Is $A+A^{\perp}$, dense in $H$ for any $A$.

Question

Let assume $H$ is a Hilbert space and $A \subset H$ is a linear subspace. Also assume $A^{\perp}$ denotes the orthonormal complemet of $A$. Is this true that: $$A+A^{\perp}$$ is dense in $H$?

From a classical Theorem, it is well known that if $A$ is closed linear subspace, then $$A+A^{\perp}=H.$$ How about non-closed linear subsets?

I thank you for any hint or comment.

Answer 1

I suppose you are asuming that $A$ is a linear subspace. In that case any $x \in H$ can be written as $a+b$ with $a \in \overline {A}$ and $b \in A^{\perp}$. [This is because $A$ and $\overline {A}$ have the same orthogonal complement]. Now it is obvious that $x =\lim a_n +b$ for some sequence $a_n $ in $A$ and some $b \in A^{\perp}$.

Answer 2

Yes it is dense. Just observe that $A^\perp$ is closed so by the standard theorem $A^\perp + A^{\perp\perp}=H$. But $A^{\perp\perp}$ is the closure of $A$. Therefore $\bar A + A^\perp=H$ and from this density follows in a straightforward manner.