Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Rudin's Real and Complex Analysis Chapter 3 Exercise 4 is:

Assume that $\varphi$ is a continuous real function on $(a,b)$ s.t. $$\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$$ for all $x,y\in(a,b)$. Prove that $\varphi$ is convex.

The conclusion does not follow if continuity is omitted from the hypotheses.

My question is, is there some way to explicitly construct a counterexample such that $\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$ for all $x,y\in(a,b)$, but $\varphi$ is not convex?

share|cite|improve this question
up vote 15 down vote accepted

I'm not sure there is a constructive example, but here is a construction using the Axiom of Choice:

$\mathbb{R}$ is a vector space over $\mathbb{Q}$. Choose (using AC) a basis $\{r_i\}_{i\in I}$ for $\mathbb{R}$, and consider the linear transformation $h$ that swaps the coefficients for two particular basis vectors $r_1$ and $r_2$. Then $h$, being a linear transformation, preserves addition and multiplication by $1/2$ (since $1/2 \in \mathbb Q$), but is everywhere discontinuous. In fact the range of $h$ on any open interval is dense in $\mathbb R$. (Prove this!)

Now consider $\phi(x) = h(x)^2$. Since $x^2$ is convex and $h$ preserves addition and halving, $\phi$ must satisfy the inequality. On the other hand, $\phi$ cannot be convex since the image of any open interval is dense in $\mathbb R_+$.

share|cite|improve this answer
Or, consider the only $\mathbb Q$-linear function $\phi$ such that $\phi(r_1)=r_1$ and $\phi(r_i)=0$ for every $i\ne1$. – Did Oct 9 '11 at 8:45
Yes, that's probably easier to prove the lemma for. (I added a squaring in my answer in order to demonstrate that the function could satisfy a strict inequality whenever $x\ne y$). – Henning Makholm Oct 9 '11 at 14:18

Any midpoint-convex function on $\mathbb R$ that is not convex must not be Lebesgue measurable. According to a result of Solovay, there are models of Zermelo-Fraenkel set theory without Axiom of Choice in which there are no non-measurable functions ${\mathbb R} \to {\mathbb R}$. Thus there is no way to explicitly construct your counterexample.

share|cite|improve this answer

It is worth mentioning that the functions with the above property are often called midpoint convex or Jensen convex. It can be shown that every midpoint convex function is rational convex.

Every non-continuous solution of Cauchy equation would work gives a counterexample. (Note that the proof of existence of such solutions uses Axiom of Choice. Most frequently, it is proved using Hamel basis.) It is basically the same example as Henning gave, but without composing the solution with $x\mapsto x^2$.

See e.g. William F. Donoghue: Distributions and Fourier transforms, p.11. (This books was among the first google books results when searching for "midpoint convex" "not convex" function.)

Interestingly, while googling (in order to add some useful links) I stumbled upon blogs of some math.stackexchange users:

A natural question is what can be said without Axiom of Choice. Although I do not know the answer to this, I'll at least add link to a mathoveflow question Are there any non-linear solutions of Cauchy’s equation $(f(x+y)=f(x)+f(y))$ without assuming the Axiom of Choice?, where it is explained that at least some form of AC is needed to obtain a non-continuous solution of Cauchy functional equation.

EDIT: I see that in the meantime the question of necessity of AC was answered by Robert Israel.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.