# P-convex hull
## Introduction

For $p$-norms are a generalization of norms. The definition requires the notion of (absolute) $p$-convex hull (see Köthe 1966).

## Definition: p-convex

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:
    $\forall _{\displaystyle x,y\in M;\lambda ,\mu \geq 0}:\lambda ^{p}+\mu ^{p}=1\,\Longrightarrow \,\lambda x+\mu y\in M$

## Definition: absolute p-convex

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:
    $\forall _{\displaystyle x,y\in M}:|\lambda |^{p}+|\mu |^{p}\leq 1\,\Longrightarrow \,\lambda x+\mu y\in M$

## Definition: p-convex hull

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}.
    ${\mathcal {C}}_{p}(M):=\displaystyle \bigcap _{\stackrel {{\widetilde {M}}\supseteq M}{{\widetilde {M}}\,p-convex}}{\widetilde {M}}$

## Definition: absolute p-convex hull

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}.
    $\Gamma _{p}(M):=\displaystyle \bigcap _{\stackrel {{\widetilde {M}}\supseteq M}{{\widetilde {M}}\,absolute\,p-convex}}{\widetilde {M}}$

## Lemma: Display of the absolutely p-convex hull

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:
    $\Gamma _{p}(M)=\left\{\sum _{j=1}^{n}\alpha _{j}x_{j}\,:\,n\in \mathbb {N} \wedge x_{j}\in M\wedge \sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1\right\}=:{\widehat {M}}$

## Proof

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

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

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 αx+βy∈M$^^${\textstyle \alpha x+\beta y\in {\widehat {M}}}.

#### Proof Part 2.1 - Absolute p-convex

Let x, y∈M$^^${\textstyle x, y\in {\widehat {M}}} now have x, y∈M$^^${\textstyle x, y\in {\widehat {M}}} the following representations:
* $\displaystyle x=\sum _{i=1}^{m}\alpha _{i}x_{i}$ with $\displaystyle \sum _{i=1}^{m}|\alpha _{i}|^{p}\leq 1$
* $\displaystyle y=\sum _{i=1}^{n}\beta _{i}y_{i}$ with $\displaystyle \sum _{i=1}^{n}|\beta _{i}|^{p}\leq 1$.
Now we have to show that the absolute $p$-convex combination is an element of ${\widehat {M}}$, i.e. αx+βy∈M$^^${\textstyle \alpha x+\beta y\in {\widehat {M}}}

#### proof-part-2.2-absolutely-p-convex

${\widehat {M}}$ is absolutely p{\textstyle p}-convex, because it holds with |α|$^p$+|β|$^p$≤1{\textstyle |\alpha |^{p}+|\beta |^{p}\leq 1}:
    ${\begin{array}{rcl}\sum _{i=1}^{m}|\alpha \alpha _{i}|^{p}+\sum _{j=1}^{n}|\beta \beta _{j}|^{p}&=&|\alpha |^{p}\underbrace {\sum _{i=1}^{m}|\alpha _{i}|^{p}} _{\leq 1}+|\beta |^{p}\underbrace {\sum _{j=1}^{n}|\beta _{j}|^{p}} _{\leq 1}\\&\leq &|\alpha |^{p}+|\beta |^{p}\leq 1.\\\end{array}}$
This gives:
    $\alpha \cdot x+\beta \cdot y=.\alpha \sum _{i=1}^{m}\alpha _{i}x_{i}+\beta \sum _{j=1}^{n}\beta _{j}y_{j}\in {\widehat {M}}.$

#### Proof Part 2.3 - Zero Vector

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

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

#### Proof Part 3.1 - Induction over Number of Summands

Now let us show inductively via the number of summands $n$ that every element of the form
    $\sum _{j=1}^{n}\alpha _{j}x_{j}{\mbox{ with }}x_{j}\in M{\mbox{ and }}\sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1$
in a given absolutely p{\textstyle p}-convex set M$^~$⊃M{\textstyle {\widetilde {M}}\supset M} is contained.

#### Proof Part 3.2 - Induction Start

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

Now let the condition for n{\textstyle n} hold, i.e.:
    $\sum _{j=1}^{n}\alpha _{j}x_{j}\in {\widetilde {M}}{\mbox{ with }}x_{j}\in M{\mbox{ and }}\sum _{j=1}^{n}|\alpha _{j}|^{p}\leq 1.$

#### Proof Part 3.4 - Induction Step

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

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

#### Proof Part 3.6 - Constructing a p-convex combination of n summands

We now construct a sum of non-negative summands $\beta _{j}\geq 0$
    $\beta _{j}:={\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}{\mbox{ with }}\sum _{j=1}^{n}\left|\beta _{j}\right|^{p}\leq 1$

#### Proof part 3.7 - Application of the induction assumption

So let |α$_{n+1}$|<1{\textstyle |\alpha _{n+1}|<1}. The inequality
    $\sum _{j=1}^{n}\underbrace {\left|{\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}\right|^{p}} _{=|\beta _{j}|^{p}}={\frac {1}{1-|\alpha _{n+1}|^{p}}}\cdot \underbrace {\sum _{j=1}^{n}|\alpha _{j}|^{p}} _{\leq 1-|\alpha _{n+1}|^{p}}\leq 1$
Returns after induction assumption z:=.∑$_{j=1}$$^n$β$_j$⋅x$_j$=.∑$_{j=1}$$^n$ α$_j$/(ᵖ√(1−|α$_{n+1}$|$^p$)) ⋅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

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

### Proof 4

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

## Lemma: p-convex hull

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

## Proof: task for learners

Transfer the above proof analogously to the $p$-convex hull.