Estimates for integrals over two balls of different radii Let $u$ and $v$ be two nonnegative integrable functions. By a standard covering argument, if we have $$\int_{B_r}u\,dx\leq\int_{B_{3r}}v\,dx$$ for any $r>0$, then we get $$\int_{B_r}u\,dx\leq C\int_{B_{2r}}v\,dx$$ for some constant C which depends on the dimension $n$.
My question is does this imply $$\int_{B_{3r}}v\,dx\leq C\int_{B_{2r}}v\,dx$$ If not, is the last inequality true? I am trying to see if I can always bound an integral over a ball with radios $r$ by an integral over a ball with radius less than $r$ given that two integrands are nonnegative. Thanks.
 A: Your arguments don't automatically imply that 
$\int \limits_{B_{3r}} v \, dx \leq C \int \limits_{B_{2r}} v \, dx$
. In your two first equations, the radius of the ball on the left side is less than the radius of the ball on the right, so it doesn't seem possible, even by playing with the values of $r$, to get your desired inequality.
In fact, it is in general not possible to find a constant such that the inequality is true (on an unbounded domain, at least). If we take $n=1$ and $v(x) = e^x$, then we compute:
$$
\int \limits_{B_{2r}} v \, dx
=
\int \limits_{-2r}^{2r} e^x \, dx
=
e^{2r} - e^{-2r}
\quad
\text{ and similarly }
\quad
\int \limits_{B_{3r}} v \, dx
=
e^{3r} - e^{-3r}
$$
Then for any $r>0$, we have:
$$
C \geq \frac{\int \limits_{B_{3r}} v \, dx}{\int \limits_{B_{2r}} v \, dx}
=
\frac{e^{3r} - e^{-3r}}{e^{2r} - e^{-2r}}
\geq
\frac{e^{3r}-1}{e^{2r}}
\geq
e^{r} - 1
$$
Taking $r \rightarrow \infty$, we get a contradiction. There might be a constant that makes the inequality hold on a bounded domain; I'm not sure.
NOTE: I'm assuming you're fixing $v$ and trying to find $C$ independent of $r$. If you're fixing $r$ and trying to find $C$ independent of $v$, then just take $v_n = e^{nx}$.
