The Wiener-Pitt tauberian theorem

Jordan Bell
March 10, 2015

1 Introduction

For f∈L1⁢(ℝd), we write


The Riemann-Lebesgue lemma tells us that f^∈C0⁢(ℝd).

For f∈C∞⁢(ℝd) and for multi-indices α,β, write


We say that f is a Schwartz function if for all multi-indices α and β we have |f|α,β<∞. We denote by 𝒮 the collection of Schwartz functions. It is a fact that 𝒮 with this family of seminorms is a Fréchet space.

Let Vd=πd/2Γ⁢(d2+1), the volume of the unit ball in ℝd.

Lemma 1.

For 1≤p≤∞, let m be the least integer ≥d+1p. There is some Cd such that for any multi-index β,


For p=∞, the claim is true with Cd,∞=1. For 1≤p<∞, let g=∂β⁡f, which satisfies

∥g∥p =(∫|x|≤1|g⁢(x)|p⁢𝑑x+∫|x|≥1|x|d+1⁢|g⁢(x)|p⁢|x|-(d+1)⁢𝑑x)1/p

Using that the function y↦∑|α|=m|yα| is continuous Sd-1→ℝ, there is some Cd such that


This gives us

∥g∥p ≤Vd1/p⁢∥g∥∞+Vd1/p⁢sup|x|≥1⁡Cd⁢∑|α|=m|xα|⁢|g⁢(x)|


The dual space 𝒮′ with the weak-* topology is a locally convex space, elements of which are called tempered distributions. It is straightforward to check that if u:𝒮→ℂ is linear, then u∈𝒮′ if and only if there is some C and some nonnegative integers m,n such that


For 1≤p≤∞ and g∈Lp⁢(ℝd), define u:𝒮→ℂ by


For 1p+1q=1, Hölder’s inequality tells us


By Lemma 1, with m the least integer ≥d+1q,




showing that u is continuous. We thus speak of elements of Lp⁢(ℝd) as tempered distributions, and speak about the Fourier transform of an element of Lp⁢(ℝd).

Let u∈𝒟′ be a distribution. For an open set ω, we say that u vanishes on ω if u⁢(ϕ)=0 for every ϕ∈𝒟⁢(ω). Let Γ be the collection of open sets ω on which u vanishes, and let Ω=⋃ω∈Γω. Γ is an open cover of Ω, and thus there is a locally finite partition of unity ψj subordinate to Γ.11 1 Walter Rudin, Functional Analysis, second ed., p. 162, Theorem 6.20. For ϕ∈𝒟⁢(Ω), because supp⁢ϕ is compact, there is some open set W, supp⁢ϕ⊂W⊂Ω, and some m such that




For each j, 1≤j≤m, there is some ωj∈Γ such that supp⁢ψj⊂ωj, which implies supp⁢ψj⁢ϕ⊂ωj, i.e. ψj⁢ϕ∈𝒟⁢(ωj). But ωj∈Γ, so u⁢(ψj⁢ϕ)=0 and hence u⁢(ϕ)=0. This shows that Ω∈Γ, namely, Ω is the largest open set on which u vanishes. The support of u is


For u∈𝒮′ we define u^:𝒮→ℂ by


It is a fact that u^∈𝒮′.

For f:ℝd→ℂ, write fˇ⁢(x)=f⁢(-x). For ϕ∈𝒮,


2 Tauberian theory

Lemma 2.

If f∈L1⁢(Rd), ζ∈Rd, and ϵ>0, then there is some h∈L1⁢(Rd) with ∥h∥1<ϵ and some r>0 such that


It is a fact that there is a Schwartz function g such that g^⁢(ξ)=1 for |ξ|<1. For λ>0, let


which satisfies, for ξ∈ℝd,

gλ^⁢(ξ) =∫ℝde-2⁢π⁢i⁢ξ⋅x⁢e2⁢π⁢i⁢ζ⋅x⁢λ-d⁢g⁢(λ-1⁢x)⁢𝑑x

In particular, for ξ∈Vλ=Bλ-1⁢(ζ) we have gλ^⁢(ξ)=1. We also define


which satisfies, for ξ∈ℝd,


Hence, for ξ∈Vλ we have hλ^⁢(ξ)=f^⁢(ζ)-f^⁢(ξ).

