# uniformly convergent subsequence of functions

Given $$\{f_n\}_{n=1}^{\infty}\subset C([0,1])$$ is a family of continuous functions and $$f_n(x)$$ is differentiable on $$(0,1]$$ with $$|f'_n(x)|\leq\frac{1}{\sqrt{x}} \forall n\in\mathbb{N}, x\in[0,1)$$ and $$f_n(0)=0 \forall n\in\mathbb{N}$$.

Show that $$\{f_n\}_{n=1}^{\infty}$$ has a uniformly convergent subsequence $$\{f_{n_{k}}\}_{k=1}^{\infty}$$

Attempt: tried to find if $$\{f_n\}$$ is equicontinuous and bounded for the Arzela-Ascoti theorem but could not prove it. Any hints would be welcome.

Thank you.

• In order for this to be true, we need to assume not just that $f_n'(x) \leq \frac{1}{\sqrt{x}}$ but that $|f_n'(x)| \leq \frac{1}{\sqrt{x}}$, right? – Alex Nolte Apr 29 at 4:05

The issue here is the point at $$0$$. The idea of this argument is to handle everything away from $$0$$ with Arzela-Ascoli and to handle $$0$$ with our estimate on the derivatives of the $$f_n$$.
Note that for all $$x\in [0,1]$$, the fundamental theorem of calculus gives $$|f_n(x)| = \left|f_n(0) + \int_{0}^x f_n'(x) dx\right| \leq |f_n(0)| + \int_0^x|f_n'(x)|dx \leq 0 + \int_0^x \frac{1}{\sqrt {x}}dx = 2\sqrt{x} \leq 2 ,$$ so that the $$f_n$$ are uniformly bounded. Next, for any $$\delta < 0$$, on $$[\delta, 1]$$, as $$\frac{1}{\sqrt{x}} \leq \frac{1}{\sqrt{\delta}}$$ on $$[\delta, 1]$$, the $$f_n$$ have uniformly bounded derivatives on $$[\delta,1]$$. This implies (again by the fundamental theorem of calculus) that the $$f_n$$ are equicontinuous on $$[\delta, 1]$$ for all $$\delta > 0$$.
By the above discussion, the $$\{f_n\}$$ are uniformly bounded and equicontinuous on $$[1/2, 1]$$. So by Arzela-Ascoli, there's a subsequence $$f_{2, k}$$ that converges uniformly on $$[1/2, 1]$$. Note that our assumptions on $$f_n(0)$$ and the derivatives of $$f_{n}$$ carry over to $$f_{2,k}$$. So the $$f_{2,k}$$ are uniformly bounded and equicontinuous on $$[1/3, 1]$$, hence we can take a susequence $$f_{3, k}$$ of $$f_{2,k}$$ that converges uniformly on $$[1/3, 1]$$. Repeat this process so that for all $$n > 2,$$ $$f_{n, k}$$ is a subsequence of $$f_{n-1, k}$$ that converges uniformly on $$[1/n, 1]$$. Finally, define $$\tilde{f}_n$$ to be $$f_{n, n}$$ for all $$n \geq 2$$.
We claim that the $$\tilde{f}_n$$ converge uniformly on $$[0,1]$$, and do this by showing that they are Cauchy with respect to the sup norm. So let $$\epsilon > 0$$. There is some $$N \in \mathbb{N}$$ so that $$4/\sqrt{N} < \epsilon$$. Furthermore, as $$(\tilde{f}_m)_{m > N}$$ is a subsequence of $$f_{N, k}$$, the $$\tilde{f}_m$$ must converge uniformly on $$[\frac{1}{n}, 1]$$. In particular, they are Cauchy with respect to the sup norm. So there is some $$N' > N$$ so that for all $$m, m' > N'$$ and $$x \in [\frac{1}{n},1]$$, $$|\tilde{f}_m(x) - \tilde{f}_{m'}(x)| < \epsilon$$. On the other hand, as $$N' > N$$, for all $$x\in [0,\frac{1}{n}]$$ and $$m, m' > N'$$, $$|\tilde{f}_m(x)| \leq \int_0^x |\tilde{f}_m'(x)|dx \leq \int_0^\frac{1}{N} \frac{1}{sqrt{x}}dx = \frac{2}{\sqrt{N}} \leq \frac{\epsilon}{2}.$$ The triangle inequality then shows that for all $$x \in [0,\frac{1}{N}]$$ and $$m, m' > N'$$, $$|\tilde{f}_m(x) - \tilde{f}_{m'}(x)| \leq \epsilon.$$ So this holds for all $$x \in [0,1]$$.
We conclude that $$\tilde{f}_k$$ is a Cauchy sequence of continuous functions on $$[0,1]$$ with respect to the sup norm, hence converges uniformly on $$[0,1]$$.