Range criterion for self-adjointness

Preparatory identity. Let $S$ be densely defined and symmetric, let $z=a+ib$ with $b\mathrm{\ne }0$ and let $x\in D(S)$. Expanding $\mathrm{\parallel }(S-z)x{\mathrm{\parallel }}^2$ and using that $⟨Sx,x⟩$ is real by [A1] gives, as in the proof of Resolvent of a self-adjoint operator: nonreal resolvents and the estimate, $\mathrm{\parallel }(S-z)x{\mathrm{\parallel }}^2=\mathrm{\parallel }(S-a)x{\mathrm{\parallel }}^2+b^2\mathrm{\parallel }x{\mathrm{\parallel }}^2\ge b^2\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{.}$

(1) implies (2): if $T=T^{\ast }$ then $T$ is closed by [A1]. If $z$ is nonreal and $v\in \ker (T^{\ast }-z)$, then $v\in D(T^{\ast })=D(T)$ and $z\mathrm{\parallel }v{\mathrm{\parallel }}^2=⟨Tv,v⟩=\bar{z}\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by [A1], so $v=0$.

(4) implies (5) is immediate, since $\pm i$ are nonreal.

(5) implies (1): let $v\in D(T^{\ast })$. Since $\mathrm{ran}(T-i)=H$, choose $w\in D(T)$ with $(T-i)w=(T^{\ast }-i)v$. Then $(T^{\ast }-i)(v-w)=0$, because $T^{\ast }w=Tw$ for $w\in D(T)$; and by [A2] with $z=-i$ the kernel of $T^{\ast }-i$ is $\mathrm{ran}(T+i{)}^{\perp }=\{0\}$. Hence $v=w\in D(T)$. So $D(T^{\ast })\subseteq D(T)$, and with $T\subseteq T^{\ast }$ this gives $T=T^{\ast }$.

(1) implies (6) by [A4]. Conversely (6) implies (4): if $\sigma (T)\subseteq \mathbb{R}$ then every nonreal $z$ lies in $\rho (T)$, so $z-T$ is surjective and $\mathrm{ran}(T-z)=H$.

Closed range for closed symmetric $S$. If in addition $S$ is closed, then $\mathrm{ran}(S-z)$ is closed for nonreal $z$: given $(S-z)x_n\to y$, step 1.1 makes $(x_n)$ Cauchy with limit $x$, so $S{x}_n\to zx+y$, and closedness of $S$ gives $x\in D(S)$ with $Sx=zx+y$, that is $y=(S-z)x$.

(5) implies (3): by 1.1 with $S=T$ and $z=-i$ the map $T+i$ satisfies $\mathrm{\parallel }(T+i)x\mathrm{\parallel }\ge \mathrm{\parallel }x\mathrm{\parallel }$, so its inverse on its range is bounded by $1$; since the range is $H$ by (5), $-i\in \rho (T)$ because $-i-T=-(T+i)$, and $T$ is closed by [A5]. The two kernels vanish by [A2] and (5).

(2) implies (4): for nonreal $z$, [A2] and (2) give $\mathrm{ran}(T-z{)}^{\perp }=\ker (T^{\ast }-\bar{z})=\{0\}$, while (2) and step 2.1 show $\mathrm{ran}(T-z)$ is closed; hence the range is all of $H$ by [A3].

(3) implies (5): by [A2], $\mathrm{ran}(T\mp i{)}^{\perp }=\ker (T^{\ast }\pm i)=\{0\}$, and by step 2.1 applied to the closed $T$ with $z=\mp i$ the two ranges are closed; hence they equal $H$ by [A3]. So (3) and (5) are equivalent.

Collecting: $(1)\Rightarrow (2)\Rightarrow (4)\Rightarrow (5)\Rightarrow (1)$ by steps 1.2, 3.1, 1.3 and 1.4; $(5)\Rightarrow (3)$ and $(3)\Rightarrow (5)$ by steps 2.2 and 3.2; and $(1)\Rightarrow (6)\Rightarrow (4)$ by step 1.5. Thus all six statements are equivalent. The final clause follows because only nonreality of the parameters was used, so $i\lambda$ with $\lambda >0$ may replace $i$.