Proving $\lim_{t\to+\infty}(I + tS)^{-1}f = P_{N(S)}f$ for all $f\in H$ if $S\in \mathcal{L}(H)$ and $(Su, u) \geq 0$ $\forall u\in H$

Question

I'm having trouble filling in the details of a step in one of the proposed solutions to exercise 5.20 from Brezis' functional analysis book. The exercise is as follows ($H$ denotes a Hilbert space with inner product $(\cdot, \cdot )$, while its norm is denoted by $|\cdot |$; finally, $\mathcal{L}(H)$ is the set of bounded linear operators on $H$):

Exercise 5.20 Assume that $S\in \mathcal{L}(H)$ satisfies $(Su, u) \geq 0$ $\forall u\in H$.

1. Prove that $N(S) = R(S)^{\perp}$.

2. Prove that $I + tS$ is bijective for every $t > 0$

3. Prove that $\lim_{t\to+\infty}(I + tS)^{-1}f = P_{N(S)}f$ for all $f\in H$.

I'm having trouble proving part 3. One of the (two) solutions proposed by the author is to first consider the cases $f\in N(S)$ and $f\in R(S)$. The idea is that every $f\in H$ can be written as $f = f_1 + f_2$ with $f_1 = P_{N(S)}f$ and $f_2 = P_{\overline{R(S)}}f$. Now, I was able to fill in all the details of the solution except for this: I haven't been able to show that if 3 holds for every $f\in R(S)$ (and every $f\in N(S)$, in case that helps), then it must hold for every $\tilde{f} \in \overline{R(S)}$. According to Brezis, one can prove this with a density argument.

Here's my attempt: suppose $\tilde{f} \in \overline{R(S)}$ and take a sequence $(f_n)_{n\in\mathbb{N}}$ in $R(S)$ with $f_n \to \tilde{f}$. Then for all $t > 0$ and $n\in\mathbb{N}$ we have: $$|(I + tS)^{-1}\tilde{f} - P_{N(S)}\tilde{f}| \leq |(I + tS)^{-1}\tilde{f} - (I + tS)^{-1}f_n| + |(I + tS)^{-1}f_n - P_{N(S)}f_n| + |P_{N(S)}f_n - P_{N(S)}\tilde{f}|$$

The first summand goes to zero as $n\to\infty$ for every fixed $t$ (by continuity of each $(I + tS)^{-1}$, which follows from the bounded inverse theorem). The second summand goes to zero as $t\to\infty$ for every fixed $n$ (since 3 holds for the elements of $R(S)$). The third summand is no larger than $|f_n - \tilde{f}|$, which goes to zero as $n\to\infty$.

I would need something stronger than what I currently have to show the LHS goes to zero as $t\to\infty$. For example, it would be sufficient to show that there is a constant $M > 0$ such that $|(I + tS)^{-1}(x)| \leq M|x|$ $\forall x\in \overline{R(S)}$ $\forall t > 0$, because, then, the first summand would be bounded above by $M|\tilde{f} - f_n|$, thus eliminating the dependence on $t$. I can't see how to prove this last bound, though.

Does anybody have a suggestion?

Answer 1

For any $f \in H$, and $0 \le t$, $$ \|f\|^{2} \le (f,f)+t(Sf,f) = ((I+tS)f,f) \le \|(I+tS)f\|\|f\|. $$ Therefore, $$ \|f\| \le \|(I+tS)f\|,\;\;\; t \ge 0,\; f \in H. $$ Presumably you have already show that $I+tS$ is invertible in $\mathcal{L}(H)$. If not, it does follow from the above that that $(I+tS)$ is injective with dense range because $\mathcal{R}(I+tS)^{\perp}=\mathcal{N}(I+tS)=\{0\}$; and the range is closed because the inverse is bounded. So $(I+tS)^{-1}\in\mathcal{L}(H)$ with $$ \|(I+tS)^{-1}g\|\le \|g\|,\;\;\; g \in H, $$ which gives you the desired uniform bound $\|(I+tS)^{-1}\|\le 1$ for all $t \ge 0$.