P-convex hull

## Contents

* 1 Introduction
* 2 Definition: p-convex
* 3 Definition: absolute p-convex
* 4 Definition: p-convex hull
* 5 Definition: absolute p-convex hull
* 6 Lemma: Display of the absolutely p-convex hull
* 7 Proof
* 7.1 Proof part 1
* 7.2 Proof part 2
* 7.2.1 Proof Part 2.1 - Absolute p-convex
* 7.2.2 proof-part-2.2-absolutely-p-convex
* 7.2.3 Proof Part 2.3 - Zero Vector
* 7.3 Proof part 3
* 7.3.1 Proof Part 3.1 - Induction over Number of Summands
* 7.3.2 Proof Part 3.2 - Induction Start
* 7.3.3 Proof Part 3.3 - Induction Precondition
* 7.3.4 Proof Part 3.4 - Induction Step
* 7.3.5 Proof Part 3.5 - Induction Step
* 7.3.6 Proof Part 3.6 - Constructing a p-convex combination of n summands
* 7.3.7 Proof part 3.7 - Application of the induction assumption
* 7.3.8 Proof Part 3.8 - Induction Step
* 7.4 Proof 4
* 8 Lemma: p-convex hull
* 9 Proof: task for learners
* 10 See also
* 11 References
* 12 Page Information
* 12.1 Wiki2Reveal

## Introduction[edit | edit source]

For  p {\displaystyle p} -norms are a generalization of norms. The definition requires the notion of (absolute)  p {\displaystyle p} -convex hull (see Köthe 1966[1]).

## Definition: p-convex[edit | edit source]

Let  M {\textstyle M} be a subset of a vector space  V {\textstyle V} and  0 < p ≤ 1 {\textstyle 0<p\leq 1} , then  M {\textstyle M} is called  p {\textstyle p} -convex if  M {\textstyle M} fulfills the following property:

∀ x , y ∈ M ; λ , μ ≥ 0 : λ p + μ p = 1 ⟹ λ x + μ y ∈ M {\displaystyle \forall _{\displaystyle x,y\in M;\lambda ,\mu \geq 0}:\lambda ^{p}+\mu ^{p}=1\,\Longrightarrow \,\lambda x+\mu y\in M}

## Definition: absolute p-convex[edit | edit source]

Let  M {\textstyle M} be a subset of a vector space  V {\textstyle V} and  0 < p ≤ 1 {\textstyle 0<p\leq 1} , then  M {\textstyle M} is said to be absolutely  p {\textstyle p} -convex if  M {\textstyle M} fulfills the following property:

∀ x , y ∈ M : | λ | p + | μ | p ≤ 1 ⟹ λ x + μ y ∈ M {\displaystyle \forall _{\displaystyle x,y\in M}:|\lambda |^{p}+|\mu |^{p}\leq 1\,\Longrightarrow \,\lambda x+\mu y\in M}

## Definition: p-convex hull[edit | edit source]

The  p {\textstyle p} -convex hull of the set  M {\textstyle M} (label:  C p ( M ) {\textstyle {\mathcal {C}}_{p}(M)} ) is the intersection over all  p {\textstyle p} -convex sets containing  M {\textstyle M} .

C p ( M ) := ⋂ M ~ p − c o n v e x M ~ ⊇ M M ~ {\displaystyle {\mathcal {C}}_{p}(M):=\displaystyle \bigcap _{\stackrel {{\widetilde {M}}\supseteq M}{{\widetilde {M}}\,p-convex}}{\widetilde {M}}}

## Definition: absolute p-convex hull[edit | edit source]

The absolutely  p {\textstyle p} -convex hull of the set  M {\textstyle M} (label:  Γ p ( M ) {\textstyle \Gamma _{p}(M)} ) is the section over all absolutely  p {\textstyle p} -convex sets containing  M {\textstyle M} .

