Brezis' exercise 5.21.3

Question

I'm trying to solve exercise 5.21.3 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 $\lim_n \sigma_n (f) = \pi_{N(I-T)} f$.

Could you have a check on my below attempt?

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1. Let $x \in H$. Then $$ \langle (I-T) x, x \rangle = |x|^2 - \langle T x, x \rangle \ge|x|^2 - |Tx||x| \ge (1-\|T\|)|x|^2 \ge 0. $$ Then $I-T$ is a positive operator and thus $N(I-T) = R(I-T)^\perp$.

2. There is $x\in H$ such that $f=x-Tx$. Then $Tf = Tx-T^2 x, \ldots, T^{n-1} f = T^{n-1} x - T^n x$. It follows that $f+Tf+T^2f+\cdots+T^{n-1}f =x-T^n x$. We have $|x-T^n x| \le (1+\|T^n\|) |x| \le (1+\|T\|^n)|x| \le 2|x|$. Then $|\sigma_n (f)| \le 2|x|/n$.

3. Let $f \in N(I-T)$. Then $f-Tf=0$. Then $f=Tf=\cdots=T^{n-1} f$. Then $\sigma_n (f) = f = \pi_{N(I-T)} f$. If $f \in R(I-T)$ then $\sigma_n (f) \to 0$ by (2.). 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 $|f-f_\varepsilon| < \varepsilon$. Then $$ \begin{align} |\sigma_n (f) - \sigma_n (f_\varepsilon)| &= \frac{|(f-f_\varepsilon)+T(f-f_\varepsilon)+T^2f+\cdots+T^{n-1} (f-f_\varepsilon)|}{n} \\ &\le \frac{|f-f_\varepsilon| (1+\|T\|+ \cdots + \|T^{n-1}\|)}{n} \\ &\le \frac{|f-f_\varepsilon| (1+\|T\|+ \cdots + \|T\|^{n-1})}{n} \\ &\le |f-f_\varepsilon|. \end{align} $$

Then $$ \begin{align} & |\sigma_n (f) - \pi_{N (I-T)} f| \\ \le & |\sigma_n (f)-\sigma_n (f_\varepsilon)| + |\sigma_n (f_\varepsilon) - \pi_{N(I-T)} f_\varepsilon| + |\pi_{N(I-T)} f_\varepsilon - \pi_{N(I-T)} f| \\ \le & |\sigma_n (f_\varepsilon) - \pi_{N(I-T)} f_\varepsilon| + (1+\|\pi_{N(I-T)}\|) |f-f_\varepsilon|. \end{align} $$

It follows that $\limsup_n |\sigma_n (f) - \pi_{N (I-T)} f| \le (1+\|\pi_{N(I-T)}\|) \varepsilon$. 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 $f_2=\pi_{\overline{R(I-T)}} f$.