# 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

## Statement

Let [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] be a Hilbert space[Note 1] and let [ilmath]A\in[/ilmath][ilmath]\big(\mathcal{P}(X)-\{\emptyset\}\big)[/ilmath] then:

• If we both of have the following:
1. [ilmath]A[/ilmath] is a closed set with respect to [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] considered as a topological space[Note 2]
2. [ilmath]A[/ilmath] is a convex set with respect to [ilmath](X,[/ilmath][ilmath]\mathbb{K} [/ilmath][ilmath])[/ilmath][Note 3] considered as a vector space
• Symbolically: [ilmath]\forall a,b\in A\forall\lambda\in[0,1]\subset\mathbb{R}\big(\subseteq\mathbb{K}\big)[a+\lambda(b-a)\in A][/ilmath]
• Then:
• There exists a unique [ilmath]p\in A[/ilmath] such that $\Vert x-p\Vert\eq\mathop{\text{Inf} }_{a\in A}\Big(\Vert x-a\Vert\Big)$
• Symbolically: [ilmath]\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][/ilmath] - furthermore such a [ilmath]p[/ilmath] is unique[Note 4]
• In totality: [ilmath]\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][/ilmath]

## Proof

Caution:This proof requires the axiom of choice
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### Uniqueness claim

Suppose [ilmath]p[/ilmath] and [ilmath]p'[/ilmath] both satisfy this property and that [ilmath]p\neq p'[/ilmath], then:

• [ilmath]\Vert p-p'\Vert^2 \le 2\Vert x-p\Vert^2+2\Vert x-p'\Vert^2 - 4I^2[/ilmath]
• But [ilmath]\Vert x-p\Vert\eq I[/ilmath] by definition of [ilmath]p[/ilmath] and [ilmath]\Vert x-p'\Vert\eq I[/ilmath] by definition of [ilmath]p'[/ilmath]
• So [ilmath]\Vert p-p'\Vert^2 \le 2I^2 + 2I^2 - 4I^2\eq 0[/ilmath]
• But [ilmath]\forall x\in X[\Vert x\Vert\ge 0][/ilmath]
• So we have [ilmath]\Vert p-p'\Vert\eq 0[/ilmath]
• But by definition of norm: [ilmath]\forall x\in X[\Vert x\Vert\eq 0\iff x\eq 0][/ilmath]
• So [ilmath]p-p'\eq 0[/ilmath], this means [ilmath]p\eq p'[/ilmath]
• This contradicts that [ilmath]p\neq p'[/ilmath], so we see [ilmath]p[/ilmath] must be unique.