Exercise 5.20.1 in Brezis' Functional Analysis is stated as follows
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 $N(S) = R(S)^\perp$.
At page 45 of the same book, we have
Corollary 2.18. Let $E, F$ be Banach spaces. Let $A: D(A) \subset E \rightarrow F$ be an unbounded linear operator with $D(A)$ a vector subspace of $E$. Let $A$ be densely defined, i.e., $D(A)$ is dense in $E$. Let $A$ be closed, i.e., its graph is closed in $E \times F$. If $A^*: D(A^*) \subset F^* \rightarrow E^*$ is the adjoint operator of $A$, then $$ \begin{aligned} N(A) & =R\left(A^{*}\right)^{\perp}, \\ N\left(A^{*}\right) & =R(A)^{\perp}, \\ N(A)^{\perp} & \supset \overline{R\left(A^{*}\right)}, \\ N\left(A^{*}\right)^{\perp} & =\overline{R(A)} . \end{aligned} $$
Let $S$ be the operator in Exercise 5.20.1 and $S^*$ its adjoint. Then Corollary 2.18. implies that $N(S) = N(S^*)$ and $R(S)^\perp = R(S^*)^\perp$.
- Is it true that $R(S) = R(S^*)$?
- Is there a geometric meaning of the condition $\langle Su, u \rangle \ge 0 \quad \forall u \in H$?
Thank you so much for your elaboration!
Here is an example that $R(S)\ne R(S^*)$. Let $H=l^2$ and $T$ the right-shift operator. Then $T^*$ is the left-shift and $T^*T=I$. Define $S:=I-T$. Then $$ \langle Su,u\rangle = \langle (I-\frac12 T - \frac12 T^*)u,u\rangle = \frac12 \langle (I- T^*)(I- T)u,u\rangle = \|(I-R)u\|^2. $$ It is easy to check that $N(S)=N(S^*)=\{0\}$.
Let $e_1 = (1,0,\dots)$. Then $S^*e_1 = e_1$, but $e_1 \not\in R(S)$. If $Sv = e_1$ it follows inductively that $v=(1,1,\dots)$, which is not in $l^2$.
The claim $R(S)=R(S^*)$ is true if $R(S)$ (or equivalently $R(S^*)$) is closed.
The condition $\langle Su,u\rangle\ge0$ for all $u$ is equivalent to $\langle (S+S^*)u,u\rangle\ge0$, which means that the self-adjoint (or symmetric) part $\frac12(S+S^*)$ of $S$ is positive semidefinite.