Γ p ( M ) := ⋂ M ~ a b s o l u t e p − c o n v e x M ~ ⊇ M M ~ {\displaystyle \Gamma _{p}(M):=\displaystyle \bigcap _{\stackrel {{\widetilde {M}}\supseteq M}{{\widetilde {M}}\,absolute\,p-convex}}{\widetilde {M}}}

## Lemma: Display of the absolutely p-convex hull[edit | edit source]

Let  M {\textstyle M} be a subset of a vector space  V {\textstyle V} over the body  K {\textstyle \mathbb {K} } and  0 < p ≤ 1 {\textstyle 0<p\leq 1} , then the absolute  p {\textstyle p} -convex hull of  M {\textstyle M} can be written as follows:

Γ p ( M ) = { ∑ j = 1 n α j x j : n ∈ N ∧ x j ∈ M ∧ ∑ j = 1 n | α j | p ≤ 1 } =: M ^ {\displaystyle \Gamma _{p}(M)=\left\\{\sum _{j=1}^{n}\alpha _{j}x_{j}\,:\,n\in \mathbb {N} \wedge x_{j}\in M\wedge \sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1\right\\}=:{\widehat {M}}}

## Proof[edit | edit source]

3 subassertions are shown, where (1) and (2) gives  Γ p ( M ) ⊆ M ^ {\displaystyle \Gamma _{p}(M)\subseteq {\widehat {M}}} and (3) gives the subset relation  M ^ ⊆ Γ p ( M ) {\displaystyle {\widehat {M}}\subseteq \Gamma _{p}(M)} .

* (Proof part 1) M ⊂ M ^ {\textstyle M\subset {\widehat {M}}} ,
* (Proof part 2) M ^ {\textstyle {\widehat {M}}} is absolutely  p {\textstyle p} -convex and.
* (Proof part 3) M ^ {\textstyle {\widehat {M}}} is contained in any absolutely  p {\textstyle p} -convex set  M ~ ⊃ M {\textstyle {\widetilde {M}}\supset M} .

### Proof part 1[edit | edit source]

M ⊂ M ^ {\textstyle M\subset {\widehat {M}}} , because  M = { α x x : α x = 1 ∧ x ∈ M } ⊂ M ^ {\textstyle M=\\{\alpha _{x}x:\alpha _{x}=1\wedge x\in M\\}\subset {\widehat {M}}}

### Proof part 2[edit | edit source]

Now let  x , y ∈ M ^ {\textstyle x,y\in {\widehat {M}}} and  | α | p + | β | p ≤ 1 {\textstyle |\alpha |^{p}+|\beta |^{p}\leq 1} be given. One must show that  α x + β y ∈ M ^ {\textstyle \alpha x+\beta y\in {\widehat {M}}} .

#### Proof Part 2.1 - Absolute p-convex[edit | edit source]

Let  x , y ∈ M ^ {\textstyle x,y\in {\widehat {M}}} now have  x , y ∈ M ^ {\textstyle x,y\in {\widehat {M}}} the following representations:

* x = ∑ i = 1 m α i x i {\displaystyle \displaystyle x=\sum _{i=1}^{m}\alpha _{i}x_{i}} with  ∑ i = 1 m | α i | p ≤ 1 {\displaystyle \displaystyle \sum _{i=1}^{m}|\alpha _{i}|^{p}\leq 1}
* y = ∑ i = 1 n β i y i {\displaystyle \displaystyle y=\sum _{i=1}^{n}\beta _{i}y_{i}} with  ∑ i = 1 n | β i | p ≤ 1 {\displaystyle \displaystyle \sum _{i=1}^{n}|\beta _{i}|^{p}\leq 1} .

Now we have to show that the absolute  p {\displaystyle p} -convex combination is an element of  M ^ {\displaystyle {\widehat {M}}} , i.e.  α x + β y ∈ M ^ {\textstyle \alpha x+\beta y\in {\widehat {M}}}

#### proof-part-2.2-absolutely-p-convex[edit | edit source]

M ^ {\displaystyle {\widehat {M}}} is absolutely  p {\textstyle p} -convex, because it holds with  | α | p + | β | p ≤ 1 {\textstyle |\alpha |^{p}+|\beta |^{p}\leq 1} :

