SVANTE JANSON

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COMPLEX METHOD INTERPOLATION DEFINED

USING A HALF-PLANE OR FULL DISC

SVANTE JANSON

Department of Mathematics, Uppsala University, PO Box 480, S-751 06 Uppsala, Sweden E-mail address: svante.janson@math.uu.se

The standard denition of the complex interpolation space [X⁰;X¹] due to Calderon [

2

^{] uses (}^X⁰⁺^X¹)-valued analytic functions in the strip^fz: 0<Rez <1^g, see below for details. Is it possible to use only functions that are analytic in the half-plane^fz: 0<Rez^g?

There is a variant of this question which arises since, as shown by Cwikel [

3

^],

the space [X⁰;X¹] may also be dened using analytic functions in the annulus

fz :R¹ <^jz^j< R⁰^g. Is it possible to use only functions that are analytic in the entire disc^fz:^jz^j< R⁰^g?

The main purpose of this paper is to show that, in general, these questions have a negative answer, even if we supposeX¹X⁰. This is done by an explicit counter example.

Section 1 contains some denitions and an equivalence theorem showing that the two questions above are equivalent.

The counter example is given in Section 2. The reason for considering this particular example is given in Section 3. Section 4 contains some additional results.

Some of these concern special cases where the answer to the above question is positive. One such case is when (X⁰;X¹) is a couple of Banach lattices withX¹ X⁰. Section 5 is an appendix written by Michael Cwikel which presents another case where the answer is positive.

Acknowledgment. I thank Michael Cwikel for posing the questions studied here, as well as for encouraging me, many years later, to write up my counter example.

1. Preliminaries

We introduce some notation. IfX andY are Banach spaces, thenX =Y means that the spaces contain the same elements and that the norms are equivalent (but 1991Mathematics Subject Classication. Primary 46B70, Secondary 42A16.

Key words and phrases. Complex interpolation, Minimal interpolation functor, Orbit space.

This paper is in nal form and no version of it will be submitted for publication elsewhere.

1

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not necessarily equal); similarly,XY means that the inclusion is continuous (but not necessarily isometric). Cwill be used to denote unspecied constants (possibly depending on some parameters that are kept xed); the meaning may change from one occurrence to the next.

Suppose that (X⁰;X¹) is a Banach couple. We dene ^FS =^FS(X⁰;X¹) as the space of all functions F from the closed strip S =^f0 Rez 1^ginto X⁰+X¹ that are bounded and continuous on S and analytic on the interior S, and such that t ^7! F(j+it) is a bounded and continuous map of the real line into Xj for j= 0;1. We let

kF^k^FS = sup_j

=0;¹ sup

1<t<¹^kF(j+it)^kX^j:

The complex interpolation space [X⁰;X¹] is dened for 0< <1 by [X⁰;X¹]=^fF() :F ²^FS^g;

equipped with the natural quotient norm ^kx^k = inf^fkF^k^F^S : F() = x^g, cf.

Calderon [

2

] and Bergh and Lofstrom [

1

^].

Remark. It is customary to impose also the condition thatF(j+it)^!0 in Xj

ast^!¹, which is convenient for some purposes but not for ours; it is easily seen that this yields the same interpolation space. (On the other hand, the continuity condition on the boundary is essential, see [

4

^].)

For the half-plane version, we let ^FH =^FH(X⁰;X¹) be the space of bounded continuous functions on the closed half-plane H=^fz: Rez0^gthat are analytic on the open half-plane H = ^fz : Rez > 0^g and such that the restriction to S belongs to^FS. We regard^FH as a subspace of ^FS; it is easily seen that ^FH is a closed subspace of^FS, and thus a Banach space. We dene, again for 0< <1,

C⁺(X⁰;X¹) =^fF() :F ²^FH(X⁰;X¹)^g;

this is a Banach space with the natural quotient norm. It should be clear thatC⁺ is an interpolation method.

