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On vector-valued inequalities for Sidon sets and sets of interpolation

arXiv:math/9209216v1 [math.FA] 25 Sep 1992On vector-valued inequalities for Sidon sets and sets of interpolationN.J. Kalton1University of Missouri-ColumbiaAbstract.

Let E be a Sidon subset of the integers and suppose X is a Banach space. ThenPisier has shown that E-spectral polynomials with values in X behave like Rademachersums with respect to Lp−norms.

We consider the situation when X is a quasi-Banachspace. For general quasi-Banach spaces we show that a similar result holds if and only if Eis a set of interpolation (I0-set).

However for certain special classes of quasi-Banach spaceswe are able to prove such a result for larger sets. Thus if X is restricted to be “natural”then the result holds for all Sidon sets.

We also consider spaces with plurisubharmonicnorms and introduce the class of analytic Sidon sets.1. Introduction.Suppose G is a compact abelian group.

We denote by µG normalized Haar measureon G and by Γ the dual group of G. We recall that a subset E of Γ is called a Sidon set ifthere is a constant M such that for every finitely nonzero map a : E →C we haveXγ∈E|a(γ)| ≤M maxg∈G |Xγ∈Ea(γ)γ(g)|.We define ∆to be the Cantor group i.e. ∆= {±1}N. If t ∈∆we denote by ǫn(t)the n-th co-ordinate of t. The sequence (ǫn) is an example of a Sidon set.

Of course thesequence (ǫn) is a model for the Rademacher functions on [0, 1]. Similarly we denote theco-ordinate maps on TN by ηn.Suppose now that G is a compact abelian group.

If X is a Banach space, or moregenerally a quasi-Banach space with a continuous quasinorm and φ : G →X is a Borel mapwe define ∥φ∥p for 0 < p ≤∞to be the Lp−norm of φ i.e. ∥φ∥p = (RG ∥φ(g)∥pdµG(g))1/pif 0 < p < ∞and ∥φ∥∞= ess supg∈G ∥φ(g)∥.1Research supported by NSF-grant DMS-89016361

It is a theorem of Pisier [12] that if E is a Sidon set then there is a constant M sothat for every subset {γ1, . .

., γn} of E, every x1, . .

., xn chosen from a Banach space Xand every 1 ≤p ≤∞we have(*)M −1∥nXk=1xkǫk∥p ≤∥nXk=1xkγk∥p ≤M∥nXk=1xkǫk∥p.Thus a Sidon set behaves like the Rademacher sequence for Banach space valued func-tions. The result can be similarly stated for (ηn) in place of (ǫn).

Recently Asmar andMontgomery-Smith [1] have taken Pisier’s ideas further by establishing distributional in-equalities in the same spirit.It is natural to ask whether Pisier’s inequalities can be extended to arbitrary quasi-Banach spaces. This question was suggested to the author by Asmar and Montgomery-Smith.

For convenience we suppose that every quasi-Banach space is r-normed for somer < 1 i.e.the quasinorm satisfies ∥x + y∥r ≤∥x∥r + ∥y∥r for all x, y; an r-norm isnecessarily continuous. We can then ask, for fixed 0 < p ≤∞for which sets E inequality(*) holds, if we restrict X to belong to some class of quasi-Banach spaces, for some constantM = M(E, X).It turns out Pisier’s results do not in general extend to the non-locally convex case.In fact we show that if we fix r < 1 and ask that a set E satisfies (*) for some fixed pand every r-normable quasi-Banach space X then this condition precisely characterizessets of interpolation as studied in [2], [3], [4], [5], [8], [9], [13] and [14].

