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Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$ (possibly unbounded, densely defined with domain $\mathcal{D}(H)$) and let $S$ be a closed subspace of $\mathcal{H}$, correspondingly let $P:\mathcal{H}\to\mathcal{H}$ be the orthogonal projection onto $S$. Note that in general $A$ does not leave $S$ invariant.

The operator $PAP$ on $\mathcal{H}$ is defined on the domain $\mathcal{D}(PHP)=(\mathcal{D}(H)\cap S)\oplus S^\perp$, dense in $\mathcal{H}$: it is zero on the sector $S^\perp$ and ``leaves'' the sector $S$ invariant in the sense that it obviously maps an element of $\mathcal{D}(H)\cap S$ into $S$. The restriction $\hat{A}$ of $PHP$ to $S$ is therefore the densely defined operator on the Hilbert space $S$ with domain $\mathcal{D}(S)=\mathcal{D}(H)\cap S$.

Question 1: Is $\hat{A}$ self-adjoint (as an operator on $S$)? I cannot find a prove or a counter-example.

Question 2: If $A$ is only assumed to be essentially self-adjoint on $\mathcal{H}$, can one say that $\hat{A}$ is essentially self-adjoint on $S$?

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In general there is no reason to assume ${\cal D}(A) \cap S$ contains anything nonzero. But let's assume it's dense. The answer to question 1 is still no.

Example: $\mathcal H = \ell^2$, $(Ax)_i = (-1)^i i^2 x_i$, with domain ${\mathcal D}(A) = \{x: \sum_i i^4 |x_i|^2 < \infty\}$. $S = \{x: x_{2i+1} = \dfrac{2i}{2i+1} x_{2i}\ \text{for all}\ i\}$.
Then ${\mathcal D}(A) \cap S$ contains all $x \in S$ with only finitely many nonzero entries, so it's dense in $S$. But if $x \in {\mathcal D}(A) \cap S$, $$\eqalign{(A x)_{2i} &= (2i)^2 x_{2i}\cr (Ax)_{2i+1} &= - (2i+1)^2 x_{2i+1} = -(2i+1)(2i) x_{2i} = - \dfrac{2i+1}{2i} (Ax)_{2i}}$$ so $Ax \in S^\perp$. Thus $P A P = 0$. On the domain ${\mathcal D}(A) \cap S$ this is not self-adjoint.

EDIT: to support my claim that there is no reason to assume ${\cal D}(A) \cap S$ contains anything nonzero. Consider $\mathcal H = \ell^2$, $(Ax)_i = i x_i$, so $\mathcal D(A) = \{x: \sum_i i^2 |x_i|^2 < \infty\}$, and $S = \{x: x_{2 k} = x_k/2 \ \text{for all}\ k \in \mathbb N\}$. Note that $S$ is closed and infinite-dimensional. But if $x \in S$, $x_{2^j k} = 2^{-j} x_k$ and $$ \sum_{i} i^2 |x_i|^2 \ge \sum_{j=1}^\infty (2^j k)^2 |x_{2^j k}|^2 = \sum_{j=1}^\infty k^2 |x_k|^2 = \infty \ \text{unless}\ x_k = 0$$ so $\mathcal D(A) \cap S = \{0\}$.

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  • $\begingroup$ Thanks a lot for your interest. I miss the main point though: $PAP$ can indeed well be zero on $S$, but isn't the zero-operator self-adjoint? Also, doesn't the fact that $\mathcal{D}(A)$ is dense in $\mathcal{H}$ forces $\mathcal{D}(A)\cap S$ to be dense in $S$? $\endgroup$ – user192047 Nov 12 '14 at 22:08
  • $\begingroup$ The zero operator is self-adjoint on all of $S$, but not on a proper subspace. And no, the intersection of a dense subspace with a closed subspace is not necessarily dense in the closed subspace. I'll edit to provide an example. $\endgroup$ – Robert Israel Nov 12 '14 at 22:40
  • $\begingroup$ Thanks for your so effective counter-example (as well as for T.A.E.'s one). Still, if one indeed assumes (or can prove independently) that $\mathcal{D}(A)\cap S$ is dense in $S$, then your example says that the zero operator is not self-adjoint on the dense domain (proper subspace) $\mathcal{D}(A)\cap S$ of $S$, but its closure, the zero operator on the whole $S$, is. In other words, the restriction $\hat{A}$ is only essentially self-adjoint. Do you think there exist cases in which this failure is more serious, namely the restriction $\hat{A}$ is not even essentially self-adjoint? $\endgroup$ – user192047 Nov 13 '14 at 9:31
  • $\begingroup$ Clever example in your EDIT. $\endgroup$ – DisintegratingByParts Nov 15 '14 at 12:24
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Let $A : \mathcal{D}(A) \subset \mathcal{H}\rightarrow \mathcal{H}$ be densely defined and selfadjoint. Choose any $x \notin\mathcal{D}(A)$ and let $S$ be the span of the single vector $x$. Then $\mathcal{D}(A)\cap S=\{0\}$, which means that the domain of your proposed operator is $\{0\}\oplus S^{\perp}$ which is not dense in $\mathcal{H}$.

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