How to prove $\|u\|_\infty\le\|u\|_1+\|\nabla u\|_1+\|\partial^2_{xy}u\|_1$ on the square $[0,1]^2$

Question

let $u(x,y)$ is continuous on $\Omega$, $$\Omega=\{(x,y)|0\le x\le 1,0\le y\le 1\}$$ and $$\dfrac{\partial u}{\partial x},\dfrac{\partial u}{\partial y},\dfrac{\partial^2 u}{\partial x\partial y}$$ are absolute integrable

show that $$\sup_{(x,y)\in\Omega}|u(x,y)|\le\int\int_{\Omega}\left(|u(x,y)|+\left|\dfrac{\partial u}{\partial x}\right|+\left|\dfrac{\partial u}{\partial y}\right|\right)dxdy+\int\int_{\Omega}\left|\dfrac{\partial^2u}{\partial x\partial y}\right|dxdy$$

I know proof this inequality $$\sup_{x\in[0,1]}|u(x)|\le\int_{0}^{1}|u(x)|dx+\int_{0}^{1}|u'(x)|dx$$

But for this I can't

Answer 1

I use repeatedly the inequality you can prove:

For each $y \in [0,1]$, $$ \sup_{x\in[0,1]} \left|\frac{\partial f}{\partial y}(x,y)\right| \le \int_{x=0}^1 \left|\frac{\partial f}{\partial y}(x,y)\right| + \left|\frac{\partial^2 f}{\partial x\partial y}(x,y)\right| \, dx .$$ Therefore $$ \sup_{x\in[0,1]} \int_{y=0}^1 \left|\frac{\partial f}{\partial y}(x,y)\right| \, dy \le \int_{y=0}^1 \sup_{x\in[0,1]} \left|\frac{\partial f}{\partial y}(x,y)\right| \, dy \le \int_{x=0}^1 \int_{y=0}^1 \left|\frac{\partial f}{\partial y}(x,y)\right| + \left|\frac{\partial^2 f}{\partial x\partial y}(x,y)\right| \, dy \, dx .$$ Also $$ \sup_{x,y \in [0,1]} |u(x,y)| \le \sup_{x\in[0,1]} \int_{y=0}^1 |u(x,y)| + \left|\frac{\partial f}{\partial y}(x,y)\right| \, dy ,$$ and $$ \sup_{x\in[0,1]} \int_{y=0}^1 |u(x,y)| \, dy \le \int_{x=0}^1 \int_{y=0}^1 |u(x,y)| \, dy + \frac{\partial}{\partial x}\left(\int_{y=0}^1 |u(x,y)| \, dy\right) \, dx $$ $$ \le \int_{x=0}^1 \int_{y=0}^1 |u(x,y)| \, dy + \left(\int_{y=0}^1 \left|\frac{\partial u}{\partial x}(x,y)\right| \, dy\right) \, dx .$$

In the last line, I cheated a bit by bringing the differential under the integral sign, and then citing the result $\frac{\partial}{\partial x} |f(x)| \le \left| \frac{\partial f}{\partial x}(x) \right|$.

But you could fix this, by first showing the result when $u$ is smooth, and then using a density argument. Also the last inequality really requires $f(x) \ne 0$, but by giggling around a bit, I think you can get around it.

Or you could put the arguments into a different sequential order, and maybe things would drop out more smoothly.

Get back to me if the details at the end don't work out. I have to go to bed now.