We recall that E iscalled a set of interpolation (set of type (I0)) if it has the property that every f ∈ℓ∞(E)(the collection of all bounded complex functions on E) can be extended to a continuousfunction on the Bohr compactification bΓ of Γ.However, in spite of this result, there are specific classes of quasi-Banach spaces forwhich (*) holds for a larger class of sets E. If we restrict X to be a natural quasi-Banachspace then (*) holds for all Sidon sets E. Here a quasi-Banach space is called natural if itis linearly isomorphic to a closed linear subspace of a (complex) quasi-Banach lattice Ywhich is q-convex for some q > 0, i.e. such that for a suitable constant C we have∥(nXk=1|yk|q)1/q∥≤C(nXk=1∥yk∥q)1/qfor every y1, .

. ., yn ∈Y.

Natural quasi-Banach spaces form a fairly broad class includingalmost all function spaces which arise in analysis.The reader is referred to [6] for adiscussion of examples. Notice that, of course, the spaces Lq for q < 1 are natural so that,in particular, (*) holds for all p and all Sidon sets E for every 0 < p ≤∞.

The case p = q2

here would be a direct consequence of Fubini’s theorem, but the other cases, includingp = ∞are less obvious.A quasi-Banach lattice X is natural if and only if it is A-convex, i.e. it has an equivalentplurisubharmonic quasi-norm.

Here a quasinorm is plurisubharmonic if it satisfies∥x∥≤Z 2π0∥x + eiθy∥dθ2πfor every x, y ∈X. There are examples of A-convex spaces which are not natural, namelythe Schatten ideals Sp for p < 1 [7].

Of course, it follows that Sp cannot be embeddedin any quasi-Banach lattice which is A-convex when 0 < p < 1. Thus we may ask forwhat sets E (*) holds for every A-convex space.

Here, we are unable to give a precisecharacterization of the sets E such that (*) holds. In fact we define E to be an analyticSidon set if (*) holds, for p = ∞(or, equivalently for any other 0 < p < ∞), for everyA-convex quasi-Banach space X.

We show that any finite union of Hadamard sequencesin N ⊂Z is an analytic Sidon set. In particular a set such as {3n}∪{3n +n} is an analyticSidon set but not a set of interpolation.

However we have no example of a Sidon set whichis not an analytic Sidon set.We would like to thank Nakhle Asmar, Stephen Montgomery-Smith and David Growfor their helpful comments on the content of this paper.2. The results.Suppose G is a compact abelian group and Γ is its dual group.

Let E be a subset ofΓ. Suppose X is a quasi-Banach space and that 0 < p ≤∞; then we will say that E hasproperty Cp(X) if there is a constant M such that for any finite subset {γ1, .

. ., γn} of Eand any x1, .

. ., xn of X we have (∗) i.e.M −1∥nXk=1xkǫk∥p ≤∥nXk=1xkγk∥p ≤M∥nXk=1xkǫk∥p.

(Note that in contrast to Pisier’s result (*), we here assume p fixed.) We start by observingthat E is a Sidon set if and only if E has property C∞(C).

It follows from the results of Pisier[12] a Sidon set has property Cp(X) for every Banach space X and for every 0 < p < ∞.See also Asmar and Montgomery-Smith [1] and Pelczynski [11].Note that for any t ∈∆we have that ∥P ǫk(t)xkǫk∥p = ∥P xkǫk∥p. Now any realsequence (a1, .

. ., an) with max |ak| ≤1 can be written in the form ak = P∞j=1 2−jǫk(tj)and it follows quickly by taking real and imaginary parts that the there is a constantC = C(r, p) so that for any complex a1, .

