Gelfand's spectral radius formula

Remarks

The maximum exists because $\sigma(a)$ is compact and, by the nonemptiness of the spectrum, nonempty. Lower bound: by the spectral mapping theorem for polynomials, $\lambda^n \in \sigma(a^n)$ for $\lambda \in \sigma(a)$, and every $\mu \in \sigma(b)$ satisfies $\abs{\mu} \le \norm{b}$; so $r(a)^n \le \norm{a^n}$ for every $n$. Upper bound: for $\abs{\lambda} > r(a)$ the resolvent $(\lambda 1 - a)^{-1}$ is holomorphic, and its Laurent series $\sum_{n \ge 0} a^n \lambda^{-n-1}$ converges there; applying functionals and the uniform boundedness principle gives $\limsup_n \norm{a^n}^{1/n} \le r(a)$.

Example: on $C([0,1])$ with the sup norm, $\norm{f^n} = \norm{f}^n$, so the formula gives $r(f) = \norm{f} = \max \abs{f}$, the largest absolute value of a point of the range. For a nilpotent element, $a^n = 0$ for large $n$, so $r(a) = 0$ and $\sigma(a) = \set{0}$, even if $\norm{a}$ is large. For matrices this is Gelfand's formula.

Going deeper

For a student

The formula shows that $r(a)$ depends only on the norms of the powers of $a$. So it does not change when the spectrum is computed in a closed subalgebra containing $a$ and $1$. In a C*-algebra, the C* identity gives $\norm{a^{2^k}} = \norm{a}^{2^k}$ for self-adjoint $a$, so $r(a) = \norm{a}$; this is a key step in the commutative Gelfand–Naimark theorem. It also shows that the Gelfand transform satisfies $\norm{\hat a}_\infty = r(a)$ in a commutative unital Banach algebra.