For any integrable function $f$, consider the following properties:
$$ \underline{V_0 \text{ property:}} \, \, \lim_{r \to 0} \, \sup_{x \in \Bbb R^n} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy = 0;$$
$$ \underline{V_\infty \text{ property:}} \, \, \lim_{r \to \infty} \, \sup_{x \in \Bbb R^n} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy = 0,$$
where $0 < \lambda < n$ and $1 \leqslant p < \infty$ are fixed elements. My goal is to show that the function
$$ f_\epsilon(x) = \begin{cases} |x|^{\frac{\lambda - n}{p} + \epsilon} \quad \text{ if } |x| \leqslant 1 \\ |x|^{\frac{\lambda-n}{p}} \quad \, \, \, \, \text{ if } |x| \geqslant 1 \end{cases},$$ where $\epsilon > 0$, has $V_0$ property but doesn't have $V_\infty$ property. Usually, when I encounter this kind of functions, my intuition is to check whether I am dealing with a radially decreasing function, since this allows me to ignore $\sup_{x \in \Bbb R^n},$ allowing me to deal with the problem using only the integral over $B(0,r).$ In this case, I am quite sure that the function in question isn't radially decreasing and thus this strategy can't be applied. The only other thing that came to my mind is to deal with the integral over $B(x,r)$ directly, which lead me to
$$ \int_{B(x,r)} |f_\epsilon(y)|^p \, dy = \int_{B(x,r)} |y|^{\frac{\lambda - n}{p} + \epsilon} \, \chi_{B(0,1)}(y) \, dy + \int_{B(x,r)} |y|^{\frac{\lambda-n}{p}} \, \chi_{\Bbb R^n \setminus B(0,1)}(y) \, dy$$
which equals
$$ \int_{B(x,r) \cap B(0,1)} |y|^{\frac{\lambda - n}{p} + \epsilon} \, dy + \int_{B(x,r) \setminus B(0,1)} |y|^{\frac{\lambda-n}{p}} \, dy.$$
But after this I am unable to proceed. I don't think my new strategy is really usefull, so I am mainly looking for where to start. Thanks for any help in advance.
By the coarea formula, we find that \begin{align*} \int_{B(x,r)} \vert f_\epsilon \vert^p\,dy &= \int_0^r\int_{\partial B(x,s)} \vert f_\epsilon \vert^p \,d\mathcal{H}^{n-1}ds \\ &=\int_0^r \int_{\partial B(x,s)} \vert y \vert^{\lambda - n + p\epsilon} \mathbf{1}_{B(0,1)}(y) + \vert y \vert^{\lambda - n}\mathbf{1}_{B(0,1)^c}(y)\,d\mathcal{H}^{n-1}ds \\ &= \int_0^r \underbrace{\mathcal{H}^{n-1}(\partial B(x,s))}_{ = c s^{n-1}}\Big( \vert s \vert^{\lambda - n + p\epsilon}\mathbf{1}_{(0,1)}(s) + \vert s \vert^{\lambda - n}\mathbf{1}_{(0,1)^c}(s)\Big)ds \\ &= c\int_0^r s^{\lambda - 1 + p\epsilon}\mathbf{1}_{(0,1)}(s) + s^{\lambda - 1}\mathbf{1}_{(0,1)^c}(s)ds. \end{align*} Now, to prove that $V_0$ is satisfied, consider $r\in(0,1)$ and $x\in\mathbb{R}^n$. Then $$ r^{-\lambda}\int_{B(x,r)} \vert f_\epsilon\vert^p\,dy = cr^{-\lambda} \int_0^r s^{\lambda - 1 + p\epsilon}\,ds = c\frac{r^{p\epsilon}}{\lambda + p\epsilon} \rightarrow 0 $$ as $r\rightarrow\infty$. To prove that $V_\infty$ is not satisfied, let $r\in (1,\infty)$ and $x\in\mathbb{R}^n$ be arbitrary. Then \begin{align*} r^{-\lambda}\int_{B(x,r)} \vert f_\epsilon\vert^p\,dy &= cr^{-\lambda} \int_0^1 s^{\lambda - 1 + p\epsilon}\,ds + cr^{-\lambda}\int_0^r s^{\lambda - 1}\,ds \\ &= \frac{c}{r^{\lambda}(\lambda + p\epsilon)} + \frac{c}{\lambda} \rightarrow \frac{c}{\lambda} \end{align*} as $r\rightarrow\infty$.