real analysis - A convex function is differentiable at all but countably many points

Let $f:\Bbb R\to\Bbb R$ be a convex function. Then $f$ is differentiable at all but countably many points.

It is clear that a convex function can be non-differentiable at countably many points, for example $f(x)=\int\lfloor x\rfloor\,dx$.

I just made this theorem up, but my heuristic for why it should be true is that the only possible non-differentiable singularities I can imagine in a convex function are corners, and these involve a jump discontinuity in the derivative, so since the derivative is increasing (where it is defined), you get an inequality like $f'(y)-f'(x)\ge \sum_{t\in(x,y)}\omega_t(f')$, where $\omega_t(f')$ is the oscillation at $t$ (limit from right minus limit from left) and the sum is over all real numbers between $x$ and $y$. Since the sum is convergent (assuming that $x\le y$ are points such that $f$ is differentiable at $x$ and $y$ so that this makes sense), there can only be countably many values in the sum which are non-zero, and at all other points the oscillation is zero and so the derivative exists. Thus there are only countably many non-differentiable points in the interval $(x,y)$, so as long as there is a suitable sequence $(x_n)\to-\infty$, $(y_n)\to\infty$ of differentiable points, the total number of non-differentiable points is a countable union of countable sets, which is countable.

Furthermore, I would conjecture that the set of non-differentiable points has empty interior-of-closure, i.e. you can't make a function that is non-differentiable at the rational numbers, but as the above discussion shows there are still a lot of holes in the proof (and I'm making a lot of unjustified assumptions regarding the derivative already being somewhat well-defined). Does anyone know how to approach such a statement?

Answer

The set of points of nondifferentiability can be dense.

But you correctly conjectured that it is at most countable. First, convexity implies that for $s
$$ \frac{f(u)-f(s)}{u-s}\leq\frac{f(t)-f(v)}{t-v} \tag{1}$$
Sketch of the proof of (1). First, use the 3-point convexity definition to show this for $u=v$. When $u $$ \frac{f(u)-f(s)}{u-s}\leq\frac{f(t)-f(u)}{t-u}\leq\frac{f(t)-f(v)}{t-v}\qquad \Box$$

From (1) it follows that $f$ has one-sided derivatives ${f}_-'$ and $f_+'$ at every point, and they satisfy
$$
{f}_-'(x)\le f_+'(x)\le {f}_-'(y)\le f_+'(y), \quad x$$

For every point where ${f}_-'(x)< f_+'(x)$, pick a rational number $q$ such that ${f}_-'(x)< q< f_+'(x)$. Inequality (2) implies all these rationals are distinct. Therefore, the set of points of nondifferentiability is at most countable.