Suppose
(1) A sequence of continuous functions$f_n(x)$ converges to a continuous function $f(x)$ pointwise on some set $I$, and
(2) a sequence $\{x_n\}$ converges to $x \in I$
(****) Is it true that $f_n(x_n)$ converges to $f(x)$? (****)
The way I try and break it down is to say that:
$|f_n(x_n) - f(x)| \le |f_n(x_n)-f(x_n)| + |f(x_n)-f(x)|$
The second term can be made arbitrarily small since $f$ is continuous (For $\epsilon > 0$ $\exists \delta$ such that $|x_n-x| < \delta \implies |f(x_n)-f(x)| < \epsilon/2$), and this holds for all $n$ large enough because of the convergence of the $x_n$ to $x$.
However, I have no idea how to bound the first term without making the assumption of uniform convergence. My question (****) comes from the idea that if I create two indexes ($f_n(x_m)$) and let $n \rightarrow \infty$, I get that $f_n(x_m) \rightarrow f(x_m)$ and subsequently let $m \rightarrow \infty$, it converges to $f(x)$.
If I take the limit of the $x_m$ followed by the limit of the $f_n(x)$, it once again converges to $f(x)$, so it would make sense for my answer to (****) to be yes, but I can't prove it. A counterexample would be helpful if it's false. Thanks for any help you can provide!
Not true. In fact this property is equivalent to uniform convergence. So it is easy to give counterexamples. On $[0,1]$ we can construct a sequence of continuous functions $f_n$ such that $f_n(x)=0$ for $x >\frac 1 n$, $f_n(0)=0$ and $f_n(\frac 1 {2n})=1$. [Think graphically]. Then $f_n \to 0$ pointwise and $\frac 1 {2n} \to 0$ but $f_n(\frac 1 {2n})=1$ for all $n$.