Moreover, it follows easily, e.g. by applying linear functionals in (X⁰+X¹), that ifF ²^FH(X⁰;X¹) andz²H, thenF(z) is given by the integral^R¹¹F(it)Pz(t)dt, wherePz(t) denotes the appropriate Poisson kernel. Sincet^!F(it) by assumption is a bounded continuous map intoX⁰, this integral converges inX⁰, andz^7!F(z) is a bounded continuous map ofH intoX⁰. Hence

FH(X⁰;X¹) =^FH(X⁰;X⁰^\X¹) and thus

C⁺(X⁰;X¹) =C⁺(X⁰;X⁰^\X¹): (1.1) It is obvious that

C⁺(X⁰;X¹)[X⁰;X¹] (1.2)

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and (1.1) thus implies

C⁺(X⁰;X¹)[X⁰;X⁰^\X¹]: (1.3) The rst question given in the introduction is whether equality holds in (1.2).

We see from (1.1) that it is natural to consider the case X¹ X⁰ only; this is equivalent to asking whether equality always holds in (1.3). A counter example is given in the next section, but rst we introduce the annulus and disc versions of the denitions above.

LetR⁰ andR¹ be two xed real numbers with 0< R¹< R⁰, and deneR= R⁰¹ R¹, 0< <1. We consider the annulusA=^fz:R¹<^jz^j< R⁰^gand the disc D=^fz:^jz^j< R⁰^g.

We dene ^FA =^FA(X⁰;X¹) as the space of all bounded, continuous functions F from A into X⁰+X¹ that are analytic on A, and such that t^7! F(R_je^it) is a continuous map of the real line intoXj forj= 0;1; we let

kF^k^F^A= sup_j

=0;¹ sup

jz^j=R^j^kF(z)^kX^j:

We also dene^FD =^FD(X⁰;X¹) as the space of all bounded continuous functions from D into X⁰+X¹ that are analytic on D and such that the restriction to Abelongs to ^FA(X⁰;X¹). We regard^FD as a subspace of^FA, and we use the subspace norm;^FDis a closed subspace and thus a Banach space. It is easily seen, as for^FH above, that^FD(X⁰;X¹) =^FD(X⁰;X⁰^\X¹).

Cwikel [

3

] showed that the complex method may be dened using analytic functions in an annulus; more precisely,

[X⁰;X¹]=^fF(R) :F ²^FA(X⁰;X¹)^g:

Cwikel's method also shows the corresponding result for the half-plane and disc.

Proposition 1. For every Banach couple (X⁰;X¹),

C⁺(X⁰;X¹) =^fF(R) :F ²^FD(X⁰;X¹)^g:

Proof. Let= ln(R⁰=R¹)>0. First, ifF ²^FD, thenG(z) =F(R⁰e ^z)²^FH, and thusF(R) =G()²C⁺(X⁰;X¹).

Conversely, if F ² ^FH, let F¹(z) = ^e ^(z⁽_z⁾⁾ ¹²F(z). Then F¹ ² ^FH, with F¹() =F(), and^kF¹(z)^kX⁰⁺X¹ C=(1 +^jz^j²). Hence the sum

F²(z) = ^X¹

k⁼ ¹F¹(z+ 2ik=)

converges for all z ² H; moreover,F² ² ^FH and F²() = F¹() = F(), because F¹(+ 2ik=) = 0 when k ⁶= 0, and^kF²(z)^kX⁰⁺X¹ C=(1 + Rez). Since F² is

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periodic with period 2i=, we may dene G(z) =F²^ln(R⁰=z)

; 0<^jz^jR⁰;

regardless of the branch of the logarithm. ThenGis an analytic (X⁰+X¹)-valued function in the punctured discDⁿ^f0^g, and since^kG(z)^kX⁰⁺X¹ ^!0 asz^!0, the origin is a removable singularity and if we deneG(0) = 0,Gbecomes analytic in D. It is easily seen thatG²^FD andG(R) =F²() =F(), which completes the

proof.

Consequently, the two questions in the introduction are equivalent.

