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Complex interpolation of compact operators mapping into the couple (FL^{\infty},FL_{1}^{\infty})

If (A_0,A_1) and (B_0,B_1) are Banach couples and a linear operator T from A_0 + A_1 to B_0 + B_1 maps A_0 compactly into B_0 and maps A_1 boundedly into B_1, does T necessarily also map [A_0,A_1]_s compactly into [B_0,B_1]_s for s in (0,1)? After 42 years this question is still not answered, not even in the case where T is also compact from A_1 to B_1. But affirmative answers are known for many special choices of (A_0,A_1) and (B_0,B_1). Furthermore it is known that it would suffice to resolve this question in the special case where (B_0,B_1) is the special couple (l^\infty(FL^\infty), l^\infty(FL^\infty_1)). Here FL^\infty is the space of all sequences which are Fourier coefficients of bounded functions, FL^\infty_1 is the weighted space of all sequences (a_n) such that (e^n a_n) is in FL^\infty, and thus B_0 and B_1 are the spaces of bounded sequences of elements in these spaces (i.e., they are spaces of doubly indexed sequences). We provide an affirmative answer to this question in the related but simpler case where (B_0,B_1) is the special couple (FL^\infty,FL^\infty_1).


Introduction
This paper deals with the following question, which originated with Alberto Calderón's study [3] of complex interpolation spaces, and has been open for the 42 years which have elapsed since the writing of [3].
Question C: Suppose that A 0 and A 1 are compatible Banach spaces, i.e., they form a Banach pair, and that so are B 0 and B 1 . Suppose that T : A 0 + A 1 → B 0 + B 1 is a linear operator such that T : A 0 → B 0 compactly and T : A 1 → B 1 boundedly. Does it follow that T maps the complex interpolation space [A 0 , A 1 ] θ into the compact interpolation space [B 0 , B 1 ] θ compactly for each θ ∈ (0, 1)?
Affirmative answers have been obtained over the years for a considerable number of particular cases of Question C. See, for example, [3], [16], [17], [5], [6], [9], and [11]. More recent studies of other aspects of this question can be found in [4], [12], [13] and [19]. We also refer to [10] and the website [7] for some general remarks and further details about this question and related questions, including a remarkable "almost counterexample" found by Fedor Nazarov.
Although we still cannot answer Question C, we can, apparently for the first time in nearly thirteen years, i.e., since the submission of [11], further enlarge the family of partial affirmative answers.
Our main result in this paper, Theorem 14, immediately and obviously implies that the answer to Question C is yes in one more particular case, namely where (A 0 , A 1 ) is an arbitrary Banach couple, and (B 0 , B 1 ) is the special couple of sequence spaces (F L ∞ , F L ∞ 1 ) of Fourier coefficients which was introduced and studied in [14].
To the best of our knowledge, this particular case of Calderón's problem cannot be resolved via any other methods or cases treated in previous partial solutions of the problem.
We believe that there ought to be a simpler proof of Theorem 14 than the one given here. One of the main underlying ideas is the use of Lusin's theorem, which leads us to the intermediate result Theorem 4. Intuitively it seems rather clear that Theorem 4 implies Theorem 14. But, unfortunately, we have not found any way so far to bypass the lengthy explanation which we need for the technical details of this implication.
We thank Fedor Nazarov for many interesting discussions and enlightening comments about Question C in general.

Some preliminaries
As usual we let T denote the torus {z ∈ C : |z| = 1}. Define the σ-algebra T of measurable subsets of T to consist of all images of Lebesgue measurable subsets of [0, 2π) under the bijection φ(t) = e it and define µ to be the measure on these sets defined by taking µ(E) to be the Lebesgue measure of φ −1 (E). In other words, µ is simply arc length measure on T.
The distance d(z 1 , z 2 ) between two points z 1 and z 2 of T will be taken to be the length of the shortest arc in T joining them, i.e., d(z 1 , z 2 ) = d(1, z 2 /z 1 ) = |t| where t is the unique number in (−π, π] which satisfies z 2 /z 1 = e it . For each z ∈ T and each r ∈ (0, π], we let Γ(z, r) be the closed arc in T of length 2r centred at z, i.e., Γ(z, r) = ze it : |t| ≤ r = {w ∈ T : d(w, z) ≤ r}.
