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?


1 Answer 1


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!

  • $\begingroup$ 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 $||f(x-t_1)-f(x-t_2)||_p<\epsilon$? $\endgroup$
    – JnPS
    Aug 12, 2020 at 0:42
  • $\begingroup$ 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 $[-1, 1]$. $\endgroup$
    – Ryan
    Aug 12, 2020 at 2:14
  • $\begingroup$ 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. $\endgroup$
    – JnPS
    Aug 13, 2020 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.