. ., an and any r-normed space X we have∥nXk=1akxkǫk∥p ≤C∥a∥∞∥nXk=1xkǫk∥p.3

From this it follows quickly that ∥Pnk=1 xkηk∥p is equivalent to ∥Pnk=1 xkǫk∥p. In partic-ular we can replace ǫk by ηk in the definition of property Cp(X).We note that if E has property Cp(X) then it is immediate that E has propertyCp(ℓp(X)) and further that E has property Cp(Y ) for any quasi-Banach space finitelyrepresentable in X (or, of course, in ℓp(X)).For a fixed quasi-Banach space X and a fixed subset E of Γ we let PE(X) denote thespace of X-valued E-polynomials i.e.

functions φ : G →X of the form φ = Pγ∈E x(γ)γwhere x(γ) is only finitely nonzero. If f ∈ℓ∞(E) we define Tf : PE(X) →PE(X) byTf(Xx(γ)γ) =Xf(γ)x(γ)γ.We then define ∥f∥Mp(E,X) to the operator norm of Tf on PE(X) for the Lp−norm (andto be ∞is this operator is unbounded).Lemma 1.

In order that E has property Cp(X) it is necessary and sufficient that thereexists a constant C such that∥f∥Mp(E,X) ≤C∥f∥∞for all f ∈ℓ∞(E).Proof: If E has property Cp(X) then it also satisfies (*) for (ηn) in place of (ǫn) for asuitable constant M. Thus if f ∈ℓ∞(E) and φ ∈PE(X) then∥Tfφ∥p ≤M 2∥f∥∞∥φ∥p.For the converse direction, we consider the case p < ∞. Suppose {γ1, .

. ., γn} is afinite subset of E. Then for any x1, .

. ., xnC−pZTN ∥nXk=1xkηk∥pdµTN = C−pZTNZG∥nXk=1xkηk(s)γk(t)∥pdµTN(s)dµG(t)≤ZG∥nXk=1xkγk∥pdµG≤CpZTNZG∥nXk=1xkηk(s)γk(t)∥pdµTN(s)dµG(t)≤CpZTN ∥nXk=1xkηk∥pdµTN.This estimate together with a similar estimate in the opposite direction gives the conclu-sion.

The case p = ∞is similar.4

If E is a subset of Γ, N ∈N and δ > 0 we let AP(E, N, δ) be the set of f ∈ℓ∞(E)such that there exist g1, . .

., gN ∈G (not necessarily distinct) and α1, . .

., αN ∈C withmax1≤j≤N |αj| ≤1 and|f(γ) −NXj=1αjγ(gj)| ≤δfor γ ∈E.The following theorem improves slightly on results of Kahane [5] and M´ela [8]. Perhapsalso, our approach is slightly more direct.

We write Bℓ∞(E) = {f ∈ℓ∞(E) : ∥f∥∞≤1}.Theorem 2. Let G be a compact abelian group and let Γ be its dual group.

Suppose Eis a subset of Γ. Then the following conditions on E are equivalent:(1) E is a set of interpolation.

(2) There exists an integer N so that Bℓ∞(E) ⊂AP(E, N, 1/2). (3) There exists M and 0 < δ < 1 so that if f ∈Bℓ∞(E) then there exist complex numbers(cj)∞j=1 with |cj| ≤Mδj and (gj)∞j=1 in G withf(γ) =∞Xj=1cjγ(gj)for γ ∈E.Proof: (1) ⇒(2).

It follows from the Stone-Weierstrass Theorem thatTE ⊂∪∞m=1AP(E, m, 1/5).Let µ = µTE. Since each AP(E, m, 1/5)∩TE is closed it is clear that there exists m so thatµ(AP(E, m, 1/5) ∩TE) > 1/2.

Thus if f ∈TE we can find f1, f2 ∈AP(E, m, 1/5) ∩TEso that f = f1f2. Hence f ∈AP(E, m2, 1/2).

This clearly implies (2) with N = 2m2. (2) ⇒(3).

We let δ = 2−1/N and M = 2. Then given f ∈Bℓ∞(E) we can find (cj)Nj=1and (gj)Nj=1 with |cj| ≤1 ≤Mδj and|f(γ) −NXj=1cjγ(gj)| ≤1/2for γ ∈E.

Let f1(γ) = 2(f(γ) −PNj=1 cjγ(gj)) and iterate the argument. (3) ⇒(1): Obvious.5

Theorem 3. Suppose G is a compact abelian group, E is a subset of the dual group Γand that 0 < r < 1, 0 < p ≤∞.

