# Projecting self-adjoint operator onto closed subspace

Let $H$ be a complex Hilbert space and let $(Q, D(Q))$ be a closed, densely defined, positive semidefinite, Hermitian quadratic form on $H$.

(That is, $D(Q)$ is a dense subspace of $H$, $Q$ maps $D(Q) \times D(Q)$ to $\mathbb{C}$, $Q$ is linear in its first variable and conjugate-linear in its second, $Q(x,y) = \overline{Q(y,x)}$, $Q(x,x) \ge 0$, and $D(Q)$ is a Hilbert space under the inner product $\langle x,y \rangle + Q(x,y)$.)

Let $(A, D(A))$ be the generator of $Q$, i.e. the unique self-adjoint operator such that $D(A) \subset D(Q)$ and for all $x \in D(A)$, $y \in D(Q)$, we have $Q(x,y) = \langle Ax, y \rangle$.

Suppose now that $K$ is a closed subspace of $H$, and that $D(Q) \cap K$ is dense in $K$. Then the restriction $(Q, D(Q) \cap K)$ is a densely defined, positive semidefinite, Hermitian form on $K$, and it is not hard to check that it is also closed. Let $(B, D(B))$ be its generator (a self-adjoint operator on $K$).

Question. Is it true that

1. $D(B) = D(A) \cap K$; and

2. $B$ is the restriction to $D(B)$ of $P_K A$, where $P_K$ is the orthogonal projection onto $K$.

The main difficulties seem to be that $P_K$ may not map $D(Q)$ into $D(Q)$, and that $A$ may not map $D(A) \cap K$ into $K$. But I also cannot come up with a counterexample.

I have a specific example in mind, but I am trying to figure out whether this is true on general principles, or requires more specific knowledge.

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Assume there is a non-zero $v \in \mathcal{D}(Q)\setminus \mathcal{D}(A)$, and let $K$ be the one-dimensional subspace spanned by $v$. Then $K\cap \mathcal{D}(Q)=K$, and $K$ is closed in $H$. The restriction of the form $Q$ to $K$ is finite-dimensional, and positive-semidefinite. $B$ is trivially defined, but $\mathcal{D}(B)\ne\mathcal{D}(A)\cap K=\{0\}$.
The above works so long as $\mathcal{D}(Q)\ne \mathcal{D}(A)$. If $\mathcal{D}(Q)=\mathcal{D}(A)$, then $A$ is less interesting.
Nice and simple, thanks. Of course if $D(Q) = D(A)$ then $A$ is bounded and everything is trivial. There may be further updates on this question (or new questions) as I figure out more relevant information about my specific situation. –  Nate Eldredge Jul 6 '14 at 4:11