and~’,
are inverses of oneanother ;
i. e., is a linear
homeomorphism
of L2(R, E)
onto itself. The norms of5
and~’
areequal
and cannot be less than one. The latter fact one seesby considering f
= ug with u E E and g a scalarL2
function. If the normof is one,
then 7fi
is anisometry.
It is clear that is extendable if E islinearly homeomorphic
to a Hilbert space.The
proofs
of both Theorem 1 and Theorem 2 arehinged
on a lemmawhich will be
proved
next. -LEMMA. 1. Let
‘~
be extendablefor
E. Let k be apositive integer,
n2 , ..., nk distinctintegers,
and ’ltt, U2(not necessarily distinct)
ele-ments
of
E. Thenlchere c denotes tlae
of’ ~’,
PROOF : Let N be a
positive integer.
Define a functionIN by
where y is the characteristic function of the open interval
(0, 1). fN is
asimple
^^
function,
its Fourier transformfN
isand
fN
E L2(R, E).
Theextendability
ofy
is not needed to show this :.11
and its factor in the above formula is bounded. The
L2
norm of~~
IN is given by
Consider now the
function g
definedby
g is
periodic
ofperiod
one, andLipschitz
continuous.Consequently,
it hasa Fourier series which converges to it
everywhere,
and the convergence is absolute and uniform([11]
Vol.I).
Denote itscomplex
Fourier coef6- cientsby
- oo n oo.By
theLebesgue
dominated convergencetheorem
(2)
becomes .The
Riemann.Lebesgue
lemma for Fourierintegrals
shows that for N appro-aching infinity
all the terms in the above series whichhave n =4= 0,
tendto zero. Since each of the
integrals
in this series is less than orequal
toone in absolute value, and
since 03A3|cn I
converges, it follows that(3)
tendsto co . * However
The lemma now follows from this
by observing
thatis
independent
ofN,
and that7
is extendable.PROOF oF THEOREM 1.
Let u, v
E ~ In Lemma 1 set k =31 ut = U
u2 = u3 = v ; n~ =
0,
n2=1, n3
= -1. SinceJ is
assumed to be anisometry (1)
becomesReplacing v by - v
in thisformula, adding
the two identities thus obtai- ned andfinally defining
afunction 1p
of the real variable Aby
one obtains
1p is
readily
seen to be a convex,non-negative,
and even function of A.An easy
computation
shows that theintegrand
of(6)
issymmetric
withrespect
to thepoint
0 =1/4 . This,
the evenness, and theperiodicity
of theintegrand
allow to rewrite(6)
in the formLet t be
non-negative ; introducing tv
instead of v in(7),
one findsthat
satisfies the
integral equation
Computation
shows that ~o(1)
=8 ~~ ro I~
12 is a solution of(8). By
thechange
of variable t cos
2~c 9 = Vs (8)
transforms into theequivalent equation
Note
that 11’,
"Po’ and theirtransforms 9?
and arelocally integrable
on[0, oo).
To show that 990 is theonly
solution of(9)
inthis class of functions
and, hence,
"Po theonly
solution of(8),
recall thefollowing
fact([12],
p.323):
Iff and g
arelocally integrable
on[0, oo) and g
is notidentically
zero, thenimplies
thatf (x)
= 0 at almost every x in[0,
Since the kernel of(9)
is
locally integrable,
it now follows from the above theorem and the linea-rity
of(9)
that 4po is itsonly
solution.Introducing
this information into(5),
one obtains
Setting A= -
2 in(10)
onefinally
hasi.e.,
the norm of E satisfies theparallelogram identity ;
in otherwords,
itis an inner
product
norm. Theorem 1 isproved.
Some further
preparation
is needed before theproof
of Theorem 2 canbe taken up. A Schauder basis of a Banach space .E is an unconditional basis if the series
giving
theexpansion
of anarbitrary
element of the space isunconditionally convergent, i.e.,
ifrearrangement
does not affect its convergence or sum. It is known[5]
that a sequence of elements of .~ which span a densesubspace
is an unconditional basis of B if andonly
if there exists M>
0 such that for any finite subsetsA,
B of theintegers
such that Ba A,
and any choice of y one hasThe
following
lemma will also be needed.LEMMA 2. Let ..4. be a
finite 8et, (Ua)aeA
afamily of
elementsof E,
{k«)«E d a farrtily of
scalars. ThenThis
,lemma
has been announced in a moregeneral
context in[8].
