The Wiener-Pitt tauberian theorem

1 Introduction

For $f\in L^1({ℝ}^d)$, we write

$$\widehat{f}(\xi )={\int }_{{ℝ}^d}f(x)e^{-2\pi i\xi \cdot x}𝑑x,\xi \in {ℝ}^d.$$

The Riemann-Lebesgue lemma tells us that $\widehat{f}\in C_0({ℝ}^d)$.

For $f\in C^{\mathrm{\infty }}({ℝ}^d)$ and for multi-indices $\alpha ,\beta$, write

$${|f|}_{\alpha ,\beta }=\underset{x\in {ℝ}^d}{sup}|x^{\alpha }({\partial }^{\beta }f)(x)|.$$

We say that $f$ is a Schwartz function if for all multi-indices $\alpha$ and $\beta$ we have . 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 $V_d=\frac{{\pi }^{d/2}}{\mathrm{\Gamma }\left(\frac{d}{2}+1\right)}$, the volume of the unit ball in ${ℝ}^d$.

Lemma 1.

For $\mathrm{1}\mathrm{\le }p\mathrm{\le }\mathrm{\infty }$, let $m$ be the least integer $\mathrm{\ge }\frac{d\mathrm{+}\mathrm{1}}{p}$. There is some $C_d$ such that for any multi-index $\beta$,

$${\parallel {\partial }^{\beta }f\parallel }_p\le V_d^{1/p}{|f|}_{0,\beta }+C_d{V}_d^{1/p}\sum _{|\alpha |=m}{|f|}_{\alpha ,\beta },f\in 𝒮.$$

Proof.

For $p=\mathrm{\infty }$, the claim is true with $C_{d,\mathrm{\infty }}=1$. For , let $g={\partial }^{\beta }f$, which satisfies

${\parallel g\parallel }_p$ $={\left({\displaystyle {\int }_{|x|\le 1}}{|g(x)|}^p𝑑x+{\displaystyle {\int }_{|x|\ge 1}}{|x|}^{d+1}{|g(x)|}^p{|x|}^{-(d+1)}𝑑x\right)}^{1/p}$

$\le {\left({\parallel g\parallel }_{\mathrm{\infty }}^p{V}_d+\underset{|x|\ge 1}{sup}\left({|x|}^{d+1}{|g(x)|}^p\right){\displaystyle {\int }_{|x|\ge 1}}{|x|}^{-(d+1)}𝑑x\right)}^{1/p}$

$={\left({\parallel g\parallel }_{\mathrm{\infty }}^p{V}_d+\underset{|x|\ge 1}{sup}\left({|x|}^{d+1}{|g(x)|}^p\right){\displaystyle {\int }_1^{\mathrm{\infty }}}\left({\displaystyle {\int }_{S^{d-1}}}{|r\gamma |}^{-(d+1)}𝑑\sigma (\gamma )\right)r^{d-1}𝑑r\right)}^{1/p}$

$={\left({\parallel g\parallel }_{\mathrm{\infty }}^p{V}_d+\underset{|x|\ge 1}{sup}\left({|x|}^{d+1}{|g(x)|}^p\right)\cdot V_d{\displaystyle {\int }_1^{\mathrm{\infty }}}{r}^{-2}𝑑r\right)}^{1/p}$

$=V_d^{1/p}{\left({\parallel g\parallel }_{\mathrm{\infty }}^p+\underset{|x|\ge 1}{sup}\left({|x|}^{d+1}{|g(x)|}^p\right)\right)}^{1/p}$

$\le V_d^{1/p}{\parallel g\parallel }_{\mathrm{\infty }}+V_d^{1/p}\underset{|x|\ge 1}{sup}\left({|x|}^{\frac{d+1}{p}}|g(x)|\right)$

$\le V_d^{1/p}{\parallel g\parallel }_{\mathrm{\infty }}+V_d^{1/p}\underset{|x|\ge 1}{sup}\left({|x|}^m|g(x)|\right).$

