# proof of basic criterion for self-adjointness

1. 1.

$(1\implies 2)$ If $A$ is self-adjoint and $Ax=ix$, then

 $i\|x\|^{2}=(ix,x)=(Ax,x)=(x,A^{*}x)=(x,Ax)=(x,ix)=\overline{(ix,x)}=-i\|x\|^{2},$

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.

$(2\implies 3)$ If $2$ holds, then $\{0\}=\operatorname{Ker}(A^{*}\pm i)^{*}=\operatorname{Ker}(A\mp i)^{*}=% \operatorname{Ran}(A\mp i)^{\perp}$, so that $\operatorname{Ran}{A\mp i}$ is dense in $\mathscr{H}$. Also, since $A$ is symmetric, for $x\in D(A)$,

 $\|(A+i)x\|^{2}=\|Ax\|^{2}+\|x\|^{2}+(Ax,ix)+(ix,Ax)=\|Ax\|^{2}+\|x\|^{2}$

because $(Ax,ix)=(x,iA^{*}x)=(x,iAx)=-(ix,Ax)$. Hence $\|x\|\leq\|(A+i)x\|$, 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 $y\in\operatorname{Ran}(A+i)$, so that $\operatorname{Ran}(A+i)$ is closed (and similarly, $\operatorname{Ran}(A-i)$ is closed. Thus $\operatorname{Ran}(A\pm i)=\mathscr{H}$.

3. 3.

$(3\implies 1)$ 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, $(A+i)x=(A^{*}+i)x=(A-i)^{*}x$, so that $(A^{*}-i)(x-y)=0$. But since $\operatorname{Ker}(A^{*}-i)=\operatorname{Ran}(A+i)^{\perp}=\{0\}$, it follows that $x=y$, so that $y\in D(A)$. Hence $D(A)=D(A^{*})$, and therefore $A$ is self-adjoint.

Title proof of basic criterion for self-adjointness ProofOfBasicCriterionForSelfadjointness 2013-03-22 14:53:05 2013-03-22 14:53:05 Koro (127) Koro (127) 5 Koro (127) Proof msc 47B25