Given $f'_n$ is less than 5 in $[0,1]$ and $f_n(x)$ converges pointwise to $f(x)$ for every x prove that $f_n(x)$ uniformly converges to $f(x)$

Question

Is my proof valid?

Let $\epsilon > 0$ Since $f'_n$ is bounded in the interval for all $n$, then they are uniformly continuous in that interval, so there exists a $\delta$ such that for all $x$, $y$ in $[0,1]$ if $|x-y|<\delta$ then

$$|f_n(x)-f_n(y)| < \epsilon / 3$$

$f_n$ converge pointwise, then (using Cauchy’s theorem for pointwise convergence) there exists an $N$ such that for all $N<n<m$:

$$|f_m(x)-f_n(x)| < \epsilon / 3$$

Then (using Cauchy’s theorem for uniform convergence) for all $N<n<m$ and all $x$ in $[0,1]$

$|f_m(x)-f_n(x)| = |f_m(x)-f_n(x)+f_m(y)-f_m(y)+f_n(y)-f_n(y)| < ... < \epsilon$

Alternate proofs are also welcomed

Answer 1

That proof is wrong. You proved that each $f_n$ is uniformly continuous (by the way: being continuous is enough for that). And they are. But that does not mean that, for each $\varepsilon>0$, there is a $\delta>0$ such that, for each $x\in[-1,1]$ and each $n\in\Bbb N$, $|x-y|<\delta\implies\bigl|f_n(x)-f_n(y)\bigr|<\varepsilon$. It just means that, for each $n\in\Bbb N$, there is a $\delta_n$ that will work for $f_n$.

Besides, the statement is false. Just consider $f_n(x)=\left(\frac{x+1}2\right)^n$. Each $f_n'$ is bounded, the sequence $(f_n)_{n\in\Bbb N}$ converges pointwise, but the convergence is not uniform.