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Let $f:\mathbb{R}^2\to\mathbb{R}$ is $C^2$, $L>0$, and $D$ be the closed unit disc in the plane. Also, suppose $f_{xx}^2+2f_{xy}+f_{yy}\leq L$ and $f(0,0)=f_x(0,0)=f_y(0,0)=0$. Show that$\int\int_Df(x,y)dxdy\leq\frac{\pi\sqrt{L}}{4}$.

Since $D$ is connected, I think that $f=0$ on the interior of $D$, so we only need to consider the values $f$ takes on the unit circle. Also, I see that $\Delta f\leq\sqrt{L}$, and so the divergence theorem gives $\int_{\partial D}\nabla f\cdot\hat{n}dl=\int\int_D\Delta fdxdy\leq\pi\sqrt{L}$, but I'm not really sure how to proceed from here. Any suggestions?

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    Why is $f=0$ on the interior of $D$? – 2017-02-05
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    @AdamHughes Maybe that's not correct, but I thought that since $f_x(0,0)=f_y(0,0)=0$, $f$ is constant along any line segment connected to $(0,0)$. But every point in the interior of $D$ can be connected by a line segment to $(0,0)$ – 2017-02-05
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    Indeed this is not necessarily true. It's not even true for a line segment, let $f(x,y)=ax^3$ with $a\le \sqrt{L}/6$. Then $f_{xx}^2+2f_{xy}+f_{yy}^2=f_{xx}^2= 36a^2x^2\le 36a^2$ (since $|x|\le 1$) satisfies the hypotheses but clearly $f$ is not constant, nor is $f$ constant along the line segment $y=0, -1\le x\le 1$. – 2017-02-05

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