An approximate eigenvalue for $ T \in B(X) $.

Problem 1

For this part, we shall assume $ X $ to be a Banach space.

Let $ T \in B(X) $. Then any eigenvalue of $ T $ is clearly an approximate eigenvalue.

Next, assume that $ \lambda \in \mathbb{C} $ is an approximate eigenvalue of $ T $, and let $ (x_{n})_{n \in \mathbb{N}} $ be a sequence of unit vectors of $ X $ such that $$ \lim_{n \to \infty} (T - \lambda I)(x_{n}) = \mathbf{0}_{X}. $$

By way of contradiction, assume that $ T - \lambda I $ is invertible in $ B(X) $. Then \begin{align} \lim_{n \to \infty} x_{n} &= \lim_{n \to \infty} (T - \lambda I)^{-1} \left( (T - \lambda I)(x_{n}) \right) \\ &= (T - \lambda I)^{-1} \left( \lim_{n \to \infty} (T - \lambda I)(x_{n}) \right) \quad (\text{As $ (T - \lambda I)^{-1} $ is continuous.}) \\ &= {(T - \lambda I)^{-1}}(\mathbf{0}_{X}) \\ &= \mathbf{0}_{X}. \end{align} This, however, is impossible because $ \| x_{n} \|_{X} = 1 $ for all $ n \in \mathbb{N} $. The assumption about the invertibility of $ T - \lambda I $ is therefore false, so we conclude that $ \lambda \in {\sigma_{B(X)}}(T) $.

Problem 2

For this part, we shall assume $ X = \mathcal{H} $ to be a Hilbert space.

Let $ \lambda \in {\sigma_{B(\mathcal{H})}}(T) $. If $ \lambda $ is an approximate eigenvalue of $ T $, then we are done; otherwise suppose that $ \lambda $ is not an approximate eigenvalue. Then $ T - \lambda I $ is bounded from below, i.e., there exists a $ c \in \mathbb{R}_{>0} $ such that $$ (\diamondsuit) \quad \forall x \in \mathcal{H}: \quad \| (T - \lambda I)(x) \|_{\mathcal{H}} \geq c \| x \|_{\mathcal{H}}. $$

Claim 1: $ \text{Range} \left( T^{*} - \overline{\lambda} I \right) $ is a dense linear subspace of $ \mathcal{H} $.

Proof: Clearly, $ (\diamondsuit) $ implies that $ \text{Ker}(T - \lambda I) = \{ \mathbf{0}_{\mathcal{H}} \} $, which yields \begin{align} \overline{\text{Range} \left( T^{*} - \overline{\lambda} I \right)} &= \overline{\text{Range}((T - \lambda I)^{*})} \\ &= (\text{Ker}(T - \lambda I))^{\perp} \\ &= (\mathbf{0}_{\mathcal{H}})^{\perp} \\ &= \mathcal{H}. \end{align} As $ \text{Range} \left( T^{*} - \overline{\lambda} I \right) $ is a linear subspace of $ \mathcal{H} $, we are done. $ \quad \spadesuit $

Claim 2: $ \text{Range} \left( T^{*} - \overline{\lambda} I \right) = \mathcal{H} $.

Proof: We shall first prove that $ \text{Range}(T - \lambda I) $ is closed in $ \mathcal{H} $. Let $ (x_{n})_{n \in \mathbb{N}} $ be a sequence in $ \mathcal{H} $ such that $ ((T - \lambda I)(x_{n}))_{n \in \mathbb{N}} $ converges to some $ y \in \mathcal{H} $. We then see by $ (\diamondsuit) $ that $ (x_{n})_{n \in \mathbb{N}} $ is a Cauchy sequence in $ \mathcal{H} $, which must have a limit $ x $ thanks to the completeness of $ \mathcal{H} $. As such, $$ y = \lim_{n \to \infty} (T - \lambda I)(x_{n}) = (T - \lambda I) \left( \lim_{n \to \infty} x_{n} \right) = (T - \lambda I)(x). $$ Therefore, $ y \in \text{Range}(T - \lambda I) $, which proves that $ \text{Range}(T - \lambda I) $ is closed in $ \mathcal{H} $.

Applying the Closed Range Theorem, we find that $ \text{Range} \left( T^{*} - \overline{\lambda} I \right) $ is also closed in $ \mathcal{H} $. By Claim $ 1 $, we therefore conclude that $ \text{Range} \left( T^{*} - \overline{\lambda} I \right) = \mathcal{H} $. $ \quad \spadesuit $

As $ \lambda \in {\sigma_{B(\mathcal{H})}}(T) $, we have $ \overline{\lambda} \in {\sigma_{B(\mathcal{H})}}(T^{*}) $. Then as $ T^{*} - \overline{\lambda} I $ is surjective (by Claim $ 2 $), it follows from the Bounded Inverse Theorem that $ T^{*} - \overline{\lambda} I $ cannot be injective. Therefore, $ \overline{\lambda} $ is an eigenvalue of $ T^{*} $.

Working backwards now, let $ \lambda \in \mathbb{C} $. If $ \lambda $ is an approximate eigenvalue of $ T $, then by Problem $ 1 $, we have $ \lambda \in {\sigma_{B(\mathcal{H})}}(T) $. If $ \overline{\lambda} $ is an eigenvalue of $ T^{*} $, then $ \overline{\lambda} \in {\sigma_{B(\mathcal{H})}}(T^{*}) $, which implies that $ \lambda \in {\sigma_{B(\mathcal{H})}}(T) $.

Conclusion: Let $ \mathcal{H} $ be a Hilbert space and $ T \in B(\mathcal{H}) $. Then $ \lambda \in {\sigma_{B(\mathcal{H})}}(T) $ if and only if $ \lambda $ is an approximate eigenvalue of $ T $ or $ \overline{\lambda} $ is an eigenvalue of $ T^{*} $.