"composition" of "pointwise convergent sequences of functions"

Question

Intuitively, if $f_n\to f$ as $n\to\infty$ and $g^{(n)}_i\to f_n$ as $i\to\infty$, can we get $g_j\to f$ as $j\to\infty$?
Formally,

Let $\{f_n\}_n$ be a sequence of functions from $\mathbb{R}^d$ to $\overline{\mathbb{R}}$, the extended real line. Let $f$ be its pointwise limit, i.e. for each $x\in\mathbb{R}^d$ and each $\varepsilon>0$, there exists $N\in\mathbb{Z}^+$ such that $|f_n(x)-f(x)|<\varepsilon$ for all $n\geq N$. Each $f_n$ is the pointwise limit of sequence $\{g^{(n)}_i\}_i$. Can we construct a sequence of $g$'s converging pointwise to $f$? If not, what additional conditions do we need?

As usual, any help is appreciated:)

Answer 1

An old question, but maybe someone likes to read about an answer somewhere in the future.

This is a basic fact about topology, the property you are asking for is an axiom of a topology defined through nets (generalized sequences).

So in a topological space $(X,\tau)$, we assume that around $x\in X$ we a have a countable basis of neighbourhoods and we take a converging sequence $x_n\to x$ and a sequence of converging sequences $y_m^n\to x_n$ for all $n\in\mathbb{N}$.

Now for any open neighbourhood $U$ of $x$ we have that there exists $N_U\in\mathbb{N}$ such that $$ x_n\in U\quad\forall n>N_U. $$ Since $U$ is open we get a natural number $M^n_U$ for all $n>N_U$ such that $$ y^n_m\in U\quad\forall m>M^n_U. $$

So now by taking $U_k$ a basis for the neighbourhoods of $x$, for any $k\in\mathbb{N}$ we can just take any $$ y^n_m\in U_k\quad m>M^n_{U_k},\, n>N_{U_k}, $$ and get the desired sequence convergin to $x$.

Given the set of functions $\mathrm{Set}(\mathbb{R}^d,\overline{\mathbb{R}})$, the pointwise convergence is the finest topology that makes all the evaluations continuous, and thus you can apply the above to the topological space $\mathrm{Set}(\mathbb{R}^d,\overline{\mathbb{R}})$.