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

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.

You might find this little article usable.

http://www.math.pku.edu.cn/teachers/fanhj/courses/fl3.pdf

• I just took a look through it, but didn't see anything all that relevant. In particular, most of the focus is on operators from an infinite dimensional Hilbert space back to itself. Still, thanks for the resource. – Mike F Aug 15 '13 at 23:48
• Did you see the closabilty criterion after Definition 2.2.7? – ncmathsadist Aug 16 '13 at 0:32
• I'm aware of this criterion. In fact, I had it in mind when I checked $\int$ is not closable above! +1, by the way. I thought I had already voted, but I guess I did not. – Mike F Aug 16 '13 at 0:36
• Actually, hang on I think this criterion is very relevant. Posting something... – Mike F Aug 16 '13 at 1:15

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.

• A nice follow-up question might be "if $\varphi$ is an unbounded (hence not closable) densely-defined linear functional, does it follow that $\ker(\varphi)$ is dense?" – Mike F Aug 16 '13 at 1:43
• Mike, the kernel of a densely defined unbounded functional is always dense. – Martin Argerami Aug 16 '13 at 16:22

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 $\ker\varphi$.

• (+1), thanks Martin. To point out a corollary: it follows that, if $\varphi$ is as in your answer above, then $\overline{ \operatorname{graph}(\varphi)} = X \times \mathbb{C}$. Since $\varphi \neq 0$, it is surjective. Given any $\lambda \in \mathbb{C}$, the plane $\{ x \in \operatorname{dom}( \varphi ): \varphi(x) = \lambda \}$ is just as dense as $\ker(\varphi)$, in fact is a translate of $\ker(\varphi)$. The assertion follows. – Mike F Aug 16 '13 at 17:10