$T$ is tempered distribution that is harmonic,then $T$ is polynomial 
QUESTION. How do I show that if $T$ is a tempered distribution that is harmonic, then $T$ is a polynomial?

Any help is greatly appreciated.
 A: The Fourier transform is well-defined over the space ${\mathcal S}'(R^d)$ of tempered distributions into itself. If $T$ is harmonic, that is $\Delta T=0$, then $|\xi|^2\hat T=0$. This tells you that the support of $\hat T$ is $\{0\}$ (unless $T=0$ of course). Using the topology of ${\mathcal S}'(R^d)$, you know that the restriction of $\hat T$ over an open ball, say the unit ball, is of finite order: there exist an integer $n$ and a finite number $C$ such that
$$|\langle\hat T,\phi\rangle|\le C\sum_{|\alpha|\le n}\sup_B|\partial^\alpha\phi(x)|.$$
This shows that $\hat T|_B$ extends uniquely to the space ${\mathcal C}^n_K(B)$. Because the support of $\hat T$ is $\{0\}$, its kernel contains the subspace $X$ of those $\phi$ which vanish in a neighbourhood of $0$. By continuity, $\hat T$ vanishes over $\bar X$, which is nothing but the finite dimensional subspace defined by $\partial^\alpha\phi(0)=0$. By elementary linear algebra, there follows that $\hat T$ is a linear combination of the forms
$$\langle\lambda_\alpha,\phi\rangle=\partial^\alpha\phi(0),$$
for $|\alpha|\le n$. By inverse Fourier transform, you obtain that $T$ is a polynomial.
Of course, this proof works for every solution $T$ of a linear PDE with constant coefficients $P(\nabla)T=0$, whenever $P(i\xi)\ne0$ for every real vector $\xi\ne0$. This is an ellipticity assumption.
