Disclaimer. I have asked this question a month ago on MSE (click here to access the original post) and even bountied it. I got an answer on MSE, but unfortunately I don't feel like it has enough quality (or, at least, I wasn't able to fully understand the reasoning presented there). Hence, I've decided to repost the question here on MathOverflow with the goal of obtaining a more clarifying answer. If this is not correct to do, please let me know and I can remove this question.
Context. Let $1 \leqslant p < \infty$ and $0 < \lambda < n$, where $n \in \mathbb N$ is a fixed integer that stands for the dimension of the euclidian space $\mathbb R^n$. In everything that follows, assume that we are dealing with the usual Lebesgue measure on $\mathbb R^n$.
Question. Let $\psi \in C_c^\infty(\mathbb R^n)$ be a function from the class $C^\infty(\mathbb R^n)$ with compact support such that $\chi_{B(0,1)} < \psi < \chi_{B(0,2)}$. Moreover, in $\mathbb R^n$, define the function
$$ f(x) := |x|^{\frac{\lambda - n}{p}}(1 - \psi(x)), $$
for every $x \in \mathbb R^n \setminus \{0\}$ (the value of $f$ at the origin can be whatever we wish, since we are dealing with Lebesgue measure and singletons are sets of measure zero). Prove that $f$ has the following properties:
- $$ \lim_{r \to \infty} \sup_{x \in \mathbb R^n} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy \neq 0 $$
and
- $$ \lim_{\xi \to 0} \, \sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f(y-\xi) - f(y)|^p \, dy = 0. $$
My attempt. I was able to deal with property $1.$ (for details about this, check the MSE post that I've linked above), but I can't seem to find a way to work with the second property. There is an idea that was presented at the answer in MSE that might be usefull. More precisely, after expanding $|f(y-\xi) - f(y)|^p$, we obtain
$$ |f(y-\xi) - f(y)|^p = |(|y-\xi|^{\frac{\lambda - n}{p}} - |y|^{\frac{\lambda - n}{p}})(1-\psi(y-\xi)) - |y|^{\frac{\lambda - n}{p}}(\psi(y-\xi) - \psi(y)))|^p. $$
After this conclusion, the answer on MSE seems to get a bit messy (at least, in my personal opinion).
Thanks for any help in advance.
