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 ${\textstyle M}$ be a subset of a vector space ${\textstyle V}$ and ${\textstyle 0<p\leq 1}$, then ${\textstyle M}$ is called ${\textstyle p}$-convex if ${\textstyle M}$ 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 ${\textstyle M}$ be a subset of a vector space ${\textstyle V}$ and ${\textstyle 0<p\leq 1}$, then ${\textstyle M}$ is said to be absolutely ${\textstyle p}$-convex if ${\textstyle M}$ 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 ${\textstyle p}$-convex hull of the set ${\textstyle M}$ (label: ${\textstyle {\mathcal {C}}_{p}(M)}$) is the intersection over all ${\textstyle p}$-convex sets containing ${\textstyle M}$.

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

Definition: absolute p-convex hull

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

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

Lemma: Display of the absolutely p-convex hull

Let ${\textstyle M}$ be a subset of a vector space ${\textstyle V}$ over the body ${\textstyle \mathbb {K} }$ and ${\textstyle 0<p\leq 1}$, then the absolute ${\textstyle p}$-convex hull of ${\textstyle M}$ 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) ${\textstyle M\subset {\widehat {M}}}$,
• (Proof part 2) ${\textstyle {\widehat {M}}}$ is absolutely ${\textstyle p}$-convex and.
• (Proof part 3) ${\textstyle {\widehat {M}}}$ is contained in any absolutely ${\textstyle p}$-convex set ${\textstyle {\widetilde {M}}\supset M}$.

Proof part 1

${\textstyle M\subset {\widehat {M}}}$, because ${\textstyle M=\{\alpha _{x}x:\alpha _{x}=1\wedge x\in M\}\subset {\widehat {M}}}$

Proof part 2

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

Proof Part 2.1 - Absolute p-convex

Let ${\textstyle x,y\in {\widehat {M}}}$ now have ${\textstyle x,y\in {\widehat {M}}}$ 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. ${\textstyle \alpha x+\beta y\in {\widehat {M}}}$

proof-part-2.2-absolutely-p-convex

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

${\begin{array}{rcl}\sum _{i=1}^{m}|\alpha \alpha _{i}|^{p}+\sum _{j=1}^{n}|\beta \beta _{j}|^{p}&=&|\alpha |^{p}\underbrace {\sum _{i=1}^{m}|\alpha _{i}|^{p}} _{\leq 1}+|\beta |^{p}\underbrace {\sum _{j=1}^{n}|\beta _{j}|^{p}} _{\leq 1}\\&\leq &|\alpha |^{p}+|\beta |^{p}\leq 1.\\\end{array}}$

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

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

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 ${\textstyle p}$-convex set ${\textstyle {\widetilde {M}}\supset M}$ is contained.

Proof Part 3.2 - Induction Start

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

Proof Part 3.3 - Induction Precondition

Now let the condition for ${\textstyle n}$ 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 ${\textstyle n+1}$, the assertion follows as follows:

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

Proof Part 3.5 - Induction Step

If ${\textstyle \alpha _{n+1}=1}$, 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 ${\textstyle |\alpha _{n+1}|<1}$. 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 ${\textstyle \displaystyle z:=.\sum _{j=1}^{n}\beta _{j}\cdot x_{j}=.\sum _{j=1}^{n}{\frac {\alpha _{j}}{\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}}\cdot x_{j}\in {\widetilde {M}}}$.

Proof Part 3.8 - Induction Step

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

$${\widetilde {M}}\ni \left({\sqrt[{p}]{1-|\alpha _{n+1}|^{p}}}\right)z+\alpha _{n+1}x_{n+1}=\sum _{j=1}^{n}\alpha _{j}x_{j}+\alpha _{n+1}x_{n+1}=.\sum _{j=1}^{n+1}\alpha _{j}x_{j}.$$

Proof 4

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

Lemma: p-convex hull

Let ${\textstyle M}$ be a subset of a vector space ${\textstyle V}$ over the body ${\textstyle \mathbb {K} }$ and ${\textstyle 0<p\leq 1}$, then the ${\textstyle p}$-convex hull of ${\textstyle M}$ 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.

See also

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

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