If $x_n→ x$ in $\ell^2$ then $\lim_{n→\infty}f(x_n)=f(x)$ if $f$ is continous?

Question

Let $(x_n)_{n\in \mathbb N} \subset \ell^2$ with component wise limit $x \in \ell^2$. Let be $f:\ell^2→\ell^2$ is continous.

Is it true that:

$$\lim_{n→\infty}f(x_n)=f(\lim_{n→\infty} x_n)=f(x)? \tag C$$

In my opinion, it is not true since continuity claims $\lim_{n→ \infty}x_n=x$ so convergence with respect to a norm and not only component-wise.

Any help appreciated!

Answer 1

This isn't true. For example you can use $f(x) = x$, which is continuous, and let $x_n$ be the sequence equal to $1$ in the $n$th entry and zero elsewhere. Then $x_n$ converges to the zero sequence componentwise, but not in $l^2$.

Answer 2

In any metrizable topological space, sequential continuity is equivalent to topological continuity!

The space $l_{2}$ is a normed space, hence metrizable with

$d(x,y)= \left\|x-y \right\|$. Therefore if $f$ is continuous, it is

also sequentially continuous which immediately gives the result!!