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

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,

• Yes, $f\ast g \in \mathcal{S}(\mathbb{R})$. If you know that it is $C^\infty$, and $\partial^\alpha (f\ast g) = f\ast \partial^\alpha g$, then you only need to check the decay. If $\operatorname{supp} f \subset [-K,K]$, then for $\lvert x\rvert > 2K$, you have $$\left\lvert \int_\mathbb{R} f(y) g(x-y)\,dy\right\rvert \leqslant 2K\lVert f\rVert_\infty \sup \left\{ \lvert g(z)\rvert : \lvert z\rvert \geqslant \frac{\lvert x\rvert}{2}\right\},$$ and the latter is bounded by $\frac{C_m}{\lvert x\rvert^m}$ for every $m$. – Daniel Fischer Jul 21 '14 at 12:40
• @DanielFischer: This could easily be an answer, not just a comment. Can you post it as such? – Giuseppe Negro Jul 21 '14 at 12:53
• @GiuseppeNegro Now I can. I had to go out buy some milk, so I hadn't the time to write an answer before. – Daniel Fischer Jul 21 '14 at 13:12
• @GiuseppeNegro Posted. – Daniel Fischer Jul 21 '14 at 13:35
• I had to buy milk, but I couldnt because I had to write an asnwer – clark Jul 21 '14 at 13:38

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.

• Indeed, Thanks a lot; – Inquisitive Jul 22 '14 at 4:41

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.