Translation and differentiation symbols

The symbol $m(\xi )=e^{-2\pi ia\cdot \xi }$ is continuous, and [F10] with $x=0$ and $y=-2\pi a\cdot \xi$ gives $\mathrm{\mid }m(\xi )\mathrm{\mid }=1$ for every $\xi$; hence $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. The symbol $\sigma (\xi )=2\pi i{\xi }_j$ is smooth with $\mathrm{\mid }\sigma (\xi )\mathrm{\mid }=2\pi \mathrm{\mid }{\xi }_j\mathrm{\mid }\le 2\pi (1+\mathrm{\mid }\xi \mathrm{\mid })$, so all its derivatives are polynomially bounded, and $\mathrm{\mid }\sigma (t{e}_j)\mathrm{\mid }=2\pi \mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$ shows it is unbounded.

For $f\in \mathcal{S}$: [F2] gives ${\tau }_{a}f\in \mathcal{S}$ with $\widehat{{\tau }_{a}f}(\xi )=m(\xi )\widehat{f}(\xi )$ at every $\xi$, and $f\in L^1$ by [F3], so the $L^1$ case of [F1] gives the distributional identity $\mathcal{F}{u}_{{\tau }_{a}f}=u_{m\widehat{f}}$. Likewise [F7] gives ${\mathrm{\partial }}_{j}f\in \mathcal{S}$ with $\mathcal{F}({\mathrm{\partial }}_{j}f)=\sigma \widehat{f}$ and hence $\mathcal{F}{u}_{{\mathrm{\partial }}_{j}f}=u_{\sigma \widehat{f}}$.

Fix $N>0$ and put $\eta =(N+2)e_j$. The singleton $K=\{\eta \}$ is compact, $U=B(\eta ,1)\subseteq \{\xi \in {\mathbb{R}}^n:{\xi }_j>N\}$ is bounded and open, and $K\subseteq U$; [F11] gives a smooth ${\psi }_N$ with $0\le {\psi }_N\le 1$, ${\psi }_N(\eta )=1$ and $\mathrm{supp}{\psi }_N\subseteq U$. Its support is closed and bounded, hence compact by [F13], so ${\psi }_N\in C_c^{\mathrm{\infty }}$. Then ${\psi }_N\mathrm{\ne }0$, and ${\psi }_N\in \mathcal{S}$ by [F12]; by [F1] the class $f_N:={\mathcal{F}}^{-1}{\psi }_N$ lies in $\mathcal{S}$ with $\mathcal{F}{f}_N={\psi }_N$.

Since $m$ is bounded and measurable by step 1.1, [F4] gives $\mathcal{S}\subseteq D_m$ and $T_{m}f={\mathcal{F}}^{-1}(u_{m\widehat{f}})$ for $f\in \mathcal{S}$; substituting step 1.2, $T_{m}f={\mathcal{F}}^{-1}\left(\mathcal{F}{u}_{{\tau }_{a}f}\right)=u_{{\tau }_{a}f}$ by [F8]. Thus the multiplier with symbol $m$ equals the translate ${\tau }_a$ on the Schwartz core, so $m$ is the multiplier symbol of ${\tau }_a$.

Since $\sigma$ is smooth and polynomially bounded by step 1.1, $\sigma \widehat{f}\in \mathcal{S}$ for $f\in \mathcal{S}$; the distributional derivative identity [F6] gives $\mathcal{F}({\mathrm{\partial }}_j{u}_f)=\sigma \mathcal{F}{u}_f=\sigma u_{\widehat{f}}=u_{\sigma \widehat{f}}$, using [F1] and [F9] for the middle identifications, while step 1.2 gives $\mathcal{F}{u}_{{\mathrm{\partial }}_{j}f}=u_{\sigma \widehat{f}}$. Injectivity of $\mathcal{F}$ on ${\mathcal{S}}^{\mathrm{\prime }}$ [F8] yields ${\mathrm{\partial }}_j{u}_f=u_{{\mathrm{\partial }}_{j}f}$: on Schwartz classes the distributional derivative is the classical derivative. Consequently $T_{\sigma }f={\mathcal{F}}^{-1}(u_{\sigma \widehat{f}})={\mathcal{F}}^{-1}\left(\mathcal{F}{u}_{{\mathrm{\partial }}_{j}f}\right)=u_{{\mathrm{\partial }}_{j}f}$ by [F8], so the Schwarz-core multiplier with symbol $\sigma =2\pi i{\xi }_j$ is exactly ${\mathrm{\partial }}_j$.

