Bounded linear operators that commute with translation

Question

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.

Answer 1

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

Answer 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.

Answer 3

A bounded linear operator $ T: L^2 \to L^2$ and commutes with translation

Known result from Riesz Representation theorem see here:$\int Tf(s)u(s) ds=\int f(s)T^*u(s)ds$ written as $<Tf,u>=<f,T^*u>$, where $ T^*$ is the linear adjoint operator of $ T $ , $ u \in L^2$

now for $f(s)$ translated by $x$ ,$f(s+x)$

we have $\int Tf(s+x)u(s)ds=\int f(s+x)T^*u(s)ds$

Now take Let $u_{\epsilon}=\phi_\epsilon(s)=\phi(s/\epsilon)\epsilon^{-1}$ where $\phi(s)$ is normalized gaussian function with zero mean.

$\int Tf(s+x)u_{\epsilon}(-s) ds=\int Tf(s+x)u_{\epsilon}(s)ds=\int f(s+x)T^*u_{\epsilon}(s)ds$

$H(s)=Tf(s)$ , by translation commutation $H(s+x)=Tf(s+x)$

$R(-s)=T^*u_{\epsilon}$

we have $\lim_{\epsilon \to 0} H* u_{\epsilon}=\lim_{\epsilon \to 0} f*R$

$\lim_{\epsilon \to 0} H*u_{\epsilon}=H(x)$ ae

for $u \in L^1$ function, we have the following theorem regarding fourier transform:

$v(x)=g*u=\int g(s)u(x-s) ds$ for $g \in L^2$ , define $\hat{g}=lim_{n \to \infty}\int_{n}^{-n}g(s)e^{2i\pi x z} dx$ the limit exists in the sense of $L^2$. If $u \in L^1$ and $\int |g(s)||u(x-s)| ds \le P$ where $P$ is a non negative integrable function in $L^2$ then $\hat{v}=\hat{g}\hat{u}$

Proof : $g_n=g1_{[-n,n]},v_n=g_n*u$ it's known that $\hat{v_n}=\hat{g_n}\hat{u}$ By dominated convergence theorem $lim_{n \to \infty} ||v-v_n||^2=0$ and by Plancherel theorem $lim_{n \to \infty} ||v-v_m||^2= lim_{n \to \infty} ||\hat{v}-\hat{v_n}||^2=0$ This implies $lim_{n \to \infty} \hat{v_n}=\hat{v}$

Now from our equation when $u$ is a normalized Gaussian we have $|H|* |u_{\epsilon}| \le MH(x)$ where $MH(x)$ is the Hardy Littlewood Maximal function (also $ ||MH(x)||^2 \le ||H(x)||^2$) using Fourier transform $\hat{H}(z)\hat{u_{\epsilon}}(z)=\hat{f}(z)G(z)$ where $G(z)=\hat{R}$

$\lim_{\epsilon \to 0}\hat{ u_{\epsilon}}=1$

$\lim_{\epsilon \to 0}G=m(z)$

consequently we have $\hat{Hf}(z)=m(z)\hat{f}(z)$