prove that integral of laplacian squared equals to integral of sum of integral of partial deritatives squared

Question

Let f be a $C^3$ function, $f: \mathbb{R}^n \to \mathbb{R}$, f is compactly supported

prove that:

$$\int_{\mathbb{R}^n} (\Delta f) ^2 = \sum_{i,j = 1}^n \int_{\mathbb{R}^n} (\frac{\partial^2f}{ \partial x_i \partial x_j})^2$$

I tried to use green formulas but didnt really know how to handle the laplacian squared

I think it might be realated to the co-area formula by using balls

Answer 1

A physicsists idea: On the one hand $$\int dV (\Delta f)^2 = \int dV \left(\sum_i \partial_i^2 f\right)^2= \int dV \left(\sum_i \partial_i^2f\right)\left(\sum_j \partial_j^2f\right)\\=\sum_{i,j} \int dV (\partial_i^2f)(\partial_j^2 f)$$ On On the other hand $$\int dV \sum_{i,j} \left(\partial_i \partial_j f \right)^2=\int dV \sum_{i,j} \left(\partial_i \partial_j f \right) \left(\partial_i \partial_j f \right)\\ =\sum_{i,j} \int dV \left(\partial_i \partial_j f \right) \left(\partial_i \partial_j f \right)$$ Let's look at the last expression. We use partial integration to move a $\partial_i$ from the right parenthesis to the left. We will pick up a minus sign. Next, we use partial integration to move a $\partial_j$ from the left parenthesis to the right parenthesis. We will pick up another minus sign and have shown the identity.

EDIT: I guess that the surface terms from the partial integration vanish due to the compact support of $f$. The appearance of the third derivative after the first partial integration should also not be a problem, since you specified that $f$ is a $C^3$ function.