For the $f_N$ of step 1.3, Plancherel [F5], the derivative identity of [F7] and $\mathcal{F}{f}_N={\psi }_N$ give $\mathrm{\parallel }{\mathrm{\partial }}_j{f}_N{\mathrm{\parallel }}_2=\mathrm{\parallel }{\mathcal{F}}_2({\mathrm{\partial }}_j{f}_N){\mathrm{\parallel }}_2=\mathrm{\parallel }2\pi i{\xi }_j{\psi }_N{\mathrm{\parallel }}_2=2\pi \mathrm{\parallel }{\xi }_j{\psi }_N{\mathrm{\parallel }}_2$. On $\mathrm{supp}{\psi }_N$ one has ${\xi }_j>N$, so $\mathrm{\parallel }{\xi }_j{\psi }_N{\mathrm{\parallel }}_2\ge N\mathrm{\parallel }{\psi }_N{\mathrm{\parallel }}_2=N\mathrm{\parallel }{f}_N{\mathrm{\parallel }}_2$ by [F5]; hence $\mathrm{\parallel }{\mathrm{\partial }}_j{f}_N{\mathrm{\parallel }}_2\ge 2\pi N\mathrm{\parallel }{f}_N{\mathrm{\parallel }}_2$, and $f_N\mathrm{\ne }0$ since ${\psi }_N(\eta )=1$.

Since $\mathrm{\mid }m\mathrm{\mid }=1$ everywhere by step 1.1, [F4] and [F5] give, for every $g\in L^2$, $T_{m}g={\mathcal{F}}_2^{-1}(m{\mathcal{F}}_{2}g)$ and $\mathrm{\parallel }{T}_{m}g{\mathrm{\parallel }}_2=\mathrm{\parallel }m{\mathcal{F}}_{2}g{\mathrm{\parallel }}_2=\mathrm{\parallel }{\mathcal{F}}_{2}g{\mathrm{\parallel }}_2=\mathrm{\parallel }g{\mathrm{\parallel }}_2$; thus $T_m$ is an isometry of $L^2$ (indeed unitary, with inverse $T_{\bar{m}}$). By step 2.1 the operator $T_m$ agrees with ${\tau }_a$ on the dense Schwartz subspace [F5]; the $L^2$ extension of ${\tau }_a$ is unique by [F4], so that extension is this isometry and ${\tau }_a$ is an $L^2$ isometry.

Suppose a bounded linear $P:L^2\to L^2$ extended ${\mathrm{\partial }}_j$ from the Schwartz core, say $\mathrm{\parallel }Pg{\mathrm{\parallel }}_2\le C\mathrm{\parallel }g{\mathrm{\parallel }}_2$ and $Pf={\mathrm{\partial }}_{j}f$ for every $f\in \mathcal{S}$. Applying this to $f_N\in \mathcal{S}$ of step 1.3 and using step 2.3 gives $2\pi N\mathrm{\parallel }{f}_N{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\mathrm{\partial }}_j{f}_N{\mathrm{\parallel }}_2=\mathrm{\parallel }P{f}_N{\mathrm{\parallel }}_2\le C\mathrm{\parallel }{f}_N{\mathrm{\parallel }}_2$, hence $2\pi N\le C$ for every $N>0$ with $f_N\mathrm{\ne }0$, which is impossible. Therefore no bounded $L^2$ extension exists, while steps 2.1 and 2.2 identify the symbols $m$ and $\sigma$ and step 3.1 establishes the $L^2$ isometry of ${\tau }_a$; no Mihlin or other smoothness criterion was used.