Remark

Let E be a complex Banach space and let $u$ be a closed operator with dense domain on E.

1) If $u\in \mathscr{L}$ (E), its spectrum coincides with the spectrum of the element $u$ of the algebra $\mathscr{L}(E)$(I, p. 2, Definition 1).

2) Let D be the domain of $u$. For every $\lambda \in \mathbf{C}-$ Sp($u$), one has

Moreover, for $\lambda_1$ and $\lambda_2$ in the resolvent set of $u$, one has

In fact, one has

Since $1_E= (\lambda_21_E-u)\circ R(u, \lambda_2)$ and $1_D= R(u, \lambda_1)\circ (\lambda_11_E-u)$, one obtains

whence the first formula. By interchanging the roles of $\lambda_1$ and $\lambda_2$, one deduces the second formula.

3) Let $\lambda \in \mathbf{C}$. By the closed graph theorem (EVT, I, p. 19, Corollary 5), if the linear mapping of dom($u$) into E defined by $x\mapsto (\lambda 1_E-u)(x)$ is bijective, then its inverse, whose graph is closed, is continuous from E into E. Therefore $\lambda$ belongs to the resolvent set of $u$ if and only if the mapping $x\mapsto (\lambda 1_E-u)(x)$ of dom($u$) into E is bijective.

4) If $\lambda$ belongs to the spectrum of $u$, one of the following properties holds:

(i) The kernel of $\lambda 1_E-u$ is not reduced to 0; one then says that $\lambda$ is an eigenvalue of $u$, and that the dimension of Ker($\lambda 1_E-u$) is its multiplicity;

(ii) The partial operator $\lambda 1_E-u$ is injective and its image is not dense in E; one then says that $\lambda$ belongs to the residual spectrum of $u$;

(iii) The partial operator $\lambda 1_E-u$ is injective, its image is dense in E, but $\lambda 1_E-u$ is not surjective; one then says that $\lambda$ belongs to the continuous spectrum of $u$.

5) Let $\lambda$ be a complex number belonging to the resolvent set of $u$. The resolvent $R(u, \lambda )$ is an injective linear mapping of E into E; its image is the domain of $u$ and $u=\lambda 1_E-R(u, \lambda )^{-1}($cf. IV, p. 226). Conversely, this property characterizes the resolvent set and the resolvent. Precisely, let $\lambda \in \mathbf{C}$; if there exists a continuous injective linear mapping $w$ of E into E such that $u=\lambda 1_E-w^{-1}$, then $\lambda$ belongs to the resolvent set of $u$ and $w= R(u, \lambda )$.

In particular, if $v$ is a closed operator with dense domain on E, and if $\lambda \in \mathbf{C}$ is a complex number not belonging to Sp($u$)$\cup$ Sp($v$), then the equality $R(u, \lambda ) = R(v, \lambda )$ implies that $u=v$.

6) The spectrum of a closable partial operator $u$ is defined as the spectrum of its closure.

There exist closed operators whose spectrum is empty, or whose spectrum is equal to $\mathbf{C}$ (exercise 12 of IV, p. 347).

Let E be a complex Banach space and $u$ a closed operator with dense domain on E. If F is a complex Banach space and if $v: E\rightarrow F$ is an isomorphism, then one has Sp($v\circ u\circ v^{-1}$) $=$ Sp($u$) and $R(v\circ u\circ v^{-1}, \lambda ) =v\circ R(u, \lambda )\circ v^{-1}$ for every $\lambda \notin$ Sp($u$).