I want to prove the following statement:
Let $H$ be a positive self-adjoint operator, not necessarily bounded. For each real positive number $\lambda$, define $H_{\lambda} = \lambda (H+\lambda I)^{-1}$.
Prove that each $H_{\lambda}$ is a bounded self-adjoint operator and that $H_{\lambda} \to H$ as $\lambda \to \infty$ strongly (i.e. $|| H_{\lambda} \varphi - H \varphi || \to 0$ when $\lambda \to \infty$ for every $\varphi \in D(H)$, where $D(H)$ is domain.
My intuition says that mentioned statement is correct, because of the following inequality (assuming $T$ bounded): $|| (I + \lambda (T - \lambda I)^{-1} || \leq \frac{||T||}{|\lambda - ||T||}$
and when $\lambda \to 0$, $|| (I + \lambda (T - \lambda I)^{-1} || \to 0$ and so $||\lambda (T - \lambda I)^{-1}|| \to I$.
The main problem is that I don't really know how to prove this for every self-adjoint $H$ not necessarily bounded.
Thanks to for the readers.
Because $H$ is positive, then, for $\lambda > 0$, $$ \lambda\|x\|^2 \le ((H+\lambda I)x,x) \le \|(H+\lambda I)x\|\|x\|,\;\;\; x\in\mathcal{D}(H) \\ \lambda\|x\|\le \|(H+\lambda I)x\|,\;\; x\in\mathcal{D}(H) \\ \lambda\|(H+\lambda I)^{-1}y\|\le \|y\|,\;\; y \in H. $$ So $\|\lambda (H+\lambda I)^{-1}\| \le 1$ for all $\lambda > 0$. For $x\in\mathcal{D}(H)$, \begin{align} \|\lambda(H+\lambda I)^{-1}x-x\|&=\|\lambda(H+\lambda I)^{-1}-(H+\lambda I)^{-1}(H+\lambda I)x\| \\ &=\|(H+\lambda I)^{-1}Hx\| \\ &\le \frac{1}{\lambda}\|Hx\|\rightarrow 0 \mbox{ as } \lambda\rightarrow\infty. \end{align} Let $y$ be general. For any $x\in\mathcal{D}(H)$, \begin{align} \|\lambda(H+\lambda I)^{-1}y-y\| &=\|\lambda(H+\lambda I)^{-1}(y-x)+\lambda(H+\lambda I)^{-1}x-x+(x-y)\| \\ &\le \|y-x\|+\|\lambda(H+\lambda I)^{-1}x-x\|+\|x-y\| \end{align} Therefore, $$ \limsup_{\lambda\rightarrow\infty}\|\lambda(H+\lambda I)^{-1}y-y\| \le 2\|y-x\|,\;\;\; x\in \mathcal{D}(H). $$ Because the above holds for all $x\in\mathcal{D}(H)$, and because $\mathcal{D}(H)$ is dense in the underlying space, then the following holds for all $y$: $$ \lim_{\lambda\rightarrow\infty}\lambda (H+\lambda I)^{-1}y=y. $$ That gives an approximation to $H$ where, for $x\in\mathcal{D}(H)$, $$ Hx= \lim_{\lambda\rightarrow\infty}\lambda (H+\lambda I)^{-1}Hx = \lim_{\lambda\rightarrow\infty}\lambda\left\{ x-\lambda(H+\lambda I)^{-1}x\right\} $$ The operator $H_{\lambda}=\lambda\{ I-\lambda(H+\lambda I)^{-1} \}$ is symmetric and bounded, which makes it selfadjoint.