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Let $A$ be a symmetric operator satisfying $\langle \phi,A\phi\rangle\geq C\lVert \phi\rVert^{2}$ for all $\phi\in \mathcal{D}(A)$ and some $C\in \mathbb{R}$. Show that the deficiency indiecs are equal, i.e., $d_{+}(A)=d_{-}(A)$, and therefore $A$ has a self-adjoint extension.

So the idea I have is to somehow show that from the condition that $A$ is closed but I can't seem to show it, then again I don't even know if I can conclude that $A$ is closed from that condition.

Since $A$ is symmetric I know that $\mathcal{D}(A)$ is dense so then we can define the adjoint $A^*$. Now $\mathcal{D}(A^{*})=\{h\in \mathcal{H}:~\exists~ \eta~ st ~\forall ~\phi\in \mathcal{D}(A) ~ we ~have ~\langle A\phi,h\rangle =\langle \phi,\eta\rangle~ \}.$ from this we see that since $A$ is symmetric that $\mathcal{D}(A)\subset \mathcal{D}(A^{*})$ so we know that $A$ is closeable since $\mathcal{D}(A^{*})$ is dense.

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Take $\psi \in Ran(A+i)^{\perp}$, then $\langle A \psi, \psi \rangle = -i \langle \psi, \psi \rangle \geq C \|\psi\|^2$. So $\psi = 0$. This makes one deficiency index $0$. Same applies to the other one. –  Michael Nov 25 '13 at 2:32
    
Thanks for that but why is it that $\psi\in\mathcal{D}(A)$ for all $\psi\in Ran(A+i)^{\perp}$? –  lance wellton Nov 25 '13 at 3:34
    
I assumed $A$ is densely defined, this is not true? –  Michael Nov 25 '13 at 6:40
    
Yeas $A$ is densely defined but can't it happen that for some $\psi\in R(A+i)^{\perp}$ we have that $\psi\notin\mathcal{D}(A)$? But it looks like you can just take limits from in $\mathcal{D}(A)$ and you still get the same result. –  lance wellton Nov 25 '13 at 18:33

1 Answer 1

Let $\rho_A$ be the set of regular points of $A,$ i.e. $$\rho_A=\{\lambda\in\mathbb C\mid\forall\varphi\in D(A): ||(A-\lambda)\varphi||\geq c||\varphi||\ \mbox{for some}\ c=c_\lambda>0\}.$$

For every $\lambda\in\rho_A$ there is a defect index defined by $d_A(\lambda)=\dim Ran(A-\lambda)^\perp.$

Theorem (see e.g. Birman, Solomyak "Spectral Theory", 3.7., Theorem 4):

The set $\rho_A\subseteq\mathbb C$ is open, and $d_A(\lambda)$ is constant on every connected component.

Since $A$ is symmetric, $\rho_A$ contains upper and lower (open) half-planes $\mathbb C_+,\mathbb C_-.$ The semiboundedness implies that $\rho_A$ contains a point from $\mathbb R.$ Hence $\pm i$ belong to the same connected component of $\rho_A$ and $d_+(A)=d_A(i)=d_A(-i)=d_-(A).$

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