In order that E satisfies Cp(X) for every r-normablequasi-Banach space X it is necessary and sufficient that E be a set of interpolation.Proof: First suppose that E is a set of interpolation so that it verifies (3) of Theorem2. Suppose X is an r-normed quasi-Banach space.

Suppose f ∈Bℓ∞(E). Then there exist(cj)∞j=1 and (gj)∞j=1 so that |cj| ≤Mδj and f(γ) = P cjγ(gj) for γ ∈E.

Now if φ ∈PE(X)it follows thatTfφ(h) =∞Xj=1cjφ(gjh)and so∥Tfφ∥p ≤M(∞Xj=1δjs)1/s∥φ∥pwhere s = min(p, r). Thus ∥f∥Mp(E,X) ≤C where C = C(p, r, E) and so by Lemma 1 Ehas property Cp(X).Now conversely suppose that 0 < r < 1, 0 < p ≤∞and that E has property Cp(X)for every r-normable space X.

It follows from consideration of ℓ∞−products that thereexists a constant C so that for every r-normed space X we have ∥f∥Mp(E,X) ≤C∥f∥∞for f ∈ℓ∞(E).Suppose F is a finite subset of E. We define an r-norm ∥∥A on ℓ∞(F) by setting∥f∥A to be the infimum of (P |cj|r)1/r over all (cj)∞j=1 and (gj)∞j=1 such thatf(γ) =∞Xj=1cjγ(gj)for γ ∈F. Notice that ∥f1f2∥A ≤∥f1∥A∥f2∥A for all f1, f2 ∈A = ℓ∞(F).For γ ∈F let eγ be defined by eγ(γ) = 1 if γ = χ and 0 otherwise.

Then for f ∈A,with ∥f∥∞≤1,(ZG∥Xγ∈Ff(γ)eγγ∥pAdµG)1/p ≤C(ZG∥Xγ∈Feγγ∥pAdµG)1/p.But for any g ∈G ∥P γ(g)eγ∥A ≤1. Define H to be subset of h ∈G such that∥Pγ∈F f(γ)γ(h)eγ∥A ≤31/pC.

Then µG(H) ≥2/3. Thus there exist h1, h2 ∈H suchthat h1h2 = 1 (the identity in G).

Hence by the algebra property of the norm∥f∥A ≤32/pC2and so if we fix an integer C0 > 32/pC2 we can find cj and gj so that P |cj|r ≤Cr0 andf(γ) =Xcjγ(gj)6

for γ ∈F. We can suppose |cj| is monotone decreasing and hence that |cj| ≤C0j−1/r.Choose N0 so that C0P∞j=N0+1 j−1/r ≤1/2.

Thus|f(γ) −N0Xj=1cjγ(gj)| ≤1/2for γ ∈F. Since each |cj| ≤C0 this implies that Bℓ∞(F ) ⊂AP(F, N, 1/2) where N = C0N0.As this holds for every finite set F it follows by an easy compactness argument thatBℓ∞(E) ⊂AP(E, N, 1/2) and so by Theorem 2 E is a set of interpolation.Theorem 4.

Let X be a natural quasi-Banach space and suppose 0 < p ≤∞. Then anySidon set has property Cp(X).Proof: Suppose E is a Sidon set.

Then there is a constant C0 so that if f ∈ℓ∞(E) thenthere exists ν ∈C(G)∗such that ˆµ(γ) = f(γ) for γ ∈E and ∥µ∥≤C0∥f∥∞. We will showthe existence of a constant C such that ∥f∥Mp(E,X) ≤C∥f∥∞.

