Let $\{f_n\}$ be a sequence of continuously differentiable functions on $[0,1]$ with $f_n(0) = f'_n(0)$ and $|f'_n(x)| \leq 1 \ \forall x \in [0,1]$ and $n \in N$. Show that if $\lim_{n \to \infty} f_n(x) = f(x) \ \forall x \in [0,1]$, then $f$ is continuous on $[0,1]$. Does the sequence have to converge? Must there be a convergent subsequence?
I've seen similar questions to this, but they all seem to have slightly different conditions. Normally I like looking at similar problems, but the thing I'm having trouble with is the conditions. I'm just saying this because I know someone will mark this as a duplicate, but I do not believe it is if you read carefully. What I'm not sure about is how these conditions can be applied to show the things being asked. I'm just not sure how to start. If anyone could get me started it would be great!
Edit: Have continued to try this and still not having any luck. If anyone could help I'd really appreciate it! This is one of the last questions I'd like to figure out for my class!
By the Mean Value Theorem, $|f_n(x)-f_n(y)|\leq|x-y|$ for all $x,y,n$, so taking the limit $n\to\infty$ yields $|f(x)-f(y)|\leq|x-y|$ as well for all $x,y$, and $f$ is continuous (actually, Lipschitz).
For an example of a sequence of such functions which does not converge, simply take $f_n(x)=(-1)^nx^2$. But yes - there will always be a subsequence of the $f_n$ which converges uniformly to some $f$ by Arzela-Ascoli: The argument above shows that all functions $f_n$ are Lipschitz, so the family $\left\{f_n\right\}_n$ is uniformly equicontinuous, and you can use the hypothesis that $|f_n(0)|=|f_n'(0)|\leq 1$ and the Mean Value Theorem again to show that these functions are uniformly bounded, so you can apply Arzela-Ascoli.
If you have not seen the Arzela-Ascoli Theorem, you can use the argument above with the Mean Value Theorem to show that $\left\{f_n(x)\right\}_n$ is bounded for all $x$ (in fact, $|f_n(x)|\leq 2$ for all $n$ and $x\in[0,1]$). Then use this fact for rational $x$, and take a diagonal subsequence which makes $f_n(x)$ converge to some $f(x)$ for all rational $x$. This function $f$ -- defined over the rationals in $[0,1]$ -- will be Lipschitz, so it extends uniquely to a Lipschitz function on all of $[0,1]$. Using the Lipschitz property again for both $f_n$ and $f$ you can actually show that $f_n\to f$ pointwise on all of $[0,1]$.