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?.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.