Morgenstern's Theorem Morgenstern proves a $\Omega(n\log n)$ lower bound for Fourier transform in the bounded coefficient model.

Let $x=[x_1,x_2,\ldots,x_n]'$ be given vector and $F$ be Fourier transform matrix.

It is not known if $F(x)$ needs $n\log n$ operations in the unbounded coefficient model when $x_i$'s are treated generically as variables.

Supposing each $x_i$ is an integer bounded by $2^b$, can one show computing $F(x)$ still needs $n\log n$ arithmetic operations in the unbounded coefficient model?

Added:
The reason I posted was a few fold. Some questions in the bounded integer $x_i$ model are:

1) We cannot use bit level manipulation on variables $x_i$. However we can do bit level operations on integers $x_i$. Are bit level operations considered valid arithmetic operations when counting the $\Omega(n\log n)$ arithmetic operation?

2) Does one multiplication operation on a word that scales as $O(n)$ bits still count as $1$ arithmetic operation?

3) Is division a valid arithmetic operation?
Of course if you show a lower bound of $\Omega(n \log n)$ for your question, it would imply the same lower bound for the unrestricted model, which is not known.

On the other hand, you are asking whether it is possible to compute the Fourier transform where each $x_i$ is known within $b$ bits of precision using $\ll n \log n$ arithmetic operations. If true, by the fact that the Fourier transform is a unitary operation (and thus, perfectly conditioned), we would be able to compute Fourier transform of any vector within $b$ bits of precision much faster than FFT, which is certainly not known either.