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I'm trying to read Elias Stein's "Singular Integrals" book, and in the beginning of the second chapter, he states two results classifying bounded linear operators that commute (on $L^1$ and $L^2$ respectively).

The first one reads:

Let $T: L^1(\mathbb{R}^n) \to L^1(\mathbb{R}^n)$ be a bounded linear transformation. Then $T$ commutes with translations if and only if there exists a measure $\mu$ in the dual of $C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), s.t. $T(f) = f \ast \mu$ for every $f \in L^1(\mathbb{R}^n)$. It is also true that $\|T\|=\|\mu\|$.

The second one says:

Let $T:L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be bounded and linear. Then $T$ commutes with translation if and only if there exists a bounded measurable function $m(y)$ so that $(T\hat{f})(y) = m(y) \hat{f}(y)$ for all $f \in L^2(\mathbb{R}^n)$. It is also true that $\|T\|=\|m\|_\infty$.

I was wondering if anyone had a reference to a proof of these two results or could explain why they are true.

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    Here is a hint for the first part. Let $\phi_\epsilon(x)=\phi(x/\epsilon)/\epsilon^n$ where $$\phi(x)=\left\{\begin{array}{cl}e^{|x|^2/(|x|^2-1)}&\text{if }|x|<1\\0&\text{if }|x|\ge1\end{array}\right.$$ $\|T\phi_\epsilon\|_{L^1}$ is bounded as $\epsilon\to0$, so there is a sequence $\{\epsilon_k\}$ and a measure $\mu$ so that $T\phi_{\epsilon_k}\to\mu$ weakly in $L^1$.2011-09-03
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    @robjohn: it may be worth pointing out that the only thing you need is that you have an approximate unit in in the convolution algebra $L^1$ consisting of continuous functions of compact support. Making it explicit doesn't help much, I think. But it's a great hint!2011-09-03
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    @Theo: Yes, you are right. There is nothing special about the $\phi$ I gave, other than it is positive, has compact support, and is in $L^1$. Originally, I started out writing up a proof using this $\phi$, but due to time constraints, I posted a hint instead. When I get time later, I may finish the proof.2011-09-03
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    Great book. For a reference check out Grafakos' "Classical Fourier Analysis". (It is not quite what you want but it is a similar result).2011-09-03
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    For the second one the definition should be $(Tf)^\wedge=m\hat{f}$ (otherwise $T$ is just multiplication by $m$ rather than a Fourier multiplier operator).2011-09-03

3 Answers 3

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For the first one:

Let $\phi_\epsilon(x)=\phi(x/\epsilon)/\epsilon^n$ where $$ \phi(x)=\left\{\begin{array}{cl}c\;e^{|x|^2/(|x|^2-1)}&\text{if }|x|<1\\0&\text{if }|x|\ge1\end{array}\right. $$ and $c$ is chosen so that $\int_{\mathbb{R}^n}\phi(x)\;\mathrm{d}x=1$.

$\|T\phi_\epsilon\|_{L^1}$ is bounded as $\epsilon\to0$, so there is a sequence $\{\epsilon_k\}$ and a measure $\mu$ so that $T\phi_{\epsilon_k}\to\mu$ weakly in $L^1$.

Since $T$ is continuous, linear, and commutes with translation, $$ \begin{align} f*\mu(x) &=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;T\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\\ &=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;T(\phi_{\epsilon_k}(x-y))\;\mathrm{d}y\\ &=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}T(f(y)\;\phi_{\epsilon_k}(x-y))\;\mathrm{d}y\\ &=\lim_{k\to\infty}\;T\left(\int_{\mathbb{R}^n}f(y)\;\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\right)\\ &=T\left(\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\right)\\ &=Tf(x) \end{align} $$ For the second one the definition should be $(Tf)^\wedge=m\hat{f}$ (otherwise $T$ is just multiplication by $m$ rather than a Fourier multiplier operator). Let $\psi(x)=e^{-\pi x^2}$ so that $\hat{\psi}=\psi$. $$ \begin{align} \psi(\xi)\;(Tf)^\wedge(\xi) &=(\psi*Tf)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}\psi(x-y)\;Tf(y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}\psi(y)\;Tf(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}T^{\;*}\psi(y)\;f(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=(T^{\;*}\psi)^\wedge(\xi)\hat{f}(\xi) \end{align} $$ Let $m(\xi)=(T^{\;*}\psi)^\wedge(\xi)/\psi(\xi)$, then we have $$ (Tf)^\wedge(\xi)=m(\xi)\hat{f}(\xi) $$ and therefore, $\|m\|_{L^\infty}=\|T\|_{L^2}$.

