# How bad can pointwise convergence in $C$ be?

$$\newcommand{\R}{\mathbb R}$$Consider the following construction. For real $$u$$, let $$\begin{equation} f(u):=\frac{2u^2}{1+u^4}, \end{equation}$$ so that the function $$f\colon\R\to\R$$ is continuous, $$0\le f\le1=f(\pm1)$$, and $$f(u)\to0=f(0)$$ as $$|u|\to\infty$$.
Let $$Q$$ be any countable dense subset of $$\R$$. Let $$(w_q)_{q\in Q}$$ be any family of (strictly) positive real numbers such that $$\sum_{q\in Q}w_q<\infty$$. Finally, for each natural $$n$$ and all real $$x$$, let $$\begin{equation} g_n(x):=\sum_{q\in Q}w_q\, f(n(x-q)). \end{equation}$$

The latter series converges uniformly in $$x$$, and hence the function $$g_n$$ is continuous. Moreover, by dominated convergence, $$g_n\to0$$ pointwise (as $$n\to\infty$$). Take now any nonempty open interval $$I\subset\R$$. Then $$r\in I$$ for some $$r\in Q$$. Moreover, $$\{r+1/n,r-1/n\}\cap I\ne\emptyset$$ eventually -- that is, for all large enough $$n$$ (depending on $$I$$). So, $$\begin{equation} \liminf_n\,\sup_I g_n\ge w_r \liminf_n f(n(r\pm1/n-r))=w_r>0. \end{equation}$$

Thus, we have a sequence of continuous functions $$g_n$$ converging to $$0$$ pointwise on $$\R$$ while for any nonempty open interval $$I\subset\R$$ we have $$\liminf\limits_n\,\sup\limits_I g_n>0$$.

This brings us to

Question: Does there exist a sequence of continuous functions $$g_n\colon\R\to\R$$ converging to $$0$$ pointwise on $$\R$$ such that for any nonempty open interval $$I\subset\R$$ we have $$\liminf\limits_n\,\sup\limits_I g_n\ge1$$?

(Of course, here we can replace $$\ge1$$ by $$=1$$.)

Proposition: Suppose that $$X$$ is a topological space where the Baire category theorem holds and $$g_{n}:X\rightarrow[0,\infty]$$ for each natural number $$n$$. Suppose that $$\overline{\lim}_{n}\sup_{I}g_{n}\geq 1$$ for each non-empty open set $$I$$. Then there is a point $$x\in X$$ such that $$\overline{\lim}_{n}g_{n}(x)\geq 1$$.
Proof: If $$0<\delta<1$$ and $$n$$ is a natural number, then let $$U_{n,\delta}=g_{n}^{-1}(\delta,\infty)$$. Then each $$U_{n,\delta}$$ is open. Let $$O_{n,\delta}=\bigcup_{k=n}^{\infty}U_{k,\delta}$$. Then $$O_{n,\delta}$$ is open and dense. Therefore, $$\bigcap_{m,n}O_{n,1-1/m}$$ is dense by the Baire category theorem. If $$x\in\bigcap_{m,n}O_{n,1-1/m}$$, then for each $$m>0$$, we have $$x\in O_{n,1-1/m}$$ for all $$n$$. Therefore, since $$x\in O_{n,1-1/m}$$, we know that $$x\in U_{k,1-1/m}$$ for infinitely many $$k$$. Thus, $$g_{k}(x)>1-1/m$$ for infinitely many $$k$$. Therefore, $$\overline{\lim}_{n\rightarrow\infty}g_{n}(x)\geq 1-1/m$$, so since $$m$$ is arbitrary, we conclude that $$\overline{\lim}_{n\rightarrow\infty}g_{n}(x)\geq 1$$ for $$x\in\bigcap_{m,n}O_{n,1-1/m}$$. Q.E.D.