View source for 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

{{Stub page|grade=A*|msg=Link to it and demote.
* Demote to grade B when a picture of my proof on paper has been included - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 04:07, 8 April 2017 (UTC)}}
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==Statement==
Let {{M|((X,}}[[K (field)|{{M|\mathbb{K} }}]]{{M|),\langle\cdot,\cdot\rangle)}} be a [[Hilbert space]] and let {{M|L\subseteq X}} be a [[vector subspace]] of the [[vector space]] {{M|(X,\mathbb{K})}}, then{{rW2014LNFARS}}:
* {{M|L^\perp}} - the [[orthogonal complement]] of {{M|L}} - is a (topologically) [[closed set]] of {{M|X}}<ref group="Note">With {{M|X}} considered with the [[topology induced by the metric]] {{M|d}}, which is the [[metric induced by the norm]] {{M|\Vert\cdot\Vert}} which is the [[norm induced by the inner product]] {{M|\langle\cdot,\cdot\rangle}}</ref>
* {{M|X\eq \overline{L} \oplus L^\perp}} - where {{M|\overline{A} }} denotes the {{link|closure|topology}} of a set {{M|A\in\mathcal{P}(X)}} and {{M|\oplus}} the [[direct sum of vector spaces]]
** This means {{M|\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]}}<ref group="Note">That's just:
* {{M|\forall x\in X\exists x_1\in\overline{L}\exists x_2\in L^\perp[x\eq x_1+x_2]}} where the {{M|x_1}} and {{M|x_2}} are [[unique]]
** Note that if we want {{M|x\eq x_1+x_2}} then {{M|x_2\eq x-x_1}} and as a [[vector space]] is an [[Abelian group]] over addition, {{M|x_2}} is unique if {{M|x_1}} is</ref>
==Proof==
We have some precursors which we use here:
# [[The orthogonal complement of a vector subspace is a vector subspace]]
# [[The orthogonal complement of a vector subspace is a topologically closed set]]
# [[The closure of a linear subspace of a normed space is a linear subspace]]
# [[A subspace of a vector space over the real or complex numbers is a convex set]] and
# [[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]]

'''Proof:'''
* Let {{M|x\in X}} 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:
**:* {{MM|\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 {{M|x_1}} is [[unique]].
** Define {{M|x_1\in\overline{L} }} to be the unique point such that {{M|\Vert x-x_1\Vert\eq\mathop{\text{Inf} }_{a\in \bar{L} }\big(\Vert x-a\Vert\big)}}
*** Now {{M|x_2\eq x-x_1}} is ''uniquely'' determined by {{M|x}} and {{M|x_1}} (as a [[vector space]] is an (additive) [[Abelian group]])<ref group="Note">{{XXX|Expand this}}</ref>
*** If {{M|x_2\in L^\perp}} we'd be done, we will show this.
**** Suppose {{M|x_2\notin L^\perp}}, that means {{M|\exists z\in\overline{L}[\langle x-x_1,z\rangle\neq 0]}}
***** Define {{M|z\in\overline{L} }} to be such that {{M|\langle x_2,z\rangle\neq 0}}
{{Requires proof|See{{rW2014LNFARS}} - 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 {{M|x_1'\in\overline{L} }} such that {{M|x_2':\eq x-x_1'\in L^\perp}}? I don't think the uniqueness part is shown! [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 04:07, 8 April 2017 (UTC)}}
==Notes==
<references group="Note"/>
==References==
<references/>
{{Theorem Of|Functional Analysis|Analysis}}