$\newcommand{\R}{\mathbb R}$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 $=1$.)


1 Answer 1


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.

  • $\begingroup$ Thank you. I had thought for a minute about trying a Baire category argument, and then inexplicably forgot about it. $\endgroup$ Aug 30, 2021 at 16:33

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