Intereting Posts

Is there any diffeomorphism from A to B that $f(A)=B$?
Can a free group over a set be constructed this way (without equivalence classes of words)?
Fourier series for $\sin^2(x)$
Checking if the Method of Moments and Maximum Likelihood Estimators are biased and/or consistent
Expected number of coin tosses to land n heads
Sum over the powers of the roots of unity $\sum \omega_j^k$
If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$
Non-finitely generated, non-projective flat module, over a polynomial ring
number of subsets of even and odd
Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$
Determinant identity: $\det M \det N = \det M_{ii} \det M_{jj} – \det M_{ij}\det M_{ji}$
The homology groups of $T^2$ by Mayer-Vietoris
If gcd$(a,b) = 1$, then I want to prove that $\forall c \in \mathbb{Z}$, $ax + by = c$ has a solution in integers $x$ and $y$.
Difference between Gentzen and Hilbert Calculi
Proof that the series expansion for exp(1) is a Cauchy sequence

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?

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- find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges
- Examples for proof of geometric vs. algebraic multiplicity

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.

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You might find this little article usable.

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.

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 $\ker\varphi$.

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