Show that the only tempered distributions which are harmonic are the the harmonic polynomials Let $d\geq 1$. Using the Fourier transform, show that the only tempered distribution $\lambda \in\mathcal{S}(\mathbb{R}^d)^*$ which are harmonic (by which we mean that $\Delta \lambda=0$ in the sense of distribution) are the harmonic polynomials. Here $\Delta=\sum_{j=1}^d \frac{\partial^2}{\partial x_j^2}$ is the Laplacian.
Suppose  that $\lambda$ is a tempered distribution such that $\Delta \lambda=0$. Then apply the Fourier transform to both sides and notice that $\mathcal{F}(\frac{\partial}{\partial x_j}\lambda)=2\pi i\xi_j \mathcal{F}\lambda$, we have$\sum_{j=1}^d (2\pi i\xi_j)^2\mathcal{F}\lambda=0$, that is $(\sum_{j=1}^d \xi_j^2 )\mathcal{F}\lambda=0$. I get stuck here, how to find all tempered distributions such that  multiplied by the polynomial $\sum_{j=1}^d \xi_j^2$ is zero? If $d=1$, I know that $\mathcal{F}\lambda$ should be the linear combination of $\delta$ and its derivative. I have no idea about  $d\geq 2$?
 A: Uhm, there is a result that roughly states that if $(x-x_0)^m f(x)=0$ with $f$ a distribution, then $f$ is given by $f=\sum_{k=0}^m c_k \delta^{(k)}(x-x_0)$. I can't find the reference right now, and I may be writing a slightly inexact formula, but that's the spirit of it. In your case, you should be able to conclude that $\mathcal{F}\lambda = c_0\delta(x) + c_1\delta'(x)$, which is obviously a tempered distribution. From here, you should be able to invert the Fourier transform to obtain $\lambda$. It looks like $\lambda$ is going to be a linear function.
A: Hint: For a tempered distribution $u$ with support $\{ 0\}$, it is possible to show that $u=\sum\limits_{|\alpha| \leq a} c_{\alpha} \partial^{\alpha} \delta_{0}$ for some $a \in\mathbb{N}_{0}$ and coefficients $c_{\alpha}$. As $\Vert \xi \Vert^{2}\mathcal{F}\lambda=0$ implies that $\text{supp}(\mathcal{F}\lambda)=\{ 0\}$, we end up with $$\mathcal{F}\lambda=\sum\limits_{|\alpha| \leq a} c_{\alpha} \partial^{\alpha}\delta_{0}.$$
To determine the requirements on the cofficients, consider $$0=\Vert \xi \Vert^{2}(\mathcal{F}\lambda)(\phi)=\sum\limits_{|\alpha| \leq a} c_{\alpha}   (-1)^{|\alpha|}\delta_{0}( \partial^{\alpha}( \Vert \xi \Vert^{2} \phi)).$$