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    I have two simple questions here: 1. Isn't the argument rather that you have a sequence $\phi_{\epsilon_k}$ that converges weak$^{\ast}$ in $(C_0)^{\ast}$ and do your computation for $f$ continuous with compact support first and then extend the equality $f \ast \mu = Tf$ for $f \in C_c$ to all of $L^1$ by continuity of both sides? 2. You say "so there is a sequence ... and a measure $\mu$ such that $\phi_{\epsilon_k} \to \mu$ weakly in $L^1$". What result do you have in mind here? I've never seen such a statement and re-proved variants by hand as needed. Could you give me a reference, please?2011-09-04
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    @robjohn: I was going to ask the same question as Theo. Also, how do you go from $\int f(y) T(\phi_{\epsilon_k}(x-y)) dy = \int T(f(y)) \phi_{\epsilon_k}(x-y) dy$? Wouldnt it be $T^*$?2011-09-04
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    @Theo: $L^1$ is a subset of the signed Radon measures, $\mathcal{M}$, which is the dual of $C_c$. $\mathcal{M}$ is weakly-* compact in that any bounded sequence $\{\mu_k\}$ in $\mathcal{M}$ contains a subsequence which converges weakly-* to some $\mu\in\mathcal{M}$ (that is when tested against any $f\in C_c$). One reference is [Weak and weak star convergences](http://jpmandallena.pagesperso-orange.fr/weak-convergence.pdf).2011-09-04
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    @user1736: your parentheses do not match mine. I am simply using the linearity and translation invariance of $T$ (that is, $T$ is operating on $x$, and $y$ is just translation that can be moved outside $T$).2011-09-04
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    @robjohn: Thanks for the reference, but you're essentially saying the same as I did in my comment. However, you want to take the dual of $C_0$ (functions vanishing at infinity) not of $C_c$ (functions of compact support--which you usually don't equip with the sup-norm but with a different topology). By the way, Theorem 7.1 in the linked notes is wrong in that the space $C_c$ is *not* complete wrt the sup-norm. I do object to your usage of "converges weakly in $L^1$", then (which would involve duality with $L^\infty$). But it's merely a technical detail and I'm happy with your argument.2011-09-04
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    @Theo: I had originally written $C_0$, but the reference I cited used $C_c$, so I thought that perhaps I had misremembered. Of course you are correct, and $C_c$ should be changed to $C_0$ in my last comment. My weak vs weak-* notation was a little sloppy, too.2011-09-04
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    @robjohn: Oh, I think I can see what you mean. So you're also just using linearity to pull $T$ outside of the integral, and to pull the limit inside then, right?2011-09-04
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    @user1736: linearity to pull $T$ outside, and continuity to pull the limit inside. Yes.2011-09-04
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    @robjohn Does your solution for part (ii) show both directions?2017-02-26
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    @user412674: my answer shows that if $T$ is a bounded linear operator on $L^2$ that commutes with translations, then $T$ is a Fourier multiplier operator with a bounded measurable function. The other direction follows pretty easily by [Plancherel's Theorem](https://en.wikipedia.org/wiki/Plancherel_theorem).2017-02-26
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Let me just give a literature reference, in case that's what you are interested in. The classical paper on translation invariant operators on Lebesgue spaces is Hormander's 1960 Acta paper "Estimates for translation invariant operators in $L^p$ spaces". If you have access to MathSciNet, here's the mref.

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Have a look at Fourier Multipliers in "classical fourier analysis by Loukas Grafakos". If you are interested in it's application in the proof of Carleson-Hunt theorem look at "modern fourier analysis by loukas grafakos".