# Showing a convolution is uniformly continuous, proof check

Here's the set-up: Let $$f\in L^{\infty}(\mathbb{R})$$ and uniformly continuous, and let $$g\in L^1(\mathbb{R})$$ Show that the convolution $$f*g(t)= \int_{-\infty}^{\infty}f(t-s)g(s)ds$$ is uniformly continuous on $$\mathbb{R}$$

Here's my reasoning so far:

Let $$\epsilon>0$$ be given. Since $$f$$ is uniformly continuous choose $$\delta$$ corresponding to $$\lambda=\dfrac{\epsilon}{\|{g}\|_1}$$, i.e. $$|t_1-t_2|< \delta \implies |f(t_1)-f(t_2)|<\lambda$$

$$|f*g(t_1)-f*g(t_2)| = \bigg| \int_{-\infty}^{\infty}f(t_1-s)g(s)-f(t_2-s)g(s) ds\text{ }\bigg|$$ $$\leq \int_{-\infty}^{\infty}|f(t_1-s)g(s)-f(t_2-s)g(s)| ds \leq \int_{-\infty}^{\infty}|g(s)||f(t_1-s)-f(t_2-s)|ds$$ Now note that $$|t_1-t_2|=|(t_1-s)-(t_2-s)|<\delta \implies |f(t_1-s)-f(t_2-s)|<\lambda$$ so we can therefore substitute making the inequality become $$\int_{-\infty}^{\infty}|g(s)||f(t_1-s)-f(t_2-s)|ds < \int_{-\infty}^{\infty}|g(s)|\lambda ds = \|g\|_1\lambda = \epsilon$$ Doesn't this therefore show that the convolution is uniformly continuous? The $$\delta$$ depends only on $$\epsilon$$ since $$f$$ was uniformally continuous. I never used the fact that $$f\in L^{\infty}(\mathbb{R})$$, and this seems suspiciously too simple to be correct, where is the flaw?

I think this is the general idea, but it might be missing some subtleties. On the very last line I think you need to use Hölder's inequality for $$p=1$$ and $$q=\infty$$. You can also consider it "sup'ing out" $$|f(t_1-s)-f(t_2-s)|$$. Since $$f$$ is $$L^\infty$$ and uniformly continuous, you can find $$\delta>0$$ small enough such that $$\|f(t_1-s)-f(t_2-s)\|_\infty<\lambda$$. Specifically,

$$\int_{-\infty}^\infty |g(s)||f(t_1-s)-f(t_2-s)| \leq \int_{-\infty}^\infty |g(s)|\|f(t_1-s)-f(t_2-s)\|_\infty \\ =\|f(t_1-s)-f(t_2-s)\|_\infty \|g\|_1.$$

I hope this is helpful!

• I think I saw a different question that asked me to prove something similar, is it a general result that $f \in L^p$ and shifting the function by two fixed real numbers, you can always find a $\delta$ such that $||f(x-t_1)-f(x-t_2)||_p<\epsilon$?
– JnPS
Aug 12, 2020 at 0:42
• I wouldn't say it's a general thing. You need the right conditions. Basically you need a little wiggle room, especially for $\| \cdot \|_p$ norm. Likewise, you'll need some kind of continuity, probably uniform continuity. Bare minimum I'd think you'd need convergence in $L^p$. Take for instance the function $f(x)=1$ for $x \in [-1, 1]$ and $f(x)=0$ for any other $x$. Note that $f \in L^p$ for all $p \geq 1$. However, we run into issues at the endpoints of $[-1, 1]$.
– Ryan
Aug 12, 2020 at 2:14
• But for that function, (I worked out some examples on paper), $||f(x-t_1)-f(x-t_2)||_p$ I believe is just $2^{1/p}|t_1-t_2|$ so as long as $|t_1-t_2|<\frac{\epsilon}{2^{1/p}}$ the p-norm of the difference will be less than epsilon. The support of the difference of the two shifted functions is just two disjoint intervals, right? So as long as I shrink those intervals the norm will shrink as well, yet that function isn't continuous.
– JnPS
Aug 13, 2020 at 3:12