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I'm a bit rusty in my Functional Analysis and couldn't solve this question:

Let $X$ be a Banach space (over either $\mathbb{R}$ or $\mathbb{C}$) and $X^*$ its dual space. Show that, if $A:D(A) \subset X \to X$ a closed and densely defined operator, then the annihilator of $D(A^*) \subset X^*$ is $\{0\}$.

The annihilators are defined as:

$\bullet$ If $M \subset X$, the annihilator of $M$ is $M^\perp := \{x^* \in X^* : \langle x,x^* \rangle = 0,\forall x \in M\}$

$\bullet$ If $M^*\subset X^*$, the annihilator of $M^*$ is $(M^*)^\perp := \{x \in X : \langle x,x^* \rangle = 0,\forall x^* \in M^*\}$

My intuition says that I should somehow include $A$ and $A^*$ in the definitions, so I can use the facts that both $A$ and $A^*$ are closed. It was also given in the question that $(M^\perp)^\perp = \overline{M}$, so it might appear somewhere in the proof.

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As $A$ is closed and densely defined we know that $A^*$ is closed and is densely defined in the weak* topology. Let $x\in D(A^*)^\perp$ and pick any $f\in X^*$. As $D(A^*)$ is weak* dense there is some $(f_n)\subset D(A^*)$ such that $f_n\xrightarrow{}f$ in the weak* topology. This means that $f_n(x)\to f(x)$, but $f_n(x)=0$ for all $n\in\mathbb N$, so $f(x)=0$. As this is true for arbitrary $f\in X^*$ it follows that $x\in (X^*)^\perp=\{0\}$ (because the dual space of a Banach space separates the points of the space).

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  • $\begingroup$ For $A^*$ to be densely defined, don't we need that $X$ is reflexive? $\endgroup$ – AspiringMathematician Mar 22 at 11:58
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    $\begingroup$ @AspiringMathematician Yes for $A^*$ to be densely defined in the norm topology, but it is always densely defined in the weak* topology. It is a well known result that the weak* closure of $D(A^*)$ is $X^*$. $\endgroup$ – K.Power Mar 22 at 12:00

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