Using that the function $y\mapsto {\sum }_{|\alpha |=m}|y^{\alpha }|$ is continuous $S^{d-1}\to ℝ$, there is some $C_d$ such that

$${|x|}^m\le C_d\sum _{|\alpha |=m}|x^{\alpha }|,x\in {ℝ}^d.$$

This gives us

${\parallel g\parallel }_p$ $\le V_d^{1/p}{\parallel g\parallel }_{\mathrm{\infty }}+V_d^{1/p}\underset{|x|\ge 1}{sup}{C}_d{\displaystyle \sum _{|\alpha |=m}}|x^{\alpha }||g(x)|$

$=V_d^{1/p}{\parallel {\partial }^{\beta }f\parallel }_{\mathrm{\infty }}+C_d{V}_d^{1/p}{\displaystyle \sum _{|\alpha |=m}}\underset{|x|\ge 1}{sup}|x^{\alpha }({\partial }^{\beta }f)(x)|$

$\le V_d^{1/p}{|f|}_{0,\beta }+C_d{V}_d^{1/p}{\displaystyle \sum _{|\alpha |=m}}{|f|}_{\alpha ,\beta }.$

∎

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

$$|u(f)|\le C\sum _{|\alpha |\le m,|\beta |\le n}{|f|}_{\alpha ,\beta },f\in 𝒮.$$

For $1\le p\le \mathrm{\infty }$ and $g\in L^p({ℝ}^d)$, define $u:𝒮\to ℂ$ by

$$u(f)={\int }_{{ℝ}^d}f(x)g(x)𝑑x,f\in 𝒮.$$

For $\frac{1}{p}+\frac{1}{q}=1$, Hölder’s inequality tells us

$$|u(f)|\le {\parallel fg\parallel }_1\le {\parallel g\parallel }_p{\parallel f\parallel }_q.$$

By Lemma 1, with $m$ the least integer $\ge \frac{d+1}{q}$,

$${\parallel f\parallel }_q\le V_d^{1/q}{|f|}_{0,0}+C_d{V}_d^{1/q}\sum _{|\alpha |=m}|f_{\alpha ,0}|.$$

Therefore,

$$|u(f)|\le C_{g,d,q}\sum _{|\alpha |\le m,|\beta |\le 0}{|f|}_{\alpha ,\beta },$$

showing that $u$ is continuous. We thus speak of elements of $L^p({ℝ}^d)$ as tempered distributions, and speak about the Fourier transform of an element of $L^p({ℝ}^d)$.

Let $u\in {𝒟}^{\prime }$ be a distribution. For an open set $\omega$, we say that $u$ vanishes on $\omega$ if $u(\varphi )=0$ for every $\varphi \in 𝒟(\omega )$. Let $\mathrm{\Gamma }$ be the collection of open sets $\omega$ on which $u$ vanishes, and let $\mathrm{\Omega }={\bigcup }_{\omega \in \mathrm{\Gamma }}\omega$. $\mathrm{\Gamma }$ is an open cover of $\mathrm{\Omega }$, and thus there is a locally finite partition of unity ${\psi }_j$ subordinate to $\mathrm{\Gamma }$. For $\varphi \in 𝒟(\mathrm{\Omega })$, because $\mathrm{supp}\varphi$ is compact, there is some open set $W$, $\mathrm{supp}\varphi \subset W\subset \mathrm{\Omega }$, and some $m$ such that

$${\psi }_1(x)+\mathrm{\cdots }+{\psi }_m(x)=1,x\in W.$$

Then

$$u(\varphi )=u(\varphi ({\psi }_1+\mathrm{\cdots }+{\psi }_m))=u({\psi }_1\varphi )+\mathrm{\cdots }+u({\psi }_m\varphi ).$$

For each $j$, $1\le j\le m$, there is some ${\omega }_j\in \mathrm{\Gamma }$ such that $\mathrm{supp}{\psi }_j\subset {\omega }_j$, which implies $\mathrm{supp}{\psi }_j\varphi \subset {\omega }_j$, i.e. ${\psi }_j\varphi \in 𝒟({\omega }_j)$. But ${\omega }_j\in \mathrm{\Gamma }$, so $u({\psi }_j\varphi )=0$ and hence $u(\varphi )=0$. This shows that $\mathrm{\Omega }\in \mathrm{\Gamma }$, namely, $\mathrm{\Omega }$ is the largest open set on which $u$ vanishes. The support of $u$ is

$$\mathrm{supp}u={ℝ}^d\setminus \mathrm{\Omega }.$$

For $u\in {𝒮}^{\prime }$ we define $\widehat{u}:𝒮\to ℂ$ by

$$\widehat{u}(\varphi )=u(\widehat{\varphi }),\varphi \in 𝒮.$$

It is a fact that $\widehat{u}\in {𝒮}^{\prime }$.

