# Let $A:H\to H$ where $H$ is Hilbert. Why $\mathcal D(A)\subset \mathcal D(A^*)$ ? (where $\mathcal D(A)$ is the domain of $A$).

Let $$(A,\mathcal D(A))$$ an unbounded densely defined linear operator on Hilbert space $$H$$ (of infinite dimensional). Defines $$\mathcal D(A^*)=\{f\in H\mid \exists u\in H:\forall \varphi \in \mathcal D(A), \left=\left\}.$$

I's written in my notes that $$\mathcal D(A)\subset \mathcal D(A^*)$$, but I don't really understand why. I tried : let $$f\in \mathcal D(A)$$. But I don't see how to construct $$u$$ s.t. $$\left=\left<\varphi ,u\right>,$$ for all $$\varphi \in \mathcal D(A)$$. I can do it if $$H$$ has finite dimension, but how can I do with infinite dimension ?

• Is $A$ densly defined ? – Fred Jul 12 at 7:10
• @Fred: Yes thank you. I edited the question. – user657324 Jul 12 at 13:29