Prove that ${\{f_n\}}_{n\in\Bbb{N}}$ has a subsequence that converges uniformly to a continuous function on $[0,1]$ 
Consider the sequence ${\{f_n\}}_{n\in\Bbb{N}}$, where for each $n\in \Bbb{N}$ the function $f_n:[0,1] \to \Bbb{R}$ is absolutely continuous and satisfies $f_n(0)=13$ and $$\int_{[0,1]}|f_n'|^4dx \le 7$$ The integration is Lebesgue integration. Prove that ${\{f_n\}}_{n\in\Bbb{N}}$ has a subsequence that converges uniformly to a continuous function on $[0,1]$.

If $f_n$ is absolutely continuous, then $f_n$ is of bounded variation and $f_n=\int|f'_n|dx$. That is all I can think about now. Then I have no clue what to do next about this question. Could someone provide some help?
 A: 
Claim: The sequence $(f_n)_{n\in\mathbb N}$ is equicontinuous. That is, for each $x\in[0,1]$ and $\varepsilon>0$, there exists some $\delta>0$ such that if $y\in[0,1]$ and $|y-x|<\delta$, then $$|f_n(y)-f_n(x)|<\varepsilon\quad\text{for every $n\in\mathbb N$.}$$

Proof: Fix $n\in\mathbb N$. First, by Hölder's inequality, one has that for any $x\in[0,1]$ and $y\in[0,1]$ such that $x\geq y$:
\begin{align*}
\int_{[y,x]}|f_n'|\,\mathrm dt=&\,\int_{[y,x]}|f_n'|\times 1\,\mathrm dt\leq\left(\int_{[y,x]}|f_n'|^4\mathrm dt\right)^{1/4}\left(\int_{[y,x]}1^{4/3}\mathrm dt\right)^{3/4}\\
\leq&\,\left(\int_{[0,1]}|f_n'|^4\mathrm dt\right)^{1/4}\left(x-y\right)^{3/4}\leq\sqrt[4]{7(x-y)^{3}}.
\end{align*}
Second, absolute continuity implies that
$$f_n(x)=f_n(0)+\int_{[0,x]}f_n'\,\mathrm dt=13+\int_{[0,x]}f_n'\,\mathrm dt\quad\text{for every $x\in[0,1]$}.$$ Now, for any $x\in[0,1]$ and $y\in[0,1]$ such that $x\geq y$,
$$|f_n(x)-f_n(y)|=\left|\int_{[y,x]}f_n'\,\mathrm dt\right|\leq\int_{[y,x]}|f_n'|\,\mathrm dt\leq\sqrt[4]{7(x-y)^{3}}.$$ Since $n\in\mathbb N$ was arbitrary and the function $y\mapsto\sqrt[4]{7(x-y)^{3}}$ is continuous on the interval $[0,x]$ for any $x\in(0,1]$, the equicontinuity of the sequence $(f_n)_{n\in\mathbb N}$ follows. $\blacksquare$


Claim: The sequence $(f_n)_{n\in\mathbb N}$ is pointwise bounded. That is, for each $x\in[0,1]$, there exists a constant $M_x\geq 0$ such that $$|f_n(x)|\leq M_x\quad\text{for each $n\in\mathbb N$}.$$

Proof: Fix $x\in[0,1]$. For any $n\in\mathbb N$, one has that $$|f_n(x)|\leq13+\int_{[0,x]}|f_n'|\,\mathrm dt\leq13+\sqrt[4]{7x^3}.$$ Therefore, the sequence $(f_n)_{n\in\mathbb N}$ is pointwise bounded. $\blacksquare$

Given that $[0,1]$ is a compact Hausdorff space, the Arzelà–Ascoli theorem implies that the set $$\{f_n\,|\,n\in\mathbb N\}$$ is precompact in $C[0,1]$ with respect to the supremum norm. The existence of a uniformly convergent subsequence readily follows.
