R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
G IOVANNI D ORE D AVIDE G UIDETTI A LBERTO V ENNI
Complex interpolation for 2
NBanach spaces
Rendiconti del Seminario Matematico della Università di Padova, tome 76 (1986), p. 1-36
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Complex Interpolation Spaces.
GIOVANNI DORE - DAVIDE GUIDETTI - ALBERTO VENNI
(*)
SUNTO - Si definiscono
spazi d’interpolazione
tra 2Nspazi
di Banach, con il metodocomplesso.
Glispazi
sono definiti apartire
daspazi
di funzioni olomorfe sullapolistriscia
di CN che sianointegrali
di Poisson di oppor- tuni dati al bordo.Vengono
dinlostrati un teorema di densita e alcuneproprieta
checollegano
questo metodo con il metodo di Calder6n. Infine viene data una caratterizzazionedegli spazi
duali.1. Introduction.
The aim of this paper is to
give
an extensioninvolving
2N Banachspaces and N
parameters
of thecomplex
method ofinterpolation
forBanach spaces introduced
by
A. P. Calder6n in[3].
In the
subsequent
years othergeneralizations
weregiven by
severalauthors: we recall Favini
[9],
whoemploys N -~-1
spaces and N param- etersdeveloping
a definition of Lions[17].
S. G. Krein and L. I. Niko- lova[18] [14] [15]
constructedinterpolation
spaces for finite or infinite familiesdepending
on acomplex parameter.
In this connection wequote
also the papers[4] [5] [6]
in whichthey
describe ageneral theory
of
complex interpolation
for infinite families of Banach spaces withone
complex parameter varying
on asimply
ormultiply
connecteddomain.
(*) Indirizzo
degli
AA. :Dipartimento
di Matematica, Piazza di Porta S. Donato 5, 40127Bologna,
Italia.The method
involving
2N spaces and Nparameters
goes back to the papers of Fernandez who deals with the real case in[10] [11] [12],
with the
complex
case in[13],
and with their connection in[2].
In thetwo last papers
they
sketch atheory following
apattern
very similar to thepattern
ofCalder6n,
but the crucialinequality
4.4.(2)
of[2]
is wrong, as we prove in our
counterexample
5.5. It isjust
the lackof that
inequality,
without whichpart
of Calderon’stheory
cannotbe
developed,
which led us togive
a new definition of theinterpolation
spaces,
starting
fromholomorphic
functions on thepolystrip
S~T thatare not
supposed
to be continuous at theboundary
but are Poissonintegrals
of functionsbelonging
tosuitably weighted
Lq spaces.The paper is
arranged
in thefollowing
way.In §
2 we obtain someinequalities
about Poisson kernels and some technical results about Poissonintegrals. In §
3 we define theinterpolation
spaces for a com-patible family (Ai)
as the spaces of values in thepoints
ofassumed
by holomorphic
functions on(]0,1[
+iR)N
which can beexpressed
as Poissonintegrals
of functions oftype Aj),
wheree is a
weight
functionequivalent
to the Poisson kernel.In ~
4 weprove the
density
of a certain class ofgood
functions » in the function spaces introduced in theprevious
section. This result isrepeatedly employed in §
5 to show thedensity of n Aj
in theinterpolation
spacesj
and other
properties.
Inparticular
zvcstudy
the connection betweenour method of
interpolation
andCald?r6n’s;
we show that our inter-polation
spaces are embedded in iteratedinterpolation
spaces of Cal- der6n. Acounterexample
shows also that ingeneral
this inclusion is proper.In 9
6 we prove some results ofduality.
2. Poisson kernels and Poisson
integrals.
In this section we
give
somedefinitions,
deduce some estimatesabout Poisson kernels for the
strip
andstudy
someproperties
ofPoisson
integrals.
We establish thefollowing
notations which wekeep
from now on.
.N is a
positive integer,
and when a e R(a
~0) ~ is
theN-tuple (a, ... , oc).
These are the Poisson kernels for the
strip
~S’.Moreover q
E[1, oo] (unless
a more restrictive condition isrequired) and q’
is itsconjugate exponent (i.e.
The
weight
is definedby
C = 1
Remark that = 1. When X is a Banach space and
RN
Le(RN, X)
is the Banach space ofstrongly
measurable functions For nota- tional convenience we shall alsowrite Lae when q
is allowed to reach00: however in the case q == 00 it is understood that this
symbol
denotes
simply
the space L°°(without weight).
