# Bounded linear map from $\mathscr{S} \to \mathscr{S}$

Let $T$ be the operator defined by $$T: f \longmapsto f \ast \partial^{\alpha} \delta_0,$$ where $\delta_0$ denotes the dirac mass at $0$. I want to show that $T$ maps the Schwartz functions $\mathscr{S}$ to itself. My thoughts so far have been to take the Fourier transform of $T(f)$, this will give $$\widehat{T(f)} = \widehat{f}(\xi) (i \xi)^{\alpha}.$$ Then perhaps take the inverse Fourier transform, using a Fourier multiplier type of idea. But I don't know how to go any further.

By definition of the distributional derivative (which is compatible with integration by parts) $$\langle \partial^\alpha\delta,f \rangle = (-1)^{\alpha} \langle \delta,\partial^\alpha f \rangle = (-1)^{\alpha} \partial^\alpha f(0) \quad \implies \quad f \ast \partial^\alpha \delta = \partial^\alpha f$$ Note $T \ast f \in C^\infty$ for any $f \in \mathscr{S}, T\in \mathscr{S}'$ and $T \ast f \in \mathscr{S}$ whenever $T$ is a compactly supported distribution.
Also if you prefer the Fourier transform, then $g \in \mathscr{S} \implies (2 i\pi \xi)^\alpha g \in \mathscr{S}$ because $(2 i\pi \xi)^\alpha \in C^\infty$ and it is polynomially bounded as well as all its derivatives. Finally $f \in \mathscr{S} \implies \widehat{f} \in \mathscr{S}\implies (2 i\pi \xi)^\alpha \widehat{f} \in \mathscr{S} \implies \mathcal{F}^{-1}[(2 i\pi \xi)^\alpha \widehat{f}] \in \mathscr{S}$