If for $x \in A \subseteq X$,$\lim \limits_{n\to \infty}f_n(x_n) = f(x ) $ for each $\{x_n\}$ s.t $\lim \limits_{n\to \infty}x_n = x$, does it necessarily follow that $f_n \to f $ uniformly. Here X is a normed vector space over $\mathbb R$ and $f_n$s are continuous.
I know that this is true if $A$ is a compact set. In that case the key point is we can construct a sequence which has a convergent subsequnce using compactness. But spent a significant time on this without any sucsess. Any hint woupd be help me. Not intended to seek a complete answer.
Not true even when $A=X=\mathbb R$. Let $f_n(x)=\frac x n$ and $f(x)=0$. Then $x_n \to x$ implies $f_n(x_n)=\frac {x_n} n \to 0=f (x)$ (since $(x_n)$ is a bounded sequence). But $sup_x |f_n(x)-f(x)|=\sup_x |\frac x n|=\infty$ for all $n$.
If you want to get an example with each $f_n$ bounded you can take $f_n(x)$ to be piece-wise linear with $f_n(x)=\frac x n$ on $(-n,n)$, $0$ on $\mathbb R \setminus (-n-1,n+1)$.