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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.

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  • $\begingroup$ 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? $\endgroup$ – Alex Nolte Apr 29 at 4:05
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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 $[0,1]$.

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