Fundamental solution Laplace-Poisson equation

Question

Let $\Phi:\mathbb{R}^n\setminus\{0\}\rightarrow\mathbb{R}$ be the fundamental solution of the Laplace equation (see e.g. in the book of Evans). For a function $f\in\mathcal{C}_c^2(\mathbb{R}^n)$ we define the function $$u(x):=\int_{\mathbb{R}^n}\Phi(y)f(x-y)dy$$ (which finally should turn out to be a solution of the Poisson equation, but I'm not interested in this fact here). What I want to prove now is that this function $u$ is continuous. Does somebody know how to do this? I attempted to use that $f$ is uniformly continuous on $\mathbb{R}^n$ which is unfortunately not the case for $\Phi$. Hence when I let a sequence $x_n$ of vectors approach some $x$, I do not know why it is possible to interchange limit and integration. Can someone give me a hint?

Answer 1

Try $\frac{u(x+he_i)-u(x)}{h}=\int_{\mathbb{R}^n}\Phi(y)\{\frac{f(x+he_i-y)-f(x-y)}{h}\}dy$. Then $\frac{f(x+he_i-y)-f(x-y)}{h}\rightarrow\frac{\partial f}{\partial x_i}(x-y)$ uniformly(!) in $h$ and of course $u\in C^1$ implies $u\in C^0$. Similarly You can show $u\in C^2$.