interpolation sets, past and present -- and future?i0 sets appplications to sidon sets open...

I0 setsAppplications to Sidon sets

Open questionsA proof

Interpolation sets, past and present – andfuture?

Colin C. Graham

Department of MathematicsUniversity of British ColumbiaMailing address: RR#1–D156

Bowen Island BC V0N1G0 [email protected]

University of Hawaii Colloquium, 13 Feb. 2008

Colin C. Graham Interpolation sets



Abstract

A survey of the theory of interpolation sets from past to present.Indications of directions for future research.

Some of this is joint work with K. E. Hare, T. W. Körner, and L.T. Ramsey.

This is a slightly emended version has of the talk originally given and has had the Beamer.cls "pauses" removed for

better web display.




Outline

1 I0 setsPreliminariesHadamard sets are I0Characterizations of interpolation setsI0 subclasses–ε-Kronecker setsI0 subclasses–restricting the cj ’s or x(j)sClosures in Z

2 Appplications to Sidon setsDefinition and propertiesProportional results

3 Open questions4 A proof




PreliminariesHadamard sets are I0Characterizations of interpolation setsI0 subclasses–ε-Kronecker setsI0 subclasses–restricting the cj ’s or x(j)sClosures in Z

The beginning, II

Let E ⊂ R and f : E → C be bounded.

When can we extend f to be periodic on all of R?

E compact– no problem.

Thm (Mycielski 1961, Lipinski 1960) If εj is a boundedsequence of real numbers and 0 ≤ λj ∈ R with λj+1/λj ≥ q > 3then there exists a continuous periodic function with f (λj) = εjall j .

The use of an exponentially growing sequence was not asurprise.





The beginning, I

Hadamard gap theorem (1892). If 1 ≤ nj , nj+1/nj > q > 1,then f (z) =

∑j cjznj is analytic precisely in a disc.

Defn:. Hadamard (lacunary) sequence; q is the ratio.Here are 2 examples from the vast literature.

Thm. (Sidon 1927) If supx

∣∣∣∑∞j=1 cjeinj x

∣∣∣ < ∞, then∑

j |cj | < ∞.

Thm.(Banach 1930) If λk →∞, ∃ cts f (x) such that

∑k

∣∣∣∣∫ eink x f (x)dx∣∣∣∣λk

= ∞.

The∫

’s are the nthk Fourier coefficients of f , and they → 0.





More generally

As a species we are not good at telling if a sequence an, n ∈ Eis from the Fourier coefficients of a function, even if E = Z .

Therefore, work has gone into finding sets E for which, forexample, every bounded sequence is

– extendible to a periodic (or almost periodic) function

– or the Fourier coefficients of a measure (explanations if timeallows later)





Notation

Z – the integers R – the real line

T: [0, 2π] with addition mod 2π and 2π identified with 0

δx : the unit point mass at x ∈ T

AP: the (Bohr) almost periodic functions on Z:the uniform limits of polynomials p(n) =

∑n1 cjeinx(j)

Interpolation set (or I0 set): E ⊂ Z such that `∞(E) = AP|E

Z: The Bohr group: all group homomorphisms ϕ : T → T,continuous or not. Z is a compact group under topology ofpointwise convergence, and AP = C(Z).





Questions and caveats

What do I0 sets look like? Structure? Subclasses?

How “big” does E have to be to be dense in Z?

Everything has versions for locally compact groups in place ofZ, though complications arise if the group: has elements offinite order; is non-metrizable; or not σ-compact.





Hadamard sets are interpolation sets

Thm. (Hartman 1961) Not every bounded εj can beinterpolated on 2j with a continuous periodic function.

But:Thm. (Strzelecki 1964) Every Hadamard sequence is I0; that is,the interpolation can be done with almost periodic functions.

Thm. (various) If E is Hadamard, then E ∪ −E is I0.

However, polynomial growth is not fast enough:Thm. (Hartman 1961) nj = k j for 2 ≤ k is not I0.





2 warnings

:1 Non-Hadamard sets: can be I0:

10k2+j : 1 ≤ j ≤ k < ∞

is I0 but is not a finite union of Hadamard sets.

2 Union difficulties: 10k :∞k=1 and 10k + k∞k=1 are Io buttheir union is not I0. (Take a net jα → 0 in Z)

The “+k ” is essential: 3k∞k=1 ∪ 3k + 1∞k=1 is I0.





3 early characterizations

Thm. (Mycielski 1961, Hartman and Ryll-Narzewski 1963) E isI0 iff whenever E = E1 ∪ E2 is a disjoint union, there is ϕ ∈ APwith ϕ = 1 on E1 and 0 on E2.

