Assorted Questions from a proof in Evans’ book regarding the Laplace and Poisson’s Equations I’m trying to understand a proof in Evan’s book, “Partial Differential Equations.” On page 23, he states

If $u$ is the fundamental solution to Laplace’s Equation then $ u \in C^2(\mathbb{R}^n)$ and $u$ satisfies Poisson’s Equation.

I have assorted questions throughout the course of the proof which I couldn’t find the answers to anywhere on stack exchange. I’m also not sure how to google them because they’re so specific. We have 
$$u(x) = \int_{\mathbb{R}^n} \Phi(x-y) f(y) \ dy =\int_{\mathbb{R}^n} f(x-y) \Phi(y) \ dy. \ \ (1)$$
My first question arises in this equality: Why is it true? I know the reasoning above the proof in the text suggests that $u(x)$ is the convolution of the fundamental solution and $f(x),$ but why is it justified to change their arguments and let them equal one another? Next we have
$$\frac{u(x+he_i)-u(x)}{h} = \int_{\mathbb{R}^n} \Phi(y) \left[\frac{f(x+he_i-y) -f(x-y)}{h}\right]dy. \ \ (2)$$ I believe what he did here (correct me if I’m wrong) is approximate the derivatives on either side. He then writes
$$\frac{\partial u}{\partial x_i}(x) = \int_{\mathbb{R}^n} \Phi(y) \frac{\partial f}{\partial x_i} \ dy. \ \ (3)$$
He uses the fact that $\frac{f(x+he_i-y)-f(x-y)}{h} \rightarrow \frac{\partial f}{\partial x_i}(x-y)$ as $h \rightarrow 0$ which I also don’t see because at $h = 0,$ the $\frac{f(x+he_i-y)-f(x-y)}{h}$  term blows up. Once again, I tried but, given its specificity, finding the answers to these questions online is impossible. Now, he computes the second partial derivative and concludes $u \in C^2(\mathbb{R}^n).$ The other half of the proof I understand. I was able to find existing questions on this site pertaining to that. I know I asked a lot of questions, so to recap: 

  
*
  
*Why is it justified to switch the arguments of the functions $\Phi$ and $f$ in equation (1)? Why are the two integrals equal?
  
*Is it true that all Evans did in equation two is change the differential terms to approximations, or is there more to it than that? 
  
*Why is equation (3) true? Why does $\frac{f(x+he_i-y)-f(x-y)}{h} \rightarrow \frac{\partial f}{x_i}$ as $h \rightarrow 0$ even with the rational term blowing up at $h = 0
  

Thanks in advance. 
 A: *

*I believe the first case is integration by substitution:
Put $t \equiv x - y$. Then $dt = -dy$, so $\int \Phi(x-y) \cdot f(y)\, dy = \int \Phi(t)\cdot f(x-t)\, dt$. The limits of integration change orientation, as does the sign of the integral, but these changes cancel, leaving you with a positive integral and the same limits of integration. You can then rename $t$ to $y$ because it's just a placeholder integration variable.

*For the second and third case, you're exactly right about approximating the derivative. Recall the definition of the derivative: if $f : \mathbb{R}^n\rightarrow \mathbb{R}$, and $\hat{r}$ is a unit vector, then the directional derivative of $f$ in the direction $\hat{r}$ (if such a derivative exists) is defined by the limit:
$$f_{\hat{r}}(x) \equiv \lim_{h\rightarrow 0} \frac{f(x+h\hat{r})-f(x)}{h}$$
In particular, if $e_k$ is one of the coordinate directions (like [1,0,0,0,0]), you can compute the directional derivative of $u$ in the coordinate direction $e_k$ as the limit:
$$\partial_k u \equiv \lim_{h\rightarrow 0}\frac{u(x+he_k)-u(x)}{h}$$
The first part of your question gives you a formula for computing $u(x)$ in terms of an integral:
$$\partial_k u = \lim_{h\rightarrow 0}\frac{1}{h} \left[ \int \Phi(y)\cdot f(x+he_k-y)\,dy - \int \Phi(y)\cdot f(x-y)\,dy\right]$$
Because of linearity, these two integrals can be combined:
$$\partial_k u  = \lim_{h\rightarrow 0}\frac{1}{h} \int \Phi(y)\cdot \left[f(x+he_k-y)-f(x-y)\right]\,dy $$
Bringing in the constant $h$, we get
$$\partial_k u  = \lim_{h\rightarrow 0} \int \Phi(y)\cdot \frac{f(x+he_k-y)-f(x-y)}{h}\,dy $$
Now the fraction in the integral looks like the definition of "The derivative of $f$ in the direction $e_k$, evaluated at point $(x-y)$". I'm sure that formally, we have to be careful about when we can bring limits from outside the integral to inside the integral, but assuming the situation is sufficiently nice, we have:
 $$\partial_k u  =  \int \Phi(y)\cdot \lim_{h\rightarrow 0}\frac{f([x-y]+he_k)-f(x-y)}{h}\,dy = \int \Phi(y) \cdot \partial_kf (x-y) \, dy$$
It's true that the $h$ in the denominator may make the fraction blow up as $h\rightarrow 0$— but the numerator also depends on $h$. As long as the numerator and denominator change at similar rates, the limit will exist and be finite. When this happens, that finite limit value defines the value of the derivative.