2. A counter example

LetR⁰,R¹andbe given with 0< R¹< R⁰and 0< <1, and let⁰= lnR⁰, = lnR¹. ThusR=R¹⁰ R¹=e⁽¹ ⁾⁰⁺¹. We may without loss of generality assumeR= 1, and thus⁰>0,¹<0 and

(1 )⁰+¹= 0: (2.1)

LetFLdenote the space of Fourier coecients of functions in L¹(

T

):

FL=^f(f^b(n))¹_n⁼ ¹:f ²L¹(

T

⁾^g^;

this is a Banach space with the norm^k(f^b(n))¹¹^kFL=^kf^k_L¹. Dene also corresponding weighted spaces by

FL=^f(xn)¹¹: (e ⁿxn)¹¹²FL^g;

where is a real number, with ^k(xn)¹¹^kFL =^k(e ⁿxn)¹¹^kFL. Note that if (xn)¹¹²FL, then

jxn^jeⁿ^k(xk)¹¹^kFL: (2.2) It is known that [FL⁰;FL¹]=FL⁽¹ ⁾⁰⁺¹ =FL, see e.g. [

7

^].

LetX⁰=FL⁰+FL¹, and X¹=FL¹; we claim that, cf. (1.2),

C⁺(X⁰;X¹)⁽[X⁰;X¹]: (2.3) In order to see this, we rst observe that

[X⁰;X¹][FL⁰;FL¹]=FL:

On the other hand, we claim that if (xn)¹¹²C⁺(X⁰;X¹) andq >1=, then

1

X

1

jxn^jq=n <¹: (2.4) Consequently, in order to verify (2.3) it suces to show that there is a sequence (xn)¹¹²FLsuch that (2.4) fails. This can be done by explicit examples, such as 1=lnln(3 +^jn^j), cf. Zygmund [

14

], Theorem V.1.5, or by observing that otherwise

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the closed graph theorem would imply^P¹¹ ^jf^b(n)^j^q=nCfor someC <¹and all f ²L¹(

T

^{) with}^k^f^kL¹ 1, but that is false as the sequence of Fejer kernels shows.

We turn to the proof of (2.4). We use the characterization with^FDin Proposition 1. Suppose that F ² ^FD(FL⁰ +FL¹;FL¹) with ^kF^k 1. We write F(z) = (fn(z))¹¹, thus eachfn is analytic inD and continuous onD; we further expand eachfn as a Taylor series

fn(z) =^X¹

k⁼⁰ankz^k; and setank= 0 fork <0.

If^jz^j=R⁰, thenF(z)²FL⁰+FL¹, with norm1 and thus, cf. (2.2),

jfn(z)^jmax(eⁿ⁰;eⁿ¹) =

eⁿ⁰; n0; eⁿ¹; n <0: Consequently, for anyk,

ja_nkR^k⁰^j sup

jz^j=R⁰^jf_n(z)^jeⁿ⁰; n0 or, recalling thatR⁰=e⁰,

jan;n⁺m^jR⁰^{n m}eⁿ⁰ =e ^m⁰; n0; ¹< m <¹: (2.5) We now turn to z with ^jz^j = R¹. By assumption, F(R¹e^it) = (fn(R¹e^it))¹¹ belongs to the unit ball of FL¹ for every real t. Hence there exist gt ²L¹(

T

), with^kgt^kL¹1, such that

bgt(n) =e ¹ⁿfn(R¹e^it) =^X¹

k⁼⁰e ¹ⁿankR^k¹e^ikt=^X¹

k⁼⁰anke ¹⁽^{n k}⁾e^ikt: (2.6) Moreover, the mappingt^7!g_tis a continuous map

R

^!^L¹(

T

).

