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Assume that $ f: \Bbb{R}^{3} \to \Bbb{R} $ is a smooth function such that $ M \stackrel{\text{df}}{=} \left\{ \mathbf{x} \in \Bbb{R}^{3} ~ \middle| ~ f(\mathbf{x}) \ge 0 \right\} $ is a non-empty compact $ 3 $-dimensional manifold with boundary. Prove that for $ g = f^{2} $, we have $$ \int_{M} \nabla g ~ \mathrm{d}{\mathbf{x}} = 0, $$ where $ \displaystyle {\nabla g}(\mathbf{x}) = \sum_{i = 1}^{3} (\partial_{i,i} g)(\mathbf{x}) $ for all $ \mathbf{x} \in \mathbb{R}^{3} $.

I plugged $ \displaystyle \nabla g = \sum_{i = 1}^{3} \partial_{i,i} g $ and $ g = f^{2} $ into the integral to get $$ \int_{M} \left( \partial_{1,1} f^{2} + \partial_{2,2} f^{2} + \partial_{3,3} f^{2} \right) \mathrm{d}{\mathbf{x}}. $$ How should I proceed further from this? Any methodological help would be highly appreciated!

Thanks in advance!

EDIT: I attempted a solution. I am trying to write the Laplacian as divergence

$\int_M\nabla g=\int_M\partial_1g+\partial_2g+\partial_3g= \int_M \partial_1f^2+ \partial_2f^2+\partial_3f^2=\int_ \partial f^2+f^2+f^2=3\int_\partial f^2=3\cdot (\frac{f^3}{3}) $

Now, I think $(\frac{f^3}{3})=0$ because the integral is evaluated on the boundary and on the boundary, f attains $0$. $\partial$ index of the integral means the boundary of $M$

Is this true?

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  • $\begingroup$ What is the notation $D_{i,i}$? $\endgroup$ – user225318 Apr 23 '15 at 19:43
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    $\begingroup$ If $\nabla$ is supposed to be the Laplacian, then you should use Stoke's theorem. $\endgroup$ – user225318 Apr 23 '15 at 19:44
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    $\begingroup$ Do you know the statement of Stokes' theorem (or the divergence theorem)? If not, look it up. If yes, try to write the Laplacian as a divergence. $\endgroup$ – user225318 Apr 23 '15 at 19:46
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    $\begingroup$ This is really horrid notation. Please write either $\nabla^2 g$ (as most math and physics texts do) or $\Delta g$ (as most research-level math texts do). $\endgroup$ – Ted Shifrin Apr 23 '15 at 20:21
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    $\begingroup$ You really should change your notation. In any event, in physicist's otation: $$\nabla^2 g = \nabla \cdot (\nabla g)\quad\text{ and }\quad \nabla g = 2 f \nabla f = 0\;\text{ on the boundary }\; \partial M $$ $\endgroup$ – achille hui Apr 23 '15 at 21:54

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