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Let $X$ be a Banach space and let $T \in L(X)$ satisfy $\|Tx\| =\|x\|$ for each $x \in X$. Suppose that the range of $T \neq X$ and let $0 < |\lambda| < 1$. Assuming that $λ \in \rho(T)$, show that $\|(λ − T)^{−1}\| \leq 1/(1−|λ|)$. Here $\rho(T)$ is the resolvent set of $T$.

I need this intermediate step in proving that $\sigma(T)$ is the unit disk.

A.Γ.
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ronalddb89
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  • If $\sigma(T)=D$ then $|\lambda|<1$ cannot be in the resolvent set. Are you trying proof by contradiction? – A.Γ. Nov 22 '16 at 12:33
  • Yes, $\lambda \in \rho(T)$ then the above inequality should hold. I have a theorem in my lecture notes which contradicts that. – ronalddb89 Nov 22 '16 at 12:35

1 Answers1

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My theory of resolvents is a bit rusty but I'll give it a try.

Let $y$ be in the domain of $(\lambda-T)^{-1}$, define $x = (\lambda-T)^{-1}y$. Then $$\|(\lambda-T)^{-1}y\| = \|x\|.$$

Using the assumption about $T$, we know that $$\|x\| = \|Tx\| = \|(\lambda-T)x - \lambda x\| \le \|(\lambda - T)x\| + |\lambda| \|x\| = \|y\| + |\lambda|\|x\|,$$ the inequality being the triangle inequality for norm. Denoting $\|x\| = u$, we find

$$\begin{aligned} u &\le \|y\| + |\lambda|u, \\ (1 - |λ|) u &\le \|y\|, \\ u &\le \frac{\|y\|}{1-|\lambda|}, \end{aligned}$$

(note that in the last step we needed that $|λ| < 1$), so for all $y$ in the domain of $(\lambda-T)^{-1}$,

$$\|(\lambda-T)^{-1}y\| \le \frac{\|y\|}{1-|\lambda|}.$$

The rest is the density of such $y$ and the definition of a norm of an operator.

The Vee
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