In my lecture notes this well known theorem is presented in the following way:
Let $f\in C_{0}^{\infty}(\mathbb{R}^n)$
and define
$$u(x) = (f*Φ)(x) =
\int_{\mathbb{R}^n}
f (y)Φ(x− y)d y,$$
where Φ is the fundamental solution of the Laplace equation. Then $u\in C^2(\mathbb{R}^n)$
and the Poisson equation: $−∆u = f$ holds in the whole space $\mathbb{R}^n$.
Just to put it here for clarification, the fundamental solution is the function $\Phi:\mathbb{R}^n\setminus \{0\}\to \mathbb{R}$ such that:
$$\Phi(\boldsymbol x)=\begin{cases}
{\displaystyle -\frac{1}{2\pi}\ln\left(\left\vert\boldsymbol x\right\vert\right) \;,
\quad n=2}
\\[2mm]
{\displaystyle\frac{1}{n\left(n – 2\right)\alpha\left(n\right)}
\frac{1}{\left\vert\boldsymbol x\right\vert^{n-2}}\;,\quad n\geq 3}\\
\end{cases}$$
Here $\alpha(n)$ is just the volume of the unit ball in $n$ dimensions.
My question: Is it not true that the solution $u$ is not just in $C^2$, but is also smooth ($u\in C^{\infty}(\mathbb{R}^n)$)? If so how would I prove this?
Best Answer
Yes, the solution is indeed smooth. Did you prove the $C^2$ regularity in your lectures? The $C^\infty$ regularity is proved in a similar way. You change variables in the integration to get $$u(x) = \int_{\mathbb{R}^n} f(x-y)\Phi(y)\, dy,$$ and then you differentiate under the integral to get $$D^\alpha u(x) = \int_{\mathbb{R}^n} [D^\alpha f(x-y)]\Phi(y)\, dy$$ for any multi-index $\alpha$. The compact support and smoothness of $f$ is used to justify exchanging the derivative and integral (with, say, the dominated convergence theorem).