Gelfand's Spectral Radius Formula/Bounded Linear Operator

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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:

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


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:

\(\ds \paren {T - z I}^{-1}\) \(=\) \(\ds -z^{-1} \paren {I - z^{-1} T}^{-1}\)
\(\ds \) \(=\) \(\ds -z^{-1} \sum_{N \mathop \ge 0} \paren {z^{-1} T}^N\) Invertibility of Identity Minus Operator, as $\norm {z^{-1} T} < 1$
\(\ds \) \(=\) \(\ds -z^{-1} \sum_{r \mathop = 0}^{m - 1} \paren {z^{-1} T}^r \sum_{q \mathop \in \N} \paren {z^{-1} T}^{m q}\) Quotient-Remainder Theorem

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:

\(\ds \map {F_\ell} z\) \(=\) \(\ds -z^{-1} \map \ell {\paren {I - z^{-1} T}^{-1} }\)
\(\ds \) \(=\) \(\ds -z^{-1} \map \ell {- \sum_{n \mathop \ge 0} \paren {z^{-1} T^n}^n }\) Invertibility of Identity Minus Operator
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds -\sum_{n \mathop \ge 0} \dfrac {\map \ell {T^n} }{z^{n + 1} }\)

Thus for all $\delta > 0$ and $m \in \N$:

\(\ds \oint _{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \map {F_\ell} z z^m \rd z\) \(=\) \(\ds \oint _{\cmod z \mathop = \norm T_{\map B T} \mathop + 1} \map {F_\ell} z z^m \rd z\) Cauchy-Goursat Theorem
\(\ds \) \(=\) \(\ds \oint_{\cmod z \mathop = \norm T_{\map B T} \mathop + 1} - \sum_{n \mathop \ge 0} \dfrac {\map \ell {T^n} }{z^{n - m + 1} } \rd z\) by $(1)$
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds -2 \pi i \; \map \ell {T^m}\) Cauchy's Residue Theorem

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:

\(\ds \norm {T^m} _{\map B T}\) \(=\) \(\ds \dfrac {-1} {2 \pi i} \oint_{\cmod z \mathop = \size {\map \sigma T} + \delta} \map {F_{\ell_m} } z z^m \rd z\) by $(2)$
\(\ds \) \(\le\) \(\ds \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \cmod {\map {F_{\ell_m} } z z^m} } \paren {\size {\map \sigma T} + \delta}\) Triangle Inequality for Contour Integrals
\(\ds \) \(=\) \(\ds \paren {\size {\map \sigma T} \mathop + \delta}^{m + 1} \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \cmod {\map {F_{\ell_m} } z} }\)
\(\ds \) \(\le\) \(\ds \paren {\size {\map \sigma T} \mathop + \delta}^{m + 1} \paren {\sup_{\cmod z \mathop = \size {\map \sigma T} \mathop + \delta} \norm {\paren {T - z I}^{-1} }_{\map B T} }\) as $\norm {\ell_m}_{ {\map B X}^\ast} = 1$

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 $\delta \to 0$.

$\blacksquare$


Sources

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