$(f_{n})_{n}$ a row of continuous function in $[0,1]$ who converges pointswise to a continuous function $f:[0,1]\to \mathbb{R}$.
Can you conclude dat for every $a \in [0,1]$ and every row $(x_{n})_{n} \in [0,1]$ who converges to $a$ :
$\lim_{n \to \infty} f_{n}(x_{n})=f(a)$ ?
I had a prove for if $f_{n}$ converges uniformly to $f$, but I can't see where it goes wrong when $f_{n}$ converges pointswise to $f$. Here my prove for $f_n$ converges uniformly.
take $a \in [0,1]$ and $(x_n)_n \in [0,1]$ randomly with $(x_n)_n \to a$. take $\epsilon_1$>0 random but fixed. Because $(x_n)_n \to a$ there exist an $n_1 \in \mathbb{N}$ so that for every $n>n_1$ , $|x_n -a|<\epsilon_1$.
Now take $\epsilon_2$>0 random. Because $(f_n)_n$ is a row of continuous functions and $|x_n -a|<\epsilon_1$ , $|f_n(x_n)-f_n(a)|<\epsilon_2/2$. Because $(f_n)_n$ converges uniformly to $f$ there exist an $n_2 \in \mathbb{N}$ so that for $x \in [0,1]$ and $n>n_2$, $|f_n(x)-f(x)|<\epsilon_2/2$.
Take now $n_0=\min\{n_1,n_2\}$. Take $n>n_0$ randomly then $|f_n(x_n)-f(a)|=|f_n(x_n)+f_n(a)-f_n(a)-f(a)|<|f_n(x_n)-f_n(a)|+|f_n(a)-f(a)|<\epsilon_2$
EDIT: as an counter example for pointwise I have
$f_n:[0,1]\to \mathbb{R}: x\to \left\{ \begin{matrix}\mbox{2nx} & 0<=x<1/2n\\ \mbox{2-2nx } & 1/2n <=x<=1/n\\ \mbox{0} & 1/n<x<=1 \end{matrix}\right.$
with $(x_n)_n=1/2n$
Now I just want to know why my prove wouldn't work if I assume pointswise convergence.
Let $f_1(x):=2x$ for $0\leqslant x\leqslant\frac{1}{2}$, and let $f_1(x):=f_1(1-x)$ for $\frac{1}{2}\leqslant x\leqslant 1$, and let $f_1(x):=0$ for $x>1$.
Now define $f_n$ on $[0,1]$ by $f_n(x):=f_1(nx)$.
Then $f_n(x)\to 0$ on $[0,1]$, so the limit function $f$ is identically $0$
However with $x_n=\frac{1}{2n}$ we have that $x_n\to 0$ but $f_n(x_n)\to 1\not=f(0)$.