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?
Best Answer
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 $\mathbb R_+$.