# Prove that $(f_n)$ has a subsequence that converge uniformly on $[0,1]$ to a continuous function f.

$C^1_0[0,1]$ 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 $f(0)=0$?

• Please tell us what is $d_u$. Jul 11 '17 at 12:56
• @JoséCarlosSantos Oh sorry I forgot to explain. $d_u$ is the uniform metric on $C^1[0,1]$.
– Y.X.
Jul 11 '17 at 12:57
• For the uniform equicontinuity of the sequence, look at how the derivatives contribute to $d$. Jul 11 '17 at 13:03

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.

• Thanks a lot for the solution! May I please ask is that the fact that we do not use the fact "$f(0)=0$" which is given in the question?
– Y.X.
Jul 11 '17 at 21:57
• @PropositionX Yes, I think that this condition is not required. However, in most applications of this kind of result, the context is PDE (partial differential equations), so then usually an initial condition is imposed on all the $f$, which would look, for example, like $f(0)=0$. That's why it's been included in the question, so that it comes directly into application in the context of PDE. Jul 12 '17 at 2:36
• Hi, Sir. Sorry to bother you again but latter I found my proof of the metric space is complete is wrong. If possible, could you please take some time to look at this question about proving completeness here? math.stackexchange.com/questions/2356115/…
– Y.X.
Jul 12 '17 at 12:43
• Ok, I will attempt the question... or actually, the answer given there is quite useful. Jul 12 '17 at 13:18
• Your hint is the mean value theorem again. Note that $f_n(x) - f_n(y) = f_n'(c)(x-y)$, where $x,y \in (0,1]$ and $c \in (x,y)$. Knowing the bounds tells us that $f_n(x)-f_n(y) \leq M \times 1$, since the derivative is bounded by $M$ , and $x \leq 1$. Hence, for all $x,y$, $f_n(x) - f_n(y)$ is bounded. Now, since we know that $f_n(0) = 0$, we know that $|f_n(x)| \leq f_n(0) + M \leq M$ by the triangle inequality, hence $f_n$ are uniformly bounded. Jul 15 '17 at 16:58