1061: For Hilbert Space, Nonempty Closed Convex Subset, and Point on Hilbert Space, There Is Unique Point on Subset Whose Distance to Point Is Minimum

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description/proof of that for Hilbert space, nonempty closed convex subset, and point on Hilbert space, there is unique point on subset whose distance to point is minimum

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Topics

About: vectors space
About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Hilbert space.
• The reader knows a definition of closed set.
• The reader admits the parallelogram law on any vectors space normed induced by any inner product.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
• The reader admits the proposition that any metric is continuous with respect to the topology induced by the metric.
• The reader admits the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
• The reader admits the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.

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Target Context

• The reader will have a description and a proof of the proposition that for any Hilbert space, any nonempty closed convex subset, and any point on the Hilbert space, there is the unique point on the subset whose distance to the point is the minimum.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V$: $\in \{\text{ the }F\text{ Hilbert spaces }\}$, with $⟨\bullet ,\bullet ⟩:V\times V\to F$, $\Vert \bullet \Vert :V\to \mathbb{R}$, and $dist:V\times V\to \mathbb{R}$
$C$: $\in \{\text{ the nonempty closed convex subsets of }V\}$
$w$: $\in V$
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Statements:
$!\mathrm{\exists }{v}_0\in C(dist(w,v_0)=inf\{dist(w,v)|v\in C\})$
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2: Note

The space's being Hilbert is crucial.

For example, when the space is the metric topological subspace of the Euclidean metric space, ${\mathbb{R}}^2$, that (the subspace) is the union of the open unit disc and the point, $(0,-2)$, the open unit disc is a nonempty closed convex subset (the intersection of the closed unit disk on ${\mathbb{R}}^2$ and the subspace), and the infimum of the distances from $(0,-2)$ is $1$, but there is no point on the closed subset whose distance to the point is $1$, in fact, there is no point on the space whose distance to the point is $1$.

Also the closed subset's being convex is crucial for Proof: at least, obviously, the uniqueness does not generally hold if the closed subset is not convex.

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3: Proof

Whole Strategy: Step 1: see that infimum of the distances, $d$, exists; Step 2: take any sequence of points on $C$ whose distances converge to $d$; Step 3: see that the sequence is a Cauchy sequence; Step 4: see that the convergence is on $C$; Step 5: see that the distance of the convergence to $w$ is $d$; Step 6: see that there is no other point on $C$ whose distance to $w$ is $d$.

Step 1:

As $\{dist(w,v)|v\in C\}$ is a subset of $\mathbb{R}$ lower bounded by $0$, $\{dist(w,v)|v\in C\}$ has the infimum $d\in \mathbb{R}$: which is a nature of $\mathbb{R}$.

Step 2:

Let us take any sequence of points on $C$, $v_1,v_2,...$, such that $d\le dist(w,v_j)<d+1/2^j$, which is possible because $d=inf\{dist(w,v)|v\in C\}$: there may be some duplications, which does not matter.

$dist(w,v_1),dist(w,v_2),...$ converges to $d$.

Step 3:

Let us see that the sequence of points is a Cauchy sequence.

$dist(v_m,v_n)=\Vert v_n-v_m\Vert =\Vert (v_n-w)-(v_m-w)\Vert$.

By the parallelogram law on any vectors space normed induced by any inner product, $\Vert (v_n-w)-(v_m-w){\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-\Vert (v_n-w)+(v_m-w){\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-\Vert v_n+v_m-2w{\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-4\Vert (v_n+v_m)/2-w{\Vert }^2$.

For any $0<ϵ$, there is an $N\in \mathbb{N}$ such that for each $N<m,n$, $\Vert v_m-w{\Vert }^2,\Vert v_n-w{\Vert }^2<d^2+{ϵ}^2/4$.

$(v_n+v_m)/2\in C$, because $C$ is convex, so, $d\le \Vert (v_n+v_m)/2-w\Vert$.

So, $\Vert v_n-v_m{\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-4\Vert (v_n+v_m)/2-w{\Vert }^2<2(2(d^2+{ϵ}^2/4))-4d^2={ϵ}^2$.

So, the sequence of points is a Cauchy sequence.

Step 4:

As $V$ is complete, there is the convergence of the sequence, $v_0\in V$.

$v_0$ is an accumulation point of $C$, because for each open ball around $v_0$, there is a $v_j\in C$ contained in the open ball, because the sequence converges to $v_0$.

By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, $v_0$ is on the closure of $C$, but the closure of $C$ is $C$, because $C$ is closed.

So, $v_0\in C$.

Step 5:

The distance map, $dist(w,v):V\to \mathbb{R}$, with $w$ fixed is a continuous map, by the proposition that any metric is continuous with respect to the topology induced by the metric and the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.

As $dist(w,v_1),dist(w,v_2),...$ converges to $d$, $dist(w,v_0)=d$, by the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.

So, there is at least 1 $v_0\in C$ such that $dist(w,v_0)=inf\{dist(w,v)|v\in C\}$.

Step 6:

Let us see that there is no other point on $C$ whose distance to $w$ is $d$.

Let us suppose there was another $v_0^{\prime }\in C$ such that $dist(w,v_0^{\prime })=d$.

$\Vert v_0-v_0^{\prime }\Vert =\Vert (v_0-w)-(v_0^{\prime }-w)\Vert$.

By the parallelogram law on any vectors space normed induced by any inner product, $\Vert (v_0-w)-(v_0^{\prime }-w){\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-\Vert (v_0-w)+(v_0^{\prime }-w){\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-\Vert (v_0+v_0^{\prime })-2w{\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-4\Vert (v_0+v_0^{\prime })/2-w{\Vert }^2$, but $\Vert v_0-w\Vert =dist(w,v_0)=d$, $\Vert v_0^{\prime }-w\Vert =dist(w,v_0^{\prime })=d$, and as $(v_0+v_0^{\prime })/2\in C$, $d\le \Vert (v_0+v_0^{\prime })/2-w\Vert$, so, $\le 2(2d^2)-4d^2=0$, which implies that $\Vert (v_0-w)-(v_0^{\prime }-w){\Vert }^2=0$, which implies that $v_0=v_0^{\prime }$.

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References

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