Problem 5

Exercise 5.5 (Spectrum is closed and bounded)   Show that the spectrum of a bounded linear operator on a complex Banach space is a non-empty compact subset of $\mathbb{C}$. Does the same hold for operator on real Banach spaces?

Proof. Examining Winter 2019 Exercise 6 shows that this problem now requires us to show only that the spectrum is non-empty. Let $B : X\to X$ be a bounded linear operator on a Banach space. Suppose the spectrum were empty. That is,

$$\sigma(B) = \{ \lambda \in \mathbb{C} \;\vert\; B-\lambda I \} = \{\}$$ (5.19)

Then $B-\lambda I$ is invertible for any complex $\lambda$, so the resolvent $R_\lambda=(B-\lambda I)^{-1}$ is defined for any $\lambda$. By the open mapping, each resolvent is bounded. Let $\phi \in \mathcal{L}(X)^*$ be a non-zero linear functional on the space of bounded opeators. Define a function $F : \mathbb{C} \to \mathbb{C}$

$$F(\lambda) = \phi(R_\lambda)$$ (5.20)

is entire with the sense of operator norm convergence. Taking the modulus we see that $\vert F(\lambda)\vert\to 0$ as $\vert\lambda\vert\to\infty$, which indicates $F\equiv 0$, so that $X=0$, a contradiction. Therefore, $B-\lambda I$ is not invertible for some $\lambda$.

The compactness does not hold in a real setting. Consider the operator

$$\int : C([0,1]) \to C([0,1])$$ (5.21)

$$f \mapsto \int_0^x f(t)dt$$ (5.22)

where $C([0,1])$ is the set of continuous real valued functions define over $[0,1]$. Let us follow a familiar derivation of the eigenvalues. Let

$$\int_0^x f(t) dt = \lambda f(x)$$ (5.23)

The very act of writing this indicates that we may differentiate on either side, so that

$$f(x) = \lambda f'(x) \iff f(x) = ke^{\lambda x}$$ (5.24)

But then the spectrum contains the real line, so it must not be compact. $\qedsymbol$