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
Revision as of 00:52, 7 April 2017 by Alec (Talk | contribs) (Created page with "__TOC__ ==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\c...")
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^{[1]}:
- If we both of have the following:
- [ilmath]A[/ilmath] is a closed set with respect to [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] considered as a topological space^{[Note 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]
- 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]}
- 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]
Proof
- 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:
The message provided is:
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.
- So [ilmath]p-p'\eq 0[/ilmath], this means [ilmath]p\eq p'[/ilmath]
- But by definition of norm: [ilmath]\forall x\in X[\Vert x\Vert\eq 0\iff x\eq 0][/ilmath]
- So we have [ilmath]\Vert p-p'\Vert\eq 0[/ilmath]
- But [ilmath]\forall x\in X[\Vert x\Vert\ge 0][/ilmath]
Notes
- ↑ Recall:
- A Hilbert space is an inner product space, [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] that is also a complete metric space when considered with the metric induced by the norm, [ilmath]d:(x,y)\mapsto\Vert x,y\Vert[/ilmath] where the norm is induced by the inner-product, [ilmath]\Vert\cdot\Vert:x\mapsto\sqrt{\langle x,x\rangle} [/ilmath]
- So: [math]d(x,y):\eq\Vert x-y\Vert:\eq\sqrt{\langle x-y,\ x-y\rangle} [/math]
- A Hilbert space is an inner product space, [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] that is also a complete metric space when considered with the metric induced by the norm, [ilmath]d:(x,y)\mapsto\Vert x,y\Vert[/ilmath] where the norm is induced by the inner-product, [ilmath]\Vert\cdot\Vert:x\mapsto\sqrt{\langle x,x\rangle} [/ilmath]
- ↑ 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.
- Where [ilmath]\mathcal{J} [/ilmath] is the topology induced by the metric [ilmath]d[/ilmath], which was induced by the norm, [ilmath]\Vert\cdot\Vert[/ilmath], which was itself induced by the inner product!
- ↑ 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
- ↑ 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
References
Categories:
- XXX Todo
- Requires choice
- Pages requiring proofs
- Theorems
- Theorems, lemmas and corollaries
- Functional Analysis Theorems
- Functional Analysis Theorems, lemmas and corollaries
- Functional Analysis
- Analysis Theorems
- Analysis Theorems, lemmas and corollaries
- Analysis
- Complex Analysis Theorems
- Complex Analysis Theorems, lemmas and corollaries
- Complex Analysis
- Real Analysis Theorems
- Real Analysis Theorems, lemmas and corollaries
- Real Analysis