Can someone show me the proof that the convolution of a compactly supported real valued function on $\mathbb{R}$ with a locally integrable function is also continuous? I feel that this is a standard analysis result, but cannot remember how to prove it. Thanks!
[Math] Convolution of compactly supported function with a locally integrable function is continuous
analysisconvolutionfunctional-analysis
Related Solutions
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.
There is a close relationship between smoothness of a function and the decay rates of its Fourier transform (and vice versa). Roughly speaking:
Some number of derivatives of $f$ existing implies some power law decay of the Fourier transform; an extra derivative speeds up the decay by a power.
Smoothness of $f$ (infinitely differentiable) gives decay of $\hat{f}$ faster than any power.
Extensibility of $f$ to an analytic function in a strip $|Im(z)|<a$ gives exponential ($e^{-at}$) decay.
Extension to an entire function of exponential type gives a Fourier transform of compact support.
See Paley-Wiener theorem for more precise statements.
In your case, you won't be able to find a compactly supported function with exponentially decaying Fourier transform, because that would require analyticity in some strip, which is clearly impossible.
Best Answer
On your comment for $L^2$, we can do it as follows:
Recall that for $1 \leq p < \infty$ translation is continuous in $L^p$, that is if $f \in L^p(\mathbb R^n)$ and $z \in \mathbb R^n$ then $\lim_{y \to 0} \|\tau_{y + z} f - \tau_z f\|_p = 0$. Here $\tau_y$ is the shift over $y$.
Okay, fine. By Young's inequality for convolutions, the convolution actually exists. So
$$\|\tau_y (f \ast g) - (f \ast g)\|_\infty = \|(\tau_y f - f) \ast g)\|_\infty \leq \|\tau f - f\|_2 \|g\|_2 \to 0 \text{ as $y \to 0$.}$$ So actually we get uniform continuity. Note that this also works for $f$ in $L^p$ and $g$ in $L^q$.