If a sequence of continuous functions converges pointwise to a continuous function on $ [a,b] $, it converges uniformly If a sequence of continuous functions converges pointwise to a continuous function on $ [a,b] $, it converges uniformly.
Looking at other theorems on the relationship between continuity and uniform convergence and how they require significant additional assumptions to assure uniform convergence, it seems like the above statement should be false in general. However I'm unable to find a counterexample. Any suggestions?
 A: $$f_n(x) = \begin{cases} nx &, x \leqslant \frac{1}{n}\\ 2 - nx &, \frac{1}{n} < x \leqslant \frac{2}{n}\\ 0 &, x > \frac{2}{n} \end{cases}$$
on the interval $[0,1]$. The sequence converges to $0$ pointwise, but the bump has constant height $1$.
A: Maybe it's a little bit far fetched but under the following relatively strong assumptions it is possible to conclude uniform convergence:
Let be $\left(f_n\right)_{n\in\mathbb{N}}$, where $f_n:[0,1]\to \mathbb{R}$, a pointwise convergent sequence of functions with $\lim\limits_{n\to\infty}f_n=f$. Further, $f_n$ and $f$ are (uniformly) continuous. Ok, now we need some strong assumptions:
We need that the sequence of functions $f_n(x)$ can be expressed as $f_n(x)=g(x,y_n)$, where $g:\mathbb{R}\times[0,1] \to \mathbb{R}$ is continuous and $y_n\to y$ is a convergent sequence. Then, a sketch of the proof of uniform convergence might be:
Define a compact set $A:=\{y_n~|~ n\in\mathbb{N}\}\cup \{y\}$. Restrict the domain of $g$ to $g:A\times[0,1] \to \mathbb{R}$ then $g$ becomes uniformly continuous which is equivalent to $f_n$ being uniformly convergent.
