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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?.

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  • $\begingroup$ Notice that $C||Au||^2$ looks conspicuously different from $C\|Au\|^2.$ I changed this to the latter, which is standard usage. $\endgroup$ – Michael Hardy Mar 26 at 0:01
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Mar 26 at 0:02
  • $\begingroup$ . . . or maybe you just translated the title yourself? $\endgroup$ – Michael Hardy Mar 26 at 0:04
  • $\begingroup$ @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). $\endgroup$ – Martin Argerami Mar 26 at 0:07
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$R(A)^{\perp} \subset N(A^{*})$: 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 $y \in N(A^{*})$.

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