2
$\begingroup$

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.

  • 1
    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
  • 0
    @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
  • 0
    @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
  • 0
    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
  • 0
    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