Convolution of compactly supported function with a locally integrable function is continuous?

Question

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!

Answer 1

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 $L^q$.

Answer 2

Here is a proof for the original form of the question with a locally integrable function. I am a little suspicious of my details near the end, especially in showing $ \| \tau_{x-y} \bar{g} - \bar{g}\|_{\hat{q}} \to 0 $, but I am quite sure this proof is salvageable. If someone can verify/correct the details, I would be grateful. Thank you.

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Let $x,y \in \mathbb{R}^N$. We must show that as $y\to x$, $$ |(f \star g)(x) - (f\star g)(y) | = | \int_{\mathbb{R}^N} f(z) ( g(x-z) - g(y-z) dz | \to 0. $$ Since $f$ has compact support, there exists $R>0$ such that $ \rm \text{supp } f \subseteq B(0,R) $. Since $f$ is $0$ outside that set, $$ \int_{\mathbb{R}^N} f(z) ( g(x-z) - g(y-z) ) dz = \int_{B(0,R)} f(z) ( g(x-z) - g(y-z) ) dz.$$ By the translation invariance of the integral we may write $$\int_{B(0,R)} f(z) ( g(x-z) - g(y-z) ) dz = \int_{B(0,R)} f(z) ( \tau_{x-y} \bar{g} (z) - \bar{g} (z) ) dz $$ where $\bar{g}(z) = g(-z)$ denotes reflection in $x$.

Let $ \| \cdot \|_{\hat{q}} $ denote the norm in $L^q( B(0,R) )$. By Minkowski's inequality, $$ \| \tau_{x-y} \bar{g} - \bar{g} \|_{\hat{q}} \leq \| \bar{g}\|_{\hat{q}} + \| \tau_{x-y} \bar{g} \|_{\hat{q}} < \infty $$ where the finiteness of both terms follows from the hypothesis $ g \in L^q_{loc} (\mathbb{R}^N) $. Thus, $ \tau_{x-y} \bar{g} - \bar{g} \in L^q( B(0,R) )$. Also, $ \| f\|_{\hat{p}}= \| f \|_{p} < \infty $ so $ f \in L^p( B(0,R) )$. Hence, by Holder's inequality we have $$ | (f\star g)(x) - (f\star g)(y) | \leq \|f \|_{\hat{p}} \| \tau_{x-y} \bar{g} - \bar{g} \|_{\hat{q}} . $$ The first factor is finite, so the continuity of $f\star g$ follows if we demonstrate that $$ \| \tau_{x-y} \bar{g} - \bar{g}\|_{\hat{q}} \to 0 $$ as $y\to x$.

Define $ h(z) = g(z) $ if $ z\in B(0,R) $, and $0$ otherwise, so that $ h \in L^q(\mathbb{R}^N) $. Then, by continuity of translation, $ \| \tau_{x-y} \bar{h} - \bar{h}\|_q \to 0$ as $y\to x.$ Since $$\| \tau_{x-y} \bar{h} - \bar{h} \|_{\hat{q}} \leq \| \tau_{x-y} \bar{h} - \bar{h}\|_{q} $$ and in $B(0,R)$, $\tau_{x-y} \bar{h}(z) - \bar{h}(z) = \tau_{x-y} \bar{g}(z) - \bar{g}(z)$, we have $$\| \tau_{x-y} \bar{g} - \bar{g}\|_{ \hat{q}} \to 0 $$ as $y\to x$. Thus, $ f\star g \in C(\mathbb{R}^N).$

Answer 3

if $f$ is continuous and if $g$ is in the set of infinitely continuous real functions, it suffices to show that the derivative, with the change of variable: $y=x+t$, passes through the integral using the definition of the derivative with limit.