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.


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).

  • $\begingroup$ For $A^*$ to be densely defined, don't we need that $X$ is reflexive? $\endgroup$ – AspiringMathematician Mar 22 at 11:58
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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