0
$\begingroup$

I'm supposed to show $f\in N(A^{*})^{\perp}$ in a proof. Here, we're dealing with a compact linear injective operator $A:X\rightarrow Y$, where both $X$ and $Y$ are Hilbert spaces. Furthermore, $A$ has dense range, i.e., $\overline{A(X)}=Y$. In the proof it is assumed $f\not\in V:=\{A\varphi\ |\ \varphi\in X,\ ||\varphi||\leq\rho\}$ for some $\rho>0$.

I have constructed the following argument:

From the information of $A$ we know the adjoint operator $A^{*}$ will be linear and injective. Therefore $N(A^{*})=\{0\}$, since for linear operators being injective means $A^{*}g=0\Rightarrow g=0$.

Since it is assumed $f\not\in V$ that implies $f\neq 0$ and therefore $f\perp N(A^{*})$, i.e., $f\in N(A^{*})^{\perp}$.

Does this make sense or do I have some mistakes/lack of arguments that $f\in N(A^{*})^{\perp}$?

Thanks in advance.

$\endgroup$
  • $\begingroup$ Why $A^*$ is injective ? $\endgroup$ – S. Cho Jan 10 at 22:29
  • $\begingroup$ @S.Cho Because $A$ has dense range. $\endgroup$ – James Jan 10 at 22:30
  • $\begingroup$ Okey, but if $N(A^*)=\{0\}$, then $N(A^{*})^{\perp}=Y$. Following what you wrote you get $f\in Y$. I think this is not important. $\endgroup$ – S. Cho Jan 10 at 22:36
  • $\begingroup$ @S.Cho The whole idea is for me to argue $f\in N(A^{*})^{\perp}$ since that implies $f\in\overline{A(X)}$ and that allows me to use a theorem to show something has a solution. $\endgroup$ – James Jan 10 at 22:40
  • $\begingroup$ What is the information on $f$, if it is just an element of $Y$ the result is obvious, since $Y=N(A^{*})^{\perp}$. If not, you need to clarify your question. $\endgroup$ – S. Cho Jan 10 at 22:43
1
$\begingroup$

I assume that you want to prove the following result: Consider the operator equation $$Au=f, u\in X, \; f\in Y, (1).$$ Then, a necessary condition for existence of a solution $u$ to $(1)$ is $f \in N(A^{*})^{\perp}$.

Assume that $(1)$ has a solution $u$. Let $v\in N(A^{*})$, we have $$(f,v)_Y=(Au, v)_Y=(u, A^* v)_Y=0,$$ Then, $f\perp v$ for all $v \in N(A^{*})$, which implies that $f \in N(A^{*})^{\perp}$.

PS: This result not require injectivity or dense range assumptions.

$\endgroup$
  • 1
    $\begingroup$ Thank you, this makes sense! $\endgroup$ – James Jan 11 at 0:15
1
$\begingroup$

Since $N(A^{*})=\{0\}$ its orthogonal complement is the entire space $Y$. So any vector $f$ belongs to it. I don't what you mean when you say ' $f\neq 0$ and therefore $f\perp (N(A^{*})$'. There is really nothing to prove here once you know that $N(A^{*})=\{0\}$.

$\endgroup$
  • $\begingroup$ Thank you for the response. I can see what you mean, I think I just made it more complicated than it needed to be. But is my way of arguing $N(A^{*})=\{0\}$ valid? $\endgroup$ – James Jan 11 at 0:12
  • 1
    $\begingroup$ @James Yes, it is valid. The fact that range of $A$ is dense implies that kernel of $A^{*}$ is $\{0\}$. $\endgroup$ – Kavi Rama Murthy Jan 11 at 0:27

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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