Let $\mathcal{H}$ be a Hilbert space, and let $N\subseteq\mathcal{H}$. I found two interesting statements (without proof):

  1. if a closed subspace $N$ is such that $N^{\perp}=\{0\}$ (which is equivalent to say that $N$ is dense in $\mathcal{H}$), then $N=\mathcal{H}$, and viceversa;
  2. a linear operator $\left(A\,;\mathcal{D}(A)\right)$ on $\mathcal{H}$ is closed if and only if, for every Cauchy sequence $\{\psi_{n}\}$ in $\mathcal{D}(A)$ such that it converges to $\psi\in\mathcal{H}$ and such that $\{A\psi_{n}\}$ converges to $\phi\in\mathcal{H}$, we have $\psi\in\mathcal{D}(A)$ and $\phi=A\psi$.

If these statements are true (and I don't know because there were no demonstrations), then an unbounded operator $A$ on $\mathcal{H}$ defined on a proper dense subspace can not be closed, because a closed operator must have a closed domain (2), and a proper dense subspace can not be closed (1).

Thus $A$ could at most be closable. If so, then there must be an extension $\overline{A}$ of $A$, with $\mathcal{D}(A)\subset\mathcal{D}(\overline{A})$, and $\mathcal{D}(\overline{A})$ must be closed because of (2). Now we have two cases: $\mathcal{D}(\overline{A})$ is the whole Hilbert space, or it is a proper closed subspace.

In the first case the closed graph theorem says that $\overline{A}$ must be bounded, but this is in constrast with the fact that we assumed $A$ to be unbounded. Therefore $\mathcal{D}(\overline{A})$ must be a proper closed subspace. In this case we must have that the proper closed subspace $\mathcal{D}(\overline{A})$ contains, as a proper subspace, the dense subspace $\mathcal{D}(A)$.

My questions are:

  1. is it possible that a proper closed subspace of an Hilbert space contains a dense subspace?
  2. if the answer to the previous question is no, then what is wrong with the construction given above?

Thank You


(1) is easier to asnwer: If $X$ is a metric space and $Y \subset X$ is a dense subset, then $\overline Y = X$. So a proper closed subspace cannot contain a dense subspace.

(2) is just a definition, it does not say that $D(A)$ is closed. It says that $A$ is called closed if and only if the graph $\{(x, Ax): x\in D(A)\}$ is closed. $D(A)$ is Not closed.

Let's have an example. Let $A : D(A) \to L^2(\mathbb R)$ be defined by

$$Af (x) = xf(x),$$

and $D(A) = \{ f\in L^2(\mathbb R): xf \in L^2(\mathbb R)\}$. Note that $D(A)$ is dense in $L^2(\mathbb R)$ as it contains all continuous function with compact support.

CLaim: $A$ is closed. Let $\phi_n \to \phi$ and $x\phi_n \to \psi$ in $L^2(\mathbb R)$. Then by passing to a subsequence if necessary, we can assume the convergence are almost everywhere. Thus we have $x\phi(x) = \psi(x)$ almost everywhere. In particular, we have $\phi \in D(A)$ and $A\phi = \psi$. Thus $A$ is closed.

Now $A$ is densely defined and closed, but $D(A)$ is not closed: the function

$$f = \min \{1, 1/|x|\} \in L^2(\mathbb R)$$

is not in $D(A)$.

(If you consider for example $f_n = \chi_{[-n, n]} f$. Then $f_n \to f$ in $L^2(\mathbb R)$. But $Af_n \in L^2(\mathbb R)$ does not converge)

  • $\begingroup$ Ok for (1). Regarding (2) I have doubts, indeed, it seems to me that the fact that $\{(\psi_{n}\,;A\psi_{n})\}\in\mathcal{G}(A)$ converges to $(\psi\,;\phi)\in\mathcal{G}(A)$ and $\phi=A\psi$, implies, because of the very definition of $\mathcal{G}(A)$, that $\psi\in\mathcal{D}(A)$, and thus $\mathcal{D}(A)$ is closed. $\endgroup$ – Ittiolo Dec 2 '14 at 11:11
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    $\begingroup$ @Ilcapitano: It might happens that $A\psi_n$ does not converges at all! $\endgroup$ – user99914 Dec 2 '14 at 11:18
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    $\begingroup$ $D(A)$ is closed if for all $\psi_n \to \psi$ and $\psi_n \in D(A)$, we have $\phi \in D(A)$. Now if you pick $\psi_n \to \psi$, you do not know what would happen for the sequence $A\psi_n$: If $A\psi_n$ converges, then $\psi \in D(A)$. If $A\psi_n$ does not converges, then $\psi \notin D(A)$. Thus $D(A)$ is not closed. $\endgroup$ – user99914 Dec 2 '14 at 11:25
  • $\begingroup$ @Ilcapitano: I added an example, please have a look. $\endgroup$ – user99914 Dec 2 '14 at 11:34
  • $\begingroup$ Thank You for your patience and disponibility. You have been clear and complete, so your answer deserves to be the "right one". $\endgroup$ – Ittiolo Dec 2 '14 at 11:38

It's not that the domain $\mathcal{D}(A)$ of a closed operator $A$ is closed in $H$, but that the graph $$\mathcal{G}(D) := \{(\xi,A\xi) \in \mathcal{H} \oplus \mathcal{H} \mid \xi \in \mathcal{D}(A)\}$$ is closed in $\mathcal{H} \oplus \mathcal{H}$. Indeed, your second statement, which is just the definition of a closed operator, actually says that if $\{(\psi_n,A \psi_n)\}$ is a Cauchy sequence in $\mathcal{G}(D)$ converging to some $(\psi,\phi) \in \mathcal{H} \oplus \mathcal{H}$, then $(\psi,\phi) \in \mathcal{G}(D)$, i.e., $\phi = A\psi$.

  • $\begingroup$ I do not get it. Indeed, it seems to me that the fact that $\{(\psi_{n}\,;A\psi_{n})\}\in\mathcal{G}(A)$ converges to $(\psi\,;\phi)\in\mathcal{G}(A)$ and $\phi=A\psi$, implies, because of the very definition of $\mathcal{G}(A)$, that $\psi\in\mathcal{D}(A)$, and thus $\mathcal{D}(A)$ is closed. $\endgroup$ – Ittiolo Dec 2 '14 at 11:07
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    $\begingroup$ The relevant situation is that both $\{\psi_n\}$ and $\{A\psi_n\}$ are simultaneously Cauchy sequences, which is a much stronger condition than just $\{\psi_n\}$ being Cauchy. $\endgroup$ – Branimir Ćaćić Dec 2 '14 at 11:20
  • $\begingroup$ I thik I understood. Not every Cauchy sequence $\{\psi_{n}\}$ is such that $\{A\psi_{n}\}$. It is so if $A$ is bounded, but if $A$ is unbounded this in general fails. This means that not all the Cauchy sequences in $\mathcal{D}(A)$ "generate" a Cauchy sequence in $\mathcal{G}(A)$. Am I right? $\endgroup$ – Ittiolo Dec 2 '14 at 11:25
  • $\begingroup$ Exactly. John's updated answer gives an excellent example. $\endgroup$ – Branimir Ćaćić Dec 2 '14 at 11:36
  • $\begingroup$ Thank You for the disponibility. $\endgroup$ – Ittiolo Dec 2 '14 at 11:39

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