Subspace of Tempered Distributions

$$\phi(\lambda D)u = \tfrac1{\lambda^n} \check \phi\left(\tfrac\cdot\lambda\right) * u := x \mapsto \int \tfrac1{\lambda^n} \check \phi\left(\tfrac y\lambda\right) u(x-y) \, dy .$$ Since $\phi$ is Schwartz, so is $\check\phi$, in particular $\check \phi \in L^p$ for all $1 \le p \le \infty$, and $$\left\|\tfrac1{\lambda^n} \check \phi\left(\tfrac\cdot\lambda\right)\right\|_p = \lambda^{-n+n/p}\left\|\check \phi\right\|_p .$$

Hence by Young's convolution inequality $$ \|\phi(\lambda D)u\|_\infty \le \lambda^{-n+n/p'}\|\check\phi\|_{p'} \|u\|_p ,$$ where $p'$ is the conjugate index to $p$. (Note $n-n/p' = n/p$.) And this converges to zero if $p < \infty$.

And of course, if $L^p$ and $L^q$ are subsets, then so is $L^p + L^q$.

Also $\phi(\lambda D)1 = \phi(0)$, where $1$ denotes the constant function. So taking any $\phi$ such that $\phi(0) \ne 0$, we see that $L^\infty \not \subset {S_{h}}'(\mathbb{R}^{n})$.

Roughly speaking, being in ${S_{h}}'(\mathbb{R}^{n})$ is saying something like the distribution tends to $0$ in the average at infinity.