Showing this function and its gradient is bounded. Let $x \in \mathbb{R}^2$ and $t \in \mathbb{R}$.  Let $v(x) \in C_c^\infty(\mathbb{R}^2)$. Lastly let 
$$u(x,t) = \frac{t}{2\pi} \int_{|y| \leq 1} \frac{v(x+ty)}{\sqrt{1-|y|^2}}dy$$ 
I want to show that for $2|x| \leq 1 +t$ 
$|u(x,t)| \leq \frac{C}{1+t} $ 
$|\nabla u(x,t)| \leq  \frac{C}{1+t}$
I'm not sure how relevant this is, but it can be shown that $$u(x,t) = \frac{t}{2\pi} \int_0^{2\pi} \int_0^{1} v(x_1 + tr\cos \theta,x_2 + tr \sin \theta) \frac{r dr}{\sqrt{1-r^2}} d\theta$$ Perhaps this is easier to work with or perhaps not. Any ideas?
 A: I have a solution to $|u(x,t)| \leq \frac {C}{1+t}$, but I left out many of the details (it'd take too long to write up). Making the substitution $z = x + ty$ in the integral helps. From there, you need to look at $t$ sufficiently large such that we are only worrying about the set where $v\neq 0$. A sketch of the solution is below:

First make the change of variables $z=x+ty$ in the integral (for fixed $x$ and $t$). Then we should have $$u(x,t) = \frac 1 {2\pi}\int_{B_t(x)} \frac{v(z)}{\sqrt{t^2 - |z-x|^2}} \, dz$$ where $B_t(x)$ is the ball of radius $t$ around the point $x$. Since $v\in C_c^\infty$, there is a ball $B_R(x_0)$ where $v=0$ outside this ball. You should be able to show that there is a $t_0$ where for $t\geq t_0$ and $|x| \leq \frac{1+t} 2$, $B_R(x_0) \subseteq B_t(x)$. Then for $x$ and $t$ satisfying these conditions, we need only evaluate the integral on $B_R(x_0)$. It is also possible to show that $|z-x| \leq (t_0 +t)/2$ for $x$ and $t$  meeting the above conditions and $z\in B_R(x_0)$. So for sufficient $x$ and $t$, we have that $$ |u(x,t)| \leq \frac{\max |v|}{2\pi} \int_{B_R(x_0)} \frac{1}{\sqrt{t^2 - (t + t_0)^2/4}}dz = \frac{\max |v| \cdot R^2}{2}\times\frac{1}{\sqrt{t^2 - (t + t_0)^2/4}}.$$ This term is $O\left(\frac{1}{1+t}\right)$. For $t<t_0$, we can see that $|u(x,t)|$ is bounded uniformly.

