Brezis' exercise 5.20.3

Question

I'm reading the solution of part (3.) of exercise 5.20 in Brezis' Functional Analysis.

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $S :H \to H$ be a bounded linear operator such that $\langle Su, u \rangle \ge 0$ for all $u \in H$. Let $N(S)$ be the kernel of $S$ and $R(S)$ the range of $S$. Then

1. $N(S) = R(S)^\perp$.
2. The map $I + t S$ is bijective for all $t>0$. Here $I:H \to H$ is the identity map.
3. $\lim_{n \to \infty} (I +n S)^{-1} f = \pi_{N(S)} f$ for all $f\in H$. Here $\pi_{N(S)} f$ is the orthogonal projection of $f$ onto $N(S)$.

Proof Let $x_n := (I +n S)^{-1} f$. Then $x_n + n S x_n =f$.

Let $f \in N(S)$. By (1.), we have $\langle Sx_n, f \rangle=0$. Then $$ \langle Sx_n, x_n + n S x_n \rangle = \langle Sx_n, x_n \rangle + n |Sx_n|^2 =0. $$

Then $Sx_n=0$ and thus $x_n=f$. Then $x_n \to f$.

Let $f \in R(S)$. Then there is $g \in S$ such that $S g=f$. Then $x_n = S(g-n x_n)$. We have $0 \le \langle S(g-n x_n), g-n x_n \rangle =\langle x_n, g-n x_n \rangle$. Then $$ |x_n | |g| \ge\langle x_n, g \rangle \ge n |x_n|^2. $$

Then $|g| \ge n |x_n|$ and thus $x_n \to 0$. By density, one can still prove that $x_n \to 0$ as $n \to \infty$ for every $f \in \overline{R(S)}$ (fill in the details).

In the general case $f \in H$, we write write $f=f_1+f_2$ with $f_1=\pi_{N(S)} f$ and $f_2=\pi_{\overline{R(S)}} f$.

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Could you confirm if I "fill in the details" correctly?

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Fix $f \in \overline{R(S)}$ and $\varepsilon >0$. There is $f_\varepsilon \in R(S)$ such that $| f - f_\varepsilon | < \varepsilon$. Let $$ x_n := (I +n S)^{-1} f_\varepsilon \quad \text{and} \quad y_n := (I +n S)^{-1} f. $$

We have proved that $x_n \to \pi_{N(S)} f_\varepsilon$. We have $x_n + n S x_n =f_\varepsilon$ and $y_n + n S y_n =f$ for all $n$. Then $$ (x_n-y_n) + nS(x_n-y_n) = f_\varepsilon- f. $$

Then $$ |x_n-y_n|^2 + 2n \langle x_n-y_n, S(x_n-y_n) \rangle + n^2 |S(x_n-y_n)|^2 = |f_\varepsilon- f|^2. $$

It follows from $\langle x_n-y_n, S(x_n-y_n) \rangle \ge 0$ that $|x_n-y_n| \le |f_\varepsilon- f|$. Then $$ \begin{align} & |y_n - \pi_{N (S)} f| \\ \le & |y_n-x_n| + |x_n - \pi_{N (S)} f_\varepsilon | + |\pi_{N (S)} f_\varepsilon - \pi_{N (S)} f| \\ \le & |f_\varepsilon- f| + |x_n - \pi_{N (S)} f_\varepsilon | + \|\pi_{N (S)}\|_{\mathcal L (H)} |f_\varepsilon- f| \\ = & (1 + \|\pi_{N (S)}\|_{\mathcal L (H)} ) |f_\varepsilon- f| + |x_n - \pi_{N (S)} f_\varepsilon |. \end{align} $$

Then $$ \limsup_n |y_n - \pi_{N (S)} f| \le (1 + \|\pi_{N (S)}\|_{\mathcal L (H)} ) \varepsilon. $$

The claim then follows by taking the limit $\varepsilon \downarrow 0$.