Here I'm defining my Fourier transform according to the convention $$ FT\left\{f(x)\right\} = \hat{f}(q) = \int_{-\infty}^{+\infty}dx\, f(x)\, e^{-i q x}\, . $$
My intuition - which may be off - tells me that this special unit-modulus form for $f(x)$ must result in some special conditions on $\hat{f}(q)$. For example, perhaps there is some special relationship between the amplitude and phase of $\hat{f}(q)$, or maybe between its real and imaginary parts.
I know that by Parseval's theorem $\hat{f}(q)$ will not be square integrable, but that's about it.
I've searched around online for information about this, but found nothing. I also note the same question asked here, which got no useful replies.
Edited to add:
I thought about this a bit, and came up with one or two things