In Theorem 1 (Solving Poisson's equation) on page 24 of Partial Differential Equations (2e) by Evans comparing equation (11) and (13) there appears to be the equality of
$$ \int_{\mathbb{R}^n \setminus B(0,\epsilon)} \Phi(y)\Delta_x f(x-y) \,dy = \int_{\mathbb{R}^n \setminus B(0,\epsilon)} \Phi(y)\Delta_y f(x-y) \,dy.$$
For context: $\Phi$ is the fundamental solution to the Laplace equation, $-\Delta u = f$ in $\mathbb{R}^n$, and $B(0,\epsilon)$ is the ball of radius $\epsilon$ centered on zero.
Where does the equality with respect to the Laplacians come from? That is why does
$$ \Delta_xf(x-y) = \Delta_yf(x-y) $$
hold?
I have seen this in another reference (page 149 of Partial Differential Equations in Action (3e) – Salsa). Is this a general property of convolution or is it something more subtle? In terms of level of understanding, explanations without reliance on measure theory would be preferred.
Best Answer
Just do the calculation. For example in 1-dimension we have that $\Delta_x = \partial_{xx}$. Using the chain rule we see that $$\Delta_x f(x-y) = \partial_x (1\cdot f'(x-y))= (1)^2 \cdot f''(x-y) = f''(x-y)$$ on the other hand we have $$\Delta_y f(x-y) = \partial_y ((-1)\cdot f'(x-y))= (-1)^2 \cdot f''(x-y) = f''(x-y).$$ The analog is true for $n$-dimensions.
To sum it up: for $\partial_y$ we get an additional minus sign from the chain rule but since we take it twice in the Laplacian, they cancel each other out.