I am reading a book with the following statement

Since $g$ has compact support $$\sum_{i=1}^d \int_{\mathbb{R}^d} \frac{\partial (fF_i g)}{x_i} dx = 0$$


  1. $\frac{dx}{dt} = F(x)$ with $x\in \mathbb{R}^d$
  2. $g: \mathbb{R}^d \rightarrow \mathbb{R}$ is continuously differentiable with compact support
  3. $f: \mathbb{R}^d \rightarrow \mathbb{R}$

The compact support means that if it is zero outside of a compact set.

How can we say the sum of integral is zero?

  • 2
    $\begingroup$ This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes. $\endgroup$ – Amey Joshi Jan 1 '19 at 7:31

The key of the reasoning is that, defined $$ V(x)=\left( \begin{matrix} fF_1 g\\ fF_2 g\\ \vdots\\ fF_{d-1} g\\ fF_d g \end{matrix} \right) \implies V\text{ has compact support in }\mathbb{R}^d $$ Since $$ \sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x = \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x, $$ then, by considering a ball $B(0,r)\in\mathbb{R}^d$ of radius $r>0$, we have that $$ \begin{split} \sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x & = \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x\\ &\triangleq\lim_{r\to\infty}\int\limits_{B(0,r)} \nabla\cdot V\, \mathrm{d}x\\ &\text{ by Gauss-Green}\\ &=\lim_{r\to\infty}\int\limits_{\partial B(0,r)} V\cdot\nu_x\, \mathrm{d}\sigma_x=0 \end{split} $$ where $\nu_x$ is the normal unit vector to $\partial B(0,r)$ in the point $x\in \partial B(0,r)$, since $V$ has compact support $\iff$ $V\cdot\nu_x=0$ for all $\nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.


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