Estimate for an integration over on a ball

Question

Let $n \in {\mathbb N}$, $x \in {\mathbb R}^{n}$ and $r>0$. Let $B(x,r)$ be a closed ball on ${\mathbb R}^{n}$ with center $x$ and radius $r$, i.e., $$ B(x,r):=\{y \in {\mathbb R}^{n}; |y-x| \leq r\}. $$ Claim: There exists a positive constant $C$ such that for any $z \in B(x,r)$, $$ \int_{B(x,r)}|y-z|^{1-n}{\rm d}y \leq C r. $$ Does this claim hold?

Answer 1

Yes. Since $B(x,r) \subset B(z,2r)$, we can extend the integral to all of $B(z,2r)$. There is a constant $s_n>0$ such that the "surface area" of an $(n-1)$-sphere in $\Bbb R^n$ equals $s_n r^{n-1}$. Therefore $$ \int_{B(x,r)} |y-z|^{1-n}\,dy \le \int_{B(z,2r)} |y-z|^{1-n}\,dy = \int_0^{2r} |r|^{1-n} \cdot s_nr^{n-1}\,dr = 2s_n r. $$