$f: \operatorname{dom}(f) \subseteq E \to F$ , $E$ and $F$ Banach then if $\operatorname{dom}(f)$ is closed, $f$ is closed

Question

I've to show the following.

Given E,F Banach spaces and $f:\operatorname{dom}(f) \subseteq E \to F$ a continuous function. Show that if $\operatorname{dom}(f)$ is closed then $f$ is a closed function ( the image of a closed set is a closed set).

I have managed to prove that if $\operatorname{dom}(f)$ is compact the result follows, but I'm stuck in the general case. I would appreciate any indication to solve it or a counterexample if the result doesn't follow.

Answer 1

It's not true, even if $E = F = \Bbb{R}$. Take, for example, $$f : \Bbb{R} \to \Bbb{R} : f(x) = \frac{1}{x^2 + 1}.$$ Then $\operatorname{dom} f = \Bbb{R}$, which is closed, but $f(\Bbb{R}) = (0, 1]$, which is not closed.