Let L 1 (T) =L 1 (T, T , µ) and L ∞ (T) =L ∞ (T, T , µ) denote the usual Lebesgue spaces of (equivalence classes of) measurable functions f : T → C normed by respectively. For each r ∈ (0, π] and z ∈ T the formula obviously holds whenever f = χ E and E ∈ T , and therefore it also holds for all f ∈ L 1 (T).
We conclude this section by recalling some standard facts related to the Lebesgue differentiation theorem, and expressing them in ways that will be convenient for our purposes here. Definition 1. For each µ-integrable function f : T → C, let Ω f denote the set of all points z ∈ T for which the limit lim r→0 It will be convenient to refer to the points of Ω f as the weak canonical Lebesgue points of f .
If two µ-integrable functions f and g coincide almost everywhere on T, then of course Ω f = Ω g and so f • (z) = g • (z) for all z ∈ T. Thus our definitions of Ω f and f • extend unambiguously to the case where f is an element of L 1 (T) rather than a single function. It will be convenient to refer to the function f • as the canonical representative of the element f . It is an obvious consequence of the Lebesgue differentiation theorem that Ω f ∈ T with µ(T\Ω f ) = 0 and that f (z) = f • (z) for almost every z ∈ T. This in turn implies that ess sup Definition 2. For each set E ∈ T , let E (d) denote the set of all points of density of E, i.e., the points z in T for which the limit lim r→0 1 2r r −r χ E (ze it )dt exists and is equal to 1, i.e., It is of course a classical result, deduced by applying the Lebesgue differentiation theorem to the function , this implies in turn that µ(E ∩ E (d) ) = µ(E) and so, of course, (2) µ(T\(E ∩ E (d) )) = µ(T\E).

The main ingredient
Our goal in this section is to prove Theorem 4 which will subsequently be the principal tool in the proof of our main result. But first we need the following lemma: Lemma 3. Let K be a relatively compact subset of L ∞ (T). Then, for each ǫ > 0, there exists a number δ = δ(ǫ) and a set U = U ǫ ⊂ T such that and, for each f ∈ K, More precisely, Proof. Since K is relatively compact, there exists some integer N and a collection of The σ-algebra T contains all the Borel sets in T, and the measure µ is finite, regular and complete on T , i.e., it satisfies conditions (b), (c) and (d) of Theorem 2.14 of [18] pp. 40-41. Thus we can apply Lusin's Theorem ( [18] Theorem 2.24 p. 55) to obtain, for each j ∈ {1, 2, ..., N }, a continuous function g j : T → C such that the set G j = {z ∈ T : f • j (z) = g j (z)} has measure µ(G j ) < ǫ/N . Since each g j is uniformly continuous, we can choose δ = δ(ǫ) to be a number with the property that Let E be the set E = T\ N j=1 G j . Then we will choose the set U to be U = E ∩ E (d) . Clearly T\E = N j=1 G j has measure µ N j=1 G j not exceeding N j=1 µ(G j ) < ǫ, and so (3) follows immediately from (2). For each point z ∈ T and each r ∈ (0, π] and each j ∈ {1, 2, ..., N }, we have If z ∈ U , and is therefore a point of density of E, then the preceding estimates show that lim sup Now suppose that f is some arbitrary element of K. Then there exists some integer j depending on f such that (6) f ∈ B(f j , ǫ/3).
Suppose that z 1 and z 2 are arbitrary points in U ∩ Ω f satisfying d(z 1 , z 2 ) < δ. For each w ∈ T we have for every w ∈ T. So, for each r ∈ (0, π], we can set w = e it and integrate the inequality (7) with respect to t on the interval [−r, r] to obtain that 1 2r In view of (5) both of the integrals in the preceding line tend to 0 as r tends to 0, and so we have shown that Finally, since z 1 and z 2 are both in Ω f , we see that This, combined with (8) establishes (4) and completes the proof of Lemma 3.
In order to be able to conveniently apply the next theorem later in the proof of our main result, we will have to distinguish, more pedantically than is usually necessary, between elements of L ∞ (T) and the functions that represent them. So here we will use bold lower case letters for elements (equivalent classes of complex valued functions) in L ∞ (T) and usual italic lower case letters for the complex valued functions in those equivalence classes.  For each δ > 0, define (11) If K is a relatively compact subset of L ∞ (T) then Proof. Let us begin by checking the almost obvious fact that the integral does not depend on our particular choices of the representatives v n of v n in the formula (9).