We observe that from the well-known
identity
it follows that
and later wc shall refer to this
inequality
even withoutexplicit
mention.We have also
In fact
PROPOSITION 2.1. T he
following
statements hold:PROOF.
(2.3), (2.4), (2.5), (2.6)
are trivial.(2.7) Obviously
and the middle term is
exactly P;(a, s) IP~ ( 2, s).
Now it is sufficient to observe that the left term is anincreasing
function of181
and theright
one is adecreasing
function of(2.8)
follows from(2.7) (with
.M’=0)
and(2.1).
Remark that from
(2.3)
it followsthat ’
i(
In the
remaining part
of thissection,
X is a fixed Banach space.PROPOSITION 2.2..Let
f
EX).
Then EJN,
a EIN, s ERN,
-
c~) f(d)
da~ exists anddefine8
a continuousfunction of
a+
is E SNRN
which is harmonic in every
couple of
variables(ak,
PROOF.
By
the estimate(2.8)
we haveThis ensures the existence of the
integral, gives
therequired growth estimate,
and allows us toemploy
the dominated convergence theorem to provecontinuity.
Moreover we can check that the function(ah, s~)
1-71-7
iPj(a, s
-a) f (a) da
has the mean valueproperty by changing
theRN
order of
integration
andexploiting
the sameproperty
ofPj.
LEMMA 2.3. Let continuous
function,
harmonic on S.Suppose
that 38 E]0,1[,
0 > 0 such that ’if a -f- isE S
PROOF.
Suppose
thatf
is real-valued. In this case we prove that Vz E Sf (z)
supf (~).
Since -f
has the sameproperties
asf,
this willEEbS
prove the statement R.
Then g
isharmonic,
and itsg.l.b.
on ~S ispositive. Vq
E R+Therefore,
if we fix z = a-E- is
E S and M islarge enough, by
themaximum
principle
on[0, 1] -[- i[- .M, MI,
Now we suppose that
f
iscomplex-valued. Obviously Re f ,
Imf
fulfilthe
assumptions
of the theorem. ThereforeFinally,
iff
has values in the Banach spaceX,
thenLEMMA 2.4. Let
f:
X be a continuousfunction
such thata
compact
subsetof Then, uniformly f or
s EH,
where
j,
k E JN and is Kronecker’ssymbol.
M’oreoverif f
isuniformly
continuous on RN and e = 1
(i.e. f
isbounded),
then the convergence isuni f orm
on RN.PROOF.
Suppose j.
Thenso that it is
enough
to show that eachintegral
in the last side of theinequality
is boundedfor ah
E I and converges to 0 when11t.
But this is shown
by
theinequality
Suppose
Then’When
a - j
the last term converges to0, uniformly
on every set onwhich f
is bounded.Having
fixed s ER+,
we can take 6 E R+ such thatthe second summand
(or
iff
isuniformly
con-tinuous on
RN).
The first summand is dominatedby
Here the second term converges
uniformly
to 0. Theref ore theproof
will be
complete
if we show that whena - j
and
have the same finite limit. For this we can suppose N = 1 and show
that
The first
equality
follows from the fact thatfor a - j ~(~ ’)
is anapproximate identity;
the second one is shownby
an easycomputation.
PROPOSITION 2.5
(see [3]
p.116).
Let continuousharmonic on every
couple of Suppose
PROOF. We
begin by supposing
that N = 1.By
lemma 2.4 andproposition 2.2,
the functionis harmonic on 8 and has a continuous extension
to S,
which vanisheson as. We can
get
our resultthrough
lemma 2.3 if we prove thatIn fact we have
This
depends only
on rx,continuously
on[0, 1] by
lemma2.4,
so thatit is bounded.
Suppose
now thatf
is a function ofN’ -~- 1
variables which satisfiesour
assumptions.
If we fix the lastvariable,
we obtain a function of .N variablessatisfying
the sameassumptions,
so that we canproceed by
induction.The aim of our next result is to show that
if fj
e.X~) Vj
E JNand
then
Vj
EIN I;
= 0.However,
y since it ispossible
thatdoes not
belong
toX),
we work in a suitablelarger
space.converges to
t
in the normof
Since
jo2(8) d8
+ from lemma 2.4 it follows that we have theRN
asserted convergence whenever
f
isuniformly
continuous and bounded.Since such functions are dense in
~’),
it suffices to show that3C>
0 such thatIn fact
so that it remains to show that
But
and direct
computation
of the lastintegral
concludes theproof.
