Let $A$ be a linear operator from $X$ to $Y$ with domain $D(A)$. I've learned the following characterization for closable operators: $A$ is closable if and only if for every sequence $(x_n) \in D(A)$ with $x_n\rightarrow 0$ in $X$ and $(Ax_n)$ convergent, it holds that $Ax_n \rightarrow 0$ in $Y$.
Suppose we have $X=L^2(\mathbb{R})$ and $Y=\mathbb{R}$. I need to prove or disprove that $Ax=\int_\mathbb{R} x dt$ with $D(A)=C_0(\mathbb{R})$ is closable. I think that it's not true (so $A$ is not closable). However, I can't find a suitable counterexample... Any help is appreciated.
You are looking for a sequence $\{f_n\}\subset C_0(\mathbb R)$ with $\|f_n\|_2\to0$ and $\int_{\mathbb R}f_n\not\to0$.
Take for instance $$f_n=\tfrac1n\,1_{[0,n]}.$$ Then $$ \|f_n\|_2=\frac{\sqrt n}n=\frac1{\sqrt n}\to0, $$ while $$ Af_n=\int_{\mathbb R}f_n=\frac nn=1. $$ The above functions are not continuous, but you can join the end points to the $x$-axis at points close to $0$ and $n$ so that the example does not change in behaviour.