I am trying to prove a statement that appears in Stein's book, Singular and integrals differentiability properties, page 95.
I have been able to demonstrate an address of the statement but the inverse failed to decipher the demonstration sketch.
I put the context of the subject a little.
Let $m$ be a bounded measurable function on $R^n$. One can then define a linear transformation $T_m$ whose domain is $L^2\cap L^p$, by the following relation between Fourier transforms
$$(T_m f)^{\wedge}(x)=m(x)\hat{f}(x),\quad f\in L^2\cap L^p$$
We shall say that $m$ is a multiplier for $L^p (1\leq p\leq \infty$) if whenever $f\in L^2\cap L^p$ then $T_m f\in L^p$ and $T_m$ is bounded, that is,
$$|T_m(f)|_{p}\leq A|f|_{p},\quad f\in L^2\cap L^p$$
If $p<\infty$, then $T_m$ has a unique bounded extension to $L^p$, wich again satisfies the same inequality. We shall also write $T_m$ for this
extension.
We denote by $\mathcal{M}_{p}$ the class of multipliers with the indicated norm.
We have already pointed out that if $m$ is a multiplier (in $\mathcal{M}_p)$, then the transformation $T_m$, wich is bounded in $L^p$, commutes
with translations.
The converse also holds: Suppose that $T$ is a bounded linear transformation on $L^p$, $p<\infty$, which commutes with translations;
then there exists an Fourier multiplier $m$ so that $T_m=T$.
I have this: Let $\tau_{h}f(x)=f(x-y), \rho_{h}f(x)=e^{ix\cdot h}f(x)$ then
\begin{align*}
T_{s}(\tau_{h}g)&=\mathcal{F}^{-1}(m(x)\mathcal{F}(\tau_{h} g))\\
&=\mathcal{F}^{-1}(m(x)\rho_{h} \mathcal{F}(g))\\
&=\mathcal{F}^{-1}(\rho_{h}m(x)\mathcal{F}(T_{s}g))\\
&=\tau_{h} T_{s}(g).
\end{align*}
therefore $T\tau_{h}=\tau_{h}T$, i.e. $T$ commute with translations.
Conversely, I can't understand the sketch of the proof.
(i) Since T commutes with translation $(Tf)\ast g=T(f\ast g$), for appropriate $f$ and $g$. Thus $Tf\ast g=f\ast Tg$.
(ii) Let $1/p+1/q=1$, and suppose that $f$ and $g$ both belong to $L^p\cap L^q$. Then the convolutions $Tf\ast g$ and $f\ast Tg$ represents continuous functions , and so the are equal at every point, in particular the origin. Hence
$$\int_{R^n}(Tf)(x)g(-x)dx=\int_{R^n}(Tg)(x)f(-x)dx$$
The usual duality argument then shows that $T$ is bounded on $L^q$, and and by the interpolation theorem (theorem 4, Chapter I ) $T$ is also bounded on $L^2$. Finally apply the Prop. in 1.4, chapter II. See also Fourier analysis, chapter I.
For (i) ,What would these $f$ and $g$ be appropriate?
For (ii) The usual duality argument is that $|T|=|T^{\ast}|$ and since $T$ is bounded in $L^p$, is $T^\ast$ also in $L^q$?
Best Answer
i) For Schwartz functions $f,g\in\mathcal{S}(\mathbb{R}^{n})$, we have \begin{align*} (Tf\ast g)^{\wedge}(\xi)&=(T f)^{\wedge}(\xi)\widehat{g}(\xi)\\ &=m(\xi)\widehat{f}(\xi)\widehat{g}(\xi)\\ &=m(\xi)\widehat{g}(\xi)\widehat{f}(\xi)\\ &=(Tg)^{\wedge}(\xi)\widehat{f}(\xi)\\ &=(Tg\ast f)^{\wedge}(\xi), \end{align*} then $Tf\ast g=Tg\ast f$. Or you could also have $(Tf\ast g)^{\wedge}(\xi)=m(\xi)\widehat{f\ast g}(\xi)=(T(f\ast g))^{\wedge}(\xi)$.
ii) For the duality, I guess is that for Schwartz functions $f$, \begin{align*} \|Tf\|_{L^{q}}&=\sup_{g\in\mathcal{S}(\mathbb{R}^{n}),\|g\|_{L^{p}}\leq 1}\left|\int g(x)Tf(x)dx\right|\\ &=\sup_{g\in\mathcal{S}(\mathbb{R}^{n}),\|g\|_{L^{p}}\leq 1}\left|\int h(-x)Tf(x)dx\right|,~~~~h(x)=g(-x)\\ &=\sup_{g\in\mathcal{S}(\mathbb{R}^{n}),\|g\|_{L^{p}}\leq 1}\left|\int Th(x)f(-x)dx\right|\\ &\leq\sup_{g\in\mathcal{S}(\mathbb{R}^{n}),\|g\|_{L^{p}}\leq 1}\|Th\|_{L^{p}}\|f\|_{L^{q}}\\ &\leq\sup_{g\in\mathcal{S}(\mathbb{R}^{n}),\|g\|_{L^{p}}\leq 1}\|T\|\|h\|_{L^{p}}\|f\|_{L^{q}}\\ &=\|T\|\|f\|_{L^{q}}. \end{align*}