It comes from estimates for wave equations.



> 
> For any $ u=u(t,x)\in C\_{0}^{\infty}(\mathbb{R}^{1+2}) $, which is a smooth compactly supported function, prove that
> $$ 
> \|u\|\_{L^{\infty}(\mathbb{R}^{1+2}\,\,)}\leq C\|\partial\_{x\_1}\square u\|\_{L^1(\mathbb{R}^{1+2}\,\,)},
> $$
> where $ \square $ is the wave operator $ \partial\_t^2-\Delta $ and $ C $ is independent of $ u $.
> 
> 
> 



> 
> Given an example that there is not constant $ C $ such that  
> 
> $$ 
> \|u\|\_{L^{\infty}(\mathbb{R}^{1+2}\,\,)}\leq C\|\partial\_t\square u\|\_{L^1(\mathbb{R}^{1+2}\,\,)}.
> $$
> 
> 
> 


Here is my try. Note that $ u $ is compactly supported, can consider $ v(t,x)=u(t+T,x) $ with $ T\gg 1 $ if necessary, I find that I can assume that $ u(0,x)=0 $ and $ \partial\_tu(0,x)=0 $. Then by using the formula of the solution for wave equations, I have
$$
u(t,x)=\int\_{0}^t\int\_{y\_1^2+y\_2^2\leq 1}\frac{sF(t-s,x\_1-sy\_1,x\_2-sy\_2)}{2\pi\sqrt{1-y\_1^2-y\_2^2}}dy\_1dy\_2ds,
$$
where $ F=\square u $. However, I cannot go on. Can you give me some references or hints?



How to prove $ \|u\|_{L^{\infty}}\leq C\|\partial_1\square u\|_{L^1} $ for any $ u\in C_0^{\infty}(\mathbb{R}^{1+2}) $?
Let $E$ be a fondamental solution of $\partial\_{x\_1}\square$. Then you have for $u$ compactly supported
$$
u=u\ast \delta=u\ast (\partial\_{x\_1}\square E)= (\partial\_{x\_1}\square u)\ast E,
$$
so that
$
\Vert u\Vert\_{L^\infty}\le \Vert \partial\_{x\_1}\square u\Vert\_{L^1}
\Vert E\Vert\_{L^\infty}
$
and
it is now enough to prove that $E$ can be chosen bounded:
take
$$
E\_0=\frac{1}{2π}H(t-\vert x\vert)(t^2-\vert x\vert^2)^{-1/2}, \quad H=\mathbf 1\_{\mathbb R\_+},
$$
the fondamental solution of $\square$ in two-space dimensions. We take
$$
E=\int\_0^{x\_1} E\_0(t, y, x\_2) dy,
$$
and noting that
$
\int\_0^a\frac{dy}{\sqrt{a^2-y^2}}
$
is bounded, you get the sought result.