If no such constant existsthen we may find a sequence En of finite subsets of E such that lim Cn = ∞where Cn isthe least constant such that ∥f∥Mp(En,X) ≤Cn∥f∥∞for all f ∈ℓ∞(En).Now the spaces Mp(En, X) are each isometric to a subspace of ℓ∞(Lp(G, X)) andhence so is Y = c0(Mp(En, X)). In particular Y is natural.

Notice that Y has a finite-dimensional Schauder decomposition.We will calculate the Banach envelope Yc of Y.Clearly Yc = c0(Yn) where Yn is the finite-dimensional space Mp(En, X) equipped withits the envelope norm ∥f∥c.Suppose f ∈ℓ∞(En).Then clearly ∥f∥∞≤∥f∥Mp(E,X) and so ∥f∥∞≤∥f∥c.Conversely if f ∈ℓ∞(En) there exists ν ∈C(G)∗with ∥ν∥≤C0∥f∥∞and such thatRγ dν = f(γ) for γ ∈En. In particular C−10 ∥f∥−1∞f is in the absolutely closed closedconvex hull of the set of functions {˜g : g ∈G} where ˜g(γ) = γ(g) for γ ∈En.

Since∥˜g∥Mp(E,X) = 1 for all g ∈G we see that ∥f∥∞≤∥f∥c ≤C0∥f∥∞.This implies that Yc is isomorphic to c0. Since Y has a finite-dimensional Schauderdecomposition and is natural we can apply Theorem 3.4 of [6] to deduce that Y = Yc isalready locally convex.

Thus there is a constant C′0 independent of n so that ∥f∥Mp(E,X) ≤C′0∥f∥∞whenever f ∈ℓ∞(En). This contradicts the choice of En and proves the theorem.We now consider the case of A-convex quasi-Banach spaces.

For this notion we willintroduce the concept of an analytic Sidon set. We say a subset E of Γ is an analytic Sidonset if E satisfies C∞(X) for every A-convex quasi-Banach space X.Proposition 5.

Suppose 0 < p < ∞. Then E is an analytic Sidon set if and only if Esatisfies Cp(X) for every A-convex quasi-Banach space X.Proof: Suppose first E is an analytic Sidon set, and that X is an A-convex quasi-Banachspace (for which we assume the quasinorm is plurisubharmonic).

Then Lp(G, X) also has7

a plurisubharmonic quasinorm and so E satisfies (1) for X replaced by Lp(G, X) and preplaced by ∞with constant M. Now suppose x1, . .

., xn ∈X and γ1, . .

., γn ∈E. Definey1, .

. ., yn ∈Lp(G, X) by yk(g) = γk(g)xk.

Thenmaxg∈G ∥nXk=1ykγk(g)∥Lp(G,X) = ∥nXk=1xkγk∥pand a similar statement holds for the characters ǫk on the Cantor group. It follows quicklythat E satisfies (1) for p and X with constant M.For the converse direction suppose E satisfies Cp(X) for every A-convex space X.Suppose X has a plurisubharmonic quasinorm.

We show that M∞(E, X) = ℓ∞(E). Infact M∞(F, X) can be isometrically embedded in ℓ∞(X) for every finite subset F of E.Thus (1) holds for X replaced by M∞(F, X) for some constant M, independent of F.Denoting by eγ the canonical basis vectors in ℓ∞(E) we see that if F = {γ1, .

. ., γn} ⊂Ethen(Z∆∥nXk=1ǫk(t)eγk∥pM∞(F,X)dµ∆(t))1/p ≤M maxg∈G ∥nXk=1γk(g)eγk∥M∞(F,X) = M.Thus the set K of t ∈∆such that ∥Pnk=1 ǫk(t)eγk∥≤31/pM has measure at least2/3.

Arguing that K.K = ∆we obtain that∥nXk=1ǫk(t)eγk∥M∞(F,X) ≤32/pM 2for every t ∈∆. It follows quite simply that there is a constant C so that for every realvalued f ∈ℓ∞(F) we have ∥f∥M∞(E,X) ≤C∥f∥∞.

