Editing Spectrum (functional analysis) (section)

==Spectrum of the adjoint operator==

Let ''X'' be a Banach space and <math>T:\,X\to X</math> a [[unbounded operator#Closed linear operators|closed linear operator]] with dense domain <math>D(T)\subset X</math>.
If ''X*'' is the dual space of ''X'', and <math>T^*:\, X^* \to X^*</math> is the [[hermitian adjoint]] of ''T'', then

:<math>\sigma(T^*) = \overline{\sigma(T)} := \{z\in\Complex : \bar{z}\in\sigma(T)\}.</math>

{{math theorem|For a bounded (or, more generally, closed and densely defined) operator ''T'', 

:<math>\sigma_{\mathrm{cp}}(T) = \overline{\sigma_{\mathrm{p}}(T^*)}</math>. 

In particular, <math>\sigma_{\mathrm{r}}(T) \subset \overline{\sigma_{\mathrm{p}}(T^*)} \subset \sigma_{\mathrm{r}}(T)\cup\sigma_{\mathrm{p}}(T)</math>.}}

{{Math proof|drop=hidden|proof=
Suppose that <math>\mathrm{Ran}(T - \lambda I)</math> is not dense in ''X''. 
By the [[Hahn–Banach theorem]], there exists a non-zero <math>\varphi \in X^*</math> that vanishes on <math>\mathrm{Ran}(T - \lambda I)</math>. 
For all ''x'' ∈ ''X'',

:<math>\langle\varphi, (T - \lambda I) x \rangle = \langle (T^* - \bar\lambda I) \varphi,x \rangle = 0.</math>

Therefore, <math>(T^*-\bar\lambda I)\varphi= 0 \in X^*</math> and <math>\bar\lambda</math> is an eigenvalue of ''T*''. 

Conversely, suppose that <math>\bar\lambda</math> is an eigenvalue of ''T*''. Then there exists a non-zero <math>\varphi \in X^*</math> such that <math>(T^* - \bar{\lambda} I) \varphi = 0</math>, i.e.

:<math>\forall x \in X,\; \langle (T^* - \bar{\lambda} I) \varphi, x \rangle = \langle \varphi,(T - \lambda I) x\rangle = 0.</math>

If <math>\mathrm{Ran}(T-\lambda I)</math> is dense in ''X'', then ''φ'' must be the zero functional, a contradiction. 
The claim is proved. 
}}

We also get  <math>\sigma_{\mathrm{p}}(T)\subset\overline{\sigma_{\mathrm{r}}(T^*)\cup \sigma_{\mathrm{p}}(T^*)}</math> by the following argument: ''X'' embeds isometrically into ''X**''. 
Therefore, for every non-zero element in the kernel of <math>T-\lambda I</math> there exists a non-zero element in ''X**'' which vanishes on <math>\mathrm{Ran}(T^* - \bar{\lambda}I)</math>. 
Thus <math>\mathrm{Ran}(T^* -\bar{\lambda} I)</math> can not be dense.

Furthermore, if ''X'' is reflexive, we have <math>\overline{\sigma_{\mathrm{r}}(T^*)}\subset\sigma_{\mathrm{p}}(T)</math>.