Cor. (H R-N 1963) E is I0 iff such E1, E2 have disjoint closuresin Z.

Thm. (Kahane 1966) E is I0 iff each f ∈ `∞(E) extends to analmost periodic function f (n) =

∑j cjenx(j), where

∑j |cj | < ∞.





2 more characterizations

Thm. (Kalton 1993) E is I0 iff for each 0 < ε there exists 1 ≤ Nsuch that for each f ∈ `∞(E) there exist x(j) ∈ T and |cj | ≤ 1,1 ≤ j ≤ N such that

supn∈E

∣∣∣∣∣∣f (n) =∑

j

cjeinx(j)

∣∣∣∣∣∣ < ε.

Kalton’s test is very useful, as it converts to a finite test.

Thm. (Bourgain 1984) E is I0 iff Bd(E) = B(E).

(Explanations later - if time)





ε-Kronecker sets, I

Def. E ⊂ Z is ε-Kronecker (ε-free, C∗-embedded in Z) if forevery ϕ : E → T there exists x ∈ T with |ϕ(n)− einx | < ε on E .

Thm. (Varopoulos 1969) If ε < 1, and E is ε-Kronecker, then Eis I0.

Thm. (Hare & CG2006) If E is Hadamard with ratio q > 3, thenE is ε-Kronecker for all ε > |1− eiπ/(q−1)|.

Thm. (ibid.) If ε <√

2, and E is ε-Kronecker, then E is I0.





ε-Kronecker sets, II

Thm.(Ibid) If 0 ≤ ε <√

2 and E is ε-Kronecker, then E = nkwith nk+1 − nk →∞

Thm. (Ibid.) There exists a√

2-Kronecker set that is not I0.





FZI0 sets

Def. E is FZI0 if every bounded Hermitian function on E is ofthe form

∑j cjeinx(j), where 0 ≤ cj and

∑cj < ∞.

(Hermitian: f (−n) = f (n) whenever n,−n ∈ E .)

Thm. (GH 2006-2008) E is FZI0 iffa) E is I0,b) 0 6∈ E , andc) E ∪ −E is I0.

There exist I0 sets for which c) fails.

Hadamard sets and ε-Kronecker sets are FZI0 (ε <√

2).





I0(U) sets, I

Let U ⊂ T.

Def. E is I0(U) if every bounded function on E can beinterpolated by a sum

∑cjeinx(j), where x(j) ∈ U and∑

|cj | < ∞.

E is I0(U) with bounded constants if there exists C such that foreach open U ⊂ T there exists a finite set F ⊂ E such that everybounded function on E\F can be interpolated by a sum∑

j cjeinx(j), where x(j) ∈ U and∑

j |cj | ≤ C.





x(j) restricted to small sets, II

Thm. (Méla 1969) Every I0 set is a finite union of I0(U) setswith bounded constants.

Thm. (GH 2005) Hadamard sets are I0(U) with boundedconstants

Thm. (GH 2005) ε-Kronecker sets are I0(U) with boundedconstants (ε <

√2)

Thm. (GH 2005) If E is I0(U) with bounded constants, thenE + F is I0 for all finite sets F . (There are I0 sets E withE ∪ (E + 1) not I0.)

Thm. (GHR 200?) Every I0 set is I0(U) for every open U.





Closures

Thm. (Ryll-Narzewski 1964, Ramsey 1980) If E is I0, then Edoes not accumulate at elements of Z.

Thm. (GHK 2006) If 0 ≤ ε < 2 and E is ε-Kronecker, then theclosure of E in Z has no interior points. (E is not always I0)

Thm.(GHK 2006) (i). If ε <√

2, and E is ε-Kronecker, then theonly Bohr cluster point of E − E in Z is 0.

(ii). If ε < 2 sin(π/(8M)), then

M︷︸︸︷E + · · ·+ E has no Bohr cluster

points in Z.




Definition and propertiesProportional results

Definition of Sidon sets

Recall Sidon’s theorem of 1927:If supx

∣∣∣∑∞j=1 cjeinj x

∣∣∣ < ∞, then∑

j |cj | < ∞.

Sidon’s Thm gave rise to a defn:E is Sidon if for all continuous f =

∑n∈E cneinx , we have∑

n∈E |cn| < ∞.





Properties

Thm. TFAE:(i). E is Sidon;(ii). For every ϕ : E → C with ϕ → 0 there exists f ∈ L1(T) withϕ(n) =

∫e−int f (t)dt , for n ∈ E .

(iii). For every bounded ϕ : E → C there is a bounded Borelmeasure µ with ϕ(n) =

∫e−intdµ(t), for n ∈ E ; and

Cor. I0 sets are Sidon.

