Prove that, if $R_\lambda$ is resolvent fuction and $\lambda>0$, then $\|R_\lambda\|\leq \frac{1}{\lambda}$

Question

Let $T$ be a linear operator defined on a limited complex Banach space $X$.

Resolvent set of T is defined as the set $ \rho (T) $ of complex numbers $\lambda$ such that the operator $ \lambda I - T $ is invertible, ie has an inverse that is a linear operator limited.

We define the resolvent function of $T$:

$$R_\lambda (T) = (\lambda I - T)^{-1} \ $$

If $\lambda>0$, prove that the norm of resolvent is bounded in this way:

$$\|R_\lambda\|\leq \frac{1}{\lambda}$$

Answer 1

That statement is wrong. As a counterexample consider $X = \mathbb{C}$ and $T = 1/2$, where we consider an element of $\mathbb{C}$ as a linear operator on $\mathbb{C}$ in the usual way. Then $$\Vert (\lambda I - T)^{-1} \Vert = |(\lambda - 1/2)^{-1}| = |(1/3 - 1/2)^{-1}| = 6 \not\leq \frac{1}{\lambda} = 3,$$ where I chose $\lambda = 1/3 > 0$.