More explicitly, we claim that if φ(ζ, w) = M n=−M ζ n f n (w) where f n (w) = v n (w) for a.e. w ∈ T and for each n, then To establish (13) we first observe that, for each fixed z ∈ T, Next we note that, for each fixed z 1 and z 2 , Integrating, and using (14), we obtain that The reverse inequality to (15) is obtained exactly analogously and so indeed we obtain (13).
We will use the usual notation ηG = {ηw : w ∈ G} for each η ∈ C and for each subset G of some complex vector space W (which will often also be simply C).
Obviously the sets E K and the quantities ρ K (δ) satisfy E ηK = ηE K and so ρ ηK (δ) = ηρ K (δ) for each η > 0. So we can suppose, without loss of generality, that the given relatively compact set K is contained in the unit ball of L ∞ (T).
Choose a positive number ǫ and let U = U ǫ and δ = δ(ǫ) be a set and a number with the properties listed in the statement of Lemma 3, corresponding to our current choices of ǫ and of the relatively compact set K. We will show that, for this choice of δ, Since ǫ can be chosen arbitrarily small, this will of course suffice to establish (12).
Let u(ζ, w) = M n=−M ζ n v n (w) be an arbitrary element of E K . In view of (13) we may assume without loss of generality that it is in fact given by the formula where, for each n, v • n is the canonical representative of v n , and of v n the element of L ∞ (T) represented by v n ).
and, for each fixed ζ ∈ T , every point in Ω is a weak canonical Lebesgue point for the function w → u(ζ, w). (Of course the set of all weak canonical Lebesgue points for this function may be strictly larger than Ω.) In view of (10) and the fact that K is a subset of the unit ball of L ∞ (T), it follows that for each fixed ζ ∈ T, we have |u(ζ, w)| ≤ 1 for almost every w ∈ T. The exceptional null set may depend on ζ, but the next simple lemma shows that the exceptional null sets for different ζ are contained in a common null set.
for some fixed integer M and all ζ, w ∈ T, where v n are some measurable functions on T, and suppose that, for each fixed ζ ∈ T, |u(ζ, w)| ≤ 1 for almost all w ∈ T. Then, for almost all w ∈ T, |u(ζ, w)| ≤ 1 holds for all ζ ∈ T.
Proof. Let, for ζ ∈ T, N ζ be the null set {w : |u(ζ, w)| > 1}. Let Q be a countable dense subset of T (for example, the set of points with rational argument), and let N = ζ∈Q N ζ . The µ(N ) = 0, and if w / ∈ N , then |u(ζ, w)| ≤ 1 for all ζ ∈ Q. However, for fixed w, u(ζ, w) is a continuous function of ζ, so the inequality holds for all ζ ∈ T.
Obviously µ(ηG) = µ(G) for every measurable set G ⊂ T and every η ∈ T. Let z 1 and z 2 be arbitrary points in T satisfying . Whenever w ∈ F , both the points z 1 w and z 2 w are in U ∩ Ω. In view of (20) we also have d(z 1 w, z 2 w) < δ. Furthermore, (by (10) and (17)), for any fixed value of w in F , the function f (z) := u(w, z) is a representative of an element of K, the points z 1 w and z 2 w are both in Ω f , and Now we need to estimate the measure of the complement of the set F . We write and µ(T\F 1 ) = µ(T\F 2 ) = µ(T\U ). We also recall that the properties listed in the statement of Lemma 3, which U must satisfy, include the condition µ(T\U ) < ǫ. From these remarks, and the fact that T\(F 1 ∩ F 1 ) = (T\F 1 ) ∪ (T\F 2 ), we can now see that Finally, by (21), (19) and (22), we obtain that Since the element u of E K and the points z 1 and z 2 in T satisfying d(z 1 , z 2 ) < δ were chosen arbitrarily, this establishes (16) and so completes the proof of Theorem 4.

Some more preliminaries
Next we recall some standard things about Alberto Calderón's complex interpolation spaces. We choose versions of the notation and definitions to suit our particular purposes. Some of the facts that we prove are implicit or even sometimes explicit in Calderón's fundamental paper [3] about these spaces, or in other papers. But we prefer to give rather full explanations here. We will assume some familiarity with the basic notions and terminology of interpolation space theory as presented, for example, in the early chapters of [1] or of [2].
