I'm trying to solve below exercise in Brezis' Functional Analysis
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 $u \in H$. Let $N(S)$ be the kernel of $S$ and $R(S)$ the range of $S$. Then $N(S) = R(S)^\perp$.
Could you verify my below attempt?
Fix $x \in N(S)$. We have $\langle S(u-x), u-x \rangle \ge 0$ for all $u\in H$. Because $Sx =0$, we get $\langle Su, u \rangle \ge \langle Su, x \rangle$ for all $u\in H$. Then $\langle S(ru), ru \rangle \ge \langle S(ru), x \rangle$ for all $u\in H$ and for all $r \in \mathbb R$.
- First, $r \langle S u, u \rangle \ge \langle Su, x \rangle$ for all $u\in H$ and for all $r > 0$. Taking the limit $r \downarrow 0$, we get $0 \ge\langle Su, x \rangle$ for all $u\in H$.
- Second, $r \langle S u, u \rangle \le \langle Su, x \rangle$ for all $u\in H$ and for all $r < 0$. Taking the limit $r \uparrow 0$, we get $0 \le \langle Su, x \rangle$ for all $u\in H$.
It follows that $\langle Su, x \rangle = 0$ for all $u\in H$ and thus $x \in R(S)^\perp$. Then $N(S) \subset R(S)^\perp$.
Let $x \in R(S)^\perp$. Then $\langle S u, x \rangle =0$ for all $u\in H$. We have $\langle S(u-x), u-x \rangle \ge 0$ for all $u\in H$. Then $\langle S(u-x), u \rangle \ge 0$ for all $u\in H$. Then $\langle Su, u \rangle \ge \langle Sx, u \rangle$ for all $u\in H$. With the same argument as above, we get $\langle Sx, u \rangle = 0$ for all $u\in H$. This implies $Sx=0$ and thus $x \in N(S)$. Then $R(S)^\perp \subset N(S)$. This completes the proof.