$\lim_{n \to \infty}f_n(x_n)=f(x)$ if $f_n \to f$ and $x_n \to x$?

Question

$f_1,f_2,...$ is a sequence of continuous functions defined on $\mathbb R$ such that $\forall x \in \mathbb R$ $\lim_{n \to \infty}f_n(x)=f(x)$, and $\{x_n\}\subseteq \mathbb R$ is a sequence converging to $x$.

Can we prove that $\lim_{n \to \infty}f_n(x_n)=f(x)$?

My attempt (motivated by Show that $f_n(x_n) \to f(x).$)

\begin{align*}|f_n(x_n)-f(x)|&=|f_n(x_n)-f(x_n)+f(x_n)-f(x)|\\ & \leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|\\ & \leq \sup_{t \in \mathbb{R}} |f_n(t)-f(t)|+|f(x_n)-f(x)|\\ & \to 0\end{align*}

To complete the proof we need to show that $\sup_{t \in \mathbb{R}} |f_n(t)-f(t)| \to 0$.

Using the definition of convergence, we know that $$\forall \epsilon\, \forall t\, \exists n^*(t) \text{ such that }|f_n(t)-f(t)|<\epsilon\quad \forall n \ge n^*(t)$$ We could complete the proof by showing that the $n*(t)$ are bounded above; that is, we need to show that there exists an integer $n^*$ such that $$n^*=\max_{t \in \mathbb R}n^*(t)$$ since then we will have $$\forall t\, \forall n \ge n^*\quad|f_n(t)-f(t)|< \epsilon$$ and therefore $$\forall n \ge n^*\quad\sup_{t \in \mathbb{R}} |f_n(t)-f(t)| \le \epsilon$$

How can we show that $n^*$ exists?

Answer 1

This is only true if $f_n(x)$ converges uniformly to $f(x)$. In fact, it will break for any situation where $f_n(x)\rightarrow f(x)$ pointwise but not uniform on an interval $[L,R]$. In other words, there exists $\epsilon>0$ s.t. for all $N>0$, there exists a point $a$ such that $|f_n(a)-f(a)|>\epsilon$. Pick a sequence of $\epsilon_n$ and increasing $N_n$. This gives a sequence of $a_n$. By Bolzano-Weierstrass, pick a convergent subsequence of $a_n$ that converges, and call it $x_n$. Now follow through the proof.

Example: $f_n(x)=2^nx$ when $0\leq x\leq 2^{-n}$, $f_n(x)=2-2^nx$ when $2^{-n}\leq x\leq 2^{-n+1}$ and 0 elsewhere. Then $f_n(x)\rightarrow 0$. Yet pick $x_n=1/2^n$. Then $x_n\rightarrow 0$ but $f_n(x_n)=1$. Essentially you have a bump which is clamped to 0 on either side of $[0,1]$, it's squishing to the left but there's always a sequences which rides on the value 1.

Answer 2

Alternatively, we may assume that the family $\{f_n\}_{n=1}^\infty$ is equicontinuous at $x$.