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I want to show that $\operatorname{Range}(A^*)^\perp \subset \operatorname{Null}(A)$ where $A:E \supset D(A) \to F$ is an unbounded closed linear operator densely defined in $E$, and $E$ and $F$ are Banach Spaces.

Notice this is trivial when $A$ is bounded. However, I am struggling to prove it with this hypothesis.

My idea so far has been to try a proof by contradiction:

Let $u \in \operatorname{Range}(A^*)^\perp$ be such that $(u,0) \notin \operatorname{Graph}(A)$. Then by Hahn Banach we can strictly separate $\{(u,0)\}$ from $\operatorname{Graph}(A)$ with a hyperplane $f \in (E \times F)^*$, say f(u,0) < \alpha < f(v,Av) \quad \forall v \in D(A).

I guess from this I should be able to find $g \in D(A^*)\subset F^*$ such that $g(Au) \neq 0$. This would contradict $u \in \operatorname{Range}(A^*)^\perp$ since $0=A^*(g)(u)=g(Au)$ for all $v\in D(A^*)$.

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    Thank you t.b., I'll have a look in Yosida. I think I am close to figuring it out on my own though. I'll try to post the full proof later.2012-04-13

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Ok, so I think I got it. If you see any problem, please let me know:

Proof:

We argue by contradiction. Let $u \in \operatorname{Range}(A^*)^\perp$ be such that $(u,0) \notin \operatorname{Graph}(A)$. Then since $\operatorname{Graph}(A)$ is closed, we can use Hahn Banach to strictly separate $\{(u,0)\}$ from $\operatorname{Graph}(A)$ with a hyperplane $f \in (E \times F)^*$, say f(u,0) < \alpha < f(v,Av) \quad \forall v \in D(A).

Since $D(A)$ is a subspace we must have $f(v,Av)=0$ for all $v\in D(A)$. Define $g(y)=f(0,y)$. Note that $\left|g(Av)\right|=\left|f(v,Av)-f(v,0)\right|=\left|f(v,0)\right|\leq \underbrace{\|f\|}_c \|v\|$ so that $g \in D(A^*)$. Note also that $g(Au)=f(u,Au)-f(u,0)>0$. But this contradicts $u\in \operatorname{Range}(A^*)^\perp$ since this means that $0=A^*(v)(u)=v(Au)$ for all $v\in D(A^*)$.

$\square$

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    Thanks. I included some minor changes as you suggested.2012-04-14