Null space of the Laplacian operator? (I guess the answer to my question is well-known in harmonic analysis, but I consider it in the framework of Schwartz distributions and in any dimension, and could not find a satisfactory answer.)
The Laplacian operator $\Delta$ has Fourier multiplier $\lVert \omega \rVert^2$. It is well-defined from $\mathcal{S}(\mathbb{R}^d)$ to itself and can be extended by duality to the space of tempered distributions $\mathcal{S}’(\mathbb{R}^d)$. 
Consider the Poisson equation on the complete space $\mathbb{R}^d$
\begin{equation}
 \Delta f = 0
\end{equation}
with $f \in \mathcal{S}’(\mathbb{R}^d)$.
Question: Is there a complete description of the space of such functions, depending on the dimension $d$?
In dimension $d=1$, of course, the Laplacian is the second derivative so that the null space is of dimension $2$. 
In dimension $d=2$, the null space has infinite dimension since it contains real parts of entire functions. But can we give an interesting description of this infinite dimensional space?
What happens in dimension $d\geq 3$?
 A: First the answer for distributions, then tempered distributions:
If $f$ is a distribution with $\Delta f=0$ then $f$ is actually a smooth function, which is to say the nullspace of the Laplacian is the space of harmonic functions.
Sketch, assuming the domain is $\Bbb R^d$: Fix $\phi\in C^\infty_c$ with $\int\phi=1$, and such that $\phi$ is radial ($\phi(x)$ depends only on $|x|$). For $\delta>0$ let $$\phi_\delta(x)=\frac1{\delta^d}\phi\left(\frac{x}{\delta}\right),$$and set $$f_\delta=\phi_\delta*f.$$Now if $f$ is any distribution it follows that $f_\delta$ is smooth and $f_\delta\to f$ in the sense of distributions as $\delta\to0$. If $\Delta f=0$ then $\Delta f_\delta=0$, so $f_\delta$ is a harmonic function. And the fun part: $\Delta f=0$ also implies $f_\delta=f$. (The last assertion is where the fact that $\phi$ is radial comes in.)
Hint for showing $f_\delta=f$: Given $\delta_1,\delta_2>0$ show that there exists $\psi\in C^\infty_c$ with $$\phi_{\delta_1}-\phi_{\delta_2}=\Delta\psi.$$

So a "harmonic distribution" is the same as a harmonic function. The harmonic tempered distributions are exactly the harmonic polynomials. (If $f\in\mathcal S'$ and $\Delta f=0$ then $\hat f$ must be supported at the origin, hence $f$ is a polynomial.)
