I'm trying to understand the proof that $u(x) = \int_{\mathbb{R}^n} \psi(x-y)f(y)dy$ is indeed a solution to $-\Delta u = f$. I can follow it if I do it in a particular dimension ie 2. However, in arbitrary dimension it is not clear to me how you bound some of the integrals in general. For example, in Evans, he arrives at the following integral and bounds it as such:
$\int_{B(0,\epsilon)} |\psi(y)|dy \le C \epsilon^2$
How does this follow in general? $\psi$ is order $\epsilon ^{2-n}$ and the volume of the ball is order $\epsilon^n$ so naively the product is $\epsilon^2$ but of course $\psi$ is singular. What are the details for this bound in general?
Edit f is assumed two be in $C^2$. But also note that what I don't understand is the bound given above where f has already been removed...
Edit 2 Also , $\psi$ is the fundamental solution, I realized I never explicitly said that.
So, we have $n\ge 3$ and $\psi(y)=c|y|^{2-n}$. Integrate in polar coordinates: $$ \int_{B(0,\epsilon)} |\psi(y)|dy = c\omega_{n-1} \int_0^\epsilon r^{2-n} r^{n-1}\,dr = c\omega_{n-1} \frac12\epsilon^2 $$ where $\omega_{n-1}$ is the area of unit sphere.
Note that the estimate is false in two dimensions: if $n=2$, then $\psi$ has logarithmic singularity, and integration of $r\log (1/r)$ leads to something with asymptotic behavior $r^{2}\log (1/r)$.