How common is it for a densely-defined linear functional to be closed?

Question

I've always held the vague belief that any densely-defined operator encountered "in nature", if it isn't bounded, is probably at least closable. But, today I noticed the following thing:

Consider the Banach space $C_0(\mathbb{R})$ of continuous, complex-valued function vanishing at $\pm \infty$ in the uniform norm. We have a densely-defined linear functional $\int : C_c(\mathbb{R}) \to \mathbb{C}$ given by Riemann integrating compactly supported functions. This functional is not closable. Indeed, choose $f \in C_c(\mathbb{R})$ with $\int f = 1$ and put $f_n(t) = (1/n) \cdot f(t/n)$. Then $f_n \to 0$ uniformly, but $\int f_n = 1$ for all $n$. Thus $(0,1)$ belongs to the closure of $\operatorname{Graph}(\int) \subset C_0(\mathbb{R}) \times \mathbb{C}$, and $\overline{\int}$ is not single-valued.

This surprised me because integration against an infinite measure is one of the most fundamental examples of an unbounded linear functional. So, if integration is not closed, how ubiquitous could closed linear functionals possibly be?

Question: Can you come up with any good "natural" examples of densely-defined linear functionals on Banach spaces which are closable? Do you have a sense for how important, or common, such examples are? Or are non-closable functionals the rule rather than the exception?

Thanks.

Answer 1

Actually I think I have an answer of sorts. Let $X$ be a Banach space and $\varphi$ a not necessarily bounded linear functional with domain $\operatorname{dom}(\varphi)$ a dense subspace of $X$.

Claim: If $\varphi$ is closed, then $\operatorname{dom}(\varphi) = X$ and $\varphi$ is bounded.

Proof: First we show that $\ker(\varphi)$ is a closed subspace of $X$. Indeed, suppose $x_n \in \ker(\varphi)$ converge to $x \in X$. We have $\lim \varphi(x_n) = \lim 0 = 0$ so, as $\varphi$ is closed, we have $\varphi(x) = 0$ i.e. $x \in \ker(\varphi)$. Now, $\operatorname{dom}(\varphi)$ is the sum of $\ker(\varphi)$ and a $1$-dimensional subspace. It follows that $\operatorname{dom}(\varphi)$ is closed in $X$ (sums of closed subspaces and finite-dimensional subspaces are closed). Since $\varphi$ is densely-definied, we have $\operatorname{dom}(\varphi) = X$. That $\varphi$ is bounded now follows either from the closed graph theorem, or the usual result that functionals with closed kernels are bounded.

Answer 2

I'll just write a short proof of what I said in the comment.

Claim: if $\varphi$ is an unbounded densely-defined linear functional on a Banach space $X$, then $\ker\varphi$ is dense.

Let $\mathcal D$ be the domain of $\varphi$ and let $x\in\mathcal D\setminus\ker\varphi$. We can assume that $\varphi(x)=1$. Since $\varphi$ is unbounded, we can find a sequence $\{x_n\}$ with $\varphi(x_n)=1$ and $x_n\to0$. Then $$ \varphi(x-x_n)=1-1=0, $$ so $\{x-x_n\}\subset\ker\varphi$. And $x-x_n\to x$, so $x$ is in the closure of $\ker\varphi$. As $\mathcal D$ is dense, so is $\ker\varphi$.