Letsbe the translation operator onL¹(

T

) given bysg(t) =g(t s) and thus (sg)^{^}(n) =e ^ins^bg(n) (2.7) Since, for anyt and",

kt⁺"gt⁺" tgt^kL¹ ^kt⁺"(gt⁺" gt)^kL¹+^kt⁺"gt tgt^kL¹

=^kg_t⁺_" g_t^k_L¹+^k_"g_t g_t^k_L¹;

it follows that t^7!tgt is a continuous map intoL¹(

T

). Thus we may dene, for m²

Z

, the Bochner integral

hm= ²¹

Z

2

e ^imttgtdt²L¹(

T

);

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with^khm^kL¹ 1. Then, by (2.7) and (2.6),

bhm(n) = ²¹

Z

2

0

e ^imte ^int^bgt(n)dt= ²¹

Z

2

0 1

X

k⁼⁰anke ¹⁽^{n k}⁾eⁱ⁽^{k m n}⁾^tdt

=an;n⁺me¹^m; n;m²

Z

:

In particular,^bhm(n) = 0 ifn+m <0. We also see that

jan;n⁺m^j=e ¹^m^j^bhm(n)^je ¹^m^khm^kL¹ e ¹^m: (2.8) Let vm be the sequence (an;n⁺m)¹_n⁼¹, and let `^q(1=n) be the sequence space

f(xn)¹¹ :^P¹¹ ^jxn^jq=n <^1g.

Consider rst the case m 0. Then ^bhm(n) = 0 when n < 0, so hm belongs to the analytic Hardy spaceH¹ L¹. By Hardy's inequality, see Zygmund [

14

^],

Theorem VII.8.7,

1

X

1

1n^jan;n⁺m^j=e ¹^m^X¹

1

n1^j^bhm(n)^jCe ¹^m^khm^kL¹Ce ¹^m and thus, using (2.8),

1

X

1

n1^jan;n⁺m^jq e ⁽^q ¹⁾¹^m^X¹

1

n1^jan;n⁺m^jCe ^q¹^m; or in the notation just introduced,

kvm^k`^q⁽¹=n⁾Ce ¹^m; m0: (2.9) Consider next the casem >0. Then^bhm(n) = 0 whenn < m, and the shifted sequence (ak m;k)¹_k⁼ ¹ is the Fourier transform of the function e ¹^me^imthm(t) inH¹. Hence Hardy's inequality yields

1

X

1

n1^jan;n⁺m^j(m+ 1)^X¹

1

n+1m^jan;n⁺m^j(m+ 1)^X¹

1

k1^jak m;k^j

(m+ 1)C^ke ¹^me^imth_m(t)^kL¹ C(m+ 1)e ¹^m: We combine this with (2.5) and obtain

1

X

1

n1^jan;n⁺m^jqe ⁽^q ¹⁾⁰^m^X¹

1

n1^jan;n⁺m^jC(m+ 1)e ⁽^q ¹⁾⁰^m¹^m: (2.10) From (1 )⁰+¹ = 0 we obtain q⁰ = q(⁰ ¹) and (q 1)⁰+¹ = (q 1)(⁰ ¹). Hence (2.10) yields

kvm^k`^q⁽¹=n⁾Cm¹^=qe ⁽ ¹^=q⁾⁽⁰ ¹⁾; m >0: (2.11)

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Since¹<0 and ( 1=q)(⁰ ¹)>0, (2.9) and (2.11) together imply

1

X

m⁼ ¹

kvm^k`^q⁽¹=n⁾<¹; and thus^P¹¹vm²`^q(1=n). But

1

X

1

vm= ^X¹

m⁼ ¹an;n⁺m

1

n⁼¹=^X¹

k⁼⁰an;k

1

n⁼¹= (fn(1))¹_n⁼¹:

This proves (2.4) for the sequence (xn)¹¹= (fn(1))¹¹=F(1) =F(R), and the proof is complete.

3. Background

The example in Section 2 solves our problem, but how was it found? The couple (FL⁰+FL¹;FL¹) may look rather unnatural at the rst sight, but the following argument shows that this is in fact the canonical (counter) example.

If Aê, Xê and A are Banach spaces such that A Aê, and J B(A;ê Xê) is a Banach space of bounded linear operators Aê ^! Xê, then the orbit of A under J is the subspace of Xê given by ^f^P¹¹ Tiai : ^P¹_i ^kTi^kJ^kai^kA <^1g (with the sum

PTiai converging in X^e). This is a Banach space with ^kx^k= inf^P_i^kTi^kJ^kai^kA

over all such representations ofx.

