Sublinear function: Difference between revisions

Revision as of 11:06, 20 May 2022

Revision as of 11:08, 20 May 2022

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====Relation to open convex sets====

{{Math theorem | name = Theorem{{sfn|Narici|Beckenstein|2011|pp=192-193}} | math_statement =

Suppose that <math>X</math> is a TVS (not necessarily [[locally convex]] or Hausdorff) over the real or complex numbers.

Then the open convex subsets of <math>X</math> are exactly those that are of the form <math display="block">z + \left\{x \in X : p(x) < 1\right\} = \left\{x \in X : p(x - z) < 1\right\}</math> for some <math>z \in X</math> and some positive continuous sublinear function <math>p</math> on <math>X.</math>

}}

{{math proof | proof=

Let <math>V</math> be an open convex subset of <math>X.</math>

If <math>0 \in V</math> then let <math>z := 0</math> and otherwise let <math>z \in V</math> be arbitrary.

Let <math>p : X \to [0, \infty)</math> be the [[Minkowski functional]] of <math>V - z</math> where <math>p</math> is a continuous sublinear function on <math>X</math> since <math>V - z</math> is convex, absorbing, and open (<math>p</math> however is not necessarily a seminorm since <math>V</math> was not assumed to be balanced).

From the properties of Minkowski functionals, it is known that <math>V - z = \{x \in X : p(x) < 1\}</math> from which

<math>V = z + \{x \in X : p(x) < 1\}</math> follows. Since <math display="block">z + \left\{x \in X : p(x) < 1\right\} = \left\{x \in X : p(x - z) < 1\right\},</math> this completes the proof. <math>\blacksquare</math>[[Q.E.D.]]

}}