So I'm in an advanced Linear Algebra class and we just moved into Hilbert spaces and such, and I'm struggling with this question.
Let $A$ be a nonempty subset of a Hilbert space $H$. Denote by $\operatorname{span}(A)$ the linear subspace of all finite linear combinations of vectors in $A$, and by $\overline{\operatorname{span}(A)}$ the
topological closure of $\operatorname{span}(A)$ with respect to $\|\cdot\|$.
Also, let $A^⊥ = \{h ∈ H : \langle h,f \rangle = 0, ∀f ∈ A\}$
and $A^{⊥⊥} = (A^⊥)^⊥$.
Use orthogonal projection to prove that
$A^{⊥⊥} =\overline{\operatorname{span}(A)}$.
The thing that trips me up is that we don't know much about $A$, like if I knew a little more, perhaps to show it's closed, then I can do the direct sum decomposition blah blah, but it also confuses me why we're using the complement of $\operatorname{span}(A)$. Is it possible to show that $\operatorname{span}(A)$ is closed, then go from there? I know it might look a bit like a duplicate, but all the questions I find don't refer to orthogonal projection at all. Any hints would be greatly appreciated!
Best Answer
Let $U=\operatorname{span}(A)$. Then it's easy to see that $$ U^{\perp}=A^{\perp} $$ It's also easy to see that $$ \overline{U}^\perp=U^{\perp} $$ (where $\overline{U}$ denotes the closure of $U$) using continuity of the inner product. Thus $$ \overline{U}=\overline{U}^{\perp\perp}=U^{\perp\perp} $$ assuming you know that, for a closed subspace $V$, $V=V^{\perp\perp}$.