In fact this is proved by writing eachsuch f with ∥f∥∞= 1 in the form f(γk) = P∞j=1 2−jǫk(tj) for a suitable sequence tj ∈∆.A similar estimate for complex f follows by estimating real and imaginary parts. Finallywe conclude that since these estimates are independent of F that ℓ∞(E) = M∞(E, X).Of course any set of interpolation is an analytic Sidon set and any analytic Sidon setis a Sidon set.

The next theorem will show that not every analytic Sidon set is a set ofinterpolation. If we take G = T and Γ = Z, we recall that a Hadamard gap sequence isa sequence (λk)∞k=1 of positive integers such that for some q > 1 we have λk+1/λk ≥q fork ≥1.

It is shown in [10] and [14] that a Hadamard gap sequence is a set of interpolation.However the union of two such sequences may fail to be a set of interpolation; for example(3n)∞n=1 ∪(3n + n)∞n=1 is not a set of interpolation, since the closures of (3n) and (3n + n)in bZ are not disjoint.8

Theorem 6. Let G = T so that Γ = Z.

Suppose E ⊂N is a finite union of Hadamardgap sequences. Then E is an analytic Sidon set.Proof: Suppose E = (λk)∞k=1 where (λk) is increasing.

We start with the observationthat E is the union of m Hadamard sequences if and only there exists q > 1 so thatλm+k ≥qmλk for every k ≥1.We will prove the theorem by induction on m. Note first that if m = 1 then E is aHadamard sequence and hence [14] a set of interpolation. Thus by Theorem 2 above, E isan analytic Sidon set.Suppose now that E is the union of m Hadamard sequences and that the theorem isproved for all unions of l Hadamard sequences where l < m. We assume that E = (λk) andthat there exists q > 1 such that λk+m ≥qmλk for k ≥1.

We first decompose E into at mostm Hadamard sequences. To do this let us define E1 = {λ1} ∪{λk : k ≥2, λk ≥qλk−1}.We will write E1 = (τk)k≥1 where τk is increasing.

Of course E1 is a Hadamard sequence.For each k let Dk = E ∩[τk, τk+1). It is easy to see that |Dk| ≤m for every k. Furtherif nk ∈Dk then nk+1 ≥τk+1 ≥qnk so that (nk) is a Hadamard sequence.

In particularE2 = E \ E1 is the union of at most m −1 Hadamard sequences and so E2 is an analyticSidon set by the inductive hypothesis.Now suppose w ∈T. We define fw ∈ℓ∞(E) by fw(n) = wn−τk for n ∈Dk.

We willshow that fw is uniformly continuous for the Bohr topology on Z; equivalently we showthat fw extends to a continuous function on the closure ˜E of E in the Bohr compactificationbZ of Z. Indeed, if this is not the case there exists ξ ∈˜E and ultrafilters U0 and U1 on Eboth converging to ξ so that limn∈U0 fw(n) = ζ0 and limn∈U1 fw(n) = ζ1 where ζ1 ̸= ζ0.We will let δ = 13|ζ1 −ζ0|.We can partition E into m sets A1, .

. .

, Am so that |Aj ∩Dk| ≤1 for each k. ClearlyU0 and U1 each contain exactly one of these sets. Let us suppose Aj0 ∈U0 and Aj1 ∈U1.Next define two ultrafilters V0 and V1 on N. V0 = {V : ∪k∈V Dk ∈U0} and V1 = {V :∪k∈V Dk ∈U1}.

We argue that V0 and V1 coincide. If not we can pick V ∈V0\V1.

Considerthe set A = (Aj0 ∩(∪k∈V Dk)) ∪(Aj1 ∩(∪k/∈V Dk)). Then A is a Hadamard sequence andhence a set of interpolation.

Thus for the Bohr topology the sets Aj0 ∩(∪k∈V Dk)) andAj1 ∩(∪k/∈V Dk)) have disjoint closures. This is contradiction since of course ξ must be inthe closure of each.

