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!
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$\begingroup$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 $V$, $V=V^{\perp\perp}$.
$\endgroup$ 8 $\begingroup$Note that $(A^\perp)^\perp=\overline{\operatorname{Span}(A)}$ is equivalent to $(A^\perp)^\perp$ being the smallest closed subspace that contains $A$.
It is obvious that $ A\subseteq (A^\perp)^\perp $ and $ (A^\perp)^\perp $ is a closed subspace of the Hilbert space $ \mathcal{H} $. It remains to prove that $ (A^\perp)^\perp $ is the smallest one. Suppose $ F $ is another closed subspace containing $ A $, then $ A^\perp\supset F^\perp $ and hence $ (A^\perp)^\perp\subset (F^\perp)^\perp $. Now it suffices to show that for a closed subspace $ F $, $ (F^\perp)^\perp\subseteq F $(thus we have $ F=(F^\perp)^\perp $). And if you know this fact then we are done.
If you do not know this fact, here comes the proof:
Since a closed subspace $ F $ of the Hilbert space $ \mathcal{H} $ is also a Hilbert space and we know there exists an orthonormal basis for $ F $, say $ \{u_\alpha\}_{\alpha\in I} $ where $ I $ is an index set. We claim that there exists an orthogonal basis $ \{v_\beta\}_{\beta\in J} $, where $ J $ is an index set, for $ \mathcal{H} $ such that $ \{u_\alpha\}_{\alpha\in I}\subseteq\{v_\beta\}_{\beta\in J} $.
To prove this fact, consider the poset $$ \mathcal{P}:=\{S\subsetneq\mathcal{H}: \{u_\alpha\}_{\alpha\in I}\subseteq S\text{ and }S\text{ is orthonormal}\}, $$
we know that $ \mathcal{P}\ne\emptyset $ and every chain has an upper bound in $ \mathcal{P} $ by taking their union, thus there exists a maximal element $ \tilde{S}\in\mathcal{P} $. If an element $ x\in\mathcal{H} $ satisfies $ \langle x,y\rangle=0 $ for any $ y\in\tilde{S} $, then by the maximality of $ \tilde{S} $, $ x $ has to be $ 0 $. Thus, by the definition of orthonormal basis of a Hilbert space, we know that $ \tilde{S} $ is an orthogonal basis for $ \mathcal{H} $. Write $ \tilde S=\{v_\beta\}_{\beta\in J} $ we know that $ \{v_\beta\}_{\beta\in J}\supseteq\{u_\alpha\}_{\alpha\in I} $.
Since every element $ x\in\mathcal{H} $ can be uniquely written as $$ x=\sum_{\alpha\in I}\langle x,u_\alpha\rangle u_\alpha+\sum_{\beta\in \{\beta\in J:v_\beta\notin\{u_\alpha\}_{\alpha\in I}\}}\langle x,v_\beta\rangle v_\beta, $$ it is clear that $ F^\perp=\operatorname{Span}(\{v_\beta\}_{\beta\in J}\setminus\{u_\alpha\}_{\alpha\in I}) $ and $ (F^\perp)^\perp=\operatorname{Span}(\{u_\alpha\}_{\alpha\in I})=F $ by the above expression and we are done.
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