If $H$ is a Hilbert space with subspace $A$ such that the orthogonal complement $A^\perp$ of $A$ is trivial, i.e. $A^\perp = \{ 0 \}$, must $A$ be dense in $H$?
Identifying a dense subspace of a Hilbert space
functional-analysishilbert-spaces
functional-analysishilbert-spaces
If $H$ is a Hilbert space with subspace $A$ such that the orthogonal complement $A^\perp$ of $A$ is trivial, i.e. $A^\perp = \{ 0 \}$, must $A$ be dense in $H$?
Best Answer
For a simple proof that this is true, use the fact that for a subspace $W$ we have $(W^{\perp})^{\perp} = \overline{W}$; thus the closure of $A$ is the whole space $H$.