# Proof $f\in N(A^{*})^{\perp}$

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 $$f\in N(A^{*})^{\perp}$$?

• Why $A^*$ is injective ? – S. Maths Jan 10 at 22:29
• @S.Cho Because $A$ has dense range. – James Jan 10 at 22:30
• 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. – S. Maths Jan 10 at 22:36
• @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. – James Jan 10 at 22:40
• 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. – S. Maths Jan 10 at 22:43

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.

• Thank you, this makes sense! – James Jan 11 at 0:15

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 $$N(A^{*})=\{0\}$$.

• 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? – James Jan 11 at 0:12
• @James Yes, it is valid. The fact that range of $A$ is dense implies that kernel of $A^{*}$ is $\{0\}$. – Kavi Rama Murthy Jan 11 at 0:27