# For any vector subspace of a Hilbert space the orthogonal complement and the closure of that subspace form a direct sum of the entire space

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• Demote to grade B when a picture of my proof on paper has been included - Alec (talk) 04:07, 8 April 2017 (UTC)

## Statement

Let [ilmath]((X,[/ilmath][ilmath]\mathbb{K} [/ilmath][ilmath]),\langle\cdot,\cdot\rangle)[/ilmath] be a Hilbert space and let [ilmath]L\subseteq X[/ilmath] be a vector subspace of the vector space [ilmath](X,\mathbb{K})[/ilmath], then[1]:

• [ilmath]L^\perp[/ilmath] - the orthogonal complement of [ilmath]L[/ilmath] - is a (topologically) closed set of [ilmath]X[/ilmath][Note 1]
• [ilmath]X\eq \overline{L} \oplus L^\perp[/ilmath] - where [ilmath]\overline{A} [/ilmath] denotes the closure of a set [ilmath]A\in\mathcal{P}(X)[/ilmath] and [ilmath]\oplus[/ilmath] the direct sum of vector spaces
• This means [ilmath]\forall x\in X\exists x_1\in\overline{L}\exists x_2\in L^\perp\big[x\eq x_1+x_2\wedge\underbrace{\forall a\in\overline{L}\forall b\in L^\perp[x\eq a+b\implies (a\eq x_1\wedge b\eq x_2)]}_\text{Uniqueness}\ \big][/ilmath][Note 2]

## Proof

We have some precursors which we use here:

Proof:

• Let [ilmath]x\in X[/ilmath] be given
• By given a Hilbert space and a non-empty, closed and convex subset then for each point in the space there is a closest point in the subset we know:
• $\exists x_1\in\overline{L}\Big[\Vert x-x_1\Vert\eq\mathop{\text{Inf} }_{a\in \overline{L} }\Big(\Vert x-a\Vert\Big)\Big]$ and furthermore such an [ilmath]x_1[/ilmath] is unique.
• Define [ilmath]x_1\in\overline{L} [/ilmath] to be the unique point such that [ilmath]\Vert x-x_1\Vert\eq\mathop{\text{Inf} }_{a\in \bar{L} }\big(\Vert x-a\Vert\big)[/ilmath]
• Now [ilmath]x_2\eq x-x_1[/ilmath] is uniquely determined by [ilmath]x[/ilmath] and [ilmath]x_1[/ilmath] (as a vector space is an (additive) Abelian group)[Note 3]
• If [ilmath]x_2\in L^\perp[/ilmath] we'd be done, we will show this.
• Suppose [ilmath]x_2\notin L^\perp[/ilmath], that means [ilmath]\exists z\in\overline{L}[\langle x-x_1,z\rangle\neq 0][/ilmath]
• Define [ilmath]z\in\overline{L} [/ilmath] to be such that [ilmath]\langle x_2,z\rangle\neq 0[/ilmath]
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See[1] - page 35 (I've annotated it) - I'm not actually sure now if it shows the decomposition is unique though, surely there could be another [ilmath]x_1'\in\overline{L} [/ilmath] such that [ilmath]x_2':\eq x-x_1'\in L^\perp[/ilmath]? I don't think the uniqueness part is shown! Alec (talk) 04:07, 8 April 2017 (UTC)

## Notes

1. With [ilmath]X[/ilmath] considered with the topology induced by the metric [ilmath]d[/ilmath], which is the metric induced by the norm [ilmath]\Vert\cdot\Vert[/ilmath] which is the norm induced by the inner product [ilmath]\langle\cdot,\cdot\rangle[/ilmath]
2. That's just:
• [ilmath]\forall x\in X\exists x_1\in\overline{L}\exists x_2\in L^\perp[x\eq x_1+x_2][/ilmath] where the [ilmath]x_1[/ilmath] and [ilmath]x_2[/ilmath] are unique
• Note that if we want [ilmath]x\eq x_1+x_2[/ilmath] then [ilmath]x_2\eq x-x_1[/ilmath] and as a vector space is an Abelian group over addition, [ilmath]x_2[/ilmath] is unique if [ilmath]x_1[/ilmath] is
3. TODO: Expand this