Spectral Radius Definition in An Invitation to Operator Theory

Question

On page 243 in 'An Invitation to Operatory Theory' by Y. A. Abramovich, C. D. Aliprantis, the spectral radius $r(T)$ of an arbitrary operator in $\mathcal{L}(X)$ is defined to be the smallest non-negative real number $r$ for which the closed disk $\{ \lambda \in \mathbb{C}: \lvert{\lambda}\rvert \leq r\}$ contains the spectrum $\sigma(T)$. That is

$$ r(T) =\sup\{\lvert \lambda \rvert : \lambda \in \sigma(T)\} = \max\{\lvert \lambda \rvert : \lambda \in \sigma(T)\}. $$

My question is how is the suprememum known to be equal to the maximum? Isn't is possible that the suprememum in the above equation might not actually be an element in $\sigma(T)$ but rather just the least possible upper bound on all elements in $\sigma(T)$, and therefore it would be incorrect to assume that the supremum was equal to the maximum?

Answer 1

Typically that notation refers to bounded linear operators, the spectrum of which is closed.

Eg see https://en.m.wikipedia.org/wiki/Spectrum_(functional_analysis)

Answer 2

For self-adjoint operators this is indeed simple, because the spectrum will be a closed subset of reals, which either is bounded, in which case the maximum is achieved, or unbounded, in which both $\sup$ and $\max$ are infinity.

You could be wondering if it could not happen that in the complex plane the spectrum is a closed set, but not so its image under $\text{abs}$. That a continuous function maps a closed set to a non-closed one can indeed happen easily, for example $\mathbb{R}$ as a whole is closed but $\arctan(\mathbb{R})$ is $(-π/2, π/2)$ which is not.

For this, imagine a sequence of lambdas which attains the supremum, that is, $$λ_n \in σ(T): \lim_{n→∞} |λ_n| = r < ∞.$$ The argument of each $λ_n$ is within a finite interval $[-π, π)$, so an infinite sequence must have a convergent subsequence. Choose this subsequence $λ_{n_k}$. Now this converges in both magnitude and argument, so $λ_∞ := \lim_{k→∞} λ_{n_k}$ exists and is an accumulation point of $σ(T)$. For the closedness of spectrum, $λ_∞$ must be an element, and its absolute value equals $r$.