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 $d\geq 2$?