Thus V0 = V1.Since both U0 and U1 converge to the same limit for the Bohr topology we can findsets H0 ∈U0 and H1 ∈U1 so that if n0 ∈H0, n1 ∈H1 then |wn1 −wn0| < δ and further|fw(n0) −ζ0| < δ and |fw(n1) −ζ1| < δ.Let V0 = {k ∈N : Dk ∩H0 ̸= ∅} and V1 = {k ∈N : Dk ∩H1 ̸= ∅}. Then V0 ∈V0and V1 ∈V1.

Thus V = V0 ∩V1 ∈V0 = V1. If k ∈V there exists n0 ∈Dk ∩H0 and9

n1 ∈Dk ∩H1. Then3δ = |ζ1 −ζ0|< |fw(n1) −fw(n0)| + 2δ= |wn1 −wn0| + 2δ< 3δ.This contradiction shows that each fw is uniformly continuous for the Bohr topology.Now suppose that X is an r-normed A-convex quasi-Banach space where the quasi-norm is plurisubharmonic.

Since both E1 and E2 are analytic Sidon sets we can introducea constant C so that if f ∈ℓ∞(Ej) where j = 1, 2 then ∥f∥M∞(Ej,X) ≤C∥f∥∞. Pick aconstant 0 < δ < 1 so that 3.41/rδ < C.Let Kl = {w ∈T : fw ∈AP(E, l, δ)}.

It is easy to see that each Kl is closed and sinceeach fw is uniformly continuous by the Bohr topology it follows from the Stone-Weierstrasstheorem that ∪Kl = T. If we pick l0 so that µT(Kl0) > 1/2 then Kl0Kl0 = T and hencesince the map w →fw is multiplicative fw ∈AP(E, l20, 3δ) for every w ∈T.Let F be an arbitrary finite subset of E. Then there is a least constant β so that∥f∥M∞(F,X) ≤β∥f∥∞. The proof is completed by establishing a uniform bound on β.For w ∈T we can find cj with |cj| ≤1 and ζj ∈T for 1 ≤j ≤l20 such that|fw(n) −l20Xj=1cjζnj | ≤3δfor n ∈E.

If ˜ζj is define by ˜ζj(n) = ζnj then of course ∥˜ζj∥M∞(E,X) = 1. Restricting to Fwe see that∥fw∥rM∞(F,X) ≤l20 + βr(3δ)r.Define F : C →M∞(F, X) by F(z)(n) = zn−τk if n ∈Dk.

Note that F is a polyno-mial. As in Theorem 5, M∞(F, X) has a plurisubharmonic norm.

Hence∥F(0)∥r ≤max|w|=1 ∥F(w)∥r ≤l20 + (3δ)rβr.Thus, if χA is the characteristic function of A,∥χE1∩F ∥rM∞(F,X) ≤l20 + (3δ)rβr.It follows that∥χE2∩F ∥rM∞(F,X) ≤l20 + (3δ)rβr + 1.10

Now suppose f ∈ℓ∞(F) and ∥f∥∞≤1. Then∥fχEj∩F ∥M∞(F,X) ≤∥fχEj∩F ∥M∞(Ej∩F,X)∥χEj∩F ∥M∞(F,X)for j = 1, 2.

Thus∥f∥rM∞(F,X) ≤Cr(1 + 2l20 + 2(3δ)rβr).By maximizing over all f this impliesβr ≤Cr(1 + 2l20 + 2(3δ)rβr)which gives an estimateβr ≤2Cr(1 + 2l20)in view of the original choice of δ. This estimate, which is independent of F, implies thatE is an analytic Sidon set.Remark: We know of no example of a Sidon set which is not an analytic Sidon set.References.1.

N. Asmar and S.J. Montgomery-Smith, On the distribution of Sidon series, ArkivMath.

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