Can I put limits together $(\lim_{t,n \rightarrow \infty} f_t(x_n) = \lim_{t \rightarrow \infty} f_t(x_t))$?

Question

I got the following problem while writing my thesis:

I have a series of functions $(f_t)_{t \in \mathbb{N}}: X \rightarrow Y_t$ on a Banach space $X$ to another $Y_t$. It is that $Y_t \rightarrow Y$ is also Banach and $Y_t \subset Y$. I also have a sequence $(x_n)_{n \in \mathbb{N}} \subset X$. The limits of the $f_t \rightarrow f:X \rightarrow Y$ (pointwise convergence) and $x_n$ are both existent. Can I do the following:

\begin{align*} \lim_{t,n \rightarrow \infty} f_t(x_n) = \lim_{t \rightarrow \infty} f_t(x_t) \end{align*}

The $(f_t)_{t \in \mathbb{N}}$ and $(x_n)_{n \in \mathbb{N}}$ have no particular dependencies.

Answer 1

This is not necessarily true in general. Consider the function $f_n=f:l^2\rightarrow l^2$ given by $f(x_1,x_2,\cdots)=\begin{cases} (x_1,x_2,\cdots) &\text{if finitely many non-zero entries}\\ 0 &\text{otherwise.}\end{cases}$ and $x_n\in l^2$ given by $x_n=(1,1/2,\cdots, 1/n,0,\cdots)$. Then $x_n\rightarrow x:=(1,1/2,\cdots)$ in $l^2$ but $\lim_{n\rightarrow\infty}\lim_{k\rightarrow\infty} f_k(x_n)= x\neq 0=f(x)=\lim_{k\rightarrow\infty}\lim_{n\rightarrow\infty}f_k(x_n)$.

Note that this counterexample only uses the fact that if $f$ is discontinuous, then $f(x_n)$ does not converge to $f(x)$.

If $f_n$ are bounded linear operators then $f_n\rightarrow f$ pointwise implies $f_n\rightarrow f$ in $\mathcal B(X,Y)$ from Banach-Steinhaus theorem so you can conclude $f_n(x_n)\rightarrow f(x)$ and $f_n(x_k)\rightarrow f(x)$.

Answer 2

This is very elementary counter-example.

$f_n:\mathbb R\to \mathbb R$ given by $f_n(x)=\begin{cases} \frac1{x-n+1} &{x>n}\\ 0 &\text{otherwise.}\end{cases}$ and $x_n=n$.

Then $\lim_{n\to\infty}f_n(x_n)=1$ is trivial, and $f_n\to0$ for pointwise convergence(if $N>x$, then $f_N(x)=0$.) And for fixed $n$, $\lim_{x\to\infty}f_n(x)=0$.

$$ \lim_{n\to\infty}\lim_{k\to\infty}f_n(x_k)=0=\lim_{k\to\infty}\lim_{n\to\infty}f_n(x_k),\ \ \ \ \lim_{n\to\infty}f_n(x_n)=1. $$