For $f:{ℝ}^d\to ℂ$, write $\check{f}(x)=f(-x)$. For $\varphi \in 𝒮$,

$$ℱ(ℱ(\varphi ))=\check{\varphi }.$$

2 Tauberian theory

Lemma 2.

If $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $\zeta \mathrm{\in }{\mathrm{R}}^d$, and $ϵ\mathrm{>}\mathrm{0}$, then there is some $h\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ with and some $r\mathrm{>}\mathrm{0}$ such that

$$\widehat{h}(\xi )=\widehat{f}(\zeta )-\widehat{f}(\xi ),\xi \in B_r(\zeta ).$$

Proof.

It is a fact that there is a Schwartz function $g$ such that $\widehat{g}(\xi )=1$ for . For $\lambda >0$, let

$$g_{\lambda }(x)=e^{2\pi i\zeta \cdot x}{\lambda }^{-d}g({\lambda }^{-1}x),x\in {ℝ}^d,$$

which satisfies, for $\xi \in {ℝ}^d$,

$\widehat{g_{\lambda }}(\xi )$ $={\displaystyle {\int }_{{ℝ}^d}}{e}^{-2\pi i\xi \cdot x}{e}^{2\pi i\zeta \cdot x}{\lambda }^{-d}g({\lambda }^{-1}x)𝑑x$

$={\displaystyle {\int }_{{ℝ}^d}}{e}^{-2\pi i\lambda \xi \cdot y}{e}^{2\pi i\lambda \zeta \cdot y}g(y)𝑑y$

$=\widehat{g}(\lambda \xi -\lambda \zeta ).$

In particular, for $\xi \in V_{\lambda }=B_{{\lambda }^{-1}}(\zeta )$ we have $\widehat{g_{\lambda }}(\xi )=1$. We also define

$$h_{\lambda }(x)=\widehat{f}(\zeta )g_{\lambda }(x)-(f*g_{\lambda })(x),x\in {ℝ}^d,$$

which satisfies, for $\xi \in {ℝ}^d$,

$$\widehat{h_{\lambda }}(\xi )=\widehat{f}(\zeta )\widehat{g_{\lambda }}(\xi )-\widehat{f*g_{\lambda }}(\xi )=\widehat{f}(\zeta )\widehat{g_{\lambda }}(\xi )-\widehat{f}(\xi )\widehat{g_{\lambda }}(\xi )=\widehat{g_{\lambda }}(\xi )(\widehat{f}(\zeta )-\widehat{f}(\xi )).$$

Hence, for $\xi \in V_{\lambda }$ we have $\widehat{h_{\lambda }}(\xi )=\widehat{f}(\zeta )-\widehat{f}(\xi )$.

For $x\in {ℝ}^d$,

$h_{\lambda }(x)$ $={\displaystyle {\int }_{{ℝ}^d}}f(y)e^{-2\pi i\zeta \cdot y}{g}_{\lambda }(x)-{\displaystyle {\int }_{{ℝ}^d}}f(y)g_{\lambda }(x-y)𝑑y$

$={\displaystyle {\int }_{{ℝ}^d}}f(y)\left(e^{-2\pi i\zeta \cdot y}{g}_{\lambda }(x)-g_{\lambda }(x-y)\right)𝑑y,$

for which

Then

${\parallel h_{\lambda }\parallel }_1$ $\le {\displaystyle {\int }_{{ℝ}^d}}\left({\displaystyle {\int }_{{ℝ}^d}}|f(y)|{\lambda }^{-d}|g({\lambda }^{-1}x)-g({\lambda }^{-1}(x-y))|𝑑y\right)𝑑x$

$={\displaystyle {\int }_{{ℝ}^d}}\left({\displaystyle {\int }_{{ℝ}^d}}|f(y)||g(u)-g({\lambda }^{-1}(\lambda u-y))|𝑑y\right)𝑑u$

