# Uniform convergence, Bounded derivative.

Let $f_n(x):[0,1] \to \Bbb R$ be a sequence of differentiable functions such that for every $n$ and every $x \in [0,1]$ we have $|f'_n(x)| \leq 1$. Also, $f_n \to f$ (pointwise convergence) in $[0,1]$. Prove that $f_n$ converges to $f$ uniformly.

I managed to prove that for every $x_0$, there is a uniform convergence in $(x_0-\delta, x_0 +\delta)$ for some $\delta > 0$.

How can I finish the proof?

• Well, how does your proof go? (Without that, it's hard to say how to build from it to get the more general statement). Jul 11 '17 at 17:14
• I used that fact that $f$ is uniformly continuous Jul 11 '17 at 17:16
• Please, provide the details of your argument. The above gives little to none. Also, what tools are you allowed to use? Do you know the dominated convergence theorem? Jul 11 '17 at 17:17

Hint. Note that $$f_n(x)=f_n(0)+\int_{0}^xf_n'(t)\,dt$$ Then $$\sup_{x\in[0,1]}|f_n(x)-f_m(x)|\leq |f_n(0)-f_m(0)|+\int_{0}^1|f_n'(t)-f'_m(t)|\,dt$$ Now by using the pointwise convergence and dominated convergence theorem show that $(f_n)_n$ is a Cauchy sequence in $C([0,1])$ with the sup norm.
You can end your proof like this: since $[0,1]$ is compact, you can find $x_1,...,x_n$ and $a_1>0,...,a_n>0$ such that $f_n$ converges uniformly of $(x_i-a_i,x_i+a_i)$ and $\bigcup_i(x_i-a_i,x_i+a_i)\cap [0,1]=[0,1]$.
For every $c>0$, there exists $N_i$ such that for every $n>N_i, x\in (x_i-a_i,x_i+a_i)$ implies that $|f_n(x)-f(x)|<c$. Take $n=Sup(N_1,...,N_n)$.
Hint: Since you managed to prove that for every $x_0 \in [0,1]$, there exists a $\delta \gt 0$ such that for all $x \in (x_0-\delta,x_0+\delta)$, $f_n \to f$ uniformly and $[0,1] \subset \cup_{x_0 \in [0,1]}(x_0-\delta,x_0+\delta)$, use the compactness of $[0,1]$ to extract a finite subcover