Double orthogonal complement of any closed subspace is it self

2.9k Views Asked by At

Let $H$ be a pre-Hilbert space such that any closed sub space $M \subset H$ has the property $M^{\bot \bot}=M$. Prove that $H$ is a Hilbert space (ie, prove that $H$ is complete)

My attempt:

As $H^*$ ($L(H,\mathbb{K})$) is complete then I want to prove that $H \cong H^*$. But I don't know how to construct an isometry between them.

I will appreciate highly who can give me some ideas

Thank in advance!