3. Definition of the
interpolation
spaces.In this section we introduce the
complex interpolation
spaces forcompatible
families ofcomplex
Banach spaces. Henceforth every vector space we consider iscomplex.
A finite
family A =
of Banach spaces is said to be com-patible
if there is a Hausdorff vectortopological space A
such thatVj
E .g(by
« - » we denotealgebraic
inclusion with continuousm
embedding;
if X and Y are normed spaces, X dl Y means that theembedding
norm does not exceed~VI) .
When A is acompatible family A(A)
is the Banach spacen A;
with norm = max and:,w jEK
jEK
-Y(A)
is the linear hull ofU Ai
with the norm11
-,. ieK
(this
too is a Banachspace). Evidently
A normed
space X
is said to be intermediate for thefamily
A if4 (A)
~ X‘ ~E(A).
When A = and B = arecompatible families, .L(A, B)
is defined as the vector space of the linearmappings T : ~{A) -~- ~(B)
such thatTIAi
EBj) Vj
E K(as usual,
when Xand Y are normed spaces,
Y)
is the space of continuous linearmappings
from X toY).
If X and Y are intermediate spaces forA,
Brespectively,
we say that.~,
Y areinterpolation
spaces withrespect
toA,
B when for everyT E L(A, B)
EL(X, Y).
From now on we consider
compatible
familiesA,
B etc. of Banachspaces whose index set is ~TN
(J = ~0, 1 ~) .
Since for acompatible family
.A and 1 q 00Aj) ~ L) (RN, 27(A)),
on the spacei we can define a linear map
T,
whose range is contained in the space of continuous functions g:SN -~ ~’(A) (see
prop.2.2), by putting
where
f
=(fj)iEJN,
a EIN,
sFrom prop. 2.6 it follows at once that is one-to- one, and we call the
image
under J offunctional ’
(with
the usual modificationif q
=cxJ)
isobviously
a Banach norm.and if
is a norm on Remark that 00) does not
depend
ona
(but
wekeep
this notation forconvenience)
and that when o0we call also the norm induced
by 5’
onPROPOSITION 3.1. For each
fixed
q E[1, o],
all the norms(a
EIN)
on,~q(A)
areequivalent, uniformly for
a incompact
subsetsof
IN.PROOF. The first statement is an obvious consequence of the ine-
quality (2.7)
with M = 0. The second one follows from H61der ine-quality
and fromPROPOSITION 3.2.
K is a
compact
subsetof
If z E
.7~,
we haveDEFINITION 3.3.
$’q(A)
is the spaceof
thef unctions f: SN-+E(A) holomorphic
andbelonging
toBy
the secondpart of
prop.3.2,
convergence in
~q(A) implies uniform
convergence oncompact
subsetsof ~SN,
withrespect
to the normof E(A),
so that is a closedsubspace
of
and hence a Banach space.-
When X is a Banach space, we write
Yq(X)
to meanNow we define
interpolation
spacesÂ(a;a).
DEFINITION 3.4.
Va E IN, ~(a; a~ _ { f (lL) ~ f
ESince the
operator f
~ is continuous fromYq(A)
toL(A) by
prop.
3.2,
we can define a Banach norm onA(a;,,) by putting
_THEOREM 3.5. an intermediate space
f or
thefamily
A.More
precisely, ~ A(a; g) ~ ~(A).
*obvious that
f c-
so that-
(when q oo)
The
proof
is evensimpler
when q = 00, since 00) ==Let x E
A(a;a)8
Thenobviously
xE L’(A)
and the estimates on thenorms follow at once from the first
part
of prop. 3.2.THEOREM 3.6.
If
A and B arecompatible f amities of
Banach spaces, thenA(a; q)
and areinterpolation
spaces withrespect
to A and B.More
precisely,
PROOF. Let x E
A(a; (1
and g = E withg(a)
= x. ThenTg
E so that Tz E and(and similarly
whenq = ~).
Asf
isarbitrary,
we obtain theinequality.
REMARKS
3.7. When X is a Banach space and
Ai == X Vj E JN,
we have4 (A)
= X =Z(A)
withequal
norms. Therefore from th. 3.5 it follows that in this case.A(a; a)
_ ~ withequal
norms.1
3.8. From prop. 3.1 we have y. whenever 1 ~ q r 00.
3.9. If 1 q 00 and
Vj
E JNAi
isreflexive,
thenAi)
is reflexive
([19]
th.5.7),
so that5T~(A), ,~ a(A.)
and arereflexive,
since
products,
closedsubspaces
andquotients
of reflexive spaces are reflexive.4. A
density
theorem.Let A be a
compatible family
of Banach spaces. We consider theholomorphic
functionsf :
such that _ lim =0,
zESN, |z|-oo
and we call the space of the restrictions to ~SN of these functions.