$={\displaystyle {\int }_{{ℝ}^d}}|f(y)|({\displaystyle {\int }_{{ℝ}^d}}|g(u)-g(u-{\lambda }^{-1}y))|du)dy.$

For each $y\in {ℝ}^d$,

$$|f(y)|({\int }_{{ℝ}^d}|g(u)-g(u-{\lambda }^{-1}y))|du)\le 2\parallel g\parallel {}_1|f(y),$$

and hence by the dominated convergence theorem,

$${\int }_{{ℝ}^d}|f(y)|({\int }_{{ℝ}^d}|g(u)-g(u-{\lambda }^{-1}y))|du)dy\to 0,\lambda \to \mathrm{\infty }.$$

Thus, there is some ${\lambda }_{ϵ}$ such that when $\lambda \ge {\lambda }_{ϵ}$. For $h=h_{{\lambda }_{ϵ}}$ and $r={\lambda }_{ϵ}^{-1}$, we have $\widehat{h}(\xi )=\widehat{f}(\zeta )-\widehat{f}(\xi )$ for $\xi \in V_{{\lambda }_{ϵ}}=B_r(\zeta )$ and , proving the claim. ∎

We remind ourselves that for $\varphi \in L^{\mathrm{\infty }}({ℝ}^d)$ and $f\in L^1({ℝ}^d)$, the convolution $f*\varphi$ belongs to $C_u({ℝ}^d)$, the collection of bounded uniformly continuous functions ${ℝ}^d\to ℂ$. We also remind ourselves that any element of $L^{\mathrm{\infty }}({ℝ}^d)$ is a tempered distribution whose Fourier transform is a tempered distribution.

Theorem 3.

If $\varphi \mathrm{\in }{L}^{\mathrm{\infty }}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $Y$ is a linear subspace of $L^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, and

$$f*\varphi =0,f\in Y,$$

then

$$Z(Y)=\bigcap _{f\in Y}\{\xi \in {ℝ}^d:\widehat{f}(\xi )=0\}$$

contains $\mathrm{supp}\mathit{}\widehat{\varphi }$.

Proof.

If $Y=\{0\}$, then $Z(Y)={ℝ}^d$, and the claim is true. If $Y$ has nonzero dimension, let $\zeta \in {ℝ}^d\setminus Z(Y)$ and let $f\in Y$ such that $\widehat{f}(\zeta )=1$; that there is such a function $f$ follows from $Y$ being a linear space. Thus by Lemma 2 there is some $h\in L^1({ℝ}^d)$ with and some $r>0$ such that

$$\widehat{h}(\xi )=1-\widehat{f}(\xi ),\xi \in B_r(\zeta );$$

because $Z(Y)$ is closed, we may take $r$ such that $B_r(\zeta )\subset {ℝ}^d\setminus Z(Y)$.

Let $\rho \in 𝒟(B_r(\zeta ))$, and let $\psi \in 𝒮$ with $\widehat{\psi }=\rho$. Define $g_0=\psi$ and $g_m=h*g_{m-1}$ for $m\ge 1$. By Young’s inequality

$${\parallel g_m\parallel }_1\le {\parallel h\parallel }_1^m{\parallel \psi \parallel }_1,$$

and because , this means that the sequence ${\sum }_{m=0}^M{g}_m$ is Cauchy in $L^1({ℝ}^d)$ so converges to some $G$, for which, as ,

$$\widehat{G}=\sum _{m=0}^{\mathrm{\infty }}\widehat{g_m}=\sum _{m=0}^{\mathrm{\infty }}\widehat{\psi }\cdot {\widehat{h}}^m=\widehat{\psi }\cdot {(1-\widehat{h})}^{-1}.$$

