Suppose that a sequence of continuous function $(f_n)$ converges pointwise to $f$ on $[0,1]$. Suppose that a sequence of continuous function $(f_n)$ converges point wise to  $f$ on $[0,1]$.
Prove that if there exists a sequence $(x_n)$ in $[0,1]$ converging to $x^*$ in $[0,1]$ such that $(f_n(x_n))$ does not converge to $f(x^*)$, then $(f_n)$ is not uniformly convergent.
 A: Hint: go back to the definition. Uniform convergence of $(f_n)_n$ to $f$ on $[0,1]$ would by definition mean that 
$$\sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert \xrightarrow[n\to\infty]{} 0$$
Suppose there exists $(x_n)_n$ as stated. Can you use this to give a lower bound on the $\sup$ (for infinitely many $n$) that would be a positive constant? (That would imply the $\sup$ cannot converge to $0$).
(See below for details)



*

*Suppose by contradiction $(f_n)_n$ does converge uniformly to $f$. Since each $f_n$ is continuous, this implies $f$ is.

*By assumption, there exists some $\varepsilon > 0$ and a subsequence $(\varphi(n))_n$ such that for all $n \geq 0$, $\lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast) \rvert \geq \varepsilon$. 

*Also, by the triangle inequality
$$\lvert f_n(x_n) - f(x_n)\rvert  \geq \lvert f_n(x_n) - f(x^\ast)\rvert - \lvert f(x^\ast) - f(x_n)\rvert
$$
for any $n$.
For the second term, as $f$ is continuous and $x_n\to x^\ast$, there exists $N \geq 0$ such that for any $n \geq N$, $\lvert f(x^\ast) - f(x_n)\rvert \leq \frac{\varepsilon}{2}$.

*Combine the two:
\begin{align}
\lvert f_{\phi(n)}(x_{\phi(n)}) - f(x_{\phi(n)})\rvert  
&\geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast)\rvert - \lvert f(x_{\phi(n)}) - f(x^\ast)\rvert \\
&\geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast)\rvert - \frac{\varepsilon}{2} \\
&\geq \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}
\end{align}
for any $n\geq N$ (which implies $\phi(n)\geq N$).

*For $n\geq N$, this implies that
$$
\sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert \geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x_{\phi(n)})\rvert \geq \frac{\varepsilon}{2}
$$
and therefore the $\sup$ cannot converge to $0$.
