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Let $f,g:\mathbb R^n \rightarrow \mathbb R^n$, $(x_1,\dots, x_n)\mapsto f,g(x_1,\dots, x_n)$ be smooth functions with compact support.

Use the following notation: for an index $\alpha = (\alpha_1,\dots, \alpha_n)$ where $\alpha_i$ is a natural number, we denote by $D^{\alpha}$ the derivative with respect to $x_1$ $\alpha_1$ times, with respect to $x_2$, $\alpha_2$ times, and so on until $x_n$, $\alpha_n$ times.

Can we then say that $$\int_{\mathbb R^n} D^{\alpha} f(x_1,\dots, x_n) g(x_1,\dots, x_n)dx_1 \cdots dx_n = (-1)^{|\alpha|} \int_{\mathbb R^n} f(x_1,\dots, x_n) D^{\alpha} g(x_1,\dots, x_n)dx_1 \cdots dx_n$$ where $|\alpha| = \sum_{i=1}^n \alpha_i$.

Thanks you!

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  • $\begingroup$ Yes, we can do this. $\endgroup$
    – Jihad
    Commented Oct 28, 2014 at 11:08

1 Answer 1

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Yes. Let $[-R,R]^n$ be a cube containing the support of $f$ and $g$. The formula is the result of integrating by parts $|\alpha|$ times in the iterated integral $$ \int_{-R}^R\Bigl(\dots\Bigl(\int_{-R}^RD^\alpha f \, g\,dx_n\Bigr)\dots\Bigr)\,dx_1. $$

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