proof of basic criterion for self-adjointness

1. $(1⟹2)$ If $A$ is self-adjoint 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. $(2⟹3)$ If $2$ holds, then $\{0\}=\mathrm{Ker}{(A^{*}\pm i)}^{*}=\mathrm{Ker}{(A\mp i)}^{*}=\mathrm{Ran}{(A\mp i)}^{⟂}$, 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, $\mathrm{Ran}(A-i)$ is closed. Thus $\mathrm{Ran}(A\pm i)=\mathscr{H}$.

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