Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\lambda \in \map \sigma T$ $\lambda$ is an approximate eigenvalue of $T$ or $\overline \lambda$ is an eigenvalue of $T^\ast$. 
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Suppose that $\lambda$ is an approximate eigenvalue of $T$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\HH$ such that:
:$\norm {x_n} = 1$ for each $n \in \N$
and:
:$\norm {T x_n - \lambda x_n} \to 0$ as $n \to \infty$. 
By Characterization of Injective Linear Transformations with Closed Range: Corollary, we either have:
:$T - \lambda I$ is not injective or $\Rng {T - \lambda I}$ is not closed.
If $\Rng {T - \lambda I}$ is not closed then $\Rng {T - \lambda I} \ne \HH$, so $T - \lambda I$ is not surjective in this case.
Either way, $\lambda \in \map \sigma T$. 
Now suppose that $\overline \lambda$ is an eigenvalue of $T^\ast$. 
That is:
:$\map \ker {T^\ast - \overline \lambda I} \ne \set { {\mathbf 0}_\HH}$
From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint, we have:
:$\map \ker {T^\ast - \overline \lambda I} = \paren {\Rng {\paren {T^\ast - \overline \lambda I}^\ast} }^\bot = \Rng {T - \lambda I}^\bot$
from Adjoint is Involutive.
So:
:$\Rng {T - \lambda I}^\bot \ne \set { {\mathbf 0}_\HH}$
From Linear Subspace Dense iff Zero Orthocomplement, we therefore have:
:$\Rng {T - \lambda I}$ is not everywhere dense.
In particular, $\Rng {T - \lambda I} \ne \HH$, so $\lambda \in \map \sigma T$.
=== Necessary Condition ===
Suppose that $\lambda \in \map \sigma T$.
From Spectrum of Adjoint of Bounded Linear Operator, we have $\overline \lambda \in \map \sigma {T^\ast}$. 
Suppose that $\lambda$ is not an approximate eigenvalue of $T$.
Then from Characterization of Injective Linear Transformations with Closed Range: Corollary, $T - \lambda I$ is injective and $\Rng {T - \lambda I}$ is closed.
From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint, we have:
:$\set { {\mathbf 0}_\HH} = \map \ker {T - \lambda I} = \paren {\Rng {T^\ast - \overline \lambda I} }^\bot$
Hence from Linear Subspace Dense iff Zero Orthocomplement, we therefore have:
:$\Rng {T^\ast - \overline \lambda I}$ is everywhere dense.
Since $\Rng {T^\ast - \overline \lambda I}$ is closed, we then have:
:$\Rng {T^\ast - \overline \lambda I} = \HH$
 $\map \ker {T^\ast - \overline \lambda I} = \set { {\mathbf 0}_\HH}$.
In this case, $T^\ast - \overline \lambda I$ is a bijective bounded linear operator.
From the Banach Isomorphism Theorem, we then have that $T^\ast - \overline \lambda I$ is invertible as a bounded linear transformation. 
This contradicts that $\lambda \not \in \map \sigma T$.
Hence we have $\map \ker {T^\ast - \overline \lambda I} \ne \set { {\mathbf 0}_\HH}$.
Hence $\overline \lambda$ is an eigenvalue of $T^\ast$.
Category:Approximate Eigenvalues (Bounded Linear Operators)
Category:Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues