# prove this step in poisson equation

$\phi(x)$ is harmonic in $C^2$

in proof for solution of poisson's equation($- \Delta u=f$) ,this step came it maybe very fundamental but I am not getting it

$u(x)= \int_{R^n}\phi(x-y)f(y)\mathrm dy = \int_{R^n}\phi(x)f(x-y)\mathrm dy$

how this transformation happened without negative sign?

and this step?

$\int_{B(0,\epsilon)}\phi(y)\Delta_xf(x-y)\mathrm dy \leq C||D^2f||_{L^{\infty}(R^n)}\int_{B(0,\epsilon)}|\phi(y)|\mathrm dy \leq C\epsilon^2|log \epsilon| for (n=2),C\epsilon^2 for (n\geq3)$

$\int_{R^n-B(0,\epsilon)}\phi(y)\Delta_xf(x-y)\mathrm dy \leq ||Df||_{L^{\infty}(R^n)}\int_{\partial B(0,\epsilon)}|\phi(y)|\mathrm dSy \leq C\epsilon|log \epsilon| for (n=2),C\epsilon for (n\geq3)$

• The first displayed equation is wrong. It would be correct if you wrote $... \int \phi(y) f(x-y) \mathrm{d}y$ instead, and can be obtained by a simple change of variables. See Wikipedia page on convolution. – Willie Wong Jun 23 '14 at 11:09