For thesake of
completeness
aproof
will be included for thespecial
case usedhere. If .E is the field of
complex
numbers(11)
iseasily proved by
separa-ting
the left hand side into its real andimaginary parts.
Let nowf
be acontinuous linear functional on E. The left hand side of
(11)
is the supre-mum of
taken over
all f’s
whose norm does not exceed one.Using (12)
for the scalarcase, one has
But if then
(13)
and(14) together
prove(12).
PROOF OF THEOREM 2 UNDER THE ASSUMPTION
i):
It is no restriction ofgenerality
also to assume that = 1 for all n. LetEo
be the densesubspace
of Espanned by
thebasis; i.e., Eo
consists of all finite linearcombinations of
en’s.
Define a linear map T :jE~ - 12 by Ten
where is the standard orthonormal basis ofl2.
. Tmapg Eo injectively
ontoa dense
subspace
of 12.k
Let now
Apply
Lemma 1 to the vectors =1,2,...,k,
j=1
and
arbitrary
distinctintegers
n, , ... , nk to obtaink k
,I =
11,
2, ...,/.,)
Lem to,
e. and to 03A3k
Let A =
( 1 ,
, 2, ... , ;i,ndapply
Lemma 2 to Zli
ej and to Zlj
ee,j=l
°
j=l
by setting
U1 =Aj
ej, =0in
the first case, and e; ,kk
=--- in the second.
lly (11)
These two
inequalities
combined with(15) give
(16)
shows that T can he extended to all of /fJ as a bounded linear map, that it has it boundedinverse,
andconsequently
a closed range. Sincel’ (Eo)
is dense inl2,
Tmaps E
ontol2, i.e.,
T is a linearhomeomorphism
ot’ E 12.
Theorem
2, assuming
ii oriii,
is a direct consequence of thefollowing
;;. Let F n,ncl G be B a bounded bilinear
form
on F x G ivhich i,g
determinig jor
both F and G.Then 7
is extendablefor
Fif
andonly
it isextendable foi-
G.OF LEMMA 3. Let
f E L2 (R,
F), g E L2(R, G).
Introduce( , )
is a bounded bilinear form on L2(R,
h’i x, L2(R, G),
and norm deter-mining
for hoth1,2 (R, F)
and L2( R, G), i.e., tor f E
EL2 (R, G)
This is
easily
checkedby
a minimal modificationof the argument given
in[9]
to prove Lemma S of that paper, which statesessentially
the same fact.7
can be defined on and ForIE c5 (R, F)
andf/ E c5 (R, G)
it isreadily
verified thatNow if
y
is extendable for F it has a boundedtranspose
withrespect
to( , ). By (17)
thistranspose
coincides on the densesubspace cS
ofL2 (B, G)
with
9, i.e., 9
definedon c5
is extendable toL2 (R, G).
Since .F and G en-ter the statement of Lemma 3
symmetrically,
this concludes theproof
ofthat lemma.
To prove Theorem 2
assuming,
say conditionii,
let be the iterated dual of B which has an unconditional basis.If 9
is extendable forE,
it isextendable for .B’’
by
Lemma3, B being
the bilinear formgiving
theduality
betwen E and E’.
steps
oneextends 9
toEBn),
this space then satis- fies condition i of Theorem 2 andis, therefore, linearly homeomorphic
to aHilbert space ; this
implies
that E itself islinearly homeomorphic
to a Hil-bert space, too. It is clear nov, how to prove Theorem 2
assuming
condi-tion iii.
_
4.
Resuls
onNon-Extendability.
In this
section,
thenon-extendability of 7
for a number ofspecific
Ba-nach spaces will be shown. These
examples
will illustratesufficiently
howother cases
might
be dealt with. The method will consist inusing
Theorem2 in combination with the
following simple
LEMMA 4. Let E and F be Banach spaces.
