Question: Let A be a closed symmetric operator that is semi-bounded from below, which means there is a $\alpha\in\mathbb R$, such that $$(Ax,x)\geq \alpha (x,x)\,,\quad\forall \,x\in D(A)\,.$$ Suppose that $n_+(A)=n_-(A)<+\infty$, where $n_\pm:=dimKer(A^*\mp iI)\,.$ Show that any self-adjoint extension of A is also semi-bounded from below.
This question is an exercise in our class “Functional analysis II”, but I have no idea how to answer it. Is there any hint or solution? Thank you!
Since $n_+(A)=n_-(A)$, we know there is a self-adjoint extension of $A$, say $A_1$. Then, we have $$D(A_1)=D(A)\oplus S,$$ where $S$ is a finite-dimensional linear space.
Suppose that $M$ is lower bound of $A$ and pick $K<M$. Then, we have $$\dim P_{(-\infty,K]}\leq\dim S,$$ where $P_{(-\infty,K]}$ is the projection-valued measure of $A$.
Otherwise, we can find $x\in D(A)\cap R(P_{(-\infty,K]}),$ so that $$(Ax,x)=\int_{\sigma(A)}\lambda \,d\|E_\lambda x\|^2\leq K\|E_{(-\infty,K]} x\|^2<M\|x\|^2,$$ contradicting with $A\geq M,$ which gives us $\dim P_{(-\infty,K]}<\infty,$ this implies that $\sigma(A_1)$ has only finitely many elements in $(-\infty,K],$ and they are eigenvalues. Therefore, $A_1$ is bounded below since there is a $M_1$ such that $\sigma(A_1)\subset [M_1,+\infty).$