Remark that
by
prop. 2.5 C5"q(A)
E[1,00].
The main result of this section is the
following
one.THEOREM 4.1.
If 1 q
oo , then is dense in,~ Q {A. ) .
We shall prove this theorem
through
a number of lemmas con-cerning
the functionsLEMMA 4.2. The above
integrals (4.1)
and(4.2)
exist in the spaces andAj respectively.
Thefunction
asdefined by f ormuta (4.2)
is
holomorphic
on CN in the normof Aj.
the
integral (4.1)
exists inf(A). Proving
the existence of(4.2)
inA;
is even
simpler,
asand
are both finite. Let .K’ be a
compact
subset of CN. Then for z E K and 8Therefore, by
the dominated convergencetheorem, f(j)n
is continuous from CN toMoreover,
if y is aclosed, piecewise
differentiable curvein
C,
when wecompute
we can
change
the order ofintegration,
so that it vanishes. This proves the lemma.PROOF. Without loss of
generality
we may consideronly
the casej
= 0. When we substitute in(4.1) f (a + is)
with itsexpression by
means of the Poisson
integral,
we canchange
the order ofintegration,
so that
We prove that in this sum every summand converges to
0, except
the one
with j
= 0 which converges to as a -+ 0.By
the domi-nated convergence theorem it is
enough
to prove thatand that
In fact the left side of
(4.3)
is dominatedby
The first
integral
iswhile the second one is
(see (2.7))
For j - 1, (4.4)
follows at once from( 2 . 6 ) . Finally
Here the second summand converges to 0
by
lemma 2.4. The firstone is dominated
by
LEMMA 4.4...For
z E CN,
does notdepend
on a E IN.
PROOF. We prove that for fixed a ==
(a2 ,
... ,aN) E IN-1, f ~a~(z)
doesnot
depend
on al. For this it isenough
to show thatdoes not
depend
on a1(the
existence of thisintegral
isguaranteed by
prop.2.2). Being holomorphic
the functionit suffices to prove that as
I MI -* + -
But this follows from prop. 2.2.
Thus we have
proved
thefollowing proposition :
PROPOSITION 4.5. E
d (A)
and thef unction f ~,a’
isholomorphic
in the norm
o f d (~.).
Lemma 4.4 allows us to write henceforth
In
insteadof j
JMoreover, by
lemmas 4.3 and 4.4and
PROOF
(recall
that cosh a c 2 cosh(a
-~3)
cosh and this proves the lemma.PROOF. is
holomorphic,
it isenough
to prove that the functionbelongs
to and that~==~((/~)).
The first statement is trivial when and is
easily proved
whenq
by
means of Minkowskiintegral inequality, through
the ex-pression
ofgiven
before lemma 4.6. For the second one we hav ePROOF
(by
Minkowskiintegral inequality)
Therefore
as it can be seen
by putting
T = nt andremembering
thatOn the other hand
Here the first summand is small when 6 is small because E E
Lq(RN, Aj).
The second one is dominatedby
and this also is small when 8 is small
by (2.2).
We have thus
proved
that the space of the restrictions to SN of the functionsf : CN ~ d (A ), holomorphic
andsatisfying
thegrowth
condition of lemma 4.6 and
belonging
to is dense in Therefore thefollowing
resultcompletes
theproof
of th. 4.1.LEMMA 4.9. Let
holomorphic function
such thatPROOF. It is obvious
that g,,
E In order to prove that it isenough
toapply
the dominated convergence theorem.5. Some
properties
of the spacesÅ(a; a) .
We prove some consequences of th. 4.1.
THEOREM 5.1.
d(A) is
dense inA(a;q).
PROOF. Trivial consequence of th. 4.1.
THEOREM 5.2.
Vj
E JN let A0j be the closureof A(A) in Aj,
with thenorm
of
so that A° _ is acompatible family.
ThenVq
E[1, oo[
, in both cases with
equal
norms.PROOF. Since
= 4(A°) (with equal norms),
we have that= and that the restriction to of the norm
of
£(A)
coincides with the restriction of thehomonymous
norm ofThen our statements follow
easily
from th. 4.1.THEOREM 5.3.
