Brezis' exercise 5.21.6

Question

I'm trying to solve exercise 5.21.6 in Brezis' Functional Analysis.

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(\mathcal L(H), \| \cdot \|)$ be the Banach space of all bounded linear operator on $H$. Let $T \in \mathcal L(H)$ with $\|T\| \le 1$. Let $f \in H$ and define for each $n \in \mathbb N^*$, $$ \begin{align} \sigma_n (f) &:= \frac{f+Tf+T^2f+\cdots+T^{n-1}f}{n}, \\ \mu_n (f) &:= \left ( \frac{I+T}{2} \right )^n f. \end{align} $$ Here $I:H \to H$ is the identity map. Our purpose is to show that $$ \lim_n \sigma_n (f) =\lim_n \mu_n (f) = \pi_{N(I-T)} f. $$ Here $\pi_{N(I-T)} f$ is the orthogonal projection of $f$ onto the kernel $N(I-T)$ of $I-T$.

1. Check that $N(I-T) = R(I-T)^\perp$ with $R(I-T)$ the range of $I-T$.
2. Assume that $f \in R(I-T)$. Prove that there is a constant $C$ such that $|\sigma_n (f)| \le C/n$ for all $n \in \mathbb N^*$.
3. Deduce that for every $f\in H$, one has $\lim_n \sigma_n (f) = \pi_{N(I-T)} f$.
4. Let $S = \frac{I+T}{2}$. Prove that $|u-Su|^2 + |Su|^2 \le |u|^2$ for all $u\in H$. Deduce that $$ \sum_{i=0}^\infty |S^i u -S^{i+1}u|^2 \le |u|^2 \quad \forall u \in H $$ and that $$ |S^n(u-Su)| \le \frac{|u|}{\sqrt{n+1}} \quad \forall u \in H, \forall n \in \mathbb N^*. $$
5. Assume $f \in R(I-T)$. Prove that there is a constant $C$ such that $|\mu_n (f)| \le C/\sqrt{n}$ for all $n \in \mathbb N^*$.
6. Deduce that for every $f\in H$, one has $\lim_n \mu_n (f) = \pi_{N(I-T)} f$.

Could you have a check on my below attempt?

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4. We have $$ \begin{align} |u|^2 &= |u-Su|^2 + 2 \langle u-Su, Su \rangle + |Su|^2 \\ &= |u-Su|^2 + \langle u - Tu, u+Tu \rangle + |Su|^2 \\ &= |u-Su|^2 + (|u|^2-|Tu|^2) + |Su|^2 \\ &\ge |u-Su|^2 + |Su|^2 \quad \text{because} \quad \|T\| \le 1.\\ \end{align} $$ We apply above inequality iteratively and get $$ \begin{align} |u|^2 &\ge |u-Su|^2 + |Su|^2 \\ &\ge |u-Su|^2 + ( |Su-S^2u|^2 + |S^2u|^2 ) \\ &\ge |u-Su|^2 + |Su-S^2u|^2 + ( |S^2u-S^3u|^2 + |S^3u|^2 ) \\ &\ge \sum_{i=0}^n |S^i u -S^{i+1}u|^2 + |S^{n+1} u|^2. \end{align} $$ It follows that $$ |u|^2 \ge \lim_n \sum_{i=0}^n |S^i u -S^{i+1}u|^2 = \sum_{i=0}^\infty |S^i u -S^{i+1}u|^2. $$ Notice that $\|S\| \le 1$, so $|S^i(u-Su)| \ge |S^n (u-Su)|$ for all $i \le n$. Then $$ |u|^2 \ge \sum_{i=0}^n |S^n (u-Su)|^2 = n |S^n (u-Su)|^2. $$

5. Let $f \in R(I-T)$. Then there is $x \in H$ such that $f=(I-T)x$. Then $$ \begin{align} \mu_n (f) &= S^n(I-T)x \\ &=2S^n(I-S)x. \end{align} $$ It follows that $|\mu_n (f)| \le \frac{2|x|}{\sqrt{n+1}}$.

6. Let $f \in N(I-T)$. Then $(I-T)f=0$. Then $S f= \frac{I-T}{2}f+ Tf=f$. Then $S^n f= f$ for all $n$. Then $\mu_n (f) = f = \pi_{N(I-T)} f$. If $f \in R(I-T)$ then $\mu_n (f) \to 0$ by (5.). By (1.), we have $0=\pi_{N(I-T)} f$. Let's prove the claim for $f \in \overline{R(I-T)}$ by density argument. Fix $\varepsilon >0$. There is $f_\varepsilon \in R(I-T)$ such that $|f-f_\varepsilon| < \varepsilon$. We have $$ \begin{align} |\mu_n (f) - \mu_n (f_\varepsilon)| &= | S^n (f-f_\varepsilon) |\\ &\le \|S^n\| |f-f_\varepsilon| \\ &\le \|S\|^n |f-f_\varepsilon| \\ &\le |f-f_\varepsilon|. \end{align} $$ Then $$ \begin{align} & |\mu_n (f) - \pi_{N (I-T)} f| \\ \le & |\mu_n (f)-\mu_n (f_\varepsilon)| + |\mu_n (f_\varepsilon) - \pi_{N(I-T)} f_\varepsilon| + |\pi_{N(I-T)} f_\varepsilon - \pi_{N(I-T)} f| \\ \le & |\mu_n (f_\varepsilon) - \pi_{N(I-T)} f_\varepsilon| + (1+\|\pi_{N(I-T)}\|) |f-f_\varepsilon|. \end{align} $$ It follows that $\limsup_n |\mu_n (f) - \pi_{N (I-T)} f| \le (1+\|\pi_{N(I-T)}\|) \varepsilon$. The claim then holds for $f \in \overline{R(I-T)}$ by taking the limit $\varepsilon \downarrow 0$. In the general case $f \in H$, we write write $f=f_1+f_2$ with $f_1=\pi_{N(I-T)} f$ and $f_2=\pi_{\overline{R(I-T)}} f$.