A strange Lipschitz function

Question

Let $n \geq 3$. Does there exist a Lipschitz function $f: \mathbb R^n \to \mathbb R$ such that the following conditions hold?

1. The origin is a weak Lebesgue point of $\nabla f$, in the sense that the following limit exists:

$$L := \lim_{\delta \to 0_+} \frac{1}{|B_\delta (0)|} \int_{B_{\delta}(0)}\nabla f.$$

2. $f$ is differentiable at $0$.

3. $\nabla f(0)$ is not equal to its Lebesgue average $L$.

Answer 1

No. Subtracting a constant and a linear function, we may assume that $f(0)=0$ and $\nabla f(0)=0$. In other words, $f(x)=o(|x|)$ for $x\to 0$. Then, fixing a coordinate direction, say, of $n$-th coordinate, we look at the $n$—th coordinate of the integral $\int_{B_\delta(0)} \nabla f$, it equals $\int_{B_\delta(0)} \partial f/\partial x_n$. We organize integration in the following way: first by $x_n$ for fixed other coordinates, then by other coordinates. On the first step each integral over $x_n$ is the difference of values of $f$ at two points in the ball, thus it is uniform $o(\delta)$. After integration by other coordinates you get $o(\delta^n)$. Thus, $n$-th coordinate of $L$ equals 0. Every other coordinate too.