Let $f_n$ be a sequence of continuous functions defined on $\{x\in \mathbb{R}^d:|x|\leq 1\}$ such that $\|f_n\|_{1}\to 0$. Further, suppose that $f_n$ is equicontinuous. I need to show that $f_n\to0$ pointwise almost everywhere.
I know what equicontinuity means: Given $\epsilon>0$, there exists $\delta>0$, such that for all $n$, if $|x-y|<\delta$. then $|f_n(x)-f_n(y)|<\epsilon$.
My difficulty with this is that I don't know what it is I must show. I don't think I need to verify the definition of pointwise convergence, since I need to show (only) almost everywhere convergence. One way I can think of is to show that $$\int\sum_{n=1}^\infty |f_n(x)|dx<\infty.$$
For then this would imply that the integrand series is finite almost everywhere which would imply that $f_n$ converges to 0 almost everywhere. I don't know how and where to use all the given hypotheses. Any hints are appreciated. Thanks!
Best Answer
Here is some idea: Let $0<\varepsilon<1$ fixed. Because $(f_n)$ is equicontinuous, we can find $0<\delta<\varepsilon$ such that $|f_n(x)-f_n(y)|<\varepsilon$ for any $n \in \mathbb{N}$ and $x,y \in B_1(0)$ satisfying $|x-y|<\delta$.
Because $(f_n)$ is convergent in $L^1$, thus it is a Cauchy's sequence in that norm. So, there is a number $N_0:=N_0(\delta(\varepsilon))\in\mathbb{N}$ such that $$ \int_{B_1(0)} |f_n(x)-f_m(x)|dx < \delta. $$
Now, consider $x_0 \in B_1(0)$ fixed and also fix $n,m >N_0$. There exists $y \in B_1(0)$ such that $|x_0-y|<\delta$ and $|f_n(y)-f_m(y)|<\delta<\varepsilon$. If not we would get a contradiction with the inequality above (why?).
We get,
$$|f_n(x_0)-f_m(x_0)|\leq |f_n(x_0)-f_n(y)|+|f_n(y)-f_m(y)|+|f_m(y)-f_m(x_0)| < 3\varepsilon.$$
I think you can conclude from here.
PS: You need to be careful with some details.