I've been asked to prove this:
If $f$ is a $C^2$ function on $\mathbb{R}^2$ with compact support, then $$\int_{\mathbb{R^2}}\Delta f = 0$$ Here, $\Delta f$ is the Laplacian defined as $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$.
I don't know any theorems related to the Laplacian, this is the first time I'm hearing of it (the question appeared in a section on Fubini's Theorem). I suppose we can begin by splitting the integral:
$$\int_{\mathbb{R^2}}\frac{\partial^2 f}{\partial x^2} + \int_{\mathbb{R^2}}\frac{\partial^2 f}{\partial y^2}$$
Should I now take a suitable rectangle $R = [a_1, b_1]\times [a_2,b_2]$ covering the support of $f$ and then applying Fubini's Theorem? I tried it but am not getting anything coherent.
How do I proceed?
Yes, by Fubini Theorem. Let $[a_1, b_1]\times [a_2, b_2]$ be large so that it covers your support. Then \begin{align} \int_{\mathbb{R^2}}\frac{\partial^2 f}{\partial x^2}&=\int_{a_2}^{b_2} \left(\int_{a_1} ^{b_1} \frac{\partial ^2 f}{\partial x^2} dx\right)dy\\ &= \int_{a_2}^{b_2} \left( \frac{\partial f}{\partial x}(b_1, y) - \frac{\partial f}{\partial x}(a_1, y) \right) dy = 0 \end{align}
Similar for $\int_{\mathbb{R^2}}\frac{\partial^2 f}{\partial y^2}$.