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Suppose $u(x,y)$ is continuous in $D=\{(x,y)| 0\le x \le 1, 0\le y \le 1\}$, $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial^2 u}{\partial x \partial y}$ is absolutely integrable, prove

$$ \sup_{(x,y) \in D} |u| \le \iint_D |u| dxdy + \iint_D \left( \left| \frac{\partial u}{\partial x} \right|+\left|\frac{\partial u}{\partial y}\right| \right) dxdy + \iint_D \left|\frac{\partial^2 u}{\partial x \partial y}\right| dxdy. $$

I have tried the Taylor formular,

$$ \sup |u|=|u(\xi_x,\xi_y)+u(x_0,y_0)-u(\xi_x,\xi_y)|=\left| u(\xi_x,\xi_y)+\frac{\partial u}{\partial x}(\xi_x,\xi_y)(x_0-\xi_x)+\frac{\partial u}{\partial y}(\xi_x,\xi_y)(y_0-\xi_y)+\frac12 \frac{\partial^2 u}{\partial x \partial y}(\eta_x,\eta_y)[(x_0-\xi_x)^2+(y_0-\xi_y)^2] \right| $$ where $u(\xi_x,\xi_y)=\iint_D u dxdy$.

but there seems to be some matters of detail.

Thanks for your help.

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  • $\begingroup$ I think you are close to the answer. Just note that $\max\{|x_0-\xi_x|,|y_0-\xi_y|,\frac{1}{2}|(x_0-\xi_x)^2+(y_0-\x_i_y)^2|\}|\le1$ when these variables are all in $D$. $\endgroup$ – Eric Yau Nov 27 '18 at 8:53
  • $\begingroup$ Thanks, but how to prove $\iint_D (|\frac{\partial u}{\partial x}(\xi_x,\xi_y)|+|\frac{\partial u}{\partial y}(\xi_x,\xi_y)|) dxdy \le \iint_D (|\frac{\partial u}{\partial x}|+|\frac{\partial u}{\partial y}|) dxdy$ and $\iint_D |\frac{\partial^2 u}{\partial x \partial y}(\eta_x,\eta_y|) dxdy \le \iint_D |\frac{\partial^2 u}{\partial x \partial y}dxdy$ ? $\endgroup$ – Fyhswdsxjj Nov 27 '18 at 11:16

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