Translation-invariant Fourier multiplier on the Schwartz core

Assume Countable Choice and let $n\ge 1$. The Fourier transform is the negative-sign $2\pi$-normalized transform of Fourier transform of a tempered distribution: on a Schwartz function it is the integral transform $\widehat{f}(\xi )={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi ix\cdot \xi }dx$, and on ${\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ it is its bilinear transpose, an automorphism with inverse ${\mathcal{F}}^{-1}$ (Fourier transform is a topological automorphism of tempered distributions).

Fix a measurable symbol $m:{\mathbb{R}}^n\to \mathbb{C}$. Define its Schwartz domain $D_m=\{f\in \mathcal{S}({\mathbb{R}}^n):m\widehat{f}\text{ is locally integrable and its regular distribution is tempered}\}\mathrm{.}$ Here $m\widehat{f}$ is the ordinary pointwise product of the measurable symbol with the Schwartz function $\widehat{f}$. For a locally integrable function $g$, its regular distribution initially means $\phi \mapsto {\int }_{{\mathbb{R}}^n}g(\xi )\phi (\xi )d\xi$ on $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Saying that this distribution is tempered means that this functional extends continuously to $\mathcal{S}({\mathbb{R}}^n)$. The extension is unique because $C_c^{\mathrm{\infty }}$ is dense in $\mathcal{S}$ (Smooth compact supports are dense in Schwartz space); we denote it by $u_g$. Local integrability alone does not guarantee an absolutely convergent integral against every Schwartz test. For $f\in D_m$ the extension exists by hypothesis, and we set $T_{m}f:={\mathcal{F}}^{-1}\left(u_{m\widehat{f}}\right)\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n),T_m:\mathcal{S}\supseteq D_m\to {\mathcal{S}}^{\mathrm{\prime }}\mathrm{.}$ The domain qualification is part of the definition: no boundedness of $T_m$, no density of $D_m$ in $\mathcal{S}$, no continuity of $m\mapsto T_m$ and no action of $T_m$ on $L^p$ is asserted. In particular this definition does not assert that $D_m=\mathcal{S}$; the zero function always belongs to $D_m$.

Translation. For $u\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ and $a\in {\mathbb{R}}^n$ define the translate ${\tau }_{a}u$ by transposition, $⟨{\tau }_{a}u,\phi ⟩:=⟨u,\phi (\cdot +a)⟩,\phi \in \mathcal{S}({\mathbb{R}}^n)\mathrm{.}$ This is a tempered distribution: translation is a continuous complex-linear endomorphism of $\mathcal{S}({\mathbb{R}}^n)$ (Basic operations are continuous on Schwartz space), so the composition is a continuous complex-linear functional (Tempered distribution). On functions, ${\tau }_{a}f(x)=f(x-a)$ for $a\in {\mathbb{R}}^n$.

Translation invariance of the domain and of the operator. If $f\in D_m$ and $a\in {\mathbb{R}}^n$, then ${\tau }_{a}f\in D_m$ and $T_m{\tau }_{a}f={\tau }_a{T}_{m}f\mathrm{.}$ Justification. Write $e_{-a}(\xi )=e^{-2\pi ia\cdot \xi }$. First, $\mathcal{F}({\tau }_{a}f)=e_{-a}\widehat{f}$: the integral formula is the translation law for the $L^1$ transform, applicable because Schwartz functions are integrable (Schwartz derivatives are integrable, Translation, modulation, linear dilation and reflection laws), and the distributional transform of the integrable function ${\tau }_{a}f$ is the regular distribution of its integral transform (Fourier transform agrees with l one and plancherel transforms). Second, with $g=m\widehat{f}$, the product $m\mathcal{F}({\tau }_{a}f)=e_{-a}g$ is locally integrable, and its regular distribution satisfies $⟨u_{e_{-a}g},\phi ⟩=⟨u_g,e_{-a}\phi ⟩$ for every compactly supported smooth test $\phi$; since $e_{-a}$ is smooth with polynomially bounded derivatives, Smooth polynomially bounded multipliers on schwartz space makes $e_{-a}{u}_g$ a tempered extension of the regular distribution of $e_{-a}g$. Uniqueness of extension gives $u_{e_{-a}g}=e_{-a}{u}_g$ on all Schwartz tests. Hence ${\tau }_{a}f\in D_m$. Third, for $u\in {\mathcal{S}}^{\mathrm{\prime }}$ one has $\mathcal{F}({\tau }_{a}u)=e_{-a}\mathcal{F}u$: pairing both sides with a Schwartz test $\phi$ and using the definition of $\mathcal{F}$ on ${\mathcal{S}}^{\mathrm{\prime }}$ gives $⟨\mathcal{F}({\tau }_{a}u),\phi ⟩=⟨{\tau }_{a}u,\mathcal{F}\phi ⟩=⟨u,{\tau }_{-a}\mathcal{F}\phi ⟩=⟨u,\mathcal{F}(e_{-a}\phi )⟩=⟨e_{-a}\mathcal{F}u,\phi ⟩,$ where the middle identity is the modulation law $\mathcal{F}(e^{-2\pi ia\cdot x}\phi )(\xi )=\widehat{\phi }(\xi +a)$ of Translation, modulation, linear dilation and reflection laws and the last identity is the definition of multiplication of a distribution by the smooth symbol $e_{-a}$ (Smooth polynomially bounded multipliers on schwartz space). Applying this to $u=T_{m}f$ and combining the three computations, $\mathcal{F}(T_m{\tau }_{a}f)=u_{e_{-a}g}=e_{-a}{u}_g=e_{-a}\mathcal{F}(T_{m}f)=\mathcal{F}({\tau }_a{T}_{m}f)\mathrm{.}$ Injectivity of $\mathcal{F}$ on ${\mathcal{S}}^{\mathrm{\prime }}$ now gives $T_m{\tau }_{a}f={\tau }_a{T}_{m}f$.

The Countable Choice hypothesis is inherited only from the cited Fourier interfaces; the transposition defining ${\tau }_a$ uses none.