Bounded variation of the partial derivatives of a convex function

Question

Let $f:\mathbb R^2\to\mathbb R$ be a convex function. For simplicity, assume that $f\in C^1$. A general theorem which can be found in the book of Evans and Gariepy says that the gradient $\nabla f$ is a function, or rather a mapping, of locally bounded variation as a function of two variables. Moreover $\frac{\partial f}{\partial x}$ is of locally bounded variation on every horizontal line as a function of one variable. This follows from the observation that the restriction of $f$ to horizontal lines is again convex and $C^1$. From the general properties (slicing) of functions of bounded variation it can be concluded that $\frac{\partial f}{\partial y}$ is of locally bounded variation on almost every horizontal line, besides being of locally bounded variation on every vertical line.

My question is: Is it true, wrong or still unknown whether $\frac{\partial f}{\partial y}$ is of locally bounded variation on every rather than on almost every horizontal line?

Answer 1

Consider $$ f(x, y) = \max \biggl\{ \frac{3 x}{2^n} - \frac{1}{4^n} + \frac{(-1)^n y}{n + 1} : n = 0, 1, 2, \ldots \biggr\} . $$ Then $f$ is clearly convex, and $$ \partial_y f(x, 0) = \frac{(-1)^n}{n + 1} \qquad \text{for } x \in [2^{-n-1}, 2^{-n}] , $$ so that $\partial_y f(x, 0)$ is a function of unbounded variation near $x = 0$.

Clearly, $f$ is not $C^1$, but it is rather clear that $f$ can be made a $C^1$ function by slightly smoothing out the edges of the graph of $f$.

Here is the plot of $f(x,0)$ (blue) and $\partial_y f(x,0)$ (yellowish) with $n$ restricted to $n \leqslant 10$:

And here is the corresponding 3-D plot of $f(x,y)$: