View source for 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

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==Statement==
Let {{M|(X,\langle\cdot,\cdot\rangle)}} be a [[Hilbert space]]<ref group="Note">Recall:
* A Hilbert space is an [[inner product space]], {{M|(X,\langle\cdot,\cdot\rangle)}} that is also a [[complete metric space]] when considered with the [[metric induced by the norm]], {{M|d:(x,y)\mapsto\Vert x,y\Vert}} where the [[norm is induced by the inner-product]], {{M|\Vert\cdot\Vert:x\mapsto\sqrt{\langle x,x\rangle} }}
** So: {{MM|d(x,y):\eq\Vert x-y\Vert:\eq\sqrt{\langle x-y,\ x-y\rangle} }}</ref> and let {{M|A\in}}[[non-empty subset|{{M|\big(\mathcal{P}(X)-\{\emptyset\}\big)}}]] then{{rW2014LNFARS}}:
* If we both of have the following:
*# {{M|A}} is a [[closed set]] with respect to {{M|(X,\langle\cdot,\cdot\rangle)}} considered as a [[topological space]]<ref group="Note">This is a topological term, now we are talking about a [[topological space]], {{M|(X,\mathcal{J})}}, and saying {{M|A}} is a [[closed set]] of that space.
** Where {{M|\mathcal{J} }} is the [[topology induced by the metric]] {{M|d}}, which was induced by the [[norm]], {{M|\Vert\cdot\Vert}}, which was itself induced by the [[inner product]]!</ref>
*# {{M|A}} is a [[convex set]] with respect to {{M|(X,}}[[K (field)|{{M|\mathbb{K} }}]]{{M|)}}<ref group="Note">Recall {{M|\mathbb{K} }} as a field means {{M|\mathbb{K}:\eq\mathbb{R} }} or {{M|\mathbb{K}:\eq\mathbb{C} }}, the reals or the complex number fields only</ref> considered as a [[vector space]]
*#* Symbolically: {{M|\forall a,b\in A\forall\lambda\in[0,1]\subset\mathbb{R}\big(\subseteq\mathbb{K}\big)[a+\lambda(b-a)\in A]}}
* Then:
** There exists a [[unique]] {{M|p\in A}} such that {{MM|\Vert x-p\Vert\eq\mathop{\text{Inf} }_{a\in A}\Big(\Vert x-a\Vert\Big)}}
*** Symbolically: {{M|\forall x\in X\exists p\in A\big[\Vert x-p\Vert\eq\mathop{\text{Inf} }_{a\in A}(\Vert x-a\Vert)\big]}} - furthermore such a {{M|p}} is [[unique]]<ref group="Note">{{XXX|We do not know whether this project will use {{M|\exists ! p}} yet. However for now the reader knows what uniqueness means, it's bog standard}}</ref>
**** In totality: {{M|\forall x\in X\exists p\in A\big[\Vert x-p\Vert\eq\mathop{\text{Inf} }_{a\in A}(\Vert x-a\Vert)\wedge\underbrace{\forall q\in A[\Vert x-q\Vert\eq\mathop{\text{Inf} }_{a\in A}(\Vert x-a\Vert)\implies p\eq q]}_\text{Uniqueness}\big]}}

==Proof==
: {{Caution|This proof requires [[the axiom of choice]][[Category:Requires choice]]}}
{{Requires proof|grade=D|msg=It only requires writing up really
<gallery>
ClosedConvexSubsetNonEmptyInHilbertSpaceProofThing.jpg|Paper proof
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}}
===Uniqueness claim===
Suppose {{M|p}} and {{M|p'}} both satisfy this property and that {{M|p\neq p'}}, then:
* {{M|\Vert p-p'\Vert^2 \le 2\Vert x-p\Vert^2+2\Vert x-p'\Vert^2 - 4I^2}}
** But {{M|\Vert x-p\Vert\eq I}} by definition of {{M|p}} and {{M|\Vert x-p'\Vert\eq I}} by definition of {{M|p'}}
** So {{M|\Vert p-p'\Vert^2 \le 2I^2 + 2I^2 - 4I^2\eq 0}}
*** But {{M|\forall x\in X[\Vert x\Vert\ge 0]}}
**** So we have {{M|\Vert p-p'\Vert\eq 0}}
***** But by definition of [[norm]]: {{M|\forall x\in X[\Vert x\Vert\eq 0\iff x\eq 0]}}
****** So {{M|p-p'\eq 0}}, this means {{M|p\eq p'}}
******* This contradicts that {{M|p\neq p'}}, so we see {{M|p}} must be unique.
==Notes==
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==References==
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{{Theorem Of|Functional Analysis|Analysis|Complex Analysis|Real Analysis|Metric Space|Topology}}