proof of basic criterion for selfadjointness

1.
$(1\u27f92)$ If $A$ is selfadjoint and $Ax=ix$, then
$$i{\parallel x\parallel}^{2}=(ix,x)=(Ax,x)=(x,{A}^{*}x)=(x,Ax)=(x,ix)=\overline{(ix,x)}=i{\parallel x\parallel}^{2},$$ so $x=0$. Similarly we prove that $Ax=ix$ implies $x=0$. That $A$ is closed follows from the fact that the adjoint of an operator is always closed.

2.
$(2\u27f93)$ If $2$ holds, then $\{0\}=\mathrm{Ker}{({A}^{*}\pm i)}^{*}=\mathrm{Ker}{(A\mp i)}^{*}=\mathrm{Ran}{(A\mp i)}^{\u27c2}$, so that $\mathrm{Ran}A\mp i$ is dense in $\mathscr{H}$. Also, since $A$ is symmetric, for $x\in D(A)$,
$${\parallel (A+i)x\parallel}^{2}={\parallel Ax\parallel}^{2}+{\parallel x\parallel}^{2}+(Ax,ix)+(ix,Ax)={\parallel Ax\parallel}^{2}+{\parallel x\parallel}^{2}$$ because $(Ax,ix)=(x,i{A}^{*}x)=(x,iAx)=(ix,Ax)$. Hence $\parallel x\parallel \le \parallel (A+i)x\parallel $, so that given a sequence ${x}_{n}\in D(A)$ such that $(A+i){x}_{n}\to y$, we have that $\{(A+i){x}_{n}\}$ is a Cauchy sequence^{} and thus $\{{x}_{n}\}$ itself is a Cauchy sequence. Hence $\{{x}_{n}\}$ converges to some $x\in \mathscr{H}$ and since $A$ is closed it follows that $x\in D(A)$ and $(A+i)x=y$. This proves that $y\in \mathrm{Ran}(A+i)$, so that $\mathrm{Ran}(A+i)$ is closed (and similarly, $\mathrm{Ran}(Ai)$ is closed. Thus $\mathrm{Ran}(A\pm i)=\mathscr{H}$.

3.
$(3\u27f91)$ Suppose $3$. If $y\in D({A}^{*})$, then there is $x\in D(A)$ such that $(A+i)x=({A}^{*}i)y$. Since $A$ is symmetric, $(A+i)x=({A}^{*}+i)x={(Ai)}^{*}x$, so that $({A}^{*}i)(xy)=0$. But since $\mathrm{Ker}({A}^{*}i)=\mathrm{Ran}{(A+i)}^{\u27c2}=\{0\}$, it follows that $x=y$, so that $y\in D(A)$. Hence $D(A)=D({A}^{*})$, and therefore $A$ is selfadjoint.
Title  proof of basic criterion for selfadjointness 

Canonical name  ProofOfBasicCriterionForSelfadjointness 
Date of creation  20130322 14:53:05 
Last modified on  20130322 14:53:05 
Owner  Koro (127) 
Last modified by  Koro (127) 
Numerical id  5 
Author  Koro (127) 
Entry type  Proof 
Classification  msc 47B25 