Suppose that $T \in BL(H)$ where $H$ is a Hilbert Space. Let $k \in \mathbb{C}$. Let $d=\text{dist}(k,W(T)) \gt 0$. Define the numerical range of $T$ as $$W(T)=\{\lambda \in \mathbb{C}: \lambda=\langle Tx,x\rangle, \|x\|=1, x \in H\}.$$ Show that $\|(kI-T)^{-1}\| \le \frac{1}{d}$.
First of all I need to see that $(kI-T)^{-1}$ makes sense. Since $\sigma(T) \subset \overline{W(T)}$ and $d=\text{dist}(k,\overline{W(T)}) \gt 0$, the inverse makes sense. Now I set out to find the solution.
My try: First of all I find relation between $W(T)$ and $W(T-kI)$. So if $\lambda \in W(T)$, then $\lambda=\langle Tx,x\rangle$ with $\|x\|=1$. Then $\langle Tx-kx,x\rangle=\lambda-k$. Hence $W(T-kI)=\{\lambda-k: \lambda \in W(T)\}$. The other way around can be proved similarly. Now, \begin{align} \|T-kI\| & = \sup_{\|x\|,\|y\| \le 1 }|\langle Tx-kx,y\rangle | \\ & \ge \sup_{\|x\|=1}|\langle Tx-kIx,x\rangle | \\ & =\sup_{\lambda \in W(T-kI)}|\lambda| \\ & =\sup_{\mu \in W(T)}{|\mu-k|} \ge d \gt 0. \end{align}
But after reaching here I am not able to go to its inverse. There might be some other way. I want to know if I can go from here
Thanks for the help!!
If $k$, $d$ are as specified, then $$ d\|x\|^2 \le |k\|x\|^2-\langle Tx,x\rangle| \\ d\|x\|^2 \le |\langle (kI-T)x,x\rangle| \le\|(kI-T)x\|\|x\| \\ d\|x\| \le \|(kI-T)x\|. $$ That's enough to give you injectivity of $kI-T$, along with a closed range. Then $$ \mathcal{R}(kI-T) = \mathcal{N}(\overline{k}I-T^{\star})^{\perp} $$ However, the first inequality also gives you $$ d\|x\|^2 \le |\overline{k\|x\|^2-\langle Tx,x\rangle}| = |\overline{k}\|x\|^2-\langle T^{\star}x,x\rangle| \\ d\|x\| \le \|(\overline{k}I-T^{\star})x\|. $$ Therefore $\mathcal{R}(kI-T)=\{0\}^{\perp}=H$. So $kI-T$ is invertible.