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 $\lim_{z\to\infty} \|f(z)\|_* = 0$.


Proof

Pick $z \in \Bbb C$ with $|z| > 2\|T\|_*$.

Then $\|T/z\|_* = \|T\|_*/|z| < 1/2$ by Operator Norm is Norm. So we have:

\(\ds \norm{ f(z) }_*\) \(=\) \(\ds \norm{ (T - zI)^{-1} }_*\)
\(\ds \) \(=\) \(\ds \norm{ (z(T/z - I))^{-1} }_*\)
\(\ds \) \(=\) \(\ds \norm{ \frac{-1}{z} (I - T/z)^{-1} }_*\) by Norm of Inverse of Constant Times Operator
\(\ds \) \(=\) \(\ds \frac{1}{\size z} \norm{ (I - T/z)^{-1} }_*\) by Operator Norm is Norm
\(\ds \) \(=\) \(\ds \frac{1}{\size z} \norm{ I + T/z + (T/z)^2 + \ldots }_*\) by Invertibility of Identity Minus Operator since $\norm{ T/z } < 1/2 < 1$
\(\ds \) \(\leq\) \(\ds \frac{1}{\size z} \sum_{k \geq 0} \norm {(T/z)^k }_*\) by Triangle Inequality
\(\ds \) \(\leq\) \(\ds \frac{1}{\size z} \sum_{k \geq 0} (\norm {T}_*/\size z)^k\) by applying Editing Operator Norm on Banach Space is Submultiplicative to each term
\(\ds \) \(\leq\) \(\ds \frac{1}{\size z} \sum_{k \geq 0} (1/2)^k\) since $\norm{ T/z }_* < 1/2$
\(\ds \) \(\leq\) \(\ds \frac{2}{\size z}.\)

Taking limits of both sides as $|z| \to \infty$, we get $\|f(z)\|_* \to 0$.


$\blacksquare$