For x∈ℝd,

hλ⁢(x) =∫ℝdf⁢(y)⁢e-2⁢π⁢i⁢ζ⋅y⁢gλ⁢(x)-∫ℝdf⁢(y)⁢gλ⁢(x-y)⁢𝑑y

for which



∥hλ∥1 ≤∫ℝd(∫ℝd|f⁢(y)|⁢λ-d⁢|g⁢(λ-1⁢x)-g⁢(λ-1⁢(x-y))|⁢𝑑y)⁢𝑑x

For each y∈ℝd,


and hence by the dominated convergence theorem,


Thus, there is some λϵ such that ∥hλ∥1<ϵ when λ≥λϵ. For h=hλϵ and r=λϵ-1, we have h^⁢(ξ)=f^⁢(ζ)-f^⁢(ξ) for ξ∈Vλϵ=Br⁢(ζ) and ∥h∥1<ϵ, proving the claim. ∎

We remind ourselves that for ϕ∈L∞⁢(ℝd) and f∈L1⁢(ℝd), the convolution f*ϕ belongs to Cu⁢(ℝd), the collection of bounded uniformly continuous functions ℝd→ℂ. We also remind ourselves that any element of L∞⁢(ℝd) is a tempered distribution whose Fourier transform is a tempered distribution.22 2 Walter Rudin, Functional Analysis, second ed., p. 228, Theorem 9.3.

Theorem 3.

If ϕ∈L∞⁢(Rd), Y is a linear subspace of L1⁢(Rd), and




contains supp⁢ϕ^.


If Y={0}, then Z⁢(Y)=ℝd, and the claim is true. If Y has nonzero dimension, let ζ∈ℝd∖Z⁢(Y) and let f∈Y such that f^⁢(ζ)=1; that there is such a function f follows from Y being a linear space. Thus by Lemma 2 there is some h∈L1⁢(ℝd) with ∥h∥1<1 and some r>0 such that


because Z⁢(Y) is closed, we may take r such that Br⁢(ζ)⊂ℝd∖Z⁢(Y).

Let ρ∈𝒟⁢(Br⁢(ζ)), and let ψ∈𝒮 with ψ^=ρ. Define g0=ψ and gm=h*gm-1 for m≥1. By Young’s inequality


and because ∥h∥1<1, this means that the sequence ∑m=0Mgm is Cauchy in L1⁢(ℝd) so converges to some G, for which, as |h^|≤∥h∥1<1,


For ξ∈supp⁢ψ^⊂Br⁢(ζ) we have h^⁢(ξ)=1-f^⁢(ξ) and so


on the other hand, for ξ∉supp⁢ψ^, ψ^⁢(ξ)=0=G^⁢(ξ)⁢f^⁢(ξ), so


which implies that ψ=G*f. Then




This is true for all ρ∈𝒟⁢(Br⁢(ζ)), which means that ϕ^ vanishes on Br⁢(ζ). This is true for any ζ∈ℝd∖Z⁢(Y), so with Ω the union of those open sets on which ϕ^ vanishes, ℝd∖Z⁢(Y)⊂Ω. Then Z⁢(Y)⊂ℝd∖Ω=supp⁢ϕ^. ∎

If X is a Banach space and M is a linear subspace of X, we define the annihilator of M as

M⟂={γ∈X*:if x∈M then ⟨x,γ⟩=0}.

It is immediate that M⟂ is a weak-* closed linear subspace of X*. If N is a linear subspace of X*, we define the annihilator of N as

N⟂={x∈X:if γ∈N then ⟨x,γ⟩=0}.

It is immediate that N⟂ is a norm closed linear subspace of the Banach space X. One proves using the Hahn-Banach theorem that (M⟂)⟂ is the norm closure of M in X.33 3 Walter Rudin, Functional Analysis, second ed., p. 96, Theorem 4.7.

We say that a subspace Y of L1⁢(ℝd) is translation-invariant if f∈Y and x∈ℝd imply that fx∈Y, where fx⁢(y)=f⁢(y-x). The following theorem gives conditions under which a closed translation-invariant subspace of L1⁢(ℝd) is equal to the entire space.44 4 Walter Rudin, Functional Analysis, second ed., p. 228, Theorem 9.4.

Theorem 4.

If Y is a closed translation-invariant subspace of L1⁢(Rd) and Z⁢(Y)=∅, then Y=L1⁢(Rd).


