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

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 $T^{*}$.

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

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 $T^{*}$.