Is the boundary of a convex subset of $\mathbb {R^2}$ always a union of a finite number of piecewise differentiable curves?

Question

A certain kind of intuition would suggest that too many non differentiable points (jagged points) would inevitably cause a few points to not be accessible from other points via a straight line. Is that correct? Is it provable?

Answer 1

In general it is false. Choose, for example, the following family of distint directions $$ \xi_j := (\cos(2\pi/j), \sin(2\pi/j)), \qquad j\geq 1, $$ and consider the intersection $K=\bigcap_j H_j$ of the half-planes $H_j$ orthogonal to $\xi_j$, containing the unit ball, and tangent to the unit ball. This set $K$ is convex (being the intersection of convex sets) but it has a countable number of faces.

Answer 2

Say $(t_j)$ is dense in $(0,1)$, $a_j>0$, $\sum a_j<\infty$, and define $f:[0,1]\to\mathbb R$ by $$f(x)=\sum a_j|x-t_j|.$$Then $f$ is convex, so the region $$C=\{(x,y):0< x< 1, f(x)<y<c\}$$is convex for a suitable $c$. But $f$ is not differentiable at $t_j$. So there's a whole arc on the boundary that does not contain any differentiable sub-arc.

Hmm, the assertion that $f$ is not differentiable at $t_j$ seems plausible. It's also easy to prove: Since all the terms are convex, if $h>0$ is small enough that $t_j\pm h\in (0,1)$ we have $$\begin{aligned}&\frac{f(t_j+h)-f(t_j)}{h}-\frac{f(t_j)-f(t_j-h)}{h} \\&\ge a_j\left(\frac{|t_j+h-t_j|-|t_j-t_j|}{h}-\frac{|t_j-t_j|-|t_j-h-t_j|}{h}\right)\\&=2a_j\not\to0\quad(h\to0).\end{aligned}$$

Answer 3

As an additional (to Rigels answer) counterexample: if $f$ is a convex function, then it's graph can be easily extended to a convex curve. In the answer to this question it is shown that the set where such a (convex) function may be not differentiable can be even dense (though still countable).