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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!

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What you stated is not true: the support of the convolution is generally about the same as the Minkowski sum of the supports of the two functions you started with. So if you convolve a function of non-compact support against another function (which has compact support), you are not going to get back a function with compact support in general. – Willie Wong Mar 8 '11 at 23:48
Yeah, whoops, got two questions confused. I will fix it now. Just want to see continuity. – Jon Beardsley Mar 9 '11 at 0:02
Doesn't this follow from the fact that translation is continuous in $L^p$ + Hölder? Oh wait, your function is only locally integrable. Should be fixable. – Jonas Teuwen Mar 9 '11 at 0:14
In the case of not locally integrable but in fact $L^2$ or something, can we get it from those facts? – Jon Beardsley Mar 9 '11 at 0:17
up vote 3 down vote accepted

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 $L^q$.

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Why can we split up the $\infty$ norm like that into the 2 norm? Thanks for this proof. – Jon Beardsley Mar 9 '11 at 0:47
It follows from Young's inequality for convolutions, like he said. You can find the result on wikipedia:'s_inequality – user1736 Mar 9 '11 at 2:22
Oh okay, thanks. Yeah, I am learning how to prove Young's inequality using Holder's, because I may need to prove that on the qual. – Jon Beardsley Mar 9 '11 at 21:20
@Jonas Sorry to bring up this old question, but it turns out I need a proof of this exact fact (the original version with a locally integrable function). I go through the usual working and the last part I have to show is that $| \tau_{x-y} g(x) - g(x) |$ can be made arbitrarily small as $y\to x$ (g is the locally integrable function by the way). Since g is only locally integrable, we don't have continuity of translation so perhaps it isn't even possible to show that. How can we fix up a proof? Thank you. – Ragib Zaman Oct 5 '11 at 16:43
@RagibZaman: I suggest that you ask a new question. – Jonas Teuwen Oct 5 '11 at 19:47

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.

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).$

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Wouldn't it be much easier to switch the rôles of $f$ and $g$? That is to say to pick $f \in L_{\rm loc}^1$ and $g \in C_c$. Then the support of $g$ is contained in some ball $\bar{B}_R$ and for $\|y\| \lt R$ the support of $\tau_y g - g$ is contained in $\bar{B}_{2R}$. Now $g$ is uniformly continuous, hence $\|\tau_y g - g\|_\infty \to 0$ as $y \to 0$ and $f$ is integrable on $\bar{B}_{2R}$, hence we're done by pulling the absolute values inside the integral in the very first equation. – t.b. Oct 9 '11 at 13:51
@t.b. I am just learning this material for the first time myself, so I'm struggling to follow your comment. If you have time, I would be appreciative if you fleshed out the details and posted it as an answer. – Ragib Zaman Oct 9 '11 at 13:55
Actually, I don't think this is necessary, because the argument would be very similar to yours (which looks fine, by the way). My suggestion amounts to the observation that in your argument you didn't really exploit the fact that $f$ is continuous, you only used that it vanishes outside the ball $B_R$ and is in $L^p$, thus yielding a stronger statement than the one my suggestion would give. All I'm saying is that it suffices to prove that for $g$ continuous and of compact support we have $\|\tau_y g - g\|_\infty \to 0$ as $y \to 0$. This implies that $\|\tau_y g - g\|_q \to 0$ for all $q$ – t.b. Oct 9 '11 at 14:21
since the support of $\tau_y g - g$ is contained in some compact set. Hence you can restrict the integral in the convolution to some ball and apply Hölder there, the same way you did. Notice also that you only exploited $(x-y) \to 0$, this gives a slight simplification in the exposition. – t.b. Oct 9 '11 at 14:25
Ahh I had been wondering why we were assuming $f$ continuous until just now when I remembered that Test functions are dense in $L^p$. I'm still getting practice in remembering all the density theorems and using the various ones in the right situations. Thanks for your help @t.b. ! – Ragib Zaman Oct 9 '11 at 14:39

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.

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