A particularly important example is when (A⁰;A¹) and (X⁰;X¹) are two Banach couples, Aê = A⁰+A¹, Xê = X⁰+X¹ and J = ^fT ² B(A;ê Xê) ^j T:(A⁰;A¹) ^! (X⁰;X¹)^gis the space of all linear operators from A⁰+A¹ to X⁰+X¹ that map A⁰ into X⁰ and A¹ into X¹. We then denote the orbit of a space A A⁰+A¹ by G(A⁰;A¹;A;X⁰;X¹). For xed A⁰, A¹ and A ⁶=^f0^g, this is an interpolation method, and it is easily seen that this is the minimal interpolation method that satisesF(A⁰;A¹)A.

Now consider the action of this interpolation method on the couple (X⁰;X⁰^\X¹).

Since

T:(A⁰;A¹)^!(X⁰;X⁰^\X¹) ⁽⁾ T:A⁰^!X⁰; T:A¹^!X⁰; T:A¹^!X¹;

() T:(A⁰+A¹;A¹)^!(X⁰;X¹); we obtain the following identity.

Proposition 2. Let(A⁰;A¹) and (X⁰;X¹) be Banach couples, andAA⁰+A¹. Then G(A⁰;A¹;A;X⁰;X⁰^\X¹) =G(A⁰+A¹;A¹;A;X⁰;X¹):

Proposition 3. Let (A⁰;A¹) be a Banach couple and A A⁰+A¹. If F is an interpolation method, then the following are equivalent.

(i) F(X⁰;X¹) G(A⁰;A¹;A;X⁰;X¹) for all Banach couples (X⁰;X¹) such thatX⁰X¹.

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(ii) F(X⁰;X⁰^\X¹)G(A⁰;A¹;A;X⁰;X⁰^\X¹) for all Banach couples (X⁰;X¹).

(iii) F(A⁰+A¹;A¹)A.

Proof. The equivalence of (i) and (ii) is clear.

By the comments before Proposition 2, (iii) holds if and only if G(A⁰+A¹;A¹;A;X⁰;X¹)F(X⁰;X¹)

for every couple (X⁰;X¹), which by Proposition 2 easily is seen to be equivalent to

(ii).

The complex method of interpolation [X⁰;X¹]can be characterized as the orbit G(FL⁰;FL¹;FL⁽¹ ⁾⁰⁺¹;X⁰;X¹) for any real ⁰ and ¹ with⁰ ⁶=¹, see [

7

]. We thus have the following corollaries.

Corollary 1. Let⁰⁶=¹and0< <1. Then, for any Banach couple (X⁰;X¹), [X⁰;X⁰^\X¹]=G(FL⁰+FL¹;FL¹;FL⁽¹ ⁾⁰⁺¹;X⁰;X¹):

Corollary 2. Let⁰⁶=¹ and0< <1 and let F be an interpolation method.

ThenF(X⁰;X¹)[X⁰;X¹] for all Banach couples(X⁰;X¹) withX⁰X¹, if and only if

F(FL⁰+FL¹;FL¹)FL⁽¹ ⁾⁰⁺¹:

Corollary 2 thus shows that if C⁺(X⁰;X¹) [X⁰;X¹] fails for any couple (X⁰;X¹), then it fails for (FL⁰+FL¹;FL¹).

4. Further comments

The interpolation method C⁺ is probably not of much practical use, but let us nevertheless give a couple of results for it. First, as another example of the propositions in Section 3, consider the -method dened by Peetre [

12

^{]. It was}

shown in [

7

] that this method, there and here denoted byG¹, can be characterized as G(c⁰;⁰;c⁰;¹;c⁰_;⁽¹ ⁾⁰⁺¹;X⁰;X¹), where c⁰; is a weighted version of c⁰ =

f(x_n)¹¹:x_n^!0 as^jn^j^!^1g.

It is easily veried (we omit this) thatC⁺(c⁰_;⁰+c⁰_;¹;c⁰_;¹)c⁰_;⁽¹ ⁾⁰⁺¹. Proposition 3 thus implies, together with (1.2), the following.

