If $g \in L^1(\mathbb{R})$ and $(f_n)$ is a sequence of measurable functions converging to $f$ a.e where $|f_n| \leq 1$, then $g * f_n \to g * f$ uniformly on every compact set where $g*f = \int_{\mathbb{R}} g(x-y)f(y)dy$.

Using Egoroff's theorem, the sequence converges almost uniformly on every compact set; however, I am having difficulty extending this to the entire set.


$f_n \to f$ almost uniformly on a set $D$ if, for every $\epsilon > 0$, there is a set $E$ such that $\mu(E) < \epsilon$ and $f_n \to f$ uniformly on $D \setminus E$.

  • $\begingroup$ What do you mean by "converges almost uniformly"? $\endgroup$ – David C. Ullrich Jan 16 '18 at 23:50
  • $\begingroup$ please clarify your question it does not correlate with the title $\endgroup$ – Guy Fsone Jan 17 '18 at 0:37
  • $\begingroup$ Sorry. I'm not sure how it does not correlate. If you are referring to the edit, that was in response to the question by David. $\endgroup$ – user2959071 Jan 17 '18 at 0:40
  • $\begingroup$ Or you meant $g\ast f_{n}\rightarrow g\ast f$ in $L^{\infty}(K)$ for every compact set $K$? $\endgroup$ – user284331 Jan 17 '18 at 0:41
  • $\begingroup$ I understood the question as being, on every compact set $K$, $g * f_n \to g * f$ uniformly. $\endgroup$ – user2959071 Jan 17 '18 at 0:47

It's clear from dominated convergence that $f_n*g\to f*g$ pointwise.

Recall that $g\in L^1$ implies that $$\lim_{h\to0}\int|g(t)-g(t+h)|\,dt=0.$$

It follows that the sequence $(f_n*g)$ is equicontinuous. And pointwise convergence plus equicontinuity implies uniform convergence (at least on compact sets).

  • $\begingroup$ Nice one. Thanks. $\endgroup$ – user2959071 Jan 17 '18 at 1:34

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.