Theorem: Gelfand's Spectral Radius Formula/Bounded Linear Operator

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

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:
{{begin-eqn}}
{{eqn | l = \paren {T - z I}^{-1}
      | r = -z^{-1} \paren {I - z^{-1} T}^{-1}
}}
{{eqn | r = -z^{-1} \sum_{N \mathop \ge 0} \paren {z^{-1} T}^N
      | c = Invertibility of Identity Minus Operator, as $\norm {z^{-1} T} < 1$
}}
{{eqn | r = -z^{-1} \sum_{r \mathop = 0}^{m - 1} \paren {z^{-1} T}^r \sum_{q \mathop \in \N} \paren {z^{-1} T}^{m q}
      | c = Quotient-Remainder Theorem
}}
{{end-eqn}}
exists, that is:
:$z \in \C \setminus \map \sigma T$
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 definition of spectral radius 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:
:$\ds \size{\map \sigma T} \ge \limsup_{n \mathop \to \infty} \paren {\norm {T^n}_{\map B X} }^{1/n}$
Let $\struct { {\map B X}^\ast, \norm {\cdot}_{ {\map B X}^\ast} }$ be the normed dual space of $\struct {\map B X, \norm \cdot_{\map B X} }$.
For $\ell \in {\map B X}^\ast$, we define a complex function $F_\ell : \C \setminus \map \sigma T \to \C$ by:
:$\map {F_\ell} z := \map \ell {\paren {T - z I}^{-1} }$
By Resolvent Mapping is Analytic, $F_\ell$ is analytic on $\C \setminus \map \sigma T$.
Moreover, if $\cmod z > \norm T_{\map B T}$, then:
{{begin-eqn}}
{{eqn | l = \map {F_\ell} z
      | r = -z^{-1} \map \ell {\paren {I - z^{-1} T}^{-1} }
}}
{{eqn | r = -z^{-1} \map \ell {- \sum_{n \mathop \ge 0} \paren {z^{-1} T^n}^n }
      | c = Invertibility of Identity Minus Operator
}}
{{eqn | r = -\sum_{n \mathop \ge 0} \dfrac {\map \ell {T^n} }{z^{n + 1} }
      | n = 1
}}
{{end-eqn}}
Thus for all $\delta > 0$ and $m \in \N$:
{{begin-eqn}}
{{eqn | l = \oint _{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \map {F_\ell} z z^m \rd z
      | r = \oint _{\cmod z \mathop = \norm T_{\map B T} \mathop + 1} \map {F_\ell} z z^m \rd z
      | c = Cauchy-Goursat Theorem
}}
{{eqn | r = \oint_{\cmod z \mathop = \norm T_{\map B T} \mathop + 1} - \sum_{n \mathop \ge 0} \dfrac {\map \ell {T^n} }{z^{n - m + 1} } \rd z
      | c = by $(1)$
}}
{{eqn | r = -2 \pi i \; \map \ell {T^m}
      | c = Cauchy's Residue Theorem
      | n = 2
}}
{{end-eqn}}
By Existence of Support Functional, for each $m \in \N$ there is an $\ell_m \in {\map B X}^\ast$ such that:
:$\norm {\ell_m}_{ {\map B X}^\ast} = 1$
:$\map {\ell_m} {T^m} = \norm {T^m}_{\map B T}$
Thus:
{{begin-eqn}}
{{eqn | l = \norm {T^m} _{\map B T}
      | r = \dfrac {-1} {2 \pi i} \oint_{\cmod z \mathop = \size {\map \sigma T} + \delta} \map {F_{\ell_m} } z z^m \rd z
      | c = by $(2)$
}}
{{eqn | o = \le
      | r = \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \cmod {\map {F_{\ell_m} } z z^m} } \paren {\size {\map \sigma T} + \delta}
      | c = Triangle Inequality for Contour Integrals
}}
{{eqn | r = \paren {\size {\map \sigma T} \mathop + \delta}^{m + 1} \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \cmod {\map {F_{\ell_m} } z} }
}}
{{eqn | o = \le
      | r = \paren {\size {\map \sigma T} \mathop + \delta}^{m + 1} \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \norm {\paren {T - z I}^{-1} }_{\map B T} }
      | c = as $\norm {\ell_m}_{ {\map B X}^\ast} = 1$
}}
{{end-eqn}}
Letting $m \to \infty$, we obtain:
:$\ds \limsup_{m \mathop \to \infty} \paren {\norm {T^m}_{\map B X} }^{1 / m} \le \size {\map \sigma T} + \delta$
The result follows by $\delta \to 0$.
{{qed}}