Selfadjoint Operator: Basic Criterion

Question

For symmetric operators one has: $$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$ How to prove this in an unveiling way?

Answer 1

Since $A$ is symmetric⁽¹⁾, we have

$$A \pm \imath \subset A^\ast \pm \imath = (A\mp\imath)^\ast.\tag{1}$$

Since $\mathcal{R}(A\pm\imath) = \mathcal{H}$ it follows that $\mathcal{R}(A^\ast \pm \imath) = \mathcal{H}$. Since $A^{\ast\ast}\subset A^\ast$, the operators $A^\ast \pm \imath$ are injective, thus bijective. But a proper suboperator of a bijective operator cannot be surjective, hence the inclusion in $(1)$ cannot be proper and in fact $A\pm\imath = A^\ast \pm \imath$, whence $A = A^\ast$.

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⁽¹⁾ If the definition of symmetry is $\langle Ax,y\rangle = \langle x, Ay\rangle$ for all $x,y\in \mathcal{D}(A)$ and not $A\subset A^\ast$, the existence of the adjoint, equivalently the denseness of $\mathcal{D}(A)$, needs to be deduced from $\mathcal{R}(A+\imath) = \mathcal{H}$ as pointed out and done by TAE. For that, one picks $x\in \mathcal{D}(A)^\perp$ and by the surjectivity, one can write $x = (A+\imath)y$ for a $y\in\mathcal{D}(A)$. Then

$$0 = \langle x,y\rangle = \langle (A+\imath)y,y\rangle = \langle Ay,y\rangle + i\lVert y\rVert^2$$

implies $y = 0$ and thus $x = 0$ since $\langle Ay,y\rangle \in\mathbb{R}$. So $\mathcal{D}(A)$ is dense, $A^\ast$ exists, and $A\subset A^\ast$.

Answer 2

Actually, you can go one step further. If $A$ is symmetric with $\mathcal{R}(A-iI)=X$, then $A$ has a dense domain. And $A=A^{\star}$ on that domain if $\mathcal{R}(A+iI)=X$ also holds.

To see that $A$ has a dense domain here, suppose that $y \perp\mathcal{D}(A)$ and write $y=(A-iI)x$. Then $$ 0 = (y,x)=((A-iI)x,x)=(Ax,x)-i(x,x). $$ Both the real and imaginary parts must be $0$, which forces $x=0$. So $\mathcal{D}(A)^{\perp}=\{0\}$, or $\mathcal{D}(A)^{c}=X$.

Next, to show that $A=A^{\star}$, suppose that $y \in \mathcal{D}(A^{\star})$. Write $(A^{\star}+iI)y=(A+iI)z$ for some $z\in\mathcal{D}(A)$ (this can be done because $A+iI$ is surjective.) Then $$ ((A-iI)x,y)=(x,(A^{\star}+iI)y)=(x,(A+iI)z)=((A-iI)x,z). $$ Because $\mathcal{R}(A-iI)=X$, then it follows that $y=z \in \mathcal{D}(A)$; in other words, $\mathcal{D}(A^{\star})\subseteq \mathcal{D}(A)$. The opposite inclusion $\mathcal{D}(A)\subseteq \mathcal{D}(A^{\star})$ follows from the symmetry of $A$. So $\mathcal{D}(A)=\mathcal{D}(A^{\star})$, which is enough to prove that $A=A^{\star}$.

Answer 3

Since $A\subseteq A^*$ (because $A$ is symmetric), it's enough to prove that $D(A^*)\subseteq D(A)$.

Take $y\in D(A^*)=D(A^*\pm\mathrm i)$.

As $(A^*\pm\mathrm i)y\in\mathcal R(A^*\pm\mathrm i)\subseteq \mathcal{H}=\mathcal R(A\pm \mathrm i)$, there exists $x\in D(A\pm\mathrm i)$ such that $$(A^*\pm\mathrm i)y=(A\pm\mathrm i)x=(A^*\pm\mathrm i)x$$ and thus $$y-x\in \ker(A^*\pm \mathrm i)=\mathcal R(A\mp\mathrm i)^{\perp}=\mathcal{H}^\perp=\{0\}.$$ So, $y=x\in D(A\pm\mathrm i)=D(A)$.