Asymptotic behavior of an integral transform Given $g\in L^2(\mathbb{R}^3)$, consider the following function ( defined for $r>0$ ):
$$c(r):=\int_{\mathbb{R}^3}\frac{g(x)}{|x|^2+r}dx$$
I'm interested in the behavior of $c(r)$ for large $r$. A simple application of Cauchy-Schwarz inequality gives
$$|c(r)|\lesssim r^{-\frac14}$$
However, I think that something better can be said about $c$. For example,
is it true that $c\in L^4(1,+\infty)$?
Thank you for any suggestions.
 A: Let $h(s)$ be the average of $g(x)$ over the sphere $|x|=s$.  Thus 
$$c(r) = \int_0^\infty \dfrac{4 \pi s^2 h(s) \; ds}{s^2 + r}$$
where $$ \int_0^\infty 4 \pi s^2 h(s)^2 \; ds \le \|g\|_2^2 < \infty$$
So Cauchy-Schwarz says
$$ c(r) \le 4 \pi \|s/(s^2+r)\|_2 \|s h(s)\|_2 \le \dfrac{\pi}{r^{1/4}} \|g\|_2 $$
with equality for $g(x) = 1/(|x|^2 + r)$.
This won't necessarily give the correct asymptotics as $r \to \infty$ for any particular $h$, but we can't improve the power of $r$ in general.  Consider  sequences $r_n \to \infty$  and $(a_n)_n \in \ell^1$, with $a_n > 0$, and
$$ h(s) = \sum_n a_n \dfrac{r_n^{1/4}}{s^2 + r_n}$$
We have $s h(s) \in L^2$ and 
$$ c(r) = \sum_n a_n {r_n}^{1/4} \int_0^\infty \dfrac{4\pi s^2 \; ds}{(s^2+r)(s^2 + r_n)} = \sum_n a_n \dfrac{2\pi^2 {r_n}^{1/4}}{\sqrt{r} + \sqrt{r_n}} $$
In particular $c(r_n) \ge \pi^2 a_n r_n^{-1/4} $.  Given $\epsilon > 0$ and $(a_n)$, we can let $r_n = a_n^{-1/\epsilon}$, and then $c(r_n) \ge \pi^2 r_n^{\epsilon - 1/4}$, so that 
$$ \limsup_{r \to \infty} c(r) r^{1/4-\epsilon} > 0$$
A: Once you asked, here goes. Let $h(r)=c(r^2)$. Then you need to look at the operator
$$
\Psi: f\mapsto \int_0^\infty f(\rho)\frac{\rho}{\rho^2+r^2}\,d\rho.
$$
Note two things: 
1) The kernel is positive and dominated by $1/(r+\rho)$ (just consider the cases $\rho\le r$ and $\rho\ge r$ separately. Hence, it is not worse than the classical Hilbert operator and it is bounded in $L^2$ (by the Schur test with $\varphi(\rho)=\rho^{-1/2}$, say).
2) If $h$ is any non-negative $L^2$ function such that $h(r)r^2$ is non-decreasing, then
$$
(\Psi h)(r)\ge\int_r^{2r}\frac{sh(s)}{s^2+r^2}\ge\int_r^{2r}\frac{r^2h(r)/s}{s^2+r^2}\ge \frac 1{10}h(r)
$$
so nothing better can be said.
The remaining change of variable should be clear.
A: I think i have a simple proof that $c\in L^4$, i give a sketch of it.
We can suppose $g$ positive (separating positive and negative part), then we can pass to its radial rearrangement (this preserves the $L^2$ norms).
Passing to polar coordinates, we get
$$c(r)=\int_0^{\infty}g(\rho)\cdot\frac{\rho^2}{\rho^2+r}d\rho$$
Since $\rho g(\rho)\in L^2(0,\infty)$, it`s natural to consider the operator
$$\Psi:f\mapsto\int_0^{\infty}f(\rho)\cdot\frac{\rho}{\rho^2+r}d\rho$$
and to show that it`s bounded from $L^2(0,\infty)$ to $L^4(0,\infty)$.
We can do this using Marcinkiewicz interpolation theorem.
For example, if $f\in L^{3/2}$, Holder inequality gives
$$|\Psi(f)(r)|\leq\Vert f\Vert_{3/2}\left\Vert\frac{\rho}{\rho^2+r}\right\Vert_3\lesssim (1+r)^{-3}\Vert f\Vert_{3/2}$$ 
This means that 
$$\Psi:L^{3/2}\rightarrow L^{3,\infty}$$
In the same way, we can show that
$$\Psi:L^{4}\rightarrow L^{8,\infty}$$
Then by Marcinkiewicz theorem
$$\Psi:L^{2}\rightarrow L^{4}$$
and we are done.
Do you think the proof it's correct? In that case i will expand it.
Thank you.