For $\xi \in \mathrm{supp}\widehat{\psi }\subset B_r(\zeta )$ we have $\widehat{h}(\xi )=1-\widehat{f}(\xi )$ and so

$$\widehat{\psi }(\xi )=\widehat{G}(\xi )(1-\widehat{h}(\xi ))=\widehat{G}(\xi )\widehat{f}(\xi );$$

on the other hand, for $\xi \notin \mathrm{supp}\widehat{\psi }$, $\widehat{\psi }(\xi )=0=\widehat{G}(\xi )\widehat{f}(\xi )$, so

$$\widehat{\psi }=\widehat{G}\cdot \widehat{f},$$

which implies that $\psi =G*f$. Then

$$\psi *\varphi =G*f*\varphi =G*0=0,$$

therefore

$$\widehat{\varphi }(\rho )=\varphi (\widehat{\rho })=\varphi ({ℱ}^2(\psi ))=\varphi (\check{\psi })={\int }_{{ℝ}^d}\psi (-x)\varphi (x)𝑑x=(\psi *\varphi )(0)=0.$$

This is true for all $\rho \in 𝒟(B_r(\zeta ))$, which means that $\widehat{\varphi }$ vanishes on $B_r(\zeta )$. This is true for any $\zeta \in {ℝ}^d\setminus Z(Y)$, so with $\mathrm{\Omega }$ the union of those open sets on which $\widehat{\varphi }$ vanishes, ${ℝ}^d\setminus Z(Y)\subset \mathrm{\Omega }$. Then $Z(Y)\subset {ℝ}^d\setminus \mathrm{\Omega }=\mathrm{supp}\widehat{\varphi }$. ∎

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

$$M^{⟂}=\{\gamma \in X^{*}:\text{if }x\in M\text{ then }⟨x,\gamma ⟩=0\}.$$

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

$${}^{⟂}N=\{x\in X:\text{if }\gamma \in N\text{ then }⟨x,\gamma ⟩=0\}.$$

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 $X$.

We say that a subspace $Y$ of $L^1({ℝ}^d)$ is translation-invariant if $f\in Y$ and $x\in {ℝ}^d$ imply that $f_x\in Y$, where $f_x(y)=f(y-x)$. The following theorem gives conditions under which a closed translation-invariant subspace of $L^1({ℝ}^d)$ is equal to the entire space.

Theorem 4.

If $Y$ is a closed translation-invariant subspace of $L^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and $Z\mathit{}\mathrm{(}Y\mathrm{)}\mathrm{=}\mathrm{\varnothing }$, then $Y\mathrm{=}{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$.

Proof.

Suppose that $\varphi \in L^{\mathrm{\infty }}({ℝ}^d)$ and $\int f\check{\varphi }=0$ for each $f\in Y$. Let $f\in Y$ and $x\in {ℝ}^d$. As $Y$ is translation-invariant, $f_{-x}\in Y$ so ${\int }_{{ℝ}^d}f(y+x)\varphi (-y)𝑑y=0$, i.e. $(f*\varphi )(x)=0$. This is true for all $x\in {ℝ}^d$, which means that $f*\varphi =0$. Theorem 3 then tells us that $\mathrm{supp}\widehat{\varphi }$ is contained in $Z(Y)$, namely, $\mathrm{supp}\widehat{\varphi }$ is empty, which means that the tempered distribution $\widehat{\varphi }$ vanishes on ${ℝ}^d$, i.e. $\mathrm{supp}\widehat{\varphi }$ is the zero element of the locally convex space ${𝒮}^{\prime }$. As the Fourier transform ${𝒮}^{\prime }\to {𝒮}^{\prime }$ is linear and one-to-one, the tempered distribution $\varphi$ is the zero element of ${𝒮}^{\prime }$, which implies that $\varphi \in L^{\mathrm{\infty }}({ℝ}^d)$ is zero. As Lebesgue measure on ${ℝ}^d$ is $\sigma$-finite, for $X$ the Banach space $L^1({ℝ}^d)$ we have $X^{*}=L^{\mathrm{\infty }}({ℝ}^d)$, with $⟨f,\gamma ⟩=\int f\gamma$. Thus $Y^{⟂}$ is the zero subspace of $L^{\mathrm{\infty }}({ℝ}^d)$, hence ${}^{⟂}(Y^{⟂})=L^1({ℝ}^d)$. This implies that $L^1({ℝ}^d)$ is equal to the closure of $Y$ in $L^1({ℝ}^d)$, and because $Y$ is closed this means $Y=L^1({ℝ}^d)$, completing the proof. ∎

Theorem 5.

