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 $H$?

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    $\begingroup$ Yes, otherwise there is some element $x$ that cannot be approximated from $A$, and by virtue, neither can the entire subspace generated by $x$ which contradicts the orthogonal complement being trivial. $\endgroup$ Sep 2, 2019 at 17:48

1 Answer 1


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 $H$.


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