Show that $\int_{B(x, 2r)} |y|^\alpha dy \lesssim \int_{B(x, r)} |y|^\alpha dy$ uniformly in $x$ and $r$ I want to show that
\begin{equation*}
\int_{B(x, 2r)} |y|^\alpha dy \lesssim \int_{B(x, r)} |y|^\alpha dy
\end{equation*}
uniformly in $x \in \mathbb R^d$ and $r > 0$ if and only if $\alpha > -d$. The forward direction follows from the fact $y \mapsto |y|^\alpha$ has a non-integrable singularity at the origin whenever $\alpha \leq -d$, however I am having trouble proving the converse.
 A: I'll treat the case $-d<\alpha <0.$ Note in this case $|y|^\alpha$ is a decreasing function of $|y|.$ Note also $|y|^\alpha$ is locally integrable on $\mathbb R^d.$
Lemma: There is a positive constant $C$ such that
$$\tag 1\int_{B(x,2)}|y|^\alpha\,dy \le C\int_{B(x,1)}|y|^\alpha\,dy\,\text{ for all } x\in \mathbb R^d.
$$
Proof: The functions
$$f(x) = \int_{B(x,1)}|y|^\alpha\,dy,\,\, g(x) = \int_{B(x,2)}|y|^\alpha\,dy $$
are positive and continuous. This follows from the local integrability of $|y|^\alpha$ and the DCT. It follows that on, say, $\{|x|\le 3\},$ $f$ has a positive minimum $m$ and $g$ has a positive maximum $M.$ Therefore
$$\frac{\int_{B(x,2)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} \le \frac{M}{m}\,\,\text{for }|x|\le 3.$$
This gives $(1)$ for $|x|\le 3.$
Now suppose $|x|>3.$ Then
$$ \int_{B(x,2)}|y|^\alpha\,dy \le  \int_{B(x,2)} (|x|-2)^\alpha\,dy =(|x|-2)^\alpha 2^d V,$$
where $V$ is the volume of the open unit ball. Similarly,
$$\int_{B(x,1)}|y|^\alpha\,dy \ge (|x|+1)^\alpha V.$$
Thus
$$\frac{\int_{B(x,2)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} \le 2^d\left (\frac{|x|-2}{|x|+1}\right )^\alpha\,\,\text{for }|x|> 3.$$
Now the function $t\to [(t-2)/(t+1)]^\alpha$ is positive and bounded on $(3,\infty).$ That implies $(1)$ holds for $|x|>3,$ and the lemma is proved.
Now suppose $x\in \mathbb R^d, r>0.$ Then the substitution $y=rz$ gives
$$\int_{B(x,r)}|y|^\alpha\,dy = r^{\alpha + d}\int_{B(x/r,1)}|z|^\alpha\,dz.$$
Similarly,
$$\int_{B(x,2r)}|y|^\alpha\,dy = r^{\alpha + d}\int_{B(x/r,2)}|z|^\alpha\,dz.$$
So
$$\tag 2 \frac{\int_{B(x,2r)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} = \frac{\int_{B(x/r,1)}|z|^\alpha\,dz}{\int_{B(x/r,2)}|z|^\alpha\,dz}.$$
By the lemma, the right side of $(2)$ is uniformly bounded for all $x,r.$
Thus we're done for the case $-d<\alpha<0.$ The proof for $\alpha \ge 0$ is much the same.
