I want to show that $$\int_\mathbb{R^n} \nabla^2\phi \mathrm{d}^nx = 0$$ provided $\nabla \phi(\mathbf{x},t) \rightarrow 0$ as $|\mathbf{x}| \rightarrow \infty$.

I have thought of maybe expanding the Laplacian but I have no idea how to evaluate $\int_\mathbb{R^n} \frac{\partial^2 \phi}{\partial x_i^2} \mathrm{d^n}x$?

I have also considered using Gauss' Theorem as follows; $$\int_\mathbb{R^n} \nabla^2\phi \mathrm{d}^nx = \lim_{r \rightarrow \infty} \int_{C_r}\nabla^2\phi \mathrm{d}^nx =\lim_{r \rightarrow \infty} \int_{\partial C_r}\nabla\phi\cdot \mathrm{d}\mathbf{S} \leq \lim_{r \rightarrow \infty}A\max_{\partial C_r }(\phi \cdot \mathbf{S}) = 0$$

Where $C_r$ is a sphere of radius $r$ centred on the origin and A is an appropriate constant. Is this correct?

I also want to prove that if we integrate over a compact space $M$ instead of $\mathbb{R^n}$ that the same holds. Although looking at my 'proof' this doesn't seem to be correct?

  • $\begingroup$ Using Gauss' Theorem should work. That's the way I like it. $\endgroup$ – md2perpe Dec 30 '17 at 22:52
  • $\begingroup$ It should also work if $\nabla\phi = 0$ on the boundary of the compact space or if it has no boundary, like a closed manifold. $\endgroup$ – md2perpe Dec 30 '17 at 22:54

This is not true. For example, let $\phi(x) = |x|^{p}$ with $0<p<1$. This is not smooth at $0$ but you can smoothen it there without changing anything for large $|x|$. Note that $|\nabla \phi(x)| = O(|x|^{p-1})$ tends to zero at infinity. On the other hand, the Laplacian of this radial function is,
$$ \Delta \phi(x) = r^{1-n}(r^{n-1}pr^{p-1})' =p r^{1-n}(r^{n+p-2})' = p(n+p-2) r^{p-2}, \quad r=|x| $$ This decays too slowly to even be integrable, as $p-2 > -2\ge -n$.

The correct statement would be to assume that $|\nabla \phi (x)|=o(|x|^{1-n})$. Then the gradient flux across the sphere of radius $r$ does tend to $0$ as $r\to \infty$, and the conclusion follows from the divergence/Gauss theorem, since the Laplacian is the divergence of the gradient.


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