The question:

(a) Let $f_n(t)=\frac{\sin(t^n)}{t^n} $ and $F_n (x)=\int_0^x f_n(t)\,dt$
for $x\in [0,1]$. Prove there exists a subsequence $\left \{ F_{n_k}\right \}$ which converges uniformly on $[0,1]$.

(b) Let $ C^{1}([0,1])$ be the set of all functions $f$ such that $f'$ is a real valued continuous map on $[0,1]$ . Prove that $ C^1([0,1])$ is a complete metric space with the metric $d_1\{f,g\}=\|f-g\|_u+\|f'-g'\|_u.$

My attempt:

(a) Let $E=[0,1]$ . If $\{f_n\} $ is pointwise bounded and equicontinuous on E, then $\{f_n\} $ contains a uniformly convergent subsequence. $f_n(t)$ is bounded since $\left | \frac{\sin(t^n)}{t^n} \right |\leq \frac 1 {t^n}$. Now it suffices to show that $f_n(t)$ is equicontinuous. I'm stuck in showing the equicontinuity.

(b) Let $E=[0,1]$ and $\{f_n\}$ be a Cauchy sequence in $ C^1(E)$.

$\forall \epsilon >0, $ $\exists N>0 $ such that $n,m\geq N\Rightarrow \|f_n-f_m\|_u+\|f_n'-f_m'\|_u<\varepsilon $

Since this is Cauchy criterion, $\{f_n\}$ converges uniformly to $f$ on $E$. Now, I want to show that $f\in C^1(E)$.

$\|f_n'-f_m'\|_u<\varepsilon $ means that $\{f_n'\}$ converges uniformly to $g$ on $E$.

Since $f_n'$ is a continuous sequence, $g$ is continuous. Then, $f'=g$. So, $f'$ is continuous and $f\in C^1(E)$. Hence $C^1[0,1] $ is a complete metric space.

I wonder whether this is a valid proof.


a) $f_n$ is bounded, but your argument is incorrect, since $1/t^n$ is not bounded on $[0,1]$. Use the fact that $|\sin x1\le|x|$ and conclude that $|f_n(x)|\le1$. Then you want to show that $\{F_n\}$ satisfies the conditions of Ascoli-Arzela's theorem. This is easy, since $$ |F_n(x)|\le\int_0^x|f_n(x)|\,dx\le x\le1\quad\forall x\in[0,1] $$ and $$ |F_n'(x)|=|f_n(x)|\le1\quad\forall x\in[0,1]. $$

b) All is correct, that is, that $f_n$ converges uniformly to some continuous $f$ and $f_n'$ converge uniformly to some continuous $g$; but you should prove that in fact $f$ is differentiable and $f'=g$.

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