Double orthogonal complement is equal to topological closure

Question

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!

Answer 1

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 $V$, $V=V^{\perp\perp}$.

Answer 2

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.