Let S be the "unit strip", i.e., S = {z ∈ C : 0 ≤ Re z ≤ 1} and let A denote the "unit annulus", A = {z ∈ C : 1 ≤ |z| ≤ e}. Calderón constructed his complex interpolation spaces in [3] via certain analytic vector valued functions defined on S. Most of the time here we prefer to use an alternative construction (cf. [8]) where S is replaced by A.
Let B = (B 0 , B 1 ) be an arbitrarily Banach couple of complex Banach spaces. For each f ∈ F A ( B) and each N ∈ N, we define the N th Fejér mean of f to be the function σ N (f ) : A → B 0 + B 1 given by the formula . Since (cf. [15] p. Definition 6. For any given subsets K 0 of B 0 and K 1 ⊂ B 1 , we let F A (K 0 , K 1 ) denote the subset of functions f ∈ F A ( B) with the additional property that f (e j+it ) ∈ K j for all t ∈ [0, 2π) and j = 0, 1.
Fact 7. Let K j be an arbitrary subset of B j for j = 0, 1. Let K j denote the closure in B j of the convex hull of K j and let K * j denote the closure in B j of the absolutely convex hull of K j . Let f be an arbitrary element of F A (K 0 , K 1 ). Then Let F L 1 be the space of complex sequences λ = {λ n } n∈Z which are Fourier coefficients of functions in L 1 (T), i.e., such that, for some function u = u λ ∈ L 1 (T), 2π 0 e −int u λ (e it )dt for all n ∈ Z.
We norm F L 1 by setting λ F L ∞ = u λ L 1 (T) . Analogously, we let F L ∞ be the space of complex sequences λ = {λ n } n∈Z which are Fourier coefficients of functions in L ∞ (T), i.e., such that the above function u λ is also in L ∞ (T). Here we use the norm λ F L ∞ = u λ L ∞ (T) . Let F C be the closed subspace of F L ∞ consisting of those sequences {λ n } n∈Z which are the Fourier coefficients of a continuous function on T, i.e., a function in C(T). (ii) If, furthermore, λ ∈ F L ∞ , then sup t∈[0,2π) P N (e it ) ≤ u λ L ∞ (T) .
(iii) If λ ∈ F C, then the convergence of the above sequence P N (e it ) N ∈N occurs for every t ∈ [0, 2π] and furthermore it is uniform on T, i.e., the sequence converges in L ∞ (T) norm.
For p = 1, ∞ and each fixed α ∈ R let F L p α be the space of complex sequences {λ n } n∈Z such that {e αn λ n } n∈Z ∈ F L p . It is normed by {λ n } n∈Z F L p α = {e αn λ n } n∈Z F L p . We also define F C α to be the closed subspace of F L ∞ α of sequences {λ n } n∈Z such that {e αn λ n } n∈Z ∈ F C.
These spaces are used extensively in [14], in particular to define the complex interpolation method as both a minimal and as a maximal method.
The continuous inclusion F L 1 θ ⊂ [ F L 1 ] θ was shown in [14], and it is not difficult to also obtain the reverse continuous inclusion. As explained in [12], the formula (28) [ F L ∞ ] θ = F C θ to within equivalence of norms can be deduced from [ F L 1 ] θ = F L 1 θ with the help of a duality argument. We need (28) crucially in this paper, and since existing proofs of it in the literature are not completely explicit, we provide a new proof below, in Appendix A. In fact there we prove the following slightly stronger result: Recall that the equality [ A] θ,A = [ A] θ holds for all Banach couples A [8], but in general the norms are equivalent only and not identical. (At least, this is proven for thick annuli in [8]; it is undoubtedly true for all annuli, but it seems that no-one has published a proof of this.) We next consider some special and natural maps on the spaces F L p α and F C α . The results in the following theorem and proposition are slight reformulations of straightforward and classical properties of analytic functions. But it seems preferable to state them explicitly in the notation needed for our purposes, and to provide explicit proofs.
Theorem 12. Suppose that the sequence λ = {λ n } n∈Z ∈ F L 1 is also an element of F L 1 1 . Then λ ∈ F C θ for all θ ∈ (0, 1). Furthermore, the limit exists for every z ∈ A • and almost every z ∈ T and almost every z ∈ eT, and the complex valued function v which it defines has the following properties: 2π 0 e −int v(e 1+it )dt = e n λ n for all n ∈ Z, (iii) for each θ ∈ (0, 1), the function e it → v(e θ+it ) is in C(T) and 1 2π 2π 0 e −int v(e θ+it )dt = e θn λ n for all n ∈ Z, and (iv) for each θ ∈ (0, 1), where the constant C θ depends only on θ.
