# 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?.

• Notice that $C||Au||^2$ looks conspicuously different from $C\|Au\|^2.$ I changed this to the latter, which is standard usage. – Michael Hardy Mar 26 at 0:01
• So Brezis's book has been translated into English. (I suppose this probably shows how non-up-to-date my information on this point is. – Michael Hardy Mar 26 at 0:02
• . . . or maybe you just translated the title yourself? – Michael Hardy Mar 26 at 0:04
• @MichaelHardy: the English version I have access to is from 2010. Meanwhile, when I took Functional Analysis in 1989 we used the Spanish translation (dated 1984). – Martin Argerami Mar 26 at 0:07

$$R(A)^{\perp} \subset N(A^{*})$$: 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 $$y \in N(A^{*})$$.