How to show $\{f_n\}_{n=1}^\infty$ has uniformly convergent subsequence on [0,1]?

Question

Let $\{f_n\}_{n=1}^\infty$ a sequence of second order differentiable functions on the interval [0,1].

If $\forall n\in \Bbb N$ $f_n(0)=f_n'(0)=0$ and for all $n\in \Bbb N$ and $x \in [0,1]$ , $|f_n''(x)|\le 1$ then show $\{f_n\}_{n=1}^\infty$ has uniformly convergent subsequence on [0,1].

Do we use arzela-ascoli which says if A is compact and $B\subset C(A;\Bbb R^m)$ is bounded and equicontinous, every sequence in B has a convergent subsequence at $C(A;\Bbb R^m).$ ?

Answer 1

Yes, you can use Arzela-Ascoli's theorem. To prove boundedness of $B = \{f_n \mid n \in \mathbb N\}$ recall that $$ f_n(x) = \int_0^x f_n'(\xi)\, d\xi = \int_0^x \int_0^\xi f_n''(\eta) \,d\eta\, d\xi $$ as $f_n(0) = f_n'(0)= 0$. For equicontinuity, prove that the $f_n$ are uniformly Lipschitz, using the mean value theorem and boundedness of $(f_n')$.

Answer 2

By Lagrange Theorem we have $$|f'_n (x) |=|f'_n (x) -f'_n (0) | =|f''_n (\xi ) ||x|\leq 1 \mbox{ for } 0\leq x\leq 1.$$ Analogously $$|f_n (x) |=|f_n (x) -f_n (0) | =|f'_n (\xi_1 ) ||x|\leq 1 \mbox{ for } 0\leq x\leq 1.$$ Hence the family $\{ f_n :n\in\mathbb{N}\}$ is uniformly bounded. Moreover $$ |f_n (u) -f_n (v) |=|f'_n (\xi_2 ) ||u-v|\leq |u-v|$$ hence the family $\{ f_n :n\in\mathbb{N}\}$ is equicontinous. So by the Arzela - Ascoli's Theorem the sequence $(f_n )$ has convergent subsequence.