Resolvent operator of a bounded linear operator

Question

If $X$ is a Banach space and $T:X\rightarrow X$ is a bounded linear operator, then how to show that $\|R_\lambda$(T)$\| \rightarrow 0$, as $|\lambda| \rightarrow \infty$.

Answer 1

Since $T-\lambda I = -\lambda( I - \lambda^{-1}T$, it is clear that $T-\lambda I$ is invertible for all $\lambda$ with $|\lambda|$ large enough, i.e., $|\lambda|>\|T\|$.

Take $x,z$ such that $(T-\lambda I)x=z$. Then $$ \|R_\lambda(T)z\|=\|x\| =\| \lambda^{-1}(z-Tx)\|\le |\lambda|^{-1}(\|z\|+\|T\|\cdot\|x\|). $$ If $|\lambda| >2 \|T\|$ then it follows $$ \frac12\|x\|\le (1-\lambda^{-1}\|T\|)\|x\| \le |\lambda|^{-1}\|z\|, $$ which is the wanted inequality.

Answer 2

For $|\lambda| > \|T\|$, $$ (T-\lambda I)^{-1} = -\frac{1}{\lambda} (I-\frac{1}{\lambda}T)^{-1}=-\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n} \\ \|(T-\lambda I)^{-1}\| \le \frac{1}{|\lambda|}\sum_{n=0}^{\infty}\frac{\|T\|^n}{|\lambda|^n}=\frac{1}{|\lambda|}\frac{1}{1-\frac{\|T\|}{|\lambda|}}=\frac{1}{|\lambda|-\|T\|}. $$