For each N ∈ N we define the function (Laurent polynomial) P N : C\{0} → C by In view of Fact 8 (i), the sequence P N (e it ) N ∈N converges a.e. to u λ (e it ) for a.e. t ∈ [0, 2π) and also in the norm of L 1 (T). For exactly the same reason, the sequence e it → P N (e 1+it ) N ∈N = e it → N n=−N 1 − |n| N +1 e (1+it)n λ n N ∈N converges a.e. on T, and also in L 1 (T) norm, to a function, (which we shall temporarily call u λ,1 (e it )) whose Fourier coefficients are {e n λ n }. For each fixed θ ∈ (0, 1) we see from (32) that the sequence of functions e it → P N (e θ+it ) converges absolutely and uniformly on T to a continuous function on T (which we shall temporarily call u λ,θ (e it )) whose Fourier coefficients are e θn λ n .
In view of the preceding remarks, it is clear that λ ∈ F C θ for each θ ∈ (0, 1). It is also clear that the function v defined as in (29) i.e., by v(z) := lim N →∞ P N (z), exists for all z ∈ A • and for almost every z on each of the circles T and eT, and it satisfies v(e it ) = u λ (e it ) and v(e 1+it ) = u λ,1 (e it ) for a.e. t ∈ [0, 2π) and v(e θ+it ) = u λ,θ (e it ) for all t ∈ [0, 2π) and all θ ∈ (0, 1). So parts (i), (ii) and (iii) of Theorem 12 have been established.
It still remains to prove part (iv). For this let us first remark that all continuous functions f : A → C which are analytic in the interior of A satisfy Cf. also the proof of the related estimate ( * * ) on p. 363 of [12].) Applying (34) to the analytic polynomials P N , we have Finally we use the fact that the above-mentioned convergence of P N (e j+it ) N ∈N to v(e j+it ) in L 1 (T) implies that lim N →∞ v(e j+it ) dt for j = 0, 1. Thus we obtain (30) from (35) by taking the limit as N tends to ∞. This completes the proof of Theorem 12.
We conclude this section with a proposition which paraphrases and expands upon some of the things in the statements and the proofs of Theorem 12 and Lemma 11. Since in our application of these things in the next section we will only need to deal with sequences λ in the smaller space F L ∞ ∩ F L ∞ 1 instead of in F L 1 ∩ F L 1 1 , we only consider such sequences here. Also, it will be convenient for us to use a slightly exotic variant (see (36)) of the formula (29) so that all the functions that we have to deal with will be defined unambiguously at every point of A if the above limit does not exist for each z ∈ A.
Then: (i) The formula (36) defines a map T from F L ∞ ∩ F L ∞ 1 into a certain class of functions on A, which is "linear almost everywhere". I.e., for each λ and ξ in F L ∞ ∩ F L ∞ 1 and each α and β in C, (37) T (αλ + βξ)(z) = αT λ(z) + βT ξ(z) for all z ∈ A • and for almost all z ∈ T ∪ eT.
Furthermore, for each λ ∈ F L ∞ ∩ F L ∞ 1 , the function v = T λ has the following properties: (iii) When α = θ ∈ (0, 1) the function v(e θ+it ) depends continuously on t and (38) can be rewritten as Proof. From Theorem 12 and its proof it is immediately clear that the right sides of (29) and of (36) coincide for all z ∈ A • and for almost all z ∈ T ∪ eT and that (keeping in mind the definitions of the spaces F L ∞ α and F C α ) properties (i), (ii), (iii) and (vii) all hold.
Property (v) is an immediate consequence of the definitions of T and of M s (cf. Definition 10).
To show (vi), we choose an arbitrary m ∈ Z and let w = T (S m λ). We have to show that This is obvious for each z ∈ A • , since, when 1 < |z| < e, we have n=−∞ z n λ n and this series is absolutely convergent. It remains to deal with z ∈ T ∪ eT. When z ∈ T, we apply part (i) of Theorem 12 to the sequence S m λ, and see that the function w = T (S m λ) must satisfy 1 2π 2π 0 e −int w(e it )dt = λ n−m for each n ∈ Z. Since we also have In view of (41) and (42) we have now established (40). This completes the proof of Proposition 13.

