Problem 6

Exercise 3.6 (Spectrum is compact)   Let $\sigma(A)\subseteq \mathbb{C}$ be the spectrum of a bounded linear operator $A : X\to X$. Then $\sigma(A)$ is compact.

Proof. To prove $\sigma(A)$ is bounded, recall the following sufficient condition for the convergence of a Neumann series which explicitly reconstructs the inverse

$$\Vert T\Vert < 1 \implies (I-T)^{-1} = \sum_{k=0}^\infty T^k$$ (3.42)

Therefore, if $I-T$ is not invertible, then $\Vert T\Vert\geq 1$. Recall the definition of $\sigma(A)$

$$\sigma(A) = \{ \lambda \in \mathbb{C} \;\vert\; A - \lambda I \}$$ (3.43)

If $A - \lambda I$ is not invertible, certainly $I-A/\lambda$ is not invertible, so that $\Vert A/\lambda\Vert\geq 1$. This implies $\vert\lambda\vert\leq \Vert A\Vert$ so that $\sigma(A)$ is bounded.

For closure, suppose $\lambda_n\to\lambda$ is a sequence satisfying $\lambda_n\in\sigma(A)$ and $\lambda\notin\sigma(A)$. The latter assumption brings into existence a bounded map $B : X\to X$ such that

$$B(A-\lambda I) = I$$ (3.44)

A little Banach algebra reveals

$$B(A-\lambda_n I) = B(A-\lambda I) - B(\lambda_n I - \lambda I)$$ (3.45)

$$= I - B(\lambda_n I - \lambda I)$$ (3.46)

If $N$ is selected so that $n\geq N$ implies $\vert\lambda_n-\lambda\vert<\frac{1}{\Vert B\Vert}$, then we may realize

$$\Vert B(\lambda_nI - \lambda I)\Vert < \Vert B\Vert/\Vert B\Vert=1$$ (3.47)

indicating $I-B(\lambda_nI - \lambda I)=B(A-\lambda_n I)$ is invertible, so that $A-\lambda_n I$ is invertible, contradicting the selection $\lambda_n\in\sigma(A)$. Therefore, $\lambda\in\sigma(A)$, so that $\sigma(A)$ is closed.

Now since $\sigma(A)$ lies in a finite-dimensional space, closed and bounded exactly prove that $\sigma(A)$ is compact. A proof that $\sigma(A)$ is non-empty is saved for Fall 2019 Exercise 5. $\qedsymbol$