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}) $?

Question

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?

Answer 1

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.