How can I show $\int_{U}\frac{\partial f}{\partial x_{i}}(x)g(x)d\lambda^{d}(x)=-\int_{U}f(x)\frac{\partial g}{\partial x_{i}}(x)d\lambda^{d}(x)$

Question

Let $U \subseteq \mathbb R^{d}$ and $f \in C_{c}^{1}(U)$ while $g \in C^{1}(U)$

Show that:

$\int_{U}\frac{\partial f}{\partial x_{i}}(x)g(x)d\lambda^{d}(x)=-\int_{U}f(x)\frac{\partial g}{\partial x_{i}}(x)d\lambda^{d}(x)$

Idea: It is difficult to know whether I am going in the correct direction.

Let $\bar{f}: \mathbb R^{d} \to \mathbb R$, where $\bar{f}(x)=f(x)$ if $x \in U$ while $\bar{f}(x)=0$ else. And

$\bar{g}: \mathbb R^{d} \to \mathbb R$, where $\bar{g}(x)=g(x)$ if $x \in U$ while $\bar{g}(x)=0$ else.

Note that:

$\int_{U}(\partial_{i}f)gd\lambda^{d}(x)+\int_{U}f(\partial_{i}g)d\lambda^{d}(x)=\int_{U}\partial_{i}(gf)d\lambda^{d}(x)=\int_{U}\partial_{i}(\bar{g}\bar{f})d\lambda^{d}(x)$

I do not know if I am on the right track?

Answer 1

You can use the Gauss integral theorem applied to the vector field $F(x) = (0, \cdots, fg, \cdots, 0)$. Then one has $$\int_U \partial_i(fg) dx = \int_U div(F) dx = \int_{\partial U} \langle F, \nu\rangle dS = 0$$ and hence $$\int_U (\partial_i f) g dx = - \int_U f (\partial_i g) dx.$$