Suppose that ϕ∈L∞⁢(ℝd) and ∫f⁢ϕˇ=0 for each f∈Y. Let f∈Y and x∈ℝd. As Y is translation-invariant, f-x∈Y so ∫ℝdf⁢(y+x)⁢ϕ⁢(-y)⁢𝑑y=0, i.e. (f*ϕ)⁢(x)=0. This is true for all x∈ℝd, which means that f*ϕ=0. Theorem 3 then tells us that supp⁢ϕ^ is contained in Z⁢(Y), namely, supp⁢ϕ^ is empty, which means that the tempered distribution ϕ^ vanishes on ℝd, i.e. supp⁢ϕ^ is the zero element of the locally convex space 𝒮′. As the Fourier transform 𝒮′→𝒮′ is linear and one-to-one, the tempered distribution ϕ is the zero element of 𝒮′, which implies that ϕ∈L∞⁢(ℝd) is zero. As Lebesgue measure on ℝd is σ-finite, for X the Banach space L1⁢(ℝd) we have X*=L∞⁢(ℝd), with ⟨f,γ⟩=∫f⁢γ. Thus Y⟂ is the zero subspace of L∞⁢(ℝd), hence (Y⟂)⟂=L1(ℝd). This implies that L1⁢(ℝd) is equal to the closure of Y in L1⁢(ℝd), and because Y is closed this means Y=L1⁢(ℝd), completing the proof. ∎

Theorem 5.

Suppose that K∈L1⁢(Rd) and that Y is the smallest closed translation-invariant subspace of L1⁢(Rd) that includes K. Y=L1⁢(Rd) if and only if


Suppose that K^⁢(ξ)≠0 for all ξ∈ℝd. As K∈Y, this implies that Z⁢(Y)=∅. Thus by Theorem 4 we get Y=L1⁢(ℝd).

Suppose that Y=L1⁢(ℝd). Then f⁢(x)=e-π⁢|x|2 belongs to Y and f^⁢(ξ)=e-π⁢|ξ|2, which has no zeros, hence Z⁢(Y)=∅. For ξ∈ℝd, define evξ:C0⁢(ℝd)→ℂ by evξ⁢(g)=g⁢(ξ), which is a bounded linear operator. The Fourier transform ℱ:L1⁢(ℝd)→C0⁢(ℝd) is a bounded linear operator, hence for each ξ∈ℝd, evξ∘ℱ:L1⁢(ℝd)→ℂ is a bounded linear operator. Hence


is a closed subspace of L1⁢(ℝd). If f∈V and x∈ℝd, then


showing that Vξ is translation-invariant. Therefore


is a closed translation-invariant subspace of L1⁢(ℝd), and because Y is the smallest closed translation-invariant subspace of L1⁢(ℝd), Y⊂V. Y⊂V implies Z⁢(V)⊂Z⁢(Y)=∅, and applying Theorem 4 we get that V=L1⁢(ℝd). But there is no ξ for which Vξ=L1⁢(ℝd), so V=L1⁢(ℝd) implies that {ξ∈ℝd:K^⁢(ξ)=0}=∅. ∎

3 Slowly oscillating functions

Let B⁢(ℝd) be the collection of bounded functions ℝd→ℂ, which with the supremum norm ∥f∥u=supx∈ℝd⁡|f⁢(x)| is a Banach algebra.

A function ϕ∈B⁢(ℝd) is said to be slowly oscillating if for each ϵ>0 there is some A and some δ>0 such that if |x|,|y|>A and |x-y|<δ, then |ϕ⁢(x)-ϕ⁢(y)|<ϵ. We now prove the Wiener-Pitt tauberian theorem; the statement supposing that a function is slowly oscillating is attributed to Pitt.55 5 Walter Rudin, Functional Analysis, second ed., p. 229, Theorem 9.7; Walter Rudin, Fourier Analysis on Groups, p. 163, Theorem 7.2.7; Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 116, Theorem 4.72; V. P. Havin and N. K. Nikolski, Commutative Harmonic Analysis II, p. 134; Edwin Hewitt and Kenneth A. Ross, Abstract Harmonic Analysis II, p. 511, Theorem 39.37.

Theorem 6 (Wiener-Pitt tauberian theorem).

If ϕ∈B⁢(Rd), K∈L1⁢(Rd), K^⁢(ξ)≠0 for all ξ∈Rd, and


then for each f∈L1⁢(Rd),

lim|x|→∞⁡(f*ϕ)⁢(x)=a⁢f^⁢(0). (1)

Furthermore, if such ϕ is slowly oscillating then

lim|x|→∞⁡ϕ⁢(x)=a. (2)

Define ψ⁢(x)=ϕ⁢(x)-a. Let Y be the set of those f∈L1⁢(ℝd) for which


It is immediate that Y is a linear subspace of L1⁢(ℝd). Suppose that fi∈Y tends to some f∈L1⁢(ℝd). As ψ∈B⁢(ℝd), f*ψ and fi*ψ belong to Cu⁢(ℝd). Then


