The theorem states that if $f \in C^{2}_{0}(\mathbb R^{n})$, then function $u(x) = \int_{\mathbb R^{n}} \phi(x-y)f(y)dy$ is the solution of $-\Delta u(x)=f(x)$ and $u \in C^{2}(\mathbb R^{n})$. ($\phi$ is the fundamental Laplace equation solution)
I have a problem with the latter statement, I don't quite understand the proof. I know I can differentiate twice under the integral sign, but don't understand the proof of continuity of the second derivative.
We have $u_{x_{i}x_{j}}(x) = \int_{\mathbb R^{n}} \phi(y)f_{x_{i}x_{j}}(x-y)dy$ for $i,j=1,2,...,n$. Now I want to prove the continuity of this partial derivative. It is done through Lebesgue dominated convergence theorem. I would do something like this:
Let $x_{0} \in \mathbb R^{n}$ and $\{x_{n}\}_{n=1}^{\infty}$ be any sequence convergent to $x_{0}$. Now we have:
$\lim_{n\to \infty} u_{x_{i}x_{j}}(x_{n}) = \lim_{n\to \infty} \int_{\mathbb R^n} \phi(y)f_{x_{i}x_{j}}(x_n -y)dy$
Now how to explain the interchange of integral sign and the limit using dominated convergence theorem? How to define a proper sequence of functions? I know that after this step it will be pretty easy to finish the proof because the support of $f$ is a compact set and $\phi$ is locally integrable, so the integral is bounded. Some definitons come to my mind as $g_n(x) := \phi(y)f(x_n -y)$ but it doesn't make sense to me as $g_n$ doesn't depend on $x$ and now I have a problem with $\phi$ as it is unbounded on any set containing $0$.
EDIT: I wonder if it could be done another way. I know that $f$ is bounded for any $y\in supp f$. Thus:
$|u_{x_{i}x_{j}}(x_n) - u_{x_{i}x_{j}}(x_0)| \le \sup_{y\in suppf}|f_{x_{i}x_{j}}(x_n -y) - f_{x_{i}x_{j}}(x_0 -y)|\int_{V} \phi(y)dy$, where $V$ is some compact set.
Now taking the limit $n\to \infty$ we get the continuity of $u_{x_{i}x_{j}}$ as the supremum converges to $0$ and $\phi$ is locally integrable. Is it OK?
I don't think there is any problem because $f$ has compact support: when $x\in K$, a compact subset of $\mathbb{R}^d$, the function $y \mapsto f_{x_i x_j}(x-y)$ vanishes outside the compact set $K - \text{supp}(f)$. It means that $$|\phi(y)f_{x_i x_j}(x-y)| \le |\phi(y)|\mathbb{1}_{K - \text{supp}(f)}(y) \|f_{x_i x_j}\|_\infty \in L^1(\mathbb{R}^d)$$ Thus, the integrated function is dominated by an integrable function uniformly when $x\in K$. The continuity follows by dominated convergence because $f\in\cal{C}^2$.
Of course, it depends on what you know about the fundamental solution, but you told us you know it is locally integrable and this is sufficient.
Remark: one often uses $\mathbb{R}^d$ instead of $\mathbb{R}^n$ in PDEs because it makes $n$ available for sequences, as you do with $x_n$.