Editing Self-adjoint operator (section)

== Spectrum of self-adjoint operators ==
{{see also|Spectrum (functional analysis)}}
Let <math>A:\operatorname{Dom}(A) \to H</math> be an unbounded operator.{{sfn|Hall|2013|pp=133,177|ps=}} The '''[[resolvent set]]''' (or '''regular set''') of <math>A</math> is defined as
: <math>\rho(A) = \left\{\lambda \in \mathbb{C}\,:\, \exist (A - \lambda I)^{-1}\;\text{bounded and densely defined}\right\}.</math>
If <math>A</math> is bounded, the definition reduces to <math>A - \lambda I</math> being [[bijective]] on <math>H</math>. The '''[[Spectrum (functional analysis)|spectrum]]''' of <math>A</math> is defined as the complement
: <math>\sigma(A) = \Complex \setminus \rho(A).</math> 
In finite dimensions, <math>\sigma(A)\subseteq \mathbb{C}</math> consists exclusively of (complex) [[eigenvalue]]s.{{sfn|de la Madrid Modino|2001|pp=95-97|ps=}} The spectrum of a self-adjoint operator is always real (i.e. <math>\sigma(A)\subseteq \mathbb{R}</math>), though non-self-adjoint operators with real spectrum exist as well.<ref>{{harvnb|Hall|2013}} Section 9.4</ref>{{sfn | Bebiano | da Providência | 2019 | p=}} For bounded ([[Normal_operator|normal]]) operators, however, the spectrum is real ''if and only if'' the operator is self-adjoint.{{sfn | Rudin | 1991 | pp=327|ps=}} This implies, for example, that a non-self-adjoint operator with real spectrum is necessarily unbounded.

As a preliminary, define <math>S=\{x \in \operatorname{Dom}A \mid \Vert x\Vert=1\},</math> <math>\textstyle m=\inf_{x\in S}  \langle Ax,x \rangle</math> and <math>\textstyle M=\sup_{x\in S} \langle Ax,x \rangle</math> with <math>m,M \in \mathbb{R} \cup \{\pm\infty\}</math>. Then, for every <math> \lambda \in \Complex </math> and every <math> x \in \operatorname{Dom}A, </math>
: <math> \Vert (A - \lambda) x\Vert \geq d(\lambda)\cdot \Vert x\Vert,</math>
where <math> \textstyle d(\lambda) = \inf_{r\in [m,M]} |r - \lambda|. </math>

Indeed, let <math>x \in \operatorname{Dom}A \setminus \{0\}.</math> By the [[Cauchy–Schwarz inequality]],
: <math>
\Vert (A - \lambda) x\Vert \geq \frac{|\langle (A - \lambda) x,x\rangle|}{\Vert x\Vert}
=\left|\left\langle A\frac{x}{\Vert x\Vert},\frac{x}{\Vert x\Vert}\right\rangle - \lambda\right| \cdot \Vert x\Vert \geq d(\lambda)\cdot \Vert x\Vert.
</math>

