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Let $f_k : [0,1] \rightarrow \mathbb{R}$ be a sequence of continuous functions. Assume that there exists $K > 0$ such that $$\int_0^1 f_k^2(t)\, dt \leq K^2$$ for all $k \in \mathbb{N}.$

For each $n \in \mathbb{N}$, define a sequence of function $(F_n)$ by $$F_n(x) = \int_0^1 f_k(t) \chi_{[0,x]}(t) \, dt.$$

Show that $F_k$ has a uniformly convergent subsequence. Let $(F_{n_k})$ be such the subsequence, and let $F$ be its limit function. Verify that for any $a \in (0,1)$ and $x \in [0,1]$, the sequence $$\int_0^1 \frac{F_{n_k}(y)}{|x-y|^a} \, dy$$ converges to $\int_0^1 \frac{F(y)}{|x-y|^a} \, dy$ as $k \rightarrow \infty.$

$\textbf{Attemp :}$ Let $X = [0,1].$ Then I can show that $\mathscr{F} \subseteq C(X) := $ space of a continuous real-valued functions on $X$, where $\mathscr{F} = \{F_n : n \in \mathbb{N}\}$.

My idea is to show that $\mathscr{F}$ is compact using Arzela-Ascoli Theorem. I can show that $\mathscr{F}$ is bounded under infinity norm $\|f\|_\infty = \sup_{x \in [0,1]} |f(x)|$ and it is also equicontinuous.

What left is to prove that $\mathscr{F}$ is closed, which I cannot do it. Any suggestion how to see that $\mathscr{F}$ is closed ? Also, the last part is exchanging limit and integral, I guess that it has something to do with dominated converegence theorem. However, I am not quite sure why $a$ and $|x-y|^a$ has to be there.

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    $\begingroup$ It is not being asserted that $\mathcal F$ is closed. You are only supposed to show that $F_n$ has a subsequence converging in $C[0,1]$; the limit need not be in $\mathcal F$. $\endgroup$ – Kavi Rama Murthy Aug 6 '18 at 23:49
  • $\begingroup$ "converges to $\int_0^1 \frac{F_n(y)}{|x-y|^a} \, dy$" You want $F$ there $\endgroup$ – zhw. Aug 7 '18 at 6:39
  • $\begingroup$ oh yes $F$, thank you. $\endgroup$ – user117375 Aug 7 '18 at 21:36
  • $\begingroup$ math.stackexchange.com/questions/323725/… might be helpful. $\endgroup$ – Wraith1995 Aug 16 '18 at 0:29

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