Show $\int_{\mathbb{R}^n}\Delta_x \Phi(x-y)f(y)dy = \int_{\mathbb{R}^n}\Delta_y \Phi(x-y)f(y)dy.$

Question

I read in an article about Laplace's equation that

$$-\int_{\mathbb{R}^n}\Delta_x \Phi(x-y)f(y)dy = -\int_{\mathbb{R}^n}\Delta_y \Phi(x-y)f(y)dy.$$

Could someone explain to me why this is? I understand that one can move the differentiation from one factor of the convolution to the other...

Answer 1

$ \nabla_x \Phi(x-y) = - \nabla_y \Phi(x-y) $ by the chain rule. Apply this identity twice.