The main result
We are finally ready to state and prove our main result.
Theorem 14. Let B = (B 0 , B 1 ) be the couple F L ∞ . Let K 0 be a relatively compact subset of the unit ball of F L ∞ and let K 1 be the unit ball of F L ∞ 1 . Let θ be a constant satisfying 0 < θ < 1 and let K θ be the set of all sequences λ ∈ F L ∞ + F L ∞ 1 of the form λ = f (e θ ) for f ∈ F A (K 0 , K 1 ). Then K θ is a relatively compact subset of F C θ .
Remark 15. As stated in the introduction, Theorem 14 immediately applies an affirmative answer to Question C in the case where (B 0 , B 1 ) is the couple of sequence spaces (F L ∞ , F L ∞ 1 ). This is because of (28) (cf. also Proposition 9). We will also use (28) in the proof of Theorem 14.
Proof. K 1 is of course closed and convex, and, since the closed convex hull of a relatively compact subset of a normed linear space is compact, we may also suppose without loss of generality that K 0 is closed and convex. It then follows from Fact 7 and (27) that the set of sequences is a dense subset of K θ in the norm of [ F L ∞ ] θ,A = F C θ . Thus we have reduced the proof of the theorem to showing that the set K θ,G is relatively compact in F C θ .
Obviously K θ,G ⊂ F L ∞ ∩ F L ∞ 1 . Let T be the map defined in (36) and let T (K θ,G ) be the collection of all functions v = T λ obtained as λ ranges over K θ,G . Let T (K θ,G )| e θ T denote the collection of all functions on the circle e θ T which are restrictions to that circle of functions in T (K θ,G ). It follows from (39) and the fact that T ξ(z) − T λ(z) = T (ξ − λ)(z) for all z ∈ e θ T and all ξ, λ ∈ F L ∞ ∩ F L ∞ 1 (cf. (37)) that, in order to show that K θ,G is relatively compact in F C θ , it suffices to show that T (K θ,G )| e θ T is a relatively compact subset of C(e θ T). By definition, K θ,G is contained in the unit ball of [ F L ∞ ] θ,A and so T (K θ,G )| e θ T is a bounded set of functions in C(e θ T). So, by the Arzelà-Ascoli theorem, it remains only to show that T (K θ,G )| e θ T is an equicontinuous set of functions. Intuitively at least, this is a rather obvious consequence of Theorem 4 and Theorem 12, in particular (30). But let us carefully go through all the steps to show it: Consider some fixed but arbitrary element ψ of T (K θ,G )| e θ T . It must be a continuous function ψ : e θ T → C which is the restriction to e θ T of some function T λ where the sequence λ is a fixed element of K θ,G . It must, by the definition of G A , be of the form for some M ∈ N, where λ n ∈ F L ∞ ∩ F L ∞ 1 for each n. (Here we are using a superscript index in λ n to avoid confusion with the previous notation λ n for individual terms of the sequence λ.) Furthermore, we have M n=−M e (j+it)n λ n ∈ K j for j = 0, 1 and for each fixed t ∈ [0, 2π). In fact, for these values of j and t, we have , K * is the set of all functions defined on T which are the restrictions to T of functions of the form T ξ for ξ ∈ K 0 ∩F L ∞ ∩F L ∞ 1 . Let K be the set of all elements (i.e., equivalence classes of functions) in L ∞ (T) which have a representative in K * . In view of (38) for α = 0 and the fact that T ξ(z) − T λ(z) = T (ξ − λ)(z) for a.e. z ∈ T and all ξ, λ ∈ F L ∞ ∩ F L ∞ 1 (cf. (37)), we can assert that K is a relatively compact subset of the unit ball of L ∞ (T). We now set v n = T (λ n ) for each n ∈ [−M, M ], (where λ n are the elements appearing in (43)) and, for each ζ ∈ A, we define u(ζ) = M n=−M ζ n v n . Correspondingly, we define u(ζ, w) = M n=−M ζ n v n (w) for each ζ and w in A. and the two norm inequalities In fact it suffices to show that (53) and (54) each hold for all finitely supported sequences λ. Since these form a dense subset of F C θ , this immediately implies that they both hold for all λ ∈ F C θ and also establishes (52).
