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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.

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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.

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