Example of a linear operator whose graph is not closed

Question

I want an example of a linear operator $T:X\to Y$ such that graph of $T$ is not closed.

My thoughts: $T$ must be unbounded. Again by closed graph theorem any unbounded linear map from a Banach space $X$ to another Banach space $Y$ should have non-closed graph. But it is not possible to define any unbounded linear map from a Banach space $X$ to another Banach space $Y$ explicitly without axiom of choice. Thus to get such an example at least one of $X$ and $Y$ must be incomplete. Could anyone please help me?

Answer 1

If $T$ is everywhere defined and closed, then it is continuous. See this.

If you mean densely-defined $T$, then it's not so hard. The (strong) derivative operator $d/dx$ defined on (a dense subset of) $L^2(-1,1)$ is closable (its closure is the weak derivative), but not closed. To see that it is not closed, Let $u_n(x)$ be a sequence of functions converging to $|x|$ suitably strongly, with $v_n(x) = u_n'(x)$ converging to the weak derivative of $|x|$, which is $x/|x| = sign(x)$. Then because $|x|$ is not in the domain of $T$, this thing is not closed.

The formula $u_n(x) = \sqrt{x^2 + n^{-1}}$ fits the bill.

Answer 2

On $L^2(R)$, consider the densely defined operator $u\mapsto \int u\,dx$, defined on $L^2\cap L^1$. This operator is neither closed nor closable. If you want the operator defined on all of X, with X incomplete, just take X to be $L^2\cap L^1$ with the $L^2$ norm.