The relation of uniform convergence and exchange of limit

Question

There are some questions that I want to ask related to section 2.2.1 part b (Poisson Equation) Theorem 1 in L.C. Evans PDE book.

So here is my problem.
Define $u(x) = \int_{\mathbb{R}^{n}}\Phi(x-y)f(y)dy$ with $\phi(x) = \begin{cases}\frac{-1}{2\pi}\log(|x|)&(n=2)\\ \frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}} &(n\geq3)\end{cases}$ and $\alpha(n)$ is the volume of unit ball in $\mathbb{R}^{n}$.

Theorem 1 (Solving Poisson's Equation)
Define $u$ as above then:
(i) $u\in C^{2}(\mathbb{R}^{n})$
(ii) $-\Delta u = f$ in $\mathbb{R}^{n}$
*Note that $f \in C^{2}_{c}(\mathbb{R}^{n})$

So, in the book Evans says that $\frac{f(x+he_{i}-y)-f(x-y)}{h}$ is uniformly convergent to $\frac{\partial f(x-y)}{\partial x_{i}}$ as $h\to 0$. Why is it? I can't see this as an obvious fact.
As for the next part, sorry for asking trivial question but why do we need uniform convergence to prove this claim below
$$\frac{\partial u}{\partial x_{i}}(x) = \int_{\mathbb{R}^{n}}\Phi(y)\frac{\partial f}{\partial x_{i}}(x-y)dy$$ What happens if the condition is only pointwise convergence? Where is the problem with that thing? Also, I have read some sources related to the proof of this claim. There is this one phrase :
”Given $h$ is small, $|\frac{f(x-y+he_{i})-f(x-y)}{h}-\frac{\partial f}{\partial x_{i}}(x-y)|$ is contained in a fixed compact set $K$."
What does it mean when it says that "it is contained in a fixed compact set $K$"?
Thank you very much and any help is much appreciated!

Answer 1

You need uniform convergence (or at least something better than pointwise convergence) to exchange a limit with an integral - for a counterexample, consider the sequence of functions $$f_n(x) = \begin{cases}n^2 & 0 \le x < 1/n \\ 0 & x \ge 1/n \end{cases}$$ on $(0,1)$. The functions converge pointwise to zero but their integrals do not converge.

As to the proof of the uniform convergence, you're right that it's not so obvious. What is obvious is pointwise convergence $$\frac{f(x + h e_i) - f(x)}h \to \frac{\partial f}{\partial x_i}(x);\tag 1$$ this is simply the definition of the partial derivative. Thus what we need to show is that the convergence is uniform in $x$.

Since $f$ is $C^1_c$, we know $f_i=\partial f/\partial x_i$ is continuous on a compact set and thus uniformly continuous. By the mean value theorem we know there is some $t \in (0,1)$ such that the LHS of $(1)$ is equal to $f_i(x+th)$; and thus $$\left|\frac{f(x + h e_i) - f(x)}h - \frac{\partial f}{\partial x_i}(x)\right|=|f_i(x+th)-f_i(x)|.$$

As $f_i$ is uniformly continuous and $|t|<1$, this converges uniformly to zero as $h \to 0$, since we are just sampling two values of $f_i$ at a distance at most $h$ apart.