Let $f: \mathbb R^n \to \mathbb R$ be a locally integrable function. Given an $\varepsilon > 0$, we say $f$ is $\varepsilon$-times Lebesgue differentiable if
$$\lim_{r \to 0} \frac{\int_{B_r (x)} |f(y) – f(x)| \, dy}{r^{n+\varepsilon}} = 0$$
for every $x \in \mathbb R^n$.
Question: Suppose $f$ is $\varepsilon$ times Lebesgue differentiable for some $\varepsilon > 0$. Does it follow that $f$ is continuous?
Note: This question is strongly related to the following problem, which was solved in the affirmative. The linked result implies, a fortiori, that if $f$ is $1 + \varepsilon$ times Lebesgue differentiable, then it is constant.
Best Answer
The answer is no. Let me present a counterexample for $n=2$.
Fix some bump function $h(x)$ supported in $[-1,1]$ and such that $|h(x)|\leq 1$ and $h(0)=1$ (though those choices don't matter much). Define $f(x,y)=h(y/x^2)$ for $x>0$ and $f(x,y)=0$ elsewhere. This function is easily seen to be smooth at all points other than $(0,0)$, and discontinuous at the origin. I claim $f$ is $\frac{1}{2}$-times Lebesgue differentiable at the origin.
$f$ is $0$ except for the region $0<|y|<x^2$, and on that region it is bounded by $1$. Therefore the integral of $|f(x,y)-f(0,0)|=|f(x,y)|$ over a ball $B_r(0)$ is bounded by the volume of the intersection of this region with the ball, which is contained inside the rectangle with vertices $(0,\pm r^2),(r,\pm r^2)$ of volume $2r^3$. We are done as $\frac{2r^3}{r^{2+1/2}}=2r^{1/2}$ tends to $0$.