# Hilbert space H is separable if and only if the dual space H* is separable

The direction (<=) is okay since I could use the fact that if the dual space H* of a normed space H is separable, then H is separable. I would like to ask how to show the other direction.

My attempt is to use the fact that Hilbert spaces are isomorphic if and only if they have the same orthonormal dimension. I know the dual space of Hilbert space is Hilbert space and dim(H)=dim(H*). But I don't know how to prove that their orthonormal dimensions are equal and I am not even sure this approach is correct or not. May I get some help?

• Do you know the Riesz representation theorem? Jun 28 at 11:20
• Do you know that Hilbert spaces are self-dual? Jun 28 at 11:35
• @jakobdt Thank you. That's a great hint. Jun 28 at 11:52
• @Hanno I don't know that it's a result of the Riesz representation theorem before asking. I would try to understand the proof. Thanks for you help. Jun 28 at 11:57

The Riesz representation theorem implements a conjugate-linear isometry $$\Phi: H \to H^*: \xi \mapsto \langle \xi\mid -\rangle$$(assuming that inner product is linear in the second factor). In particular, $$\Phi$$ is a homeomorphism of the underlying topological spaces, and thus $$H$$ is separable if and only if $$H^*$$ is separable.
The idea you provide can also be made to work, but you will essentially be reproving the Riesz representation theorem and the argument is a bit more roundabout. Concretely, if $$\{e_i: i \in I\}$$ is an orthonormal basis for $$H$$, then $$\{\langle e_i\mid-\rangle \in H^*: i \in I\}$$ is an orthonormal basis for $$H$$. Hence, $$\dim_{\operatorname{Hilb}}(H) = \dim_{\operatorname{Hilb}}(H^*).$$ Recalling that $$H$$ is separable if and only if $$\dim_{\operatorname{Hilb}}(H)\le \aleph_0$$, we also conclude the result.