Finding a uniformly convergent subsequence? I have the following problem which has been bugging me for a bit.
Suppose that $(f_n)_{n=1}^{\infty} \subset C^1([0,1])$ is a sequence of RVed functions such that $$|f_n(x)| \leq -\frac{\ln(x)}{x^2}, \: \: |f_n'(x)| \leq -\ln(x), \forall x \in [0,1], \: n \in \mathbb{N}.$$ Prove that $(f_n)$ has a uniformly convergent subsequence. Hint: Notice that $|f_n(x)| \leq |f_n(y)| + |f_n(x)-f_n(y)|$.
So I know that $f_n,f_n'$ are both bounded by decreasing functions with $f_n(1)=f_n'(1)=0$ for all $n$.
I however am not sure how to interpret the hint, other than: $$|f_n(x)| \leq |f_n(y)| + |f_n(x)-f_n(y)| \leq -\frac{\ln(y)}{y^2} - |x-y|\ln(x) + |x-y|M(y)$$ where $M(y) \rightarrow 0$ as $y \rightarrow x$, using the fact that $|f_n'(x)| \leq -\ln(x)$ and writing a difference quotient.
I also know that $Y = \{f \in C^1([0,1]) \; | \: f(1)=f'(1)=0\}$ is a closed subspace and Banach under an appropriate norm, which would allow me to simply find a uniformly Cauchy subsequence.
If anyone could provide a small $\textbf{hint}$ to point me in the right direction that would be greatly appreciated.
 A: Here are some hints on how to follow up on the suggestion made in Mason's comment.  To apply the Arzelà-Ascoli theorem you need to show that your sequence is equicontinuous  and uniformly bounded.

*

*For $\ 0\le a<b\le 1\ $, we have
\begin{align}
\left|f_n(b)-f_n(a)\,\right|&=\left|\,\int_a^bf_n'(y)\,dy\,\right|\\
&\le -\int_a^b\ln y\,dy\\
&=b-b\ln b-a+a\ln a\\
&\le (-\ln a)(\,b-a)\ ,
\end{align}
and, in particular,
$$
|f_n(b)-f_n(0)|=b-b\ln b\
$$
(where I have taken $\ a\ln a=0\ $ at $\ a=0\ $), so \begin{align}|f_n(0)|&\le|f_n(b)|+|f_n(b)-f_n(0)|\\
&\le\frac{-\ln b}{b^2}+b-b\ln b\ ,\end{align} and \begin{align} |f_n(x)|&\le|f_n(0)|+|f_n(x)-f_n(0)|\\&\le\frac{-\ln b}{b^2}+b-b\ln b+x-x\ln x\ ,\end{align} from which the uniform boundedness of your sequence follows.

*From the inequality $\ \left|f_n(b)-f_n(a)\,\right|\le (-\ln a)(\,b-a)\ $, it follows that $\ |f_n(x)-f_n(y)|\le$$-(\ln\epsilon)|x-y|\ $ for any $\ x,y\in[\epsilon,1]\ $, where $\ 0<\epsilon<1\ $. The sequence $\ \left\{f_n\right\}_{n=1}^\infty\ $ is therefore uniformly Lipschitz, and hence equicontinuous, over every such interval. But for $\ x,y\in[0,2\epsilon]\ $, we also have
\begin{align}
|f_n(x)-f_n(y)|&\le|f_n(x)-f_n(0)|+|f_n(y)-f_n(0)|\\
&\le x-x\ln x+ y-y\ln y\ ,
\end{align}
which will be arbitrarily small if $\ \epsilon\ $ is sufficiently small.  I'm fairly sure these two observations are sufficient to give you the equicontinuity of your sequence.  I haven't written out the argument in full, however, so if you have any trouble finishing it off, please let me know in the comments.

