Orthogonality Relations Exercise 2.19, Brezis' Book Functional Analysis I am studying some topics of Brezis's book and I am trying to solve this excercises:
Let $E$ a Banach space and let $A\,\colon\, D(A)\subset E \rightarrow E^*$ be a densely defined unbounded operator.
Assume that there exists a constant $C$ such that
$$
\langle Au, u \rangle \geq -C \|Au\|^2\quad \forall u\in D(A).
$$
Prove that $N(A)\subset N(A^*)$.
I was trying some solutions but, I coudn't. In fact, I search the hint in the back of the book and say:
Recall that $N(A^*)=R(A)^{\perp}$. Let $u\in N(A)$ and $v\in D(A)$; we have
$$
\langle A(u+tv),u+tv\rangle \geq -C \|A(u+tv)\|^2\quad \forall t\in\mathbb{R},
$$
which implies that $\langle A v,u\rangle=0$. Thus $N(A)\subset R(A)^\perp$.
Could anyone explain me why this inequality implies the result?.

$R(A)^{\perp} \subset N(A^{*})$: If $ \langle y, Ax \rangle=0$ for all $x \in D(A)$ then $ \langle A^{*}y, x \rangle=0$ for all $x \in D(A)$. Since $D(A)$ is dense it follows that $A^{*}y=0$ so $y \in N(A^{*})$.
