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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 $L^1$. –  robjohn Sep 3 '11 at 1:12
    
@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! –  t.b. Sep 3 '11 at 1:18
    
@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. –  robjohn Sep 3 '11 at 2:09
    
Great book. For a reference check out Grafakos' "Classical Fourier Analysis". (It is not quite what you want but it is a similar result). –  Jonas Teuwen Sep 3 '11 at 10:18
    
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). –  robjohn Sep 3 '11 at 16:26

2 Answers 2

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? –  t.b. Sep 4 '11 at 9:42
    
@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 $T^*$? –  user1736 Sep 4 '11 at 15:31
    
@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. –  robjohn Sep 4 '11 at 19:17
    
@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 $T$). –  robjohn Sep 4 '11 at 19:20
    
@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. –  t.b. Sep 4 '11 at 19:32

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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