Let $f_n$ be a sequence of $C^2[0,1]$ functions which satisfies $|f(0)|, |f'(0)| < M$ for some $M$. And $|f''_n(x)| < (1-x) ^{-a}$ for some $1<a<2$ for all $x \in (0,1)$. Show that $f_n$ has a uniformly convergent subsequence.
What I tried is to apply Arzelà Ascoli theorem. I integrated the function $(1-x) ^{1-a}$ twice to get a uniform upper bound for the functions $f_n$. But I have trouble showing equicontinuity. Since if I integrate $(1-x) ^{-a}$ once I get $c (1-x)^{1-a } $ which is a upper bound for $f'_n(x)$ but I can't get equicontinuity with this bound at the point $x=1$.
Any help is appreciated. Many thanks in advance.
By assumption, we have the uniform bound $\lvert f_n''(x)\rvert \leqslant (1-x)^{-a}$ for $x\in (0,1)$ with some $a\in (1,2)$. Since $\lvert f_n'(0)\rvert \leqslant M$ for all $n$, it follows that
\begin{align} \lvert f_n'(t)\rvert &= \biggl\lvert f_n'(0) + \int_0^t f_n''(x)\,dx\biggr\rvert\\ &\leqslant \lvert f_n'(0)\rvert + \int_0^t \lvert f_n''(x)\rvert\,dx\\ &\leqslant M + \int_0^t (1-x)^{-a}\,dx\\ &= M + \biggl[\frac{(1-x)^{1-a}}{a-1}\biggr]_0^t\\ &= M + \frac{(1-t)^{1-a}-1}{a-1}\\ &\leqslant M + \frac{(1-t)^{1-a}}{a-1} \end{align}
for all $n$ and $t \in [0,1)$. The bound is monotonically increasing in $t$, and integrable over $[0,1]$.
Therefore, whenever we have $0 \leqslant x < y \leqslant 1$, we have
\begin{align} \lvert f_n(y) - f_n(x)\rvert &= \biggl\lvert \int_x^y f_n'(t)\,dt\biggr\rvert\\ &\leqslant \int_x^y \lvert f_n'(t)\rvert\,dt\\ &\leqslant \int_x^y M + \frac{(1-t)^{1-a}}{a-1}\,dt\\ &\leqslant M(y-x) + \frac{1}{a-1}\int_{1-(y-x)}^1 (1-t)^{1-a}\,dt\\ &= M(y-x) + \biggl[\frac{(1-t)^{2-a}}{(a-1)(a-2)}\biggr]_{1-(y-x)}^1\\ &= M(y-x) + \frac{(y-x)^{2-a}}{(a-1)(a-2)}. \end{align}
This shows the equicontinuity of the $f_n$. More, it shows that we have an equi-Hölder-continuous family for the exponent $2-a$.