Proposition 4. For any Banach couple(X⁰;X¹) such that X⁰X¹, G¹(X⁰;X¹)C⁺(X⁰;X¹)[X⁰;X¹]:

In particular, this shows that if (X⁰;X¹) is a Banach couple such thatX⁰X¹

and G¹(X⁰;X¹) = [X⁰;X¹]; (4.1)

then [X⁰;X¹] may be dened usingX⁰-valued functions F that are analytic in a half-plane or disc, as dened in detail in Section 1.

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Remark. One important case when (4.1) holds, pointed out to me by Michael Cwikel, is for a couple of lattices on the same measure space (;;) i.e. when Xj =Yj(

C

) is the complexication of a Banach latticeYjof real valued measurable functions on for j = 0;1. This fact has been observed by a number of authors.

(Sometimes they impose additional conditions.) For the reader's convenience let us list some results which can be combined to immediately prove it:

(i) the continous inclusion [X⁰;X¹]Y⁰¹ Y¹(

C

) (see [

2

] section 13.6 (i) p. 125), (ii) the continuous inclusion Y⁰¹ Y¹(

C

)G²(X⁰;X¹) (see [

11

] Lemma 8.2.1 p.

453.)

The rest of these \ingredients" also hold for arbitrary Banach couples (X⁰;X¹).

(iii) the density ofX⁰^\X¹ in [X⁰;X¹]( [

2

] p. 116) and the fact that the closure ofX⁰^\X¹ inG²(X⁰;X¹) isG¹(X⁰;X¹). ([

7

] Theorem 8 p. 60.)

(iv) the continuous inclusionG¹(X⁰;X¹)[X⁰;X¹]. (See [

12

] p. 176 or [

7

^{] p. 67).}

The interpolation methodC⁺can also be represented as an orbit method. Dene P⁺(xn)¹¹ = (xn)¹⁰ , the restriction of a doubly innite sequence to non-negative indices, and letFL⁺ =P⁺(FL), equipped with the quotient norm.

Proposition 5. Let⁰> ¹. Then for any Banach couple(X⁰;X¹), C⁺(X⁰;X¹) =G(FL⁺⁰;FL⁺¹;FL⁺⁽¹ ⁾0

+¹;X⁰;X¹):

Proof. We may assume (1 )⁰+¹ = 0. We use Proposition 1, choosing R⁰=e⁰andR¹=e¹, and thusR= 1. First, suppose thatx= (xn)¹¹is a nite sequence of complex numbers, i.e. a sequence with all but nitely many elements 0. ThenF(z) = (x_nzⁿ)¹⁰ denes an entire analytic function intoFL⁺⁰^\FL⁺¹ and ifz=Rje^it,j = 0 or 1, then

kF(z)^k_FL⁺ j

=^k(xne^int)¹⁰ ^k_FL⁺ ^k(xne^int)¹¹^kFL=^k(xn)¹¹^kFL: Hence F(z) ² ^FD(FL⁺⁰;FL⁺¹) with norm ^kx^k_FL, and thus P⁺x = F(1) ² C⁺(FL⁺⁰;FL⁺¹) with norm^kx^kFL. The set of all nite sequences is dense in FL, so by continuity

FL⁺=P⁺(FL)C⁺(FL⁺⁰;FL⁺¹); which implies, by the minimality of the interpolation functorG,

G(FL⁺⁰;FL⁺¹;FL⁺;X⁰;X¹)C⁺(X⁰;X¹) for every Banach couple (X⁰;X¹).

