Give an example of compactly supported smooth function $f$ on $\mathbb{R}^2$ such that $\int_{\mathbb{R}^2} f dx_1 \wedge dx_2 = 0$, but one of the functions $f_1(x_1, x_2) = \int_{-\infty}^{x_1} f(t,x_2) dt$ or $f_2(x_1, x_2) = \int_{-\infty}^{x_2} f(x_1,t) dt$ is not compactly supported where $f_j : \mathbb{R}^2 \to \mathbb{R}$.

Okay so I think bump functions will be the key here, but my current issue is not understanding what are these $f_j$'s are. What is the intuition for the defined maps?


1 Answer 1


I'm not sure how to provide useful "intuition" for the functions $f_1$ and $f_2$, but I can describe the gist of the argument.

Consider the function $$ f(x_1, x_2) = \begin{cases} 1 & \text{if } (x_1, x_2) \in [-1, 0) \times [0, 1] \\ -1 & \text{if } (x_1, x_2) \in [0, 1] \times [0, 1] \\ 0 & \text{otherwise} \end{cases}.$$

This function is compactly supported, though it is not continuous (let alone smooth). By direct calculation, you should be able to see that $f_2$ is not compactly supported: the support of $f_2$ is $[-1,1] \times [0, \infty)$. And yet $\int f dx_1 \wedge dx_2 = 0$.

However, my $f$ doesn't solve your problem, since it's not smooth. So you need to think of an $f$ that has a similar profile to my $f$, but is smooth. Your idea of using bump functions should do the trick.


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