Suppose that $K\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ and that $Y$ is the smallest closed translation-invariant subspace of $L^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ that includes $K$. $Y\mathrm{=}{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$ if and only if

$$\widehat{K}(\xi )\ne 0,\xi \in {ℝ}^d.$$

Proof.

Suppose that $\widehat{K}(\xi )\ne 0$ for all $\xi \in {ℝ}^d$. As $K\in Y$, this implies that $Z(Y)=\mathrm{\varnothing }$. Thus by Theorem 4 we get $Y=L^1({ℝ}^d)$.

Suppose that $Y=L^1({ℝ}^d)$. Then $f(x)=e^{-\pi {|x|}^2}$ belongs to $Y$ and $\widehat{f}(\xi )=e^{-\pi {|\xi |}^2}$, which has no zeros, hence $Z(Y)=\mathrm{\varnothing }$. For $\xi \in {ℝ}^d$, define ${\mathrm{ev}}_{\xi }:C_0({ℝ}^d)\to ℂ$ by ${\mathrm{ev}}_{\xi }(g)=g(\xi )$, which is a bounded linear operator. The Fourier transform $ℱ:L^1({ℝ}^d)\to C_0({ℝ}^d)$ is a bounded linear operator, hence for each $\xi \in {ℝ}^d$, ${\mathrm{ev}}_{\xi }\circ ℱ:L^1({ℝ}^d)\to ℂ$ is a bounded linear operator. Hence

$$V_{\xi }=\{f\in L^1({ℝ}^d):\widehat{f}(\xi )=0\}=\ker ({\mathrm{ev}}_{\xi }\circ ℱ)$$

is a closed subspace of $L^1({ℝ}^d)$. If $f\in V$ and $x\in {ℝ}^d$, then

$$\widehat{f_x}(\xi )={\int }_{{ℝ}^d}f(y-x)e^{-2\pi i\xi \cdot y}𝑑y=e^{-2\pi i\xi \cdot x}\widehat{f}(\xi )=0,$$

showing that $V_{\xi }$ is translation-invariant. Therefore

$$V=\bigcap _{\widehat{K}(\xi )=0}{V}_{\xi }$$

is a closed translation-invariant subspace of $L^1({ℝ}^d)$, and because $Y$ is the smallest closed translation-invariant subspace of $L^1({ℝ}^d)$, $Y\subset V$. $Y\subset V$ implies $Z(V)\subset Z(Y)=\mathrm{\varnothing }$, and applying Theorem 4 we get that $V=L^1({ℝ}^d)$. But there is no $\xi$ for which $V_{\xi }=L^1({ℝ}^d)$, so $V=L^1({ℝ}^d)$ implies that $\{\xi \in {ℝ}^d:\widehat{K}(\xi )=0\}=\mathrm{\varnothing }$. ∎

3 Slowly oscillating functions

Let $B({ℝ}^d)$ be the collection of bounded functions ${ℝ}^d\to ℂ$, which with the supremum norm ${\parallel f\parallel }_u={sup}_{x\in {ℝ}^d}|f(x)|$ is a Banach algebra.

A function $\varphi \in B({ℝ}^d)$ is said to be slowly oscillating if for each $ϵ>0$ there is some $A$ and some $\delta >0$ such that if $|x|,|y|>A$ and , then . We now prove the Wiener-Pitt tauberian theorem; the statement supposing that a function is slowly oscillating is attributed to Pitt.

Theorem 6 (Wiener-Pitt tauberian theorem).

If $\varphi \mathrm{\in }B\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $K\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $\widehat{K}\mathit{}\mathrm{(}\xi \mathrm{)}\mathrm{\ne }\mathrm{0}$ for all $\xi \mathrm{\in }{\mathrm{R}}^d$, and

$$\underset{|x|\to \mathrm{\infty }}{lim}(K*\varphi )(x)=a\widehat{K}(0),$$

then for each $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$,

$$\underset{|x|\to \mathrm{\infty }}{lim}(f*\varphi )(x)=a\widehat{f}(0).$$ (1)

Furthermore, if such $\varphi$ is slowly oscillating then

$$\underset{|x|\to \mathrm{\infty }}{lim}\varphi (x)=a.$$ (2)

Proof.