Suppose
thatThen Vb =
(b2,
andf or 1 ,
PROOF. Vf
E we setThen it is obvious that We prove
that 3 C >
0 such thatVf E F0(A)
But
by (2.7)
and(2.8) (the
constantsdepcnding only
ona’, a").
Therefore-Here
and
by
H61derinequality
This allows us to
get
theinequality.
Let ~ E and
f
E5~-,(A)
such thatf(a’, b)
= x. Let(fn)neN
Wa sequence in
Yo(A) converging
tof
in the norm · thenThe above
inequality
shows thatb))neN
is aCallchy
sequence in i becausegfn(a", b)
=fn(a’, b),
and as andare both
continuously
embedded in.E(A),
it follows that x EA(a’,
b; q), Therefore
11
as
f
isarbitrary
REMARK 5.4. When
N=1,
ourinterpolation
spacesA(a;
q)(a
E]0, 1[, 1 ~ q ~ ~)
coincide(with equal norms)
with the space[Ao , A1]a
definedby
A. P. Calder6n in[3].
In fact it is obvious that Calder6n’s spaceAx)
is contained in our5l~(A)
and thatVf e AI)
, and this
implies
thatI
~ A(~ ; a ~ . Conversely,
let Then and the ine-quality IIIflll(a; 1)
holds(see [3],
p.117 ) . By
anargui-nent
similar to that of the
proof
of th.5.3,
we obtain theembedding A(a;
1)~ [Ao,
which proves our statementby
remark 3.8.When
.~1~~ ~ 2
the situation is no more soclear,
because of the lack of theinequality (9.4 iii)
of[3]
p.117,
as we see in thesubsequent counterexample.
As a matter of fact the assertedproof
of that ine-quality, given
in[2]
pp. 199-200 is wrong, since theargument
is basedon the false assertion
that, given
2 N boundedinfinitely
differentiable real-valuedfunctions
there exists a functionf : SN- C,
continuous on SN and
holomorphic
on such that Vs e RNVj
E INRe -;-
is)
=COUNTEREXAMPLE 5.5. We exhibit a
compatible quadruple
A = and a sequence of func tions
tn : 82-7- L(A),
continuousand bounded on
82, holomorphic
on82,
such thatfn(j
+is, k
+it)
Eand is bounded and continuous in the norm
of Ajk,
for which the normof
f ~( 2 , 2 }
in theinterpolation
space defined in[13]
and dealed with in[2]
isequal
to1,
while This proves that theinequality 4.4(2)
of[2] (corresponding
to(9.4 iii)
of[3]) fails,
andalso that the
interpolation
space of[13]
relative to thepoint (t, ~)
is different from
A(~,~;1) (actually
it iseasily
checked that it issmaller).
have the usual
meaning
and let I be the Banach space ofcomplex-valued
sequences such that --- sup + 00,1
’ n
so that loo - 1. we set
Aoo = All
=Zoo, Aol = A1o
= I. Then =Zoo, Z(A) =
Z(with equal norms).
If[A ; 2 , tJ
is theinterpolation
space defined in[13],
we have Zoo~ [A ; ~ , ~ ] .
On the otherhand,
letf :
82 -¿ 17(A)
= be a bounded continuousfunction, holomorphic
on 82and such that
f (j
+is, k
+it)
EA j,
and is bounded and continuous(with respect
to(s, t))
inAjk.
Putg(z)
=f (z, z)
Vz E S.As g
is boundedin the norm
of 1,
we canrepresent
it as the Poissonintegral
of itsboundary values,
and these values are bounded and continuous inZoo,
so that
g(z)
E Zoo VzE S (see [3]
p.116) .
Thereforef ( 2 , 1 )
E100,
andthis proves that loo =
[A; 1 2, 1 2] . Moreover,
fromit follows that ~ Therefore we have also
equality
of norms.
_
Let
a(n)
=(Kronecker’s symbol)
and 0f,: ~~--~ Z°°,,
If we
fix n
>0,
we can find E > 0 such that the lastintegrals
isand if we take n > e~ we obtain
THEOREM 5.6. Let A =
(A;);eJN
be acompatible family of
BanachPROOF. We shall prove that
vf
E,~=o(-A) b)
and this will prove the theorem
by
anargument
similar to that ofthe
proof
of th. 5.3. In fact(9.4 iii)
of[3]
and H61der
inequality)
THEOREM 5.7. Let and A ---- be a
compatible family- of
Banach spaces that E JN-1 IA,,,.
