Let $x_n := (I +n S)^{-1} f$ for all $n$. How to prove that $(x_n)$ is convergent?

Question

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space. Let $S :H \to H$ be a bounded linear operator such that $\langle Su, u \rangle \ge 0$ for all $u \in H$. Then the map $I + t S$ is bijective for all $t>0$. Here $I:H \to H$ is the identity map. Fix $f\in H$ and let $x_n := (I +n S)^{-1} f$ for all $n \in \mathbb N$.

I'm trying to prove that

Theorem $(x_n)$ is convergent in $H$.

It suffices to prove that $(x_n)$ is a Cauchy sequence. Clearly, $$ x_n+n Sx_n = f = x_m + m S x_m \quad \forall m,n \in \mathbb N. $$

Then I don't know how to move forward.

Could you elaborate how to prove above theorem?

Answer 1

Assume $x\in \ker S.$ Then $(I+nS)x=x,$ hence $x=(I+nS)^{-1}x.$ For $0\neq x\in (\ker S)^\perp$ we have $Sx\neq 0$ and $${\|(I+nS)x\|\over \|x\|}\ge n{\|Sx\|\over \|x\|}-1\underset{n\to\infty}{\longrightarrow} \infty\quad (*)$$ Let $x\in {\rm Im}\,S.$ Then $x=Sv $ for some $v.$ We have $$ y_n=(I+nS)^{-1}x=S(I+nS)^{-1}v$$ Thus $y_n\in {\rm Im}\,S\subset (\ker S)^\perp$ and $(I+nS)y_n=Sv=x.$ The formula $(*)$ implies $\|y_n\|\underset{n}{\to}0.$ By continuity we obtain $$\|(I+nS)^{-1}x\|\underset{n\to \infty}{\longrightarrow} 0,\quad x\in \overline{{\rm Im}\,S}=(\ker S)^\perp$$ Summarizing, the sequence $(I+nS)^{-1}$ tends strongly (i.e. pointwise) to the orthogonal projection onto $\ker S.$

Remark We have used the property $\ker S=({\rm Im}\,S)^\perp.$ The latter space is equal $\ker (S^T),$ so it suffices to show that $\ker (S^T)=\ker S.$ The proof is in the spoiler.

We have $$\langle (S+S^T)x,x\rangle =2\langle Sx,x\rangle \ge 0$$ Assume $Sx=0.$ Then $$\langle (S+S^T)x,x\rangle=0$$ As $S+S^T$ is positive self adjoint we get $(S+S^T)x=0.$ Hence $S^Tx=0.$