Define $\psi (x)=\varphi (x)-a$. Let $Y$ be the set of those $f\in L^1({ℝ}^d)$ for which

$$\underset{|x|\to \mathrm{\infty }}{lim}(f*\psi )(x)=0.$$

It is immediate that $Y$ is a linear subspace of $L^1({ℝ}^d)$. Suppose that $f_i\in Y$ tends to some $f\in L^1({ℝ}^d)$. As $\psi \in B({ℝ}^d)$, $f*\psi$ and $f_i*\psi$ belong to $C_u({ℝ}^d)$. Then

$${\parallel f*\psi -f_i*\psi \parallel }_u={\parallel (f-f_i)*\psi \parallel }_u={\parallel \psi \parallel }_u{\parallel f-f_i\parallel }_1.$$

There is some $i_0$ such that $i\ge i_0$ implies , and because $f_{i_0}\in Y$ there is some $M$ such that $|x|\ge M$ implies . Then for $|x|\ge M$,

$|(f*\psi )(x)|$ $\le |(f*\psi )(x)-(f_{i_0}*\psi )(x)|+|(f_{i_0}*\psi )(x)|$

$\le {\parallel \psi \parallel }_u{\parallel f-f_i\parallel }_1+|(f_{i_0}*\psi )(x)|$

showing that $f\in Y$, namely, that $Y$ is closed. Let $f\in Y$ and $x\in {ℝ}^d$. $f_x\in L^1({ℝ}^d)$, and for $y\in {ℝ}^d$,

$$(({\tau }_{x}f)*\psi )(y)=(f*\psi )(y-x),$$

and as $|y|\to \mathrm{\infty }$ we have $|y-x|\to \mathrm{\infty }$ and thus $(f*\psi )(y-x)\to 0$, hence ${\tau }_{x}f\in Y$, i.e. $Y$ is translation-invariant. Therefore $Y$ is a closed translation-invariant subspace of $L^1({ℝ}^d)$. For $x\in {ℝ}^d$,

$$(K*\psi )(x)={\int }_{{ℝ}^d}K(y)(\varphi (x-y)-a)𝑑y=(K*\varphi )(x)-a\widehat{K}(0),$$

and by hypothesis we get $(K*\psi )(x)\to 0$ as $|x|\to \mathrm{\infty }$, i.e. $K\in Y$.

Let $Y_0$ be the smallest closed translation-invariant subspace of $L^1({ℝ}^d)$ that includes $K$. On the one hand, because $Y$ is a closed translation-invariant subspace of $L^1({ℝ}^d)$ and $K\in Y$ we have $Y_0\subset Y$. On the other hand, because $\widehat{K}(\xi )\ne 0$ for all $\xi$ we have by Theorem 5 that $Y_0=L^1({ℝ}^d)$. Therefore $Y=L^1({ℝ}^d)$. This means that for each $f\in L^1({ℝ}^d)$, $(f*\psi )(x)\to 0$ as $|x|\to \mathrm{\infty }$, i.e. $(f*\varphi )(x)\to a\widehat{f}(0)$ as $|x|\to \mathrm{\infty }$, proving (1)

Assume further now that $\varphi$ is slowly-oscillating and let $ϵ>0$. There is some $A$ and some $\delta >0$ such that if $|x|,|y|>A$ and then

There is a test function $h$ such that $h\ge 0$, $h(x)=0$ for $|x|\ge \delta$, and $\widehat{h}(0)=1$. By (1),

$$\underset{|x|\to \mathrm{\infty }}{lim}(h*\varphi )(x)=a\widehat{h}(0)=a.$$

On the other hand, for $x\in {ℝ}^d$,

$\varphi (x)-(h*\varphi )(x)$ $=\widehat{h}(0)\varphi (x)-(h*\varphi )(x)$

$={\displaystyle {\int }_{{ℝ}^d}}(h(y)\varphi (x)-\varphi (x-y)h(y))𝑑y$

and so for $|x|>A+\delta$,

We have thus established that as $|x|\to \mathrm{\infty }$, (i) $(h*\varphi )(x)=a+o(1)$ and (ii) $\varphi (x)=(h*\varphi )(x)+o(1)$, which together yield $\varphi (x)=a+o(1)$, i.e. $\varphi (x)\to a$ as $|x|\to \mathrm{\infty }$, proving (2). ∎

4 Closed ideals in 𝐿¹(ℝ^d)

$L^1({ℝ}^d)$ is a Banach algebra using convolution as the product.