=Ak,1
=Bk. Then V(a, b)
c-PROOF.
By
theorem 5.6 1 Letand
put ;
.1t is an easy
computation
to checkthat g
E£ (A )
and that ==- This proves the converse
embedding.
We conclude this section
by showing that,
ingeneral,
the ill-clusion of th. 5.6 is not an
equality.
For this we need twopreliminary
lemmas.
LEMMA. 5.8. Let
.,A.o , AI, Bo , .B~
be Ba,nach spaces such thatis ac
compatible family, Ao
4-Bo
~-->Ao
+Ai ,
and suppose that300
E]0, 1[
such that
[.Ao,
~Bo.
Let T Et(Bo, Bl)
and endow kerTIAo
with thePROOF. Let
f c- Y(ker TIAo’ A1) (Calderón’s
functionspaces, see
[3]).
Thenf (it)
EBo, f (80
+it)
E R. For 0 0eo f (8 + it)
is the Poissonintegral (in Ao
+Ai)
of its values for Re z = 0 and Re z =80 .
But this is also a Poissonintegral
ofBo-valued, Bo-
bounded and continuousfunctions,
so thatf (0
-t-it)
EBo .
Moreoverf
is aBo-continuous
functionwhich,
for 00 00
coincides with itsCauchy integral
in the norm ofAo
+A,
and hence in the norm ofBo.
This proves that
Tf
isholomorphic
for 0 Re z00
and continuous for in the norm ofB1.
But -- 0b’t E R.,
so thatTf(z)
= 0 for each z such that0 ~ Re z ~ 8p .
Sincef
isarbitrary,
thisconcludes the
proof.
We recall that if vo and xi are
positive weight
sequences, then 01),
wherevo(n)
=(see [1]
th.5.5.3).
LEMMA 5.9. Let Vi: Z - R+
(j = 0,1)
benon-decreasing
sequences.We suppose that inf vo > 0 . = inf v1.
Then
10, 1 [ [Ao,
.---l1v0
withequivalent
norms..PROOF. First of all from
Ao l;o’ 11
it followsby
inter-,
It is obvious that the function
and
Thus in
general Therefore,
assince lim
ve(m)
= 0 as i t followseasily
from theassumptions.
Thism- - oo
proves that ’BIn E Z
(a">
- converges in[Ao,
as m -?- - 00.But
II
and as bothA1
and[Ao ,
are con-tinuously
embedded in we have that andHence it follows
easily
that each« finite » sequence À
belongs
to with and sowe
get
at oncethat l1v0 2- [Ao ,
COUNTEREXAMPLE 5.10. We show an
example
of acompatible quadruple
whereA(~,~ ;1)
isstrictly
smaller thanWe set with the
l1-norm,
with the natural norms.
By applying
lemma 5.9 we see that[.Aoo~
is
equal
to l1 withweight
2n/2 and that[Aol, .A1l]i is equal
to 11 withweight 2-n/2,
, so that the iteratedinterpolation
space isequal
to lx(without weight).
But4 (A)
is contained in the proper closedsubspace
of 11 1 defined
by
the condition1
and
by
th. 5.1A~2, ~ ;1~
=,4 11.COUNTEREXAMPLE 5.11. We show an
example
of acompatible quadruple A
= where for some(0, e)
C J2We define
~:Z-~C
in thefollowing
way:~oo(~)=ma.x{1~2"}~
1VIO(N)
= maxfl, 2-"1, vol(n)
= min11, 21~, vll(n)
= min{1, 2-"1.
Weput Ail
== and(in
every case with the naturalnorm). Then, by
lemma5.9,
=(voe
=and = so that
By
th.1.17.1/1
of[21]
M~Z~j In = 0),
and weknow that
[.Aol, Al1]o = ~~e~ .
Anapplication
of lemma 5.8 shows that6. Duals
In this section we
study
the duals of ourinterpolation
spaces.When we are
given
acompatible family A = (Aj);eJN
of Banachspaces, the
assumption
thatd (A)
is dense ineach Aj (which
we suppose satisfiedthroughout
thissection)
ensures that the dual spaceAi
ofA~
can beidentified,
in the usual way, with asubspace AJ
ofA (A)~’.
