$f_n$ is a sequence of real functions defined on $[0,1]$.
For all n, we have
i) $ f_n(0)=f'_n(0)=0$
ii) $f''$ is continuous, and $|f''(x)|\leq 1$ for all $x \in [0,1]$
Prove that ${f_n}$ has a subsequece converging uniformly on $[0,1]$
My Solution:
I have proved that $f_n$ is pointwise bounded.
Thus, for any $x\in [0,1]$, we can find a sequence ${x_k}$ converging to x and a subsequence of $f_n$ (denote this subsequence as $\{g_n\}$) converging for every point in $x_k$. i.e. given any $x_k$, $ \{ g_n(x_k) \}$ converges as $n\rightarrow \infty$
I want to prove that $ \{g_n\}$ converges uniformly on $[0,1]$
I am considering the following:
$|g_n(x)-g_m(x)|\leq|g_n(x)-g_n(x_k)|+|g_n(x_k)-g_m(x_k)|+|g_m(x_k)-g_m(x)|$
I think the first part and the third part con be controlled below $\epsilon$ by continuous and convergence of $x_k$ and the second part can be controlled by pointwise convergence. But when writing the proof, I found that the selection of $m$ and $n$ depends on $k$ , and that the selection of $k$ depends on $m$ and $n$.
Did I ignore anything? Or can I just say, "for any $\epsilon$ we can find $x_k$, $N_k$ such that $m, n> N_k$ implies that all conditions hold"?
Best Answer
Use Arzela-Ascoli Theorem. Since $|f_n''|\leq 1$ applying Mean-Value theorem we have, $$\frac{f_n'(x)-f_n'(0)}{x-0}=f_n''(c_{n,x})\implies |f_n'(x)|\leq x|f_n''(c_{n,x})|\leq 1,\forall n\in \Bbb N,\forall x\in [0,1].$$ Now, $$\frac{f_n(x)-f_n(0)}{x-0}=f_n'(d_{n,x})\implies |f_n(x)|\leq x|f_n'(d_{n,x})|\leq 1,\forall x\in [0,1],\forall n\in \Bbb N$$$$\implies \{f_n\}\text{ is uniformly bounded}$$ and $\forall n\in \Bbb N,\forall x,y\in [0,1]$ we have, $$\frac{f_n(x)-f_n(y)}{x-y}=f_n'(e_{n,x})\implies |f_n(x)-f_n(y)|\leq |x-y||f_n'(e_{n,x})|\leq |x-y|$$$$\implies \{f_n\}\text{ is equicontinuous family.}$$ So $\{f_n\}$ is a equicontinuos family of continuous functions which is uniformly bounded. So by Arzela-Ascoli theorem this seqnce has a uniformly convergent subsequence.