Let $E$ and $F$ be two Banach spaces and let $ A:D(A)\subset E \rightarrow F$ be a densely defined unbounded operator.
(1) Prove that $N(A')=R(A)^{\bot}$ and $N(A)\subset R(A')^{\bot}$.
(2) Let's assume A is closed, prove that $N(A)=R(A')^{\bot}$.
I have already proved (1) and I am trying to prove (2) with the Hahn-Banach theorem.
By contradiction: if $\bar{u}\in R(A')^{\bot}$ such that $[\bar{u},0]\not\in G(A)$ ( $G(A)$ is the graph of A) there exists a function $f$ and some $\alpha \in \mathbb{R}$ such that: $f([u,Au])<\alpha<f([\bar{u},0])$ $\forall u\in D(A)$ . Cause $G(A)$ is a linear space $f([u,Au])=0$ $\forall [u,Au] \in G(A)$, so $f(\bar{u},0)>0$. Is there a way to find a contradiction saying $f(\bar{u},0)$ should be equal to $0$?
Could anyone help me?
Best Answer
$u\in R(A')^\perp$ means that for every $y'\in D(A')$ and every sequence $(u_n)\subset D(A)$ with $u_n\to u$ we have $y'(Au_n)\to 0$ as $n\to\infty$.
Now, as you started, if $(u,0)\notin G(A)$, then there exists a bounded linear functional $f$ on $E\times F$ such that $f(x,Ax) = 0$ for all $x\in D(A)$ and $f(u,0) = 1$. Now, define the functional $y' : F\to\Bbb K$ by $y'(y) := f(0,y)$. Then, obviously, $y'\in F'$. But also for $x\in D(A)$ we have $$ |y'(Ax)| = |f(0,Ax)| = |f(x,Ax)-f(x,0)| = |f(x,0)|\le \|f\|\|x\|. $$ That is, $y'\in D(A')$.
So, if $(u_n)\subset D(A)$ is a sequence with $\lim_{n\to\infty}u_n = u$, then \begin{align*} 0 &= \lim_{n\to\infty}y'(Au_n) = \lim_{n\to\infty}f(0,Au_n) = \lim_{n\to\infty}f(u_n,Au_n)-f(u_n,0)\\ &= -\lim_{n\to\infty}f(u_n,0) = -f(u,0) = -1, \end{align*} a contradiction.