Thus A’ _
(A;)ieJN
is anothercompatible family
and we canput
thequestion
whether the space(A(a; a~)’,
identified with the dual ofA(a; q),
is an intermediate space for the
family
A’ . WhenN=1,
from thefact that
d (A)*= Z(A’ )
andJ (A) (see [1]
th.2. 7.1)
it followsat once that
(A(a; q»)’
is intermediate forA’,
and moreover in that caseCalder6n
[3]
show ed that it is aninterpolation
space. But as soon asN ~ 2
theequality d (A’ )
fails(see [8] §4)
and in fact thefollowing counterexample
shows that(A(a;q»)’
may also fail to be anintermediate space for the
family
A’.COUNTEREXAMPLE 6.1. We set
where
vo(n)
= min11, 2~~,
== min{1, 2’"}.
is dense in each
(see [5] appendix 1).
We set Then 0 can be con-
tinuously
extended to eachA.1k, and
so 0 Ed(A.’).
Now we fix(0, e)
E J2such that
0 + e > 1, 0 Lo
and we showthat OE (2(0,e;q))’.
To dothis,
weput
I and we show that inthis will prove our assertion.
n, m wi th and we set
converges to a~~~ -
a~-?~~, we have proved
thatIn the
sequel
we have to consider dual spaces of spaces likerLQ(RN, .X),
where X is a Banach space, a: RN- R+ is a measurableweight
function and1 ~ q
We shall assume that for 1 q o0.X))*
isisometrically isomorphic
to +(1iq’)=1)
and that
(La(~BN, .X))*
isisometrically isomorphic
to inboth cases with
respect
to theduality
These
assumptions
are fulfilled when .X’~ has theRadon-Nikodym property,
and inparticular
when X’~ is reflexive orseparable (see [7]
cor. 5 p. 117
and §6
p.118).
We shall also
employ
functions defined on thepolydisk
DN(.Z)
-=11) (or
on(aD)N)
which we obtain from functionsdefined on
(or
on(aS)N ) by
means ofchanges
of variablesgiven by
directproducts
of conformalmappings
of g into D. Moreprecisely,
let w and
/~p: ~S 2013~D
be the map definedby
Then pw
is ahomeomorphism
of 9 onto exp(2ni
Rew)~
suchthat
pw(w)
= 0 and its restriction to S is abiholomorphic
diffeo-morphism
of S onto D.When w -
(~,vl , ... ,
eSN,
we define-
..., Since exp
(2ni
Re weobtain that the N-dimensional measure of is zero.
The functions
p )
act aschanges
ofvariables, giving
riseto one-to-one
correspondances
between functions defined on SN and functions defined onDN,
and between(classes of)
functions defineda.e. on
(aS)N
and(classes of)
functions defined a.e. on(aD)N.
Moreoverthese
changes
of variables « commute » with the Poissonintegrals (on (aD)N
and on in thefollowing
sense.and 99 = Let g be
the Poissonintegral of 1jJ
on DN(here
the Poisson kernel for DN is the tensorproduct
of Poisson kernels for
D)
andf =
Then andf
is the Poisson
integral
of 99The
computations
arestraightforward
and we omit them. As a par- ticular case(with ~
+i?7
= u~ and11(p(j
+ instead of~(~ c is))
we
get
Hence,
EX),
then ELq((oD)N, X),
so that the cor-respondance
is onto.It is well-known
([20]
th.2.1.4)
that a necessary and sufficient conditionthat g
beholomorphic
on DN is that the Fourier coefficientsca
of 1jJ
are zero such that tXj 0.Thus for a
function 99
EX)
we are interested inlooking
forthe
vanishing
Fourier coefficients of itscorresponding function y
on(8D)N.
In connection withthis,
remark that ifv:
SN~ DN are directproducts
of conformalmappings
such that =v(w)
= 0(where
w E
SN), then, by
well-knownproperties
of the conformalmappings
ofthe
disk,
there areÂ1, ...,
with = 1 such that Vk =(15
=1,
... ,N) ;
therefore after we have chosen thepoint
w E SN whichwe map into
0,
the fact that a fixed Fourier coefficient vanishes does notdepend
on theproduct
of conformalmappings
weemploy.
THEOREM 6.2. Let 1 q 00,
q’
be itsconjugate exponent,
X be aBanach space and
V j
E JN . letfj X ).
Weput f(j
+is)
==== fj(s)
Vs EJN,
and g =(where
W ESN) .
T hen thefollowing
statements are
equivalent:
(a)
the Fouriercoefficient of
gis zero cx =F 0 and
]
In
particular, if
X is the dual spaceof
a Banach spaceY,
then
( a )
and(b)
areequivalent
tosuch that
99(w)
= 0PROOF. We show that
(b)
~(a)
and that(a)
=>(c). Analogously
it can be shown that
(a)
=>(b).