If <math> \lambda \notin [m,M], </math> then <math> d(\lambda) > 0, </math> and <math> A - \lambda I </math> is called ''bounded below''.
{{math theorem|Theorem|Self-adjoint operator has real spectrum}}
{{math proof| Let <math> A </math> be self-adjoint and  denote <math> R_\lambda = A - \lambda I</math> with <math>\lambda \in \Complex.</math>  It suffices to prove that <math>\sigma(A) \subseteq [m,M].</math>
# Let <math>\lambda \in \Complex \setminus [m,M].</math> The goal is to prove the existence and boundedness of <math> R_\lambda^{-1}, </math> and show that <math> \operatorname{Dom} R_\lambda^{-1} = H. </math> We begin by showing that <math>\ker R_\lambda = \{0\}</math> and <math>\operatorname{Im} R_\lambda = H.</math> {{ ordered list | list-style-type = lower-alpha
| As shown above, <math> R_\lambda </math> is bounded below, i.e. <math> \Vert R_\lambda x\Vert \geq d(\lambda)\cdot \Vert x\Vert, </math> with <math> d(\lambda) > 0. </math> The triviality of <math>\ker R_\lambda </math> follows.
| It remains to show that <math> \operatorname{Im}R_\lambda = H.</math> Indeed, {{ ordered list | list-style-type = lower-roman
  | <math> \operatorname{Im}R_\lambda </math> is closed. To prove this, pick a sequence <math> y_n = R_\lambda x_n \in \operatorname{Im}R_\lambda </math> converging to some <math> y \in H.</math> Since <math display="block"> \|x_n - x_m\| \leq \frac{1}{d(\lambda)}\|y_n - y_m\|, </math> <math> x_n </math> is [[Cauchy sequence|fundamental]]. Hence, it converges to some <math> x\in H. </math> Furthermore, <math> y_n + \lambda x_n = Ax_n </math> and <math> y_n + \lambda x_n \to y + \lambda x. </math> The arguments made thus far hold for any symmetric operator. It now follows from self-adjointness that <math> A </math> is closed, so <math> x \in \operatorname{Dom}A = \operatorname{Dom}R_\lambda, </math> <math> Ax = y+\lambda x \in \operatorname{Im}A, </math> and consequently <math> y = R_\lambda x \in \operatorname{Im}R_\lambda. </math>
  | <math> \operatorname{Im}R_\lambda</math> is dense in <math> H. </math> The self-adjointness of <math> A </math> (i.e. <math> A^*=A </math>) implies <math> R_\lambda^* = R_{\bar\lambda}</math> and thus <math> \left(\operatorname{Im}R_\lambda\right)^\perp = \ker R_{\bar{\lambda}}</math>. The subsequent inclusion <math> \bar\lambda \in \Complex \setminus [m,M] </math> implies <math> d(\bar\lambda) > 0 </math> and, consequently, <math> \ker R_{\bar\lambda} = \{0\}. </math>}}
}}
# The operator <math> R_\lambda \colon \operatorname{Dom} A \to H </math> has now been proven to be bijective, so <math> R^{-1}_\lambda </math> exists and is everywhere defined. The graph of <math> R^{-1}_\lambda </math> is the set <math> \{ (R_\lambda x,x) \mid x \in \operatorname{Dom}A \}. </math> Since <math> R_\lambda </math> is closed (because <math> A </math> is), so is <math> R^{-1}_\lambda. </math> By [[closed graph theorem]], <math> R^{-1}_\lambda </math> is bounded, so <math> \lambda \notin \sigma(A). </math>
}}
{{math theorem|Theorem|Symmetric operator with real spectrum is self-adjoint}}
{{math proof|
# <math> A </math> is symmetric; therefore <math> A \subseteq A^*</math> and <math> A - \lambda I \subseteq A^* - \lambda I</math> for every <math> \lambda \in \Complex</math>. Let <math> \sigma(A) \subseteq [m,M]. </math> If <math> \lambda \notin [m,M]</math> then <math> \bar\lambda \notin [m,M] </math> and the operators <math>\{A - \lambda I,A - \bar\lambda I\} : \operatorname{Dom}A \to H </math> are both bijective.
# <math> A - \lambda I =  A^* - \lambda I. </math> Indeed, <math> H = \operatorname{Im}(A - \lambda I) \subseteq \operatorname{Im}(A^* - \lambda I)</math>. That is, if <math> \operatorname{Dom} (A - \lambda I) \subsetneq \operatorname{Dom} (A^* - \lambda I)</math> then <math> A^* - \lambda I </math> would not be injective (i.e. <math> \ker(A^* - \lambda I) \neq \{0\} </math>). But <math> \operatorname{Im}(A - \bar\lambda I)^\perp = \ker(A^* - \lambda I)</math> and, hence, <math> \operatorname{Im}(A - \bar\lambda I) \neq H. </math> This contradicts the bijectiveness.
# The equality <math> A - \lambda I =  A^* - \lambda I </math> shows that <math> A =A^*,</math> i.e. <math> A </math> is self-adjoint. Indeed, it suffices to prove that <math> A^* \subseteq A. </math> For every <math> x \in \operatorname{Dom} A^*</math> and <math> y =A^*x, </math> <math display="block">
A^*x = y \Leftrightarrow (A^*-\lambda I)x = y - \lambda x \Leftrightarrow (A-\lambda I)x = y - \lambda x \Leftrightarrow Ax = y.
</math>
}}