TITLE: Evans Proof of Solution to Poisson's Equation
QUESTION [3 upvotes]: There already are many questions about the proof of Theorem 1 in chapter 2 of Evans' PDE on here. However, I don't think this specific question has been asked yet.
On pages 23 and 24, Evans proves that $u \in C^2(\mathbb{R}^n)$ where $u$ is defined as
$$u(x) = \int_{\mathbb{R}^n} \Phi(x-y)f(y)dy$$
and $\Phi$ is defined as
$\Phi(x) = -\frac{1}{2\pi}\log|x|$ if n = 2,
$\Phi(x) = -\frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}}$ if $n \geq 3$.
The assumption is $f \in C^2(\mathbb{R}^n)$ with compact support.
I understand up to the bottom of page 23, where for i, j = 1, 2, ... , n
$$u_{x_ix_j}(x) = \int_{\mathbb{R}^n}\Phi(y)f_{x_ix_j}(x-y)dy.$$
But what I don't understand is on page 24, he claims $u_{x_ix_j}(x)$ is continuous in x.
To prove it, let $\{x_m\}$ be a sequence which converges to $x$ for any $x$ in $\mathbb{R}^n$. We must show
$$\lim_{m \to \infty} u_{x_ix_j}(x_m) = u_{x_ix_j}(x).$$
And to this end, I think we need to show that $\Phi(y)f_{x_ix_j}(x_m-y)$ converges uniformly to $\Phi(y)f_{x_ix_j}(x-y)$, but I don't know how to do this. Would anyone be able to give me a hint on the proof, or to complete it in a different direction? Thank you very much in advance.

REPLY [1 votes]: Uniform convergence is a good tool to prove we can pass the limit inside the integral sign, but in more general contexts this tool is likely not enough. As mentioned by Pedro in the comments, you can use dominated convergence.
It goes like this: If $|f_m(x)|\leq g(x)$ for every $x$ and $m$, where $g(x)$ is integrable and $\lim_n f_n(x)$ exists for every $x$, then 
$$\lim_{m\to\infty} \int f_m(x)dx = \int \lim_{m\to\infty} f_m(x) dx.$$
Of course, the integrals above are to be taken on the domain of the functions (assumed to be the same for all of them).
The usual proof of this result usually uses Fatou's Lemma (which follows, from monotone convergence).
In the case above, we have functions $U(x)=u_{x_ix_j}(x)$ and $g(x)=f_{x_ix_j}(x)$ related by 
$$ U(x) = \int_{\mathbb{R}^n}\Phi(y)g(x-y)dy,$$
and we wish to prove $U(x)$ is continuous.
Fix $x\in\mathbb{R}^n$ and let's prove $U(x)$ is continuous in $x$. Let $(x_m)$ be a sequence converging to $x$, and 
$$f_m(y)=\Phi(y)g(x_m-y).$$
Since $(x_m)$ is a bounded sequence and $g$ is compactly supported, the support of all $f_m$ lie in some compact set $K$ (a big ball if you like it). Using the continuity of $g$, it is possible to find a constant $C>0$ (for example $C=sup \ |g(x)|$) such that 
$$|f_m(y)|\leq C|\Phi(y)|, \ \ \forall y\in K.$$
It is not hard to check that $|\Phi(y)|$ is locally integrable (that is, its integral if finite in any compact set).
The result follows from 

Dominated Convergence, and
$U(x_m) = \int_K f_m(y)dy$.
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