Moreover it is obvious that(c)
~(b).
Then =
0,
so thatWe also
call1jJ
the function on( aD)N
obtainedby
thechange
of variable~w
from thefunction j + ~ ~ pj(s)
on Sincewe have to prove that the first
integral
vanishes. Setwhere
and ba
is the a-th Fourier coefficient of Vy.By
meansof the Fourier coefficients
of g
we defineanalogously
gm.As y
is
holomorphic
on DN and =0,
we havethat ba
= 0 when a = 0and whenever min oc, 0. Therefore
k
We now remark that E
L°°((aD)N, Y),
as it is a finite sum offunctions in and that g E
Lq’((aD)N, X),
so that. Therefore Vn E N
same
theorem)
and soREMARK 6.3.
Suppose
thatf
E(1 q’ co).
Thenf
isholomorphic
if andonly
if theequivalent properties
of th. 6.2 hold Vw E ~N. Infact,
iff
isholomorphic,
then condition(a~)
is fulfilledas we have
already
remarked.Conversely,
if the condition(b)
issatisfied ‘dw E
SN,
then it is easy to check thatso that
ggf
is harmonic in eachcomplex variable zk = Ek
+iq,,.
Inparticular
if we setq;(z)
= weget
so that it is
proved
thatf
isholomorphic.
DEFINITION 6.4. Let A =
(Ai)jEJN
be acompatible family of
Banachspaces, a E
IN,
1 q 00, and letq’
be theconjugate exponent
to q. We callg(A)(a; a’)
the closedsubspace of ,~ a.(A)
whose elements are thefunctions f
withboundary
valuesfulfilling
the conditions(a)-(b) of
th. 6.2(where X -.E(A)
and w =a).
We endow with the normBy
prop. 3.2f H- f (a)
is continuous onS(~)(~~
so that we candefine a Banach space in the
following
way:From remark 6.3 it follows at once that is a closed
subspace
of
S(JL)(~~
so that ATHEOREM 6.6. > is an intermediate space
for the family A,
and
actuall y J (A) ~ .d~a ~ a~ > ~ ~(A ) Moreover, i f
A and B are com-patible families o f
Banach spaces, thenA(a;
a’) andB~a ~
a~ areinterpolation
spaces with
respect
to A and B.PROOF. From the remark
preceding
the theorem and from prop. 3.2and th. 3.5 it follows that It is
easily
checked that whenever andf c
thenhence the theorem follows at once.
As
above,
when Y is a Banach space such that is a densesubspace
of Y(we
recall that we have made thisassumption
forY = and this is not restrictive
by
th.5.2),
we denoteby
Y’ thevector
subspace
of4(A)*
which can be identified in the usual way with the dual space Y* of Y.THEOREM 6.7.
I f 1 ~ ~ ~ ,
then(.A(a; q»)’
is a closedsubspace of A’ ~a ~ g~
o)ith the same norm.PROOF. Let
q/
E(A(a;,))’
and letqJ*
be its continuous extension to Thenf
is a continuous linear functional onand
(if .~~ (A )
has the norm it has the same norm asqJ’. By
the Hahn-Banach theorem we
get
a continuous linear functional onand
by
acomposition
with T weget
a continuous linear functionalsuch that has the
norm
~~~ ’ ~~~ (~ ; ~> ,
and thatwhenever
T((f,))
E In connection with there are(j
EIN)
such that(recall
theassumptions
that we have made above and theequivalence
between the
weights o
andP,(a,.))
and11 (1J’
=J ~~y~~}~).
As the constant
A(A )-valued
functionsbelong
toFq(A),
so that =
~9’.
To show that y E~(A‘)(a;
a.~, sincewe can
verify that 1p
fulfils the condition(c)
of th. 6.2 with Y ==d (A).
Thus we have
proved
thatp’
andthat I It remains to show that Vy’ E (A(a;q))’,
Let g E such that
g(a)
=T’
and letf
E ThenREMARK 6.8.
Along
the lines of theproof
of th.6.7,
it can beproved
that the dual space
(A(~; a))’
can be obtained as the space of vectorsh(a),
where h runs over the space whose elements are the functions h E!,,,(A’)
such thatVf
E withf (a)
=0,
We note that the definition of differs from the definition of
9(J.~)~;~’)
because in the latteronly functions f
E areemployed (see 6.2(c)
and the definition6.4).
However forwe do not know any characterization in terms of Fourier coefficients