Let 7
be non-extendablefor I’ ;
then it is non-extendablefor
Eif
eitherof
thefollowing
two conditionsJrolds.
i)
E contains a closedsubspace
whiclz islinearly homeomorphic
to F.ii)
.E can belinearly
andcontinuously
onto F.The statement is obvious in Case i. In Case
ii,
it followsby duality
from Theorem 2 and Part i of Lemma 4.
a) Sequence
spaces. Theorem 2implies
that if E has an unconditional basis and is notlinearly homeomorphic
to a Hilbert space, then7
is not extend- able.Let c, c0
and1P
1 1 p oo denote the familiar sequence spaces.Except
for l°° all there spaces have obvious unconditional bases.Now l p , p #
2 is notlinearly homeorphic
to a Hilbert space([2],
p.120)
Soby
theabove
remark, 7
is not extendable#
2. Since l°° =(l1)’
and
(co)’= l1,
nonextendability
follows for 100 and co from Theorem 2. Fi-nally,
ccontains co
as a direct summand.b)
L P spaces. Let(8, I, p)
be a measure space, thenJ
is non-extend able for LP(S, E, /z),
1p
oo unless this space is finite dimensional or p = 2. It is easy to see that LP is finite dimensional if andonly
if thecollection of measurable sets of
positve
finite measure consists offinitely
many
atoms,
each of finite measure.Excluding
this trivial case, one has to deal with a measure space which containsinfinitely
many measurable sets ofpositive
finite measure. One canalways
find in such a space a sequence[An)
ofdisjoint
measurable sets which havepositive
finite measure. This isclear if the space contains
infinitely
many atoms of finite measure. Ifnot,
let
Bo
be a measurable set of finitepositive
measure which is not a finiteunion of
atoms ;
such a set existsby assumption. Ba
contains a proper subset ofpositive measure B1
such that 0 fl(Bt)
p(Bo).
LetAl
=Bo - B, ;
p
(A,)
> 0. Nowproceed inductively. If xi
denotes the characteristic func- tionof Ai,
then(,u
Xi has Lp norin one, and the closedsubspace
ofLP
spanned by
these functions islinearly
isometric to the sequence space lp . Lemma 4 anda)
nowimply that 7
is non-extendable. If(8, I, p)
is to-tally sigma-finite,
then L°° is the dual ofL1,
hence7
is non-extendable to L°°.c)
Let K be acompact
Hausdorff space, and the space of con- tinuouscomplex
valued functions on K. Let be the dual of this space,i.e.,
the set of all finitesigned
Baire measures on K.7
is non extendable forC (h’)’
and hence forC (h’).
’1°o see thislet p
be a non-zero element ofh 0. is a closed
subspace
of C(1~)’, J
is non extendable forby b). Hence, by
Lemma4,
it is non extendable to C(h.’)’.
d)
Let ek(Rn)
be the space of hounded continuous functions on Eu- clidean n-space which have bounded continuous derivatives up to and inclu-ding
those of order k. ek(Rn)
is a Banach space with the normwhere 7~ is
the, by
now, standard multi-index notation for derivatives. Re- striction to a, line maps ekcontinuously
ontoCk (R).
Restriction to the interval I =[0, 1] ]
mapscontinuously
intook (I).
It iseasily
checkedthat this map is also onto. It is known
([0],
p.184)
thatOk (I)
islinearly homeomorphic
to0 (I) (The
norm of used in[0]
is different from theone
above, however, they
areeasily
seen to beequivalent).
It followsby
Lemma 4 and
c) that F
is not extendable for ek(Rn).
e)
Toconclude,
consider thefollowing
case: Let H be acomplex
infinite dimensional Hilbert space and let denote the
algebra
of boun-ded linear
operators
on H.7
is not extendable for B(H).
To seethis,
letA E
B (H)
beself-adjoint
and such that the norm closure of thesubalgebra generated by
A is infinite dimensional.(It
suffices to take an A whosespectrum
is not a finiteset.)
Fromspectral theory
one knows that thissubalgebra
isisometrically isomorphic
to thealgebra
of continuouscomplex
valued functions on a
compact
Hausdorff space([6],
p.95), i.e.,
the Banach(H)
contains a closedsubspace
forwhich 7
is not extendable. Onecan go
slightly
further : ifC (H)
denotes the closed ideal ofB (H)
consis-ting
of thecompact
linearoperators
onH,
and if B isseparable,
it followsthat
7
is not extendable forC (H)
because([7],
p.208) B (H)
islinearly
isometric to the second dual of
De University, Chicago,
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M. M. DAY, Normed linear spaces, Springer Verlag, Berlin, 1958.[3]
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