Let {$f_{n}$} be a sequence of functions on [0, 1], such that all $f_{v}$ have continuous second derivatives. Assume that there exists $M$ such that $f_{n}(0) ≤ M, f'_{n}(0) ≤ M$ for all $n > 0$, and that for all $n > 0$ and all $x ∈ [0, 1)$,$|f''_{n}(x)| ≤\frac{1}{\sqrt{1-x}}$. Prove that then {$f_{n}$} contains a uniformly convergent subsequence.
My thought is, show {$f_{n}$} is pointwise bounded and equicontinous. Since $f_{n}(0)$ is bounded, I only need to prove they're equicontinous. But I don't know how to use the second derivative premise.
In addition, it's not homework problem, I just saw it on exam last year and stucked here for some days. Thanks in advance for any help.
I am assuming that you have $\lvert f_n(0) \rvert \le M$ and $\lvert f'_n(0) \lvert \le M$. That is, you need these bounded from top and bottom; otherwise, I'm pretty sure the property isn't true. With this modification, the following reasoning works:
Note that for any $x \in [0,1]$, $$f_n'(x) = f'_n(0) + \int^{x}_0 f''(t) dt.$$ Thus $$\lvert f_n'(x) \rvert \le \lvert f'_n(0)\rvert + \left \lvert\int^x_0 f''_n(t) dt\right \rvert \le M + \int^x_0 \lvert f''_n(t)\rvert dt \le M + \int^1_0 \frac{1}{\sqrt{1-t}}dt = M+2.$$ This is a uniform (in both $x$ and $n$) bound on $f'_n(x)$. By the mean value theorem, we have for any $x,y \in [0,1]$ and any $n \in \mathbb N$, $$\lvert f_n(x) - f_n(y) \rvert = \lvert f'_n(c)\rvert \lvert x - y \rvert \le (M+2) \lvert x-y\rvert,$$ where $c$ is some value between $x$ and $y$ (and it depends on $n$, but that doesn't matter). This shows that all of $f_n$ are Lipschitz continuous with the same Lipschitz constant, and hence the family is equicontinuous.
Of course, you also need uniform boundedness. This follows since $$\lvert f_n(x) \rvert \le \lvert f_n(0) \rvert + \lvert f_n(x) - f_n(0) \rvert \le M + (M+2) \lvert x \rvert \le 2M+2.$$
EDIT: Indeed, you need $\lvert f_n(0) \rvert \le M$ and $\lvert f'_n(0) \lvert \le M$. If you only have the bound from above, then $f_n(x) = x-n$ gives a counterexample. In this case you can take $M =1$ and you have $f_n(0), f'_n(0) \le M$. You also have $\lvert f''_n(x) \rvert \le \frac{1}{\sqrt {1-x}}$. But there is no uniformly convergent subsequence.