Alphabeta Math
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Convexity: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedaudited 2026-08-02Open item page →

The absolute-value function is convex

Example

The function f(x)=xf(x)=|x| is convex on R\mathbb R. Indeed, for 0λ10\le\lambda\le1,

λx+(1λ)yλx+(1λ)y=λx+(1λ)y,|\lambda x+(1-\lambda)y|\le|\lambda x|+|(1-\lambda)y|=\lambda|x|+(1-\lambda)|y|,

by the triangle inequality (The triangle inequality) and absolute homogeneity (Basic properties of the absolute value), which is precisely the convexity inequality (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

Remarks

Convexity does not entail differentiability at every point; the nondifferentiability of absolute value at zero is recorded in xxx \mapsto |x| is continuous everywhere and not differentiable at 00: the difference quotient equals 11 on the right and 1-1 on the left, so the two one-sided limits differ but is not a dependency of this example.

ExampleConstruction: AI-adaptedVerification: AI-generatedaudited 2026-08-02Open item page →

Finite Jensen for xx2x\mapsto x^2 gives that the square of a weighted mean is at most the weighted mean of the squares

Example

For nonnegative weights λi\lambda_i with iλi=1\sum_i\lambda_i=1 and real xix_i,

(iλixi)2iλixi2.\left(\sum_i\lambda_i x_i\right)^2\le\sum_i\lambda_i x_i^2.

The function q(t)=t2q(t)=t^2 is convex because

λx2+(1λ)y2(λx+(1λ)y)2=λ(1λ)(xy)20\lambda x^2+(1-\lambda)y^2-(\lambda x+(1-\lambda)y)^2=\lambda(1-\lambda)(x-y)^2\ge0

for 0λ10\le\lambda\le1, using nonnegativity of squares (Squares of nonzero elements are positive). Applying finite Jensen (Finite Jensen inequality for a convex function and nonnegative weights summing to one) to qq gives the displayed inequality.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

Assuming Choice, a Hamel coefficient map is midpoint convex but discontinuous and therefore not convex

Statement refuted

Every midpoint-convex function RR\mathbb R\to\mathbb R is continuous, and hence convex.

Facts & Assumptions

Counterexample

technique · contradiction
1.1

Choose a Hamel coefficient map Λ\Lambda from [L1]. Additivity gives Λ((x+y)/2)=(Λ(x)+Λ(y))/2\Lambda((x+y)/2)=(\Lambda(x)+\Lambda(y))/2, so Λ\Lambda is midpoint convex with equality.

L1L2
2.1

Suppose for contradiction that Λ\Lambda is continuous. Then [L2] gives Λ(x)=cx\Lambda(x)=cx; a nonzero ww in its kernel forces c=0c=0, whereas Λ(b)=1\Lambda(b)=1, a contradiction.

step 1.1L2assume-contra
3.1

Thus Λ\Lambda is discontinuous. By [L3], this midpoint-convex function cannot be convex.

step 1.1step 2.1L3discharge-contradiction
ExampleConstruction: AI-adaptedVerification: AI-generatedaudited 2026-08-02Open item page →

xx3x\mapsto x^3 changes from concave to convex at zero and has an inflection point there

Sources