$C^1_0[0,1]$ is the set of functions $f:[0,1]\rightarrow \Bbb R$ such that $f,f'$ are continuous on $[0,1]$ and $f(0)=0.$
The metric $d$ is given by:
$d(f,g)=\int^1_0 |f(x)-g(x)|\,\mathrm dx+\sup_{x\in[0,1]}|f'(x)-g'(x)|$
Now I aim to prove that:
Given that $(f_n)$ is a bounded sequence in $(C^1_0[0,1], d)$.
Prove that $(f_n)$ has a subsequence that converge uniformly on $[0,1]$ to a continuous function $f$.
I have managed to prove that $C^1_0[0,1]$ with the matric $d$ is a complete metric space.
And I believe that the question can be solved by using the Arzela-Ascoli theorem. I think boundedness comes from the fact that $\|f_n\|_{d_u}<\|f_n\|_d<M$ where $M\in \Bbb R$. But may I please ask how to prove the equicontinous part? How can we use the fact that $f(0)=0$?
Thanks in advance!
Equicontinuity means that for all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in [0,1]$ and for all $n$ , $|x - y| < \delta \implies |f_n(x) - f_n(y)| < \epsilon $.
It is clear that the metric $d$ has come from the norm $||f|| = \int_0^1 |f|dx + \sup_{x \in [0,1]} |f'(x)|$.
Note that since there is some large constant $M$ such that $|f_n'(x)| < M$ for all $n$,$x$, it also follows that $|f_n(x) - f_n(y)| < M|x-y|$ for all $n,x,y$. This is because of the mean value theorem,which says that $|f_n(x) - f_n(y)| \leq |f'(c)||x-y|$, where $c$ is in between $x$ and $y$. Of course, since $|f'(c)| < M$, the proposition follows.
Hence, equicontinuity also follows, so by Arzela Ascoli, you get the result.