# If operator is closed and densely defined then $D(A^*)^\perp = \{0\}$

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.

## 1 Answer

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).

• For $A^*$ to be densely defined, don't we need that $X$ is reflexive? – AspiringMathematician Mar 22 at 11:58
• @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 $X^*$. – K.Power Mar 22 at 12:00