In order to prove the converse, suppose that F ² ^FD(X⁰;X¹), and expand F as a Taylor seriesF(z) =^P¹⁰ xkz^k, withxk ²X⁰^\X¹. If (an)¹¹ ²FL⁰, then

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an=eⁿ⁰f^b(n) for somef ²L¹(

T

) and (withX⁰-valued integrals)

1

2

Z

2

0

f(e ^it)F(R⁰e^it)dt= lim_r

%1 1

2

Z

2

0

f(e ^it)F(rR⁰e^it)dt= lim_r

%1 1

X

n⁼⁰

fb(n)rⁿRⁿ⁰x_n

= lim_r

%1 1

X

n⁼⁰rⁿanxn: (4.2)

Hence the mapping T : (an)¹¹ ^! limr^%1^P¹n⁼⁰rⁿanxn is a well-dened linear mapFL⁰ ^!X⁰, and

kT((an)¹¹)^kX⁰ ^kf^k_L¹sup_t ^kF(R⁰e^it)^kX⁰ ^k(an)¹¹^kFL⁰^kF^k^FD: SinceTaobviously depends onP⁺aonly, we can also regardT as a bounded linear mapFL⁺⁰ ^!X⁰.

Similarly if (an)¹⁰ ²FL⁺¹, then an =eⁿ¹f^b¹(n), n 0, for some f¹ ² L¹(

T

⁾

and, by the same argument as in (4.2), T((an)¹⁰ ) =²¹^Z ²

0

f¹(e ^it)F(R¹e^it)dt;

with the integral convergent inX¹. It follows thatT:FL⁺¹ ^!X¹. Thus T:(FL⁺⁰;FL⁺¹)^!(X⁰;X¹);

and we see also that ^kT^k^kF^k^F^D. Furthermore, if f ²L¹(

T

), then by the same argument again,

1

2

Z

2

0

f(e ^it)F(e^it)dt=T((f^b(n))¹¹): Thus, if we letf F denote theX⁰-valued function

fF(z) = ²¹

Z

2

0

f(e ^it)F(ze^it)dt;

andY =G(FL⁺⁰;FL⁺¹;FL⁺;X⁰;X¹), then by the denition of the latter space, f F(1) =T(f^b)²Y with ^kfF(1)^kY ^kf^k_L^1kF^k^FD: (4.3) It is easily seen that f F ² ^FD(X⁰;X¹) for all f ² L¹(

T

), and that, if Kn

denotes then:th Fejer kernel, KnF ^!F in ^FD(X⁰;X¹) as n ^!¹; moreover, (fg)F=f(gF) for allf;g²L¹(

T

^{). Dene}^yn =KnKnF(1). By (4.3), yn ²Y. Furthermore,

yn ym= (Kn+Km)(Kn Km)F(1) and thus, by (4.3) for (Kn Km)F,

kyn ym^kY ^kKn+Km^kL¹^k(Kn Km)F^k^FD 2^kKnF KmF^k^FD ^!0;

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as n;m ^! ¹. Hence ^fyn^g is a Cauchy sequence in Y, so yn ^! y ² Y. But KnKnF ^!F in^FD(X⁰;X¹), and thusyn^!F(1) inC⁺(X⁰;X¹). Consequently

F(1) =y²Y, which completes the proof.

Remark. It is also possible to dene interpolation methods using harmonic functions in a half-plane or disc, see Janson and Peetre [

9

]. The resulting interpolation spaces contain the ones given by analytic functions, and are in general larger.

For harmonic interpolation, we do not know if the half-plane and and disc versions always yield the same interpolation spaces, or even if the disc interpolation spaces are independent of the ratio R⁰=R¹ as in the analytic case, cf. Proposition 1. For the disc version, with xedR⁰=e⁰ andR¹=e¹, the following analogues of the results above hold. (We omit the proofs.)

First, a somewhat more complicated version of the argument in Section 2 shows that, at least for the disc version, the harmonic interpolation space for the couple (FL⁰+FL¹;FL¹) does not contain the standard complex method space.

Furthermore, the method has an orbit description asG(A⁰;A¹;A;X⁰;X¹), whereA=^f(an)¹¹: (e ^jⁿ^jan)¹¹²FL^g.

An argument similar to the one in Section 2 shows that for the couple (A⁰;A¹), the standard complex method space does not contain the harmonic interpolation space. Consequently, even assuming X⁰ X¹, the harmonic method and the standard complex method are not comparable.

