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

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. $\left<A\varphi ,f\right>=\left<\varphi ,u\right>,$ for all $\varphi \in \mathcal D(A)$. I can do it if $H$ has finite dimension, but how can I do with infinite dimension ?

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

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