There is some i0 such that i≥i0 implies ∥f-fi∥1<ϵ, and because fi0∈Y there is some M such that |x|≥M implies |(fi0*ψ)⁢(x)|<ϵ. Then for |x|≥M,

|(f*ψ)⁢(x)| ≤|(f*ψ)⁢(x)-(fi0*ψ)⁢(x)|+|(fi0*ψ)⁢(x)|

showing that f∈Y, namely, that Y is closed. Let f∈Y and x∈ℝd. fx∈L1⁢(ℝd), and for y∈ℝd,


and as |y|→∞ we have |y-x|→∞ and thus (f*ψ)⁢(y-x)→0, hence τx⁢f∈Y, i.e. Y is translation-invariant. Therefore Y is a closed translation-invariant subspace of L1⁢(ℝd). For x∈ℝd,


and by hypothesis we get (K*ψ)⁢(x)→0 as |x|→∞, i.e. K∈Y.

Let Y0 be the smallest closed translation-invariant subspace of L1⁢(ℝd) that includes K. On the one hand, because Y is a closed translation-invariant subspace of L1⁢(ℝd) and K∈Y we have Y0⊂Y. On the other hand, because K^⁢(ξ)≠0 for all ξ we have by Theorem 5 that Y0=L1⁢(ℝd). Therefore Y=L1⁢(ℝd). This means that for each f∈L1⁢(ℝd), (f*ψ)⁢(x)→0 as |x|→∞, i.e. (f*ϕ)⁢(x)→a⁢f^⁢(0) as |x|→∞, proving (1)

Assume further now that ϕ is slowly-oscillating and let ϵ>0. There is some A and some δ>0 such that if |x|,|y|>A and |x-y|<δ then


There is a test function h such that h≥0, h⁢(x)=0 for |x|≥δ, and h^⁢(0)=1. By (1),


On the other hand, for x∈ℝd,

ϕ⁢(x)-(h*ϕ)⁢(x) =h^⁢(0)⁢ϕ⁢(x)-(h*ϕ)⁢(x)

and so for |x|>A+δ,


We have thus established that as |x|→∞, (i) (h*ϕ)⁢(x)=a+o⁢(1) and (ii) ϕ⁢(x)=(h*ϕ)⁢(x)+o⁢(1), which together yield ϕ⁢(x)=a+o⁢(1), i.e. ϕ⁢(x)→a as |x|→∞, proving (2). ∎

4 Closed ideals in 𝐿¹(ℝd)

L1⁢(ℝd) is a Banach algebra using convolution as the product.66 6 Eberhard Kaniuth, A Course in Commutative Banach Algebras, p. 25, Proposition 1.4.7.

Theorem 7.

Suppose that I is a closed linear subspace of L1⁢(Rd). I is translation-invariant if and only if I is an ideal.


Assume that I is translation-invariant and let f∈I and g∈L1⁢(ℝd). For ϕ∈I⟂⊂L∞⁢(ℝd),

⟨g*f,ϕ⟩ =∫ℝd(g*f)⁢(x)⁢ϕ⁢(x)⁢𝑑x

because fz∈I for each z∈ℝd. This shows that f*g∈⟂(I⟂). But (I⟂)⟂ is the closure of I in L1⁢(ℝd),77 7 Walter Rudin, Functional Analysis, second ed., p. 96, Theorem 4.7. and I is closed so f*g∈I, showing that I is an ideal.

Assume that I is an ideal and let f∈I and x∈ℝd. Let V be a closed ball centered at 0, and let χA be the indicator function of a set A. We have

∥fx-1μ⁢(V)⁢χx+V*f∥1 =∫ℝd|fx⁢(y)-1μ⁢(V)⁢(χx+V*f)⁢(y)|⁢𝑑y

Let ϵ>0. The map z↦fz is continuous ℝd→L1⁢(ℝd), so there is some δ>0 such that if |z|<δ then ∥fz-f0∥1<ϵ, i.e. ∥f-fz∥1<ϵ. Then let V be the closed ball of radius δ, with which

∥fx-1μ⁢(V)⁢χx+V*f∥1≤supz∈V⁡∥f-fz∥1≤ϵ. (3)

As I is an ideal and 1μ⁢(V)⁢χx+V∈L1⁢(ℝd) we have 1μ⁢(V)⁢χx+V*f∈L1⁢(ℝd), and then (3) and the fact that I is closed imply fx∈I. Therefore I is translation-invariant. ∎