Finally, the harmonic method space is included in the one given by Ovchinnikov's method u [

10

], denoted by H¹ in [

7

]. It follows that for `tame' couples, the harmonic and analytic disc methods coincide with the standard complex method.

5. Appendix:

ANOTHER CASE WHERE

C⁺(X⁰;X¹) = [X⁰;X¹]

. by Michael Cwikel

Here we consider couples (X⁰;X¹) which can be obtained by \one sided reiteration".

Theorem 5.1. Let(X⁰;X¹) be a Banach couple satisfyingX¹X⁰. Suppose that there exists another Banach spaceB such that(X⁰;B) forms a Banach couple and such that

X¹= [X⁰;B] for some²(0;1): Then C⁺(X⁰;X¹) = [X⁰;X¹] for all ²(0;1):

Proof. Let (Y⁰;Y¹) be the Banach couple obtained by settingY⁰=X¹ andY¹=

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X⁰, and let= 1 . I.e. we haveY⁰Y¹ and

Y⁰= [B;Y¹] (5:2) (We introduce Y⁰ and Y¹ and because this is the easiest way of adapting our original version of this proof, which was written without access to a denitive version of the previous sections of this paper, to the format of the notation used in those sections.)

The proof of this theorem amounts to showing that [X⁰;X¹]C⁺(X⁰;X¹) for all²(0;1). This is of course equivalent to showing that

[Y⁰;Y¹]C¹⁺(Y¹;Y⁰) for all ²(0;1): (5:3) We may suppose without loss of generality thatB^\Y¹is dense inB, since if not we can simply replaceBbyB= [B;Y¹]⁰, the closure ofB^\Y¹inB, which satises [B;Y¹] = [B;Y¹]. (Cf. [

2

] Paragraph 9.3 p. 116.) It follows immediately from (5.2) that

ka^k_Y⁰ ^ka^k¹_B ^ka^k_Y¹ for every a ² B^\Y¹. But since also ^ka^k_Y1

C^ka^k_Y0 for some constant C, we deduce that ^ka^k_Y⁰ C⁼⁽¹ ⁾^ka^k_B for all such a. Using the density of B ^\Y¹ in B we deduce that the previous estimate holds for all a²B and so obtain the continuous embedding

BY⁰: (5:4)

By [

2

] Paragraph 9.3 p. 116, the spaces [B;Y¹]¹and [Y⁰;Y¹]¹are the completions ofB and of Y⁰ respectively inY¹. SinceB =B^\Y¹ is dense inY⁰, it follows that [B;Y¹]¹ = [Y⁰;Y¹]¹. Applying [

2

] Paragraph 9.3 p. 116 once more, we have that [Y⁰;[Y⁰;Y¹]¹] = [Y⁰;Y¹] and so [Y⁰;Y¹] = [[B;Y¹];[B;Y¹]¹]. Applying the reiteration formula (see [

3

] p. 1005) to this latter space we deduce that

[Y⁰;Y¹]= [B;Y¹]⁽¹ ⁾⁺: (5:5) Let us now consider an arbitrary xed element a ² [Y⁰;Y¹]. By (5.5) it can be represented in the form a = g¹((1 )+) for some g¹ ² ^FS(B;Y¹). Now letg²(z) =g¹((1 z)+z). Then g² is a continuous and bounded B+Y¹ =Y¹ valued function on the strip Rez²[¹ ;1] and it is analytic in the interior of this strip. Furthermore g²() = a and the restrictions of g² to the lines Rez = ¹ and Rez= 0 are continuous boundedB valued and [B;Y¹]=Y⁰ valued functions respectively.

The next step will be to use the construction dened in [

3

] p. 1008. (Cf. also the proof of Proposition 1 above.) This enables us to construct fromg²a new function g³ which has all the properties listed above forg²and also the additional property thatg³(z+ 2i) =g³(z) for allz in the strip Rez²[¹ ;1] . (The functionsw(z) ande⁽^z⁾²which were used in that construction on the strip Rez²[0;1] of course also satisfy estimates of the required form on the wider strip which we need here.)