For any open interval $(a, b)\subset {\mathbb R}\,$, define a weakly convex function $f:(a, b) \rightarrow {\mathbb R}$ as one for which
$$f(q\;x_0 + (1 – q)\;x_1) \leq q\;f(x_0) + (1-q)\;f(x_1)$$
for all $x_0, x_1 \in (a, b)$, and all $q \in [0, 1] \cap {\mathbb Q}$.
Does this definition of weak convexity imply continuity (in $(a, b)$)?
My intuition says "yes", because hard as I try, I can't see how a function that satisfies the inequality above for all $x_0, x_1,$ and $q$ could have a single point of discontinuity in $(a, b)$.
My attempt to prove this implication, however, has been slippery business. A pointer to a proof would be welcome (assuming, of course, that my intuition is correct).
The book where I found this definition of weak convexity implies (though does not state outright) that weak convexity does not guarantee continuity. If this is the case, I'd love to see a counterexample.
Best Answer
Such a function cannot have a single point of discontinuity. But it can have lots of points of discontinuity.
Sorry, you will not see it. But you have to believe it, if you believe the axiom of choice.
Consider $\mathbb R$ as a vector space over $\mathbb Q$. As any vector space, it has a basis $\{v_\alpha\}$ (once again, by the axiom of choice; see Hamel basis). Pick one basis element $v_{\alpha_0}$ and define a $\mathbb Q$-linear map $f:\mathbb R\to\mathbb R$ so that $f(v_{\alpha_0})=2v_{\alpha_0}$ and $f(v_\alpha)=v_\alpha$ for $\alpha\ne \alpha_0$. Since $f$ is $\mathbb Q$-linear, it satisfies $$f(q\;x_0 + (1 - q)\;x_1) = q\;f(x_0) + (1-q)\;f(x_1)\tag{1}$$ for all $q\in \mathbb Q$ and all $x_0,x_1\in\mathbb R$. Yet, it is discontinuous at every non-zero point because both sets $\{x: f(x)=2x\}$ and $\{x:f(x)=x\}$ are dense in $\mathbb R$.