convolution of compactly supported continuous function with schwartz class function is again a Schwartz class function?

Question

Suppose $f$ is continuous function on $\mathbb R$ with compact support; and $g\in \mathcal{S}(\mathbb R),$ (Schwartz space)

My Question is: Can we expect $f\ast g \in \mathcal{S(\mathbb R)}$ ? (Bit roughly speaking, convolution of compactly supported continuous function with schwartz class function is again a Schwartz class function )

[We note that convolution of arbitrary continuous function with schwartz class function need not be Schwartz class function, for instance, see this question, ]

Thanks,

Answer 1

Yes, the convolution of an integrable function $f$ with compact support, and a Schwartz class function $g$ belongs to the Schwartz space again.

Since all derivatives of Schwartz class functions belong to the Schwartz space, in particular are bounded, the convolution

$$(f\ast g)(x) = \int f(y)g(x-y)\,dy$$

is smooth, since the dominated convergence theorem allows differentiating under the integral arbitrarily often. (Since the difference quotients of $\partial^\alpha g$ converge uniformly on $\mathbb{R}$, and the support of $f$ is compact, one can get that result also without the dominated convergence theorem.)

So only the decay remains to be checked. Choose $K > 0$ such that $\operatorname{supp} f \subset [-K,K]$. Since $g\in \mathcal{S}(\mathbb{R})$, for every $\alpha,m\in\mathbb{N}$ there is a constant $C_{\alpha,m}$ such that

$$\lvert \partial^\alpha g(x)\rvert \leqslant \frac{C_{\alpha,m}}{(1+\lvert x\rvert)^m}$$

for all $x\in\mathbb{R}$.

Then for $\lvert x\rvert \geqslant 2K$ we have

$$\begin{align} \lvert \partial^\alpha(f\ast g)(x)\rvert &= \left\lvert \int_{-K}^K f(y) \partial^\alpha g(x-y)\,dy \right\rvert\\ &\leqslant \int_{-K}^K \lvert f(y)\rvert\, \lvert \partial^\alpha g(x-y)\rvert\,dy\\ &\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{(1+\lvert x-y\rvert)^m}\,dy\\ &\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{\left(1 + \frac{\lvert x\rvert}{2}\right)^m}\,dy\\ &= \frac{2^mC_{\alpha,m}}{(2+\lvert x\rvert)^m}\int_{-K}^K\lvert f(y)\rvert\,dy\\ &\leqslant \frac{C'_{\alpha,m}}{(1+\lvert x\rvert)^m}, \end{align}$$

where $C'_{\alpha,m} = 2^m\lVert f\rVert_{L^1}C_{\alpha,m}$.

Since $(1+\lvert x\rvert)^m \partial^\alpha(f\ast g)(x)$ is continuous, it is bounded on the compact set $[-K,K]$, hence we have

$$(1+\lvert x\rvert)^m\lvert\partial^\alpha(f\ast g)(x)\rvert \leqslant \tilde{C}_{\alpha,m}$$

for all $x\in\mathbb{R}$ and some constant $\tilde{C}_{\alpha,m}$.

So $f\ast g$ is a smooth function such that $x^m\partial^\alpha(f\ast g)(x)$ is bounded for all $\alpha,m\in\mathbb{N}$, and that means precisely $f\ast g\in \mathcal{S}(\mathbb{R})$.

The generalisation to $\mathbb{R}^n$ is immediate.

Answer 2

One of the key properties of the Schwartz space is of course the fact that it is invariant under the Fourier transform. Assume this has been already verified, then the so-called convolution theorem for the Fourier transform, telling you that $$ \widehat{f \ast g} = \hat{f} \cdot \hat{g} $$ and vice versa, i.e. $$ \widehat{f \cdot g} = \hat{f} \ast \hat{g} $$ tell you that you can, equivalently, verify that pointwise products of Schwartz functions are again Schwartz functions. This can be done using essentially the Leibniz rule for differentation.

Answer 3

A late addition to the other good answers, but possibly of interest: that the convolution of a Schwartz function and a compactly-supported continuous function is again Schwartz is an example of a very general phenomenon.

Namely, let $V$ be any quasi-complete, locally convex topological vector space acted-upon continuously by a topological group $G$ (the action map $G\times V\to V$ is jointly continuous). The quasi-completeness is a generalization of completeness on metrizable spaces like Hilbert and Banach and Frechet spaces, and also includes "strict colimits of Frechet", such as spaces of distributions, and also includes the weak duals of all these. Then compactly-supported continuous functions $f$ act on $V$ by $f\cdot v=\int_G f(g)\,g\cdot v\;dg$ (with left or right Haar measure on $G$). (This is one of the first things one would prove after developing the Gelfand-Pettis vector-valued integral set-up.)

Since $\mathbb R$ acts continously on Schwartz functions by $(g\cdot f)(x)=f(g+x)$ (not hard to prove), this applies in the present situation. :)