Theorem 7.

Suppose that $I$ is a closed linear subspace of $L^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$. $I$ is translation-invariant if and only if $I$ is an ideal.

Proof.

Assume that $I$ is translation-invariant and let $f\in I$ and $g\in L^1({ℝ}^d)$. For $\varphi \in I^{⟂}\subset L^{\mathrm{\infty }}({ℝ}^d)$,

$⟨g*f,\varphi ⟩$ $={\displaystyle {\int }_{{ℝ}^d}}(g*f)(x)\varphi (x)𝑑x$

$={\displaystyle {\int }_{{ℝ}^d}}\varphi (x)\left({\displaystyle {\int }_{{ℝ}^d}}g(x-y)f(y)𝑑y\right)𝑑x$

$={\displaystyle {\int }_{{ℝ}^d}}g(z)\left({\displaystyle {\int }_{{ℝ}^d}}\varphi (x)f_z(x)𝑑x\right)𝑑z$

$={\displaystyle {\int }_{{ℝ}^d}}g(z)⟨\varphi ,f_z⟩𝑑z$

$=0,$

because $f_z\in I$ for each $z\in {ℝ}^d$. This shows that $f*g{\in }^{⟂}(I^{⟂})$. But ${}^{⟂}(I^{⟂})$ is the closure of $I$ in $L^1({ℝ}^d)$, and $I$ is closed so $f*g\in I$, showing that $I$ is an ideal.

Assume that $I$ is an ideal and let $f\in I$ and $x\in {ℝ}^d$. Let $V$ be a closed ball centered at $0$, and let ${\chi }_A$ be the indicator function of a set $A$. We have

${\parallel f_x-{\displaystyle \frac{1}{\mu (V)}}{\chi }_{x+V}*f\parallel }_1$ $={\displaystyle {\int }_{{ℝ}^d}}\left|f_x(y)-{\displaystyle \frac{1}{\mu (V)}}({\chi }_{x+V}*f)(y)\right|𝑑y$

$={\displaystyle {\int }_{{ℝ}^d}}\left|{\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_V}{f}_x(y)𝑑z-{\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_{{ℝ}^d}}{\chi }_{x+V}(z)f(y-z)𝑑z\right|𝑑y$

$={\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_{{ℝ}^d}}\left|{\displaystyle {\int }_V}f(y-x)𝑑z-{\displaystyle {\int }_V}f(y-z-x)𝑑z\right|𝑑y$

$={\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_{{ℝ}^d}}\left|{\displaystyle {\int }_V}(f(y-x)-f(y-z-x))𝑑z\right|𝑑y$

$\le {\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_V}\left({\displaystyle {\int }_{{ℝ}^d}}\left|f(y-x)-f(y-z-x)\right|𝑑y\right)𝑑z$

$={\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_V}{\parallel f_x-f_{z+x}\parallel }_1𝑑z$

$={\displaystyle \frac{1}{\mu (V)}}{\displaystyle {\int }_V}{\parallel f-f_z\parallel }_1𝑑z$

$\le \underset{z\in V}{sup}{\parallel f-f_z\parallel }_1.$

Let $ϵ>0$. The map $z\mapsto f_z$ is continuous ${ℝ}^d\to L^1({ℝ}^d)$, so there is some $\delta >0$ such that if then , i.e. . Then let $V$ be the closed ball of radius $\delta$, with which

$${\parallel f_x-\frac{1}{\mu (V)}{\chi }_{x+V}*f\parallel }_1\le \underset{z\in V}{sup}{\parallel f-f_z\parallel }_1\le ϵ.$$ (3)

As $I$ is an ideal and $\frac{1}{\mu (V)}{\chi }_{x+V}\in L^1({ℝ}^d)$ we have $\frac{1}{\mu (V)}{\chi }_{x+V}*f\in L^1({ℝ}^d)$, and then (3) and the fact that $I$ is closed imply $f_x\in I$. Therefore $I$ is translation-invariant. ∎