Differential operator and Fourier Transform

[shwedka]

Well, in this case, you have to consider your $u$ as a distribution and apply everything , i.e., Fourier and diff. operator in the sense of distributions (in $\mathcal{S}'$). The same reasoning will give that the distribution $\hat{u}$ has support in the set of zeros of the polynomial. such distribution has to be a collection of derivatives of $\delta$ distributions centred at the zeros of the polynomial, therefore, applying the inverse Fourier, again in the distributional sense, you get that $u$ is a collection of polynomials times exponents. Such functions do not tend to zero at infty.