∑ i = 1 m | α α i | p + ∑ j = 1 n | β β j | p = | α | p ∑ i = 1 m | α i | p ⏟ ≤ 1 + | β | p ∑ j = 1 n | β j | p ⏟ ≤ 1 ≤ | α | p + | β | p ≤ 1. {\displaystyle {\begin{array}{rcl}\sum _{i=1}^{m}|\alpha \alpha _{i}|^{p}+\sum _{j=1}^{n}|\beta \beta _{j}|^{p}&=&|\alpha |^{p}\underbrace {\sum _{i=1}^{m}|\alpha _{i}|^{p}} _{\leq 1}+|\beta |^{p}\underbrace {\sum _{j=1}^{n}|\beta _{j}|^{p}} _{\leq 1}\\\&\leq &|\alpha |^{p}+|\beta |^{p}\leq 1.\\\\\end{array}}}

This gives:

α ⋅ x + β ⋅ y = . α ∑ i = 1 m α i x i + β ∑ j = 1 n β j y j ∈ M ^ . {\displaystyle \alpha \cdot x+\beta \cdot y=.\alpha \sum _{i=1}^{m}\alpha _{i}x_{i}+\beta \sum _{j=1}^{n}\beta _{j}y_{j}\in {\widehat {M}}.}

#### Proof Part 2.3 - Zero Vector[edit | edit source]

0 V ∈ M ^ {\textstyle 0_{V}\in {\widehat {M}}} , because it holds  0 V = α ⋅ x {\textstyle 0_{V}=\alpha \cdot x} with  α = 0 = | α | p ≤ 1 {\textstyle \alpha =0=|\alpha |^{p}\leq 1} and any  x ∈ M {\textstyle x\in M} gets  0 V = α ⋅ x ∈ M ^ {\textstyle 0_{V}=\alpha \cdot x\in {\widehat {M}}} .

### Proof part 3[edit | edit source]

We now show that the absolutely  p {\displaystyle p} -convex hull is contained in every absolutely  p {\displaystyle p} -convex superset  M ~ {\displaystyle {\widetilde {M}}} of  M {\displaystyle M} .

#### Proof Part 3.1 - Induction over Number of Summands[edit | edit source]

Now let us show inductively via the number of summands  n {\displaystyle n} that every element of the form

∑ j = 1 n α j x j with x j ∈ M and ∑ j = 1 n | α j | p ≤ 1 {\displaystyle \sum _{j=1}^{n}\alpha _{j}x_{j}{\mbox{ with }}x_{j}\in M{\mbox{ and }}\sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1}

in a given absolutely  p {\textstyle p} -convex set  M ~ ⊃ M {\textstyle {\widetilde {M}}\supset M} is contained.

#### Proof Part 3.2 - Induction Start[edit | edit source]

For  n = 2 {\textstyle n=2} , the assertion follows via the definition of an absolutely  p {\textstyle p} -convex set  M ~ ⊃ M {\textstyle {\widetilde {M}}\supset M} .

#### Proof Part 3.3 - Induction Precondition[edit | edit source]

Now let the condition for  n {\textstyle n} hold, i.e.:

∑ j = 1 n α j x j ∈ M ~ with x j ∈ M and ∑ j = 1 n | α j | p ≤ 1. {\displaystyle \sum _{j=1}^{n}\alpha _{j}x_{j}\in {\widetilde {M}}{\mbox{ with }}x_{j}\in M{\mbox{ and }}\sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1.}

#### Proof Part 3.4 - Induction Step[edit | edit source]

For  n + 1 {\textstyle n+1} , the assertion follows as follows:

Let  x := ∑ j = 1 n + 1 α j x j {\textstyle \displaystyle x:=\sum _{j=1}^{n+1}\alpha _{j}x_{j}} and  ∑ j = 1 n + 1 | α j | p ≤ 1 {\textstyle \displaystyle \sum _{j=1}^{n+1}|\alpha _{j}|^{p}\leq 1} with  x j ∈ M {\textstyle x_{j}\in M} for all  j ∈ { 1 , … , n + 1 } {\textstyle j\in \\{1,\dots ,n+1\\}} .  x ∈ M ~ {\textstyle x\in {\widetilde {M}}} is now to be proved.

#### Proof Part 3.5 - Induction Step[edit | edit source]

If  α n + 1 = 1 {\textstyle \alpha _{n+1}=1} , then there is nothing to show, since then all  | α j | = 0 {\displaystyle |\alpha _{j}|=0} are for  j ∈ { 1 , … , n } {\displaystyle j\in \\{1,\ldots ,n\\}} .

