Continuity of $\ell^2$-norm in $\overline{\mathbb{R}}$

Question

Is the map $f:\overline{\mathbb{R}}^\mathbb{N}\to [0,\infty]$, defined by $f(x):= \sum_{n\in \mathbb{N}} x_n^2$ continuous ?

Here we use the usual conventions for the multiplication and addition in $\overline{\mathbb{R}}:=[-\infty,\infty]$, so that $f(x)<\infty$ if and only if $x\in \ell^2$.

I am considering the product topology on $\overline{\mathbb{R}}^\mathbb{N}$.

Answer 1

No, it's not even sequentially continuous, and one reason is substantially the dominated convergence theorem. Consider $\{x^n\}_{n\in\Bbb N}$, $$(x^n)_k=\begin{cases}\frac1n&\text{if }1\le k\le n^2\\0&\text{otherwise}\end{cases}.$$ Then, $x^n\to 0$ in the product topology but $\lim_{n\to\infty} f(x^n)=1$.