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Assuming Choice, a Hamel coefficient map is midpoint convex but discontinuous and therefore not convex
Statement refuted
Every midpoint-convex function is continuous, and hence convex.
Facts & Assumptions
Given: The Axiom of Choice.
Assuming Choice, a Hamel basis has a coefficient map that is additive, has for its selected basis vector, and has a nonzero kernel (Assuming the Axiom of Choice, has a Hamel basis over : there is such that every real is a finite -linear combination of elements of in exactly one way, and each basis vector carries a well-defined -linear coefficient map).
An additive function satisfying any listed regularity condition, including continuity at one point, equals (Six regularity conditions each force an additive to be : continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in ).
A convex real function is continuous at every interior point of its interval domain (A convex real function is Lipschitz on every closed bounded subinterval of the interior of its domain, hence continuous throughout the interior).
Counterexample
Choose a Hamel coefficient map from [L1]. Additivity gives , so is midpoint convex with equality.
Suppose for contradiction that is continuous. Then [L2] gives ; a nonzero in its kernel forces , whereas , a contradiction.
Thus is discontinuous. By [L3], this midpoint-convex function cannot be convex.
Depends on
- The Axiom of Choice
- Assuming the Axiom of Choice, $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$: there is $B \subseteq \mathbb{R}$ such that every real is a finite $\mathbb{Q}$-linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $\mathbb{Q}$-linear coefficient map
- Cauchy's functional equation $f(x+y) = f(x) + f(y)$, and the additive functions $\mathbb{R} \to \mathbb{R}$
- Six regularity conditions each force an additive $f : \mathbb{R} \to \mathbb{R}$ to be $x \mapsto f(1)x$: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in $\mathbb{R}^{2}$
- Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval
- A convex real function is Lipschitz on every closed bounded subinterval of the interior of its domain, hence continuous throughout the interior
Used by
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Sources
- P. Green and W. Gustin, On the gap between convexity and midpoint convexity (standard reference, not scraped)