Q : I need to prove that if $K$ is tempered distribution on $\mathbb{R}$ satisfying: \begin{equation} K = K*e^{-\pi |x|^2} \end{equation} then $K$ is first degree polynomial. mean $K(x) = Ax + b$
Remark: The question was changed. The original was to prove that if $K = K * e^{-\pi |x|^2}$ then $K$ is constant, which is false.
The first thing I did is to apply fourier transform on both sides to work with multiplication instead of convolution. and I got $\hat{K} = e^{-\pi |x|^2} \hat{K}$.
I succeeded to prove $\hat{K}$ is supported at the origin and by theorem 1.7 at page 110, from Stein and Shakarchi functional analysis(Can't find the pdf online) or theorem 6.25 at page 165 from Rudin Functional analysis: \begin{equation} \hat{K} =\sum_{|\alpha| \leq N} a_{\alpha} \partial^{\alpha}\delta \end{equation}.
Now, if I apply the inverse fourier transform I get that $K$ is a polynomial.
The solution will arise if I will prove that if $p$ is a polyomial in $\mathbb{R}^{d}$ satisfying $p*e^{-\pi |x|^2} = p$, then $p$ is constant.
It sounds true(which is not, please see the comments) but I think it is kind of "ugly" to prove and I am pretty sure that there is another way for me to continue.
Hot to continue?
Thanks :)
As I mentioned in the comment, the problem is false. In fact, let $p$ be any harmonic polynomial. Then by noting that
$$ \Phi(x,t) = \frac{1}{(4\pi t)^{d/2}}e^{-\frac{|x|^2}{4t}} $$
is the fundamental solution of the heat equation $\partial_t\Phi = \Delta\Phi$, we have
$$ \partial_t(p*\Phi) = p*(\partial_t\Phi)=p*(\Delta\Phi)=(\Delta p)*\Phi=0. $$
Together with $ p(x) = \lim_{t\to 0} (p*\Phi)(x,t) $, this implies that $p = p*\Phi$. Then plugging $t = \frac{1}{4\pi}$ proves
$$ p = p * e^{-\pi|\cdot|^2}. \tag{*} $$
I suspect that the converse is also true, since the condition $\text{(*)}$ implies that $p(x) = (p*\Phi)(x,\frac{n}{4\pi})$ for any integer $n \geq 1$. This surely sounds like another line of interesting question, although I have no good idea to prove this right now.