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

From Maths
Jump to: navigation, search


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[1]:

  • 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 [math]\Vert x-p\Vert\eq\mathop{\text{Inf} }_{a\in A}\Big(\Vert x-a\Vert\Big)[/math]
      • 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]


Caution:This proof requires the axiom of choice
Grade: D
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
It only requires writing up really

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.


  1. Recall:
  2. This is a topological term, now we are talking about a topological space, [ilmath](X,\mathcal{J})[/ilmath], and saying [ilmath]A[/ilmath] is a closed set of that space.
  3. Recall [ilmath]\mathbb{K} [/ilmath] as a field means [ilmath]\mathbb{K}:\eq\mathbb{R} [/ilmath] or [ilmath]\mathbb{K}:\eq\mathbb{C} [/ilmath], the reals or the complex number fields only
  4. TODO: We do not know whether this project will use [ilmath]\exists ! p[/ilmath] yet. However for now the reader knows what uniqueness means, it's bog standard


  1. Warwick 2014 Lecture Notes - Functional Analysis - Richard Sharp