14
$\begingroup$

This is a problem from Conway’s Functional Analysis:

Definition An approximate eigenvalue for $ T \in B(X) $ is a scalar $ \lambda $ such that there is a sequence of unit vectors $ x_{n} \in X $ such that $ T(x_{n}) - \lambda x_{n} \rightarrow 0 $.

  1. Show that any eigenvalue for $ T $ is an approximate eigenvalue for $ T $, and that any approximate eigenvalue for $ T $ lies in the spectrum of $ T $, which we denote by $ \sigma(T) $.

  2. If $ X $ is a Hilbert space, show that $ \lambda \in \sigma(T) $ if and only if either $ \lambda $ is an approximate eigenvalue for $ T $ or $ \overline{\lambda} $ is an eigenvalue for $ T^{*} $.

$\endgroup$
  • 1
    $\begingroup$ Is this homework? What part of it are you having trouble with? $\endgroup$ – Robert Israel Mar 3 '13 at 20:51
  • 1
    $\begingroup$ What are your thoughts? Eigenvalue implies approximate eigenvalue is trivial, right? What about the other assumption of 1)? $\endgroup$ – Julien Mar 3 '13 at 20:51
9
$\begingroup$

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^{*} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.