Resolvent and spectrum of a closed operator on a Banach space

Let $X$ be a Banach space over $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ (Banach space, Real and complex scalar conventions for normed spaces) and let $A:D(A)\subseteq X\to X$ be a closed linear operator: $D(A)$ is a linear subspace and $\mathrm{\Gamma }(A)=\{(x,Ax):x\in D(A)\}$ is closed in $X\times X$ with norm $\mathrm{\parallel }(x,y)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$. This extends the Hilbert-space vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores to Banach spaces. A scalar $\lambda \in \mathbb{K}$ belongs to the resolvent set $\rho (A)$ if $\lambda I-A:D(A)\to X$ is bijective and its inverse is bounded on $X$. Under Dependent Choice, boundedness of the inverse follows from bijectivity by Closed graph theorem; thus under DC this is equivalent to the bijectivity-only convention. For $\lambda \in \rho (A)$ the bounded operator $R(\lambda ,A):=(\lambda I-A{)}^{-1}\in \mathcal{B}(X)$ is the resolvent, and $\sigma (A):=\mathbb{K}\setminus \rho (A)$ is the spectrum. One has $R(\lambda ,A)X=D(A)$, $R(\lambda ,A)(\lambda I-A)y=y$ for $y\in D(A)$, and $\lambda R(\lambda ,A)x-x=AR(\lambda ,A)x$ for $x\in X$; in particular $AR(\lambda ,A)=\lambda R(\lambda ,A)-I\in \mathcal{B}(X)$. This is the Banach-space form of the Hilbert-space vocabulary Resolvent and spectrum of an unbounded operator; the shift convention $\lambda I-A$ is the same, with the closed graph theorem replacing the Hilbert-space bounded-inverse convention.

Boundedness under DC. Assume Dependent Choice. If $\lambda \in \mathbb{K}$ and $\lambda I-A$ is bijective, then $\lambda I-A$ is a closed operator: its graph is the image of the graph $\mathrm{\Gamma }(A)\subseteq X\oplus X$ (Unbounded linear operators: domain, graph and extension) under the homeomorphism $(y,z)\mapsto (y,\lambda y-z)$ of $X\oplus X$ with inverse $(u,v)\mapsto (u,\lambda u-v)$. Hence $(\lambda I-A{)}^{-1}:X\to X$ has closed graph and is everywhere defined, so the closed graph theorem (Closed graph theorem, which assumes DC) makes it bounded; this is the one place where an axiom beyond ZF enters, and it is the DC carried by the cited theorem. The Banach-space vocabulary is Banach space with the scalar convention of Real and complex scalar conventions for normed spaces; boundedness and the space $\mathcal{B}(X)$ are those of A bounded linear operator between normed spaces.

Elementary identities. Let $\lambda \in \rho (A)$ and $R:=R(\lambda ,A)$. Since $R$ is the inverse of the bijection $\lambda I-A:D(A)\to X$, its range is $D(A)$: $RX=D(A)$. It satisfies $R(\lambda I-A)y=y(y\in D(A)),(\lambda I-A)Rx=x(x\in X)\mathrm{.}$ The first identity says $R(\lambda ,A)(\lambda I-A)=I_{D(A)}$ and the second says $(\lambda I-A)R(\lambda ,A)=I_X$. Reading the second identity as $\lambda Rx-ARx=x$ and rearranging gives $\lambda R(\lambda ,A)x-x=AR(\lambda ,A)x(x\in X),$ that is $AR(\lambda ,A)=\lambda R(\lambda ,A)-I$ as everywhere-defined operators $X\to X$; in particular $AR(\lambda ,A)\in \mathcal{B}(X)$ even though $A$ itself need not be bounded. The same identities hold for every $\lambda \in \rho (A)$, and $\sigma (A):=\mathbb{K}\setminus \rho (A)$ collects the scalars for which $\lambda I-A$ fails to be bijective or has an unbounded inverse. Under DC, the latter possibility is excluded by the closed graph argument above. When $X$ is a Hilbert space this is the Banach-space form of Resolvent and spectrum of an unbounded operator, with the same shift convention $\lambda I-A$; the closed graph theorem replaces the Hilbert-space convention that the resolvent is bounded by definition.