There are Banach space reformulations of (i)-(iii).





Union Thm; quasi-independent sets, I

Thm. (Drury 1970) The union of 2 Sidon sets is Sidon.

Def. E is quasi-independent if whenever 1 ≤ N, ej = 0,±1,nj ∈ E , and

∑N1 ejnj = 0 we have ej = 0, 1 ≤ j ≤ N.

Thm.(Classical) Quasi-independent sets are Sidon.





Pisier’s Thm

Def. E is proportionally quasi-independent if there exist 0 < δsuch that for every finite F ⊂ E , there is a quasi-independentset H such that H ⊂ F and #H ≥ δ#F .

Thm. (Pisier 1982) E is Sidon iff it is proportionallyquasi-independent.

Cor. The union of 2 Sidon sets is Sidon.





Kalton’s Thm

Recall Kalton’s Thm:Thm. E is I0 iff for each 0 < ε there exists 1 ≤ N such that foreach f ∈ `∞(E) there exist x(j) ∈ T and |cj | ≤ 1, 1 ≤ j ≤ Nsuch that

supn

∣∣∣∣∣∣f (n) =N∑

j=1

cjeinx(j)

∣∣∣∣∣∣ < ε.

We say that E is I0(N, ε)

Thm. (Ramsey 1996) E is Sidon iff it is proportionally I0(N, ε)for some 1 ≤ N and 0 < ε < 1.





Other proportionality results

Thm. (GH 2008) TFAE.(i) E is Sidon.(ii). E is proportionally ε-Kronecker.(iii). E is proportionally FZI0(N, ε) for some 1 ≤ N and0 < ε < 1.(iv.) E is cofinitely proportional FZI0(N, ε, U) for all open U.

(I will save “cofinitely” for questioners.)




Questions

1 If E is ε-Kronecker and√

2 ≤ ε < 2, is E Sidon? (It’s notnecessarily I0 – G-H 2006)

2 Is every quasi-independent set I0?

3 Is every Sidon set a finite union of I0 sets?

4 Can a Sidon set be dense in Z?(Yes to #3 implies No to #4.)




Kahane’s Theorem and sketch of a proof, I

Thm. (Kahane 1966) E is I0 iff each f ∈ `∞(E) extends to analmost periodic function f (n) =

∑j cjenx(j), where

∑j |cj | < ∞.

Proof. Let Ω = z : |z| = 1E , the set of functions on E whosevalues are in z : |z| = 1. With the product topology, Ω is acompact metric space.With coordinate-wise multiplication, Ω is an abelian group. Ofcourse, the multiplication is continuous.For 1 ≤ N and 0 < ε, let Ω(N, ε) be the set of ω ∈ Ω for whichthere exist x(1), . . . , x(N) ∈ T and c1, . . . , cN ∈ C with |cj | ≤ 1and |ω(n)−

∑j cjeinx(j)| ≤ ε for all n ∈ E .




Sketch of proof, II

Thena) Ω(N, ε) is closed in Ω,

b) Ω(N, ε) ⊂ Ω(N + 1, ε) for 1 ≤ N, and

c) Ω(N, ε) · Ω(M, δ) ⊂ Ω(N + M, ε + δ + εδ). (Exercise.)




Sketch of proof, III

Since every bounded function on E extends to an AP function,and since the AP functions are uniform limits of trig polys (finitesums

∑N1 cjeinx(j)),

d)⋃

N Ω(N, ε) = Ω for all 0 < ε.

The Baire category theorem says one of the Ω(N, ε) has nonempty interior.

Since Ω is a compact group, a finite number of translates ofΩ(N, ε) cover Ω.




Sketch of proof, IV

Let ωk + Ω(N, ε), 1 ≤ k ≤ K be such translates. For each kthere exists N(k) such that ωk ∈ Ω(N(k), ε).Let M = max N(k). Then Ω ⊂ Ω(M + N, 2ε + ε2) by c).12Ω + 1

2Ω contains all functions on E bounded by 1, so we haveshownThere exists L such that for every f bounded by 1 on E , thereexists cj and x(j) such that

|f (n)−L∑1

cjeinx(j)| ≤ ε for all n ∈ E and∑

j

cj | ≤ L.

That proves Kalton’s theorem.




Sketch of proof, V

A standard iteration shows that every bounded f : E → C is ofthe form

f (n) =∞∑1

cjeinx(j), where∑

j

|cj | < ∞.

And that proves Kahane’s theorem.


interpolation sets, past and present -- and future?i0 sets appplications to sidon sets open...

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