#### Proof Part 3.6 - Constructing a p-convex combination of n summands[edit | edit source]

We now construct a sum of non-negative summands  β j ≥ 0 {\displaystyle \beta _{j}\geq 0}

β j := α j 1 − | α n + 1 | p p with ∑ j = 1 n | β j | p ≤ 1 {\displaystyle \beta _{j}:={\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}{\mbox{ with }}\sum _{j=1}^{n}\left|\beta _{j}\right|^{p}\leq 1}

#### Proof part 3.7 - Application of the induction assumption[edit | edit source]

So let  | α n + 1 | < 1 {\textstyle |\alpha _{n+1}|<1} . The inequality

∑ j = 1 n | α j 1 − | α n + 1 | p p | p ⏟ = | β j | p = 1 1 − | α n + 1 | p ⋅ ∑ j = 1 n | α j | p ⏟ ≤ 1 − | α n + 1 | p ≤ 1 {\displaystyle \sum _{j=1}^{n}\underbrace {\left|{\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}\right|^{p}} _{=|\beta _{j}|^{p}}={\frac {1}{1-|\alpha _{n+1}|^{p}}}\cdot \underbrace {\sum _{j=1}^{n}|\alpha _{j}|^{p}} _{\leq 1-|\alpha _{n+1}|^{p}}\leq 1}

Returns after induction assumption  z := . ∑ j = 1 n β j ⋅ x j = . ∑ j = 1 n α j 1 − | α n + 1 | p p ⋅ x j ∈ M ~ {\textstyle \displaystyle z:=.\sum _{j=1}^{n}\beta _{j}\cdot x_{j}=.\sum _{j=1}^{n}{\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}\cdot x_{j}\in {\widetilde {M}}} .

#### Proof Part 3.8 - Induction Step[edit | edit source]

Since  M ~ {\textstyle {\widetilde {M}}} is absolutely  p {\textstyle p} -convex, it follows with  ( 1 − | α n + 1 | p p ) p + | α n + 1 | p = 1 {\textstyle \left({\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}\right)^{p}+|\alpha _{n+1}|^{p}=1}

M ~ ∋ ( 1 − | α n + 1 | p p ) z + α n + 1 x n + 1 = ∑ j = 1 n α j x j + α n + 1 x n + 1 = . ∑ j = 1 n + 1 α j x j . {\displaystyle {\widetilde {M}}\ni \left({\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}\right)z+\alpha _{n+1}x_{n+1}=\sum _{j=1}^{n}\alpha _{j}x_{j}+\alpha _{n+1}x_{n+1}=.\sum _{j=1}^{n+1}\alpha _{j}x_{j}.}

### Proof 4[edit | edit source]

From the proof parts  ( 1 ) {\textstyle (1)} ,  ( 2 ) {\textstyle (2)} and  ( 3 ) {\textstyle (3)} together the assertion follows.  ◻ {\displaystyle \Box }

## Lemma: p-convex hull[edit | edit source]

Let  M {\textstyle M} be a subset of a vector space  V {\textstyle V} over the body  K {\textstyle \mathbb {K} } and  0 < p ≤ 1 {\textstyle 0<p\leq 1} , then the  p {\textstyle p} -convex hull of  M {\textstyle M} can be written as follows:

C p ( M ) = { . ∑ j = 1 n α j x j : n ∈ N ∧ x j ∈ M ∧ α j ∈ [ 0 , 1 ] ∧ ∑ j = 1 n α j p = 1 } {\displaystyle {\cal {C}}_{p}(M)=\left\\{.\sum _{j=1}^{n}\alpha _{j}x_{j}\,:\,n\in \mathbb {N} \wedge x_{j}\in M\wedge \alpha _{j}\in [0,1]\wedge \sum _{j=1}^{n}\alpha _{j}^{p}=1\right\\}}

## Proof: task for learners[edit | edit source]

Transfer the above proof analogously to the  p {\displaystyle p} -convex hull.

## See also[edit | edit source]

* Linear algebra
* Minkowski functional
* P-regularity
* PC-regularity
* Convex Combination
* Lemma - subadditivity p-convexity

## References[edit | edit source]

1. ↑ Gottfried Köthe (1966) Topological Vector Spaces, 15.10, pp.159-162.

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