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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 $\|T\|=\|\mu\|$.

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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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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, $\|m\|_{L^\infty}=\|T\|_{L^2}$.

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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