We use essentially the same simple reasoning as was used in [14] to show that Given an arbitrary sequence λ = {λ n } n∈Z with finite support, we introduce the sequence valued function f (z) = e n(θ−z) λ n n∈Z . Clearly f (θ) = λ, and for all real t and j = 0, 1 we have Since f (z) is an entire function taking values in a finite dimensional subspace of F L ∞ ∩ F L ∞ 1 and is bounded on S, it is clear that, for each δ > 0, the function Since δ can be chosen arbitrarily small, this gives (53). Let us also consider the function g : A → F L ∞ ∩ F L ∞ 1 defined by g(ζ) = e nθ ζ −n λ n n∈Z . Since g(e z ) = f (z) we of course have g(e θ ) = λ and also g ∈ F A (F L ∞ , F L ∞ 1 ) with norm This immediately gives us (54). We now know that (53) and (54) hold for all finitely supported sequences and so, as already explained, this establishes all of (52), (53) and (54). Now we turn to proving the inclusion and inequalities which are the reverse of (52), (53) and (54) respectively. Again we will use density properties, more explicitly, the facts that G(F L ∞ , F L ∞ 1 ) is dense in F (F L ∞ , F L ∞ 1 ) and G A (F L ∞ , F L ∞ 1 ) is dense in F A (F L ∞ , F L ∞ 1 ). Our main step will be to show that, for each g ∈ G(F L ∞ , F L ∞ 1 ), we have g(θ) ∈ F C θ and (56) g(θ) F C θ ≤ g F (F L ∞ ,F L ∞ 1 ) . Analogously, for every g ∈ G A (F L ∞ , F L ∞ 1 ) we will show that g(e θ ) ∈ F C θ and (57) g(e θ ) F C θ ≤ g F A (F L ∞ ,F L ∞ 1 ) .
If g ∈ G(F L ∞ , F L ∞ 1 ) then g(z) is a finite sum g(z) = M m=1 φ m (z)a m where each a m = {a m n } n∈Z is a sequence in the space F L ∞ ∩ F L ∞ 1 and each φ m is a scalar valued entire function which is bounded on S. Applying Theorem 12 to each sequence a m , we see that a m ∈ F C θ for each m. (Note that F L ∞ j ⊂ F L 1 j and that the proof of Theorem 12 does not use Proposition 9. In fact, we only need the simple argument at the beginning of the proof.) So obviously also g(θ) ∈ F C θ .
Given any sequence λ = {λ n } n∈Z ∈ F L ∞ and any other sequence σ = {σ n } n∈Z whose support Z σ = {n ∈ Z : σ n = 0} is a finite set, we of course have that the finite sum n∈Z λ n σ n = n∈Zσ λ n σ n = 1 2π 2π 0 f λ (e it )f σ (e −it )dt, where f λ and f σ are the functions whose sequences of Fourier coefficients are respectively λ and σ.
Applying (58), we see that, for j = 0, 1 and for all t ∈ R, |ψ σ (j + it)| ≤ 1 2π Since g(θ) ∈ F C θ we have g(θ) F L ∞ θ = g(θ) F C θ and so (56) follows from (63), (61) and (62). The proof of (57) is almost the same. The only differences are that this time the functions φ m are each of the form φ m (z) = z km for some k m ∈ Z, and we have to use the function ψ σ (z) = M m=1 φ m (z) n∈Zσ a m n σ n z n . By the maximum modulus principle for functions which are analytic on A, we have ψ σ (e θ ) ≤ max ψ σ (e j+it ) : j = 0, 1, t ∈ [0, 2π) . Analogously to (62) this last expression is bounded from above by g F A (F L ∞ ,F L ∞ 1 ) , and we obtain (57). Finally, if f is an arbitrary element of F (F L ∞ , F L ∞ 1 ) then we approximate it in F (F L ∞ , F L ∞ 1 ) norm by a sequence {g n } n∈N in G(F L ∞ , F L ∞ 1 ). Thus, by (56), {g n (θ)} n∈N is a Cauchy sequence in F C θ . Since it converges to f (θ) in the norm of F L ∞ + F L ∞ 1 it follows that f (θ) ∈ F C θ and f (θ , an exactly analogous argument using (57) shows that λ F C θ ≤ λ [F L ∞ ,F L ∞ 1 ] θ,A . (Alternatively, one could use the first case and the general fact that the inclusion [ A] θ,A → [ A] θ has norm 1, see [8].) This completes the proof of Proposition 9.