Given a sequence $f_{n}:[0,1] \rightarrow [0,1]$ of continuous functions, define $g_{n}:[0,1] \rightarrow \mathbb{R}$ by setting $g_{n}(x)=\int_{0}^{1} \frac{f_{n}(t)}{(t-x)^{1/3}}dt$; $x \in [0,1]$. Show that the sequence $g_{n}$ has a uniformly convergent subsequence.
My attempt: To prove this result, we need to show $g_{n}$ is equicontinuous and pointwise bounded on $[0,1]$.
- Equicontinuous: For given $\epsilon>0$, there exist a $\delta >0$ such that if $x,y \in [0,1]$ and $|x-y|<\delta$, then
$|g_{n}(x)-g_{n}(y)|=|\int_{0}^{1} \frac{f_{n}(t)}{(t-x)^{1/3}}dt - \int_{0}^{1} \frac{f_{n}(t)}{(t-y)^{1/3}}dt|\leq \int_{0}^{1} \frac{|f_{n}(t)|}{(t-x)^{1/3}}dt+\int_{0}^{1} \frac{|f_{n}(t)|}{(t-y)^{1/3}}dt$
Since, $f_{n}$ is continuous over compact set [0,1]. This implies $f_{n}$ is bounded over $[0,1]$. This implies there exist $M>0$ such that $|f_{n}(x)|\leq M$, for all $x \in [0,1]$.
$|g_{n}(x)-g_{n}(y)| \leq M[\int_{0}^{1} \frac{1}{(t-x)^{1/3}}dt + \int_{0}^{1} \frac{1}{(t-y)^{1/3}}dt]$
Now I solved the improper integral $\int_{0}^{1} \frac{1}{(t-x)^{1/3}}dt= 3/2[(x-1)^{2/3}-x^{2/3}]$.
I am confused here, how can I write this in terms of $\epsilon$, so I can show the equicontinuous step. Can anyone please suggest me the last step of this question? Many thanks in advance.
There is an issue with the $M$, just because $f_n$ is a sequence of continuous functions doesn't mean that it's converging to a continuous or bounded function. We don't know anything about what $f_n$ is doing. You are absolutely right in saying that each $f_n$ is bounded because it's a continuous function on a compact set, however that doesn't mean all the $f_n$'s are bounded by the same $M$.
Take for example the sequence of functions $f_n(x)=nx$ this sequence is bounded for all $n$ on say $[0,1]$ but the bound depends upon $n$ and in the pointwise limit it goes to infinity for $x\neq0$.