P-convex hull

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Introduction[edit | edit source]

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

Definition: p-convex[edit | edit source]

Let be a subset of a vector space and , then is called -convex if fulfills the following property:

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

Let be a subset of a vector space and , then is said to be absolutely -convex if fulfills the following property:

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

The -convex hull of the set (label: ) is the intersection over all -convex sets containing .

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

The absolutely -convex hull of the set (label: ) is the section over all absolutely -convex sets containing .

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

Let be a subset of a vector space over the body and , then the absolute -convex hull of can be written as follows:

Proof[edit | edit source]

3 subassertions are shown, where (1) and (2) gives and (3) gives the subset relation .

  • (Proof part 1) ,
  • (Proof part 2) is absolutely -convex and.
  • (Proof part 3) is contained in any absolutely -convex set .


Proof part 1[edit | edit source]

, because

Proof part 2[edit | edit source]

Now let and be given. One must show that .

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

Let now have the following representations:

  • with
  • with .

Now we have to show that the absolute -convex combination is an element of , i.e.


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

is absolutely -convex, because it holds with :

This gives:

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

, because it holds with and any gets .

Proof part 3[edit | edit source]

We now show that the absolutely -convex hull is contained in every absolutely -convex superset of .

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

Now let us show inductively via the number of summands that every element of the form

in a given absolutely -convex set is contained.

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

For , the assertion follows via the definition of an absolutely -convex set .

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

Now let the condition for hold, i.e.:

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

For , the assertion follows as follows:

Let and with for all . is now to be proved.

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

If , then there is nothing to show, since then all are for .

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

We now construct a sum of non-negative summands

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

So let . The inequality

Returns after induction assumption .


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

Since is absolutely -convex, it follows with

Proof 4[edit | edit source]

From the proof parts , and together the assertion follows.

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

Let be a subset of a vector space over the body and , then the -convex hull of can be written as follows:

Proof: task for learners[edit | edit source]

Transfer the above proof analogously to the -convex hull.


See also[edit | edit source]

References[edit | edit source]

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

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