Resolvent Mapping Converges to 0 at Infinity

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Theorem

Let $B$ be a Banach space.

Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.

Let $T \in \map \LL {B, B}$.

Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.

Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is such that $\ds \lim_{z \mathop \to \infty} \norm {\map f z}_* = 0$.


Proof

Pick $z \in \Bbb C$ such that $\size z > 2 \norm T_*$.

By Operator Norm is Norm:

$\norm {\dfrac T z}_* = \dfrac {\norm T_*} {\size z} < \dfrac 1 2$

Hence:

\(\ds \norm {\map f z}_*\) \(=\) \(\ds \norm {\paren {T - z I}^{-1} }_*\)
\(\ds \) \(=\) \(\ds \norm {\paren {z \paren {\dfrac T z - I} }^{-1} }_*\)
\(\ds \) \(=\) \(\ds \norm {\frac{-1} z \paren {I - \dfrac T z}^{-1} }_*\) Norm of Inverse of Constant Times Operator
\(\ds \) \(=\) \(\ds \frac 1 {\size z} \norm {\paren {I - \dfrac T z}^{-1} }_*\) Operator Norm is Norm
\(\ds \) \(=\) \(\ds \frac 1 {\size z} \norm {I + \dfrac T z + \paren {\dfrac T z}^2 + \cdots}_*\) Invertibility of Identity Minus Operator since $\norm {\dfrac T z} < \dfrac 1 2 < 1$
\(\ds \) \(\le\) \(\ds \frac 1 {\size z} \sum_{k \mathop \ge 0} \norm {\paren {\dfrac T z}^k }_*\) Triangle Inequality
\(\ds \) \(\le\) \(\ds \frac 1 {\size z} \sum_{k \mathop \ge 0} \paren {\dfrac {\norm T_*} {\size z} }^k\) Operator Norm on Banach Space is Submultiplicative
\(\ds \) \(\le\) \(\ds \frac 1 {\size z} \sum_{k \mathop \ge 0} \paren {\dfrac 1 2}^k\) since $\norm {\dfrac T z}_* < \dfrac 1 2$
\(\ds \) \(\le\) \(\ds \frac 2 {\size z}\)

Taking limits of both sides as $\size z \to \infty$:

$\norm {\map f z}_* \to 0$

$\blacksquare$

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