$\newcommand{\R}{\mathbb R}$Consider the following [construction](https://www.mathcounterexamples.net/pointwise-convergence-not-uniform-on-any-interval/). 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 $=1$.)



How bad can pointwise convergence in $C$ be?
This is just a standard application of the Baire category theorem.


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.


