Resolvent and spectrum of an unbounded operator

Let $H$ be a complex Hilbert space and $T$ a linear operator on $H$ with domain $D(T)$. Write $(z-T)x=zx-Tx$ for $x\in D(T)$. The resolvent set of $T$ is $\rho (T):=\{z\in \mathbb{C}:z-T:D(T)\to H\text{ is bijective and }(z-T{)}^{-1}\in \mathcal{B}(H)\},$ where $\mathcal{B}(H)$ is the space of bounded everywhere defined operators on $H$ (A bounded linear operator between normed spaces); the spectrum is $\sigma (T):=\mathbb{C}\setminus \rho (T)$. For $z\in \rho (T)$ the bounded operator $R_T(z):=(z-T{)}^{-1}\in \mathcal{B}(H)$ is the resolvent of $T$ at $z$. This is the library convention used throughout the page, matching the bounded resolvent of Spectrum and resolvent of a bounded operator: the shift is $zI-T$, with coefficient $-1$ on $T$.

The resolvent determines $T$ back. If $z\in \rho (T)$ and $R:=R_T(z)$, then $D(T)=\mathrm{ran}R$ and $TRy=zRy-y$ for $y\in H$; equivalently $T=zI-R^{-1}$ with $D(T)=\mathrm{ran}R$, and $R(z-T)x=x$ for $x\in D(T)$, $(z-T)Ry=y$ for $y\in H$.

Nonempty resolvent set forces closedness, without the closed graph theorem. Assume $\rho (T)\mathrm{\ne }\mathrm{\varnothing }$ and fix $z\in \rho (T)$ with $R:=R_T(z)$. Choose a bound $C\ge 0$ such that $\mathrm{\parallel }Rv\mathrm{\parallel }\le C\mathrm{\parallel }v\mathrm{\parallel }$. The map $F(v,w)=w-Rv$ is continuous, since $\mathrm{\parallel }F(v,w)-F(v^{\mathrm{\prime }},w^{\mathrm{\prime }})\mathrm{\parallel }\le \mathrm{\parallel }w-w^{\mathrm{\prime }}\mathrm{\parallel }+C\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }$. Its zero set $\mathrm{\Gamma }(R)=F^{-1}(\{0\})$ is closed. Because $D(T)=\mathrm{ran}R$ and $(z-T)Rv=v$ for every $v\in H$, the graph of $T$ is $\mathrm{\Gamma }(T)=\{(x,Tx):x\in D(T)\}=\{(Rv,zRv-v):v\in H\}=\mathrm{\Phi }(\mathrm{\Gamma }(R)),$ where $\mathrm{\Phi }(v,w):=(w,zw-v)$ is a continuous linear bijection of $H\oplus H$ with continuous inverse $\mathrm{\Psi }(u,v)=(zu-v,u)$, since $\mathrm{\Phi }(\mathrm{\Psi }(u,v))=(u,zu-(zu-v))=(u,v)$ and $\mathrm{\Psi }(\mathrm{\Phi }(v,w))=\mathrm{\Psi }(w,zw-v)=(v,w)$. A homeomorphism carries closed sets to closed sets, so $\mathrm{\Gamma }(T)$ is closed and $T$ is closed. No closed graph theorem and hence no choice principle is used here.

For the zero operator on a nonzero $H$, $\rho (0)=\mathbb{C}\setminus \{0\}$: the nonzero shifts have inverse $z^{-1}I$, whereas the zero shift is not bijective. If $H=\{0\}$, every shift is the unique bijection of the zero space, so $\rho (T)=\mathbb{C}$ and $\sigma (T)=\mathrm{\varnothing }$.