Solving Poisson Equation Proof in Lawrence C. Evans Theorem 1 page 23

Question

I am trying to understand a proof of a theorem in L.C. Evans related to Poisson Equation. However, there are some steps which I do not understand.

Here is the statement of the theorem:

Theorem 1. Let $u$ be a function defined as the following:
$$u(x) = \int_{\mathbb{R}^{n}}\Phi(x-y)f(y)dy$$
given that $\Phi(x) = \Phi(|x|) = \begin{cases} -\frac{1}{2\pi}\log|x| & (n=2) \\ \frac{1}{n(n-2)\alpha(n)} \frac{1}{|x|^{n-2}} & (n \geq 3) \end{cases}$ with $\alpha(n)$ as the volume of a unit ball in $\mathbb{R}^{n}$ and $f : \mathbb{R}^{n} \to \mathbb{R}$.
Then (i) $u \in C^{2}(\mathbb{R}^{n})$ and (ii) $-\Delta u = f$ in $\mathbb{R}^{n}$

So that is the statement of the theorem. In proving this theorem, I have several things that I can't understand quite well.

First, is this statement:
Fix $\varepsilon > 0$. Then
$\Delta u(x) = \int_{B(0,\varepsilon)}\Phi(y)\Delta_{x}f(x-y)dy + \int_{\mathbb{R}^{n}-B(0,\varepsilon)}\Phi(y)\Delta_{x}f(x-y)dy =: I_{\varepsilon}+K_{\varepsilon}$ (eq.11)
Now
$|I_{\varepsilon}| \leq C ||D^{2}f||_{L^{\infty}(\mathbb{R}^{n})}\int_{B(0,\varepsilon)}\Phi(y)dy\leq \begin{cases} C\varepsilon^{2}|\log\varepsilon|& (n=2) \\ C\varepsilon& (n \geq 3)\end{cases}$ (eq.12)
An integration by parts yields
$$J_{\varepsilon}=\int_{\mathbb{R}^{n}-B(0,\varepsilon)}\Phi(y)\Delta_{x}f(x-y)dy = -\int_{\mathbb{R}^{n}-B(0,\varepsilon)}D\Phi(y)D_{y}f(x-y)dy+\int_{\partial B(0,\varepsilon)}\Phi(y)\frac{\partial f}{\partial \nu}(x-y)dS(y)$$ $$ J_{\varepsilon} =: K_{\varepsilon} + L_{\varepsilon}\quad \text{(eq.13)}$$ with $\nu$ denoting the inward pointing unit normal along $\partial B(0,\varepsilon)$. We readily check
$|L_{\varepsilon}|\leq ||Df||_{L^{\infty}(\mathbb{R}^{n})}\int_{\partial B(0,\varepsilon)}|\Phi(y)|dS(y)\leq \begin{cases} C\varepsilon\log|\varepsilon| & (n=2) \\ C\varepsilon & (n \geq 3) \end{cases}$ (eq.14)

So, my question is, how do I get the inequality in (eq.12) and (eq.14)? Also I find it weird that in (eq.13) the book uses $\Delta_{y}$ instead of $\Delta_{x}$. Is there any explanation of why they can do it like that?

As for the last part, I also don't understand about this part:
$$K_{\varepsilon} = -\frac{1}{n\alpha(n)\varepsilon^{n-1}}\int_{\partial B(0,\varepsilon)}f(x-y)dS(y)$$ $$K_{\varepsilon}=-\,\,\mathrel{\int\!\!\!\!\!\!-}_{\partial B(0,\varepsilon)}f(y)dS(y)\to-f(x)$$ as $\varepsilon \to 0$
Why can we argue that it converges to $f$ as $\varepsilon$ goes to zero?

Thank you very much and any help is appreciated!

Answer 1

It's been a little bit since I've read Evans so I write this with caution, but if I remember correctly we use $\Delta_y$ because the dummy variable inside the integral is in y, and so it's important to make that difference to which variable you are taking the derivative of. This really only takes place because you are repeatedly integrating by parts with some of these integrals.

As for $K_\epsilon$, that second integral should be over $\partial B(x,\epsilon)$, and the reason for its limit is that (as you seem to have ommited), we assume $f \in C^2_c(\mathbb{R}^n)$(use this!), so you should be able to show for all $\eta>0$, that

$$\left|\frac{1}{n\alpha(n)\epsilon^{n-1}}\int_{\partial B(x,\epsilon)} f(y)\,dS(y) - f(x)\right|<\eta$$