From Evan's book, we know that, in $R^n$, the function $u$ defined as $$ u(x):=\int_{R^n}\Phi(y)f(x-y)dy,$$ where $f\in C^2(R^n)$ and $\Phi$ is the fundamental solutions of Laplace equation. Then we have $u\in C^2(R^n)$ as well and also we have $-\triangle u = f$.
However, my professor in addition comments that "it's possible to extend the theorem to more general $f$. For example, if $f$ is integrable, and Holder continuous, then $$ u(x)=\int_{R^n}\Phi(y)f(x-y)dy,$$ satisfies $-\triangle u = f$, at least for $n\geq 3$.
I can not even prove that $u\in C^2(R^n)$, not to mention that $-\triangle u =f$... Help is very appreciate!
It is true that $f$ does not need to be so smooth, and by using the fact that $-\Delta \Phi(x)=\delta_{0}$ i.e. the dirac delta function we may calculate $$-\Delta u(x)=\int_{\Bbb R^n}-\Delta_{x}\Phi(x-y)f(y)\,dy=\int_{\Bbb R^n}\delta_{x}f(y)\,dy=f(x).$$