"Let $T \in B(X,X)$. Prove that $||R_{\lambda}(T)|| \to 0$ as $\lambda \to \infty$."
We have that $R_{\lambda}(T)x=(T-\lambda I)^{-1}(x)$. The problem doesn't specify but the book says that they will always assume $X$ to be Banach unless otherwise stated. Since $T$ is bounded and $Dom T=X$, we know that that $R_{\lambda}(T)$ is defined i.e. $(T-\lambda I)x=y$ has a unique solution $x$.
We want to show that $||R_{\lambda}(T)||=\sup_{||x||=1}||R_{\lambda}(T)x||$ is getting closer and closer to $0$. Since $||x||=0$ iff $x=0$, I suppose we want to show that $R_{\lambda}(T)x$ is getting close to $0$ for all $x$ on the boundary of the unit ball.
I'm having trouble getting beyond what I've mentioned above.
Any help is appreciated and thanks in advance.
Edit:
Since $||\lambda x||=||(\lambda x-Tx) + Tx|| \leq ||\lambda x - Tx|| + ||Tx||$ and $||\lambda x - Tx||=||Tx-\lambda x||$, we have that $||\lambda x|| - ||Tx|| \leq ||(T-\lambda I)x||$. Then since $||\lambda x = |\lambda|\cdot||x||$ and $-||Tx|| \geq -||T||\cdot||x||$, we have:
$$(|\lambda|-||T||)\cdot||x|| \leq ||(T-\lambda I)x|| \leq ||(T-\lambda I)||\cdot||x||$$
Assuming $x \neq 0$, we have $(|\lambda|-||T||) \leq ||(T-\lambda I)||$. I'm pretty sure that for an operator $S$, $||S^{-1}|| \geq \frac{1}{||S||}$.
If it is true that $(|\lambda|-||T||) \leq \frac{1}{||(T-\lambda I)^{-1}||} \leq ||(T-\lambda I)||$, then we have $||R_{\lambda}(T)||=||(T-\lambda I)^{-1}|| \leq \frac{1}{|\lambda|-||T||}$. Then, as $\lambda \to \infty$, the LHS goes to $0$.
Is the inequality $(|\lambda|-||T||) \leq \frac{1}{||(T-\lambda I)^{-1}||}$ okay and if so, why?
For $\lvert\lambda\rvert > \lVert T\rVert$, we have a non-negative lower bound (strictly positive for $x\neq 0$) for $\lVert (T-\lambda I)x\rVert$, namely
$$\lVert (T-\lambda I)x\rVert \geqslant (\lvert\lambda\rvert - \lVert T\rVert)\lVert x\rVert.$$
Now set $x = (T-\lambda I)^{-1} y$ to obtain the inequality
$$\lVert y\rVert = \lVert(T-\lambda I)(T-\lambda I)^{-1}y\rVert \geqslant (\lvert\lambda\rvert -\lVert T\rVert)\lVert (T-\lambda I)^{-1}y\rVert$$
holdong for all $y\in X$. Dividing by $\lvert\lambda\rvert -\lVert T\rVert$ then yields
$$\lVert (T-\lambda I)^{-1} y\rVert \leqslant \frac{1}{\lvert\lambda\rvert -\lVert T\rVert}\lVert y\rVert$$
for all $y\in X$, whence
$$\lVert (T-\lambda I)^{-1}\rVert \leqslant \frac{1}{\lvert\lambda\rvert - \lVert T\rVert}\tag{$\ast$}$$
for all $\lambda$ with $\lvert\lambda\rvert > \lVert T\rVert$. It is clear from $(\ast)$ that $\lVert (T-\lambda I)^{-1}\rVert \to 0$ as $\lvert\lambda\rvert \to +\infty$.