Dense subspaces, closed subspaces and unbounded operators in Hilbert spaces Let $\mathcal{H}$ be a Hilbert space, and let $N\subseteq\mathcal{H}$.
I found two interesting statements (without proof):


*

*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;

*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:


*

*is it possible that a proper closed subspace of an Hilbert space contains a dense subspace?

*if the answer to the previous question is no, then what is wrong with the construction given above?


Thank You
 A: (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)
A: 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., $\phi = A\psi$.
