Gelfand's Spectral Radius Formula/Bounded Linear Operator

Theorem
Let $\struct {X, \norm \cdot _X}$ be a Banach space over $\C$.

Let $\map B X$ be the set of bounded linear operators on $X$.

Let $\norm \cdot_{\map B X}$ denote the operator norm on $\map B X$.

Let $T \in \map B X$.

Let $\size {\map \sigma T}$ be the spectral radius of $T$.

Then:
 * $\ds \size {\map \sigma T} = \lim_{n \to \infty} \paren {\norm{T^n}_{\map B X} }^{1/n} = \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$

Proof
Let $z \in \C$ be such that:
 * $\ds \cmod z > \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$

That is, there exists an $m \in \N_{>0}$ such that:
 * $\ds \cmod z > \paren {\norm{T^m}_{\map B X} }^{1/m}$

Then:

exists, i.e. $z \in \C \setminus \map \sigma T$.

Therefore:
 * $\forall z \in \C : \ds z \in \map \sigma T \implies \cmod z \le \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$

By we have:
 * $\ds \size{\map \sigma T} \le \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$

It remains to show:
 * $\ds \size{\map \sigma T} \ge \limsup_{n \to \infty} \paren {\norm{T^n}_{\map B X} }^{1/n}$