Countable Non-Differentiable Points on Convex Curve Boundaries

[alanlu]

I am having trouble proving the following:

Suppose that $E$ is a convex region in the plane bounded by a curve $C$. Show that $C$ has a tangent line except at a countable number of points.

$E$ is convex iff for every $x, y \in E,$ and for every $\lambda \in [0,1], (1-\lambda) x + \lambda y \in E$.

I am considering an approach where I parametrize $C$ in a fixed orientation and then look at the places where it is not differentiable, showing somehow that corners with some angular measure $a \in [0,\pi)$ are the only flavor of non-differentiable parts on this curve, and then showing that the number of corners is bounded by $\frac{2\pi}{\pi - a}$ for the largest $a$.

Any thoughts?

 

[Dick]

I think you are on the right general track. But there can be a countably infinite number of nondifferentiable points, yes? Can you show that an uncountable sum of positive numbers must be infinite?

 

[alanlu]

Ah, right, we could have a polygon(?) with angles $\pi, \frac{3\pi}{2}, \frac{7\pi}{4}...$

How do you take an uncountable sum? That sounds quite exotic! I am trying to imagine a proof by contradiction and pigeon-hole.

 

[Dick]

Ah, right, we could have a polygon(?) with angles $\pi, \frac{3\pi}{2}, \frac{7\pi}{4}...$

How do you take an uncountable sum? That sounds quite exotic! I am trying to imagine a proof by contradiction and pigeon-hole.

It's not that exotic. Suppose you are summing c_i over an uncountable index i belonging to a set I. Define I_n to be the set of all i such that c_i>1/n for n a positive integer. Then if the sum is finite, I_n must be finite for all n, right? What's the union of all of the I_n? Forgive me for not TeXing this.

 

[alanlu]

It would be all the positive values in I, so I has at most a countable subset of nonzero values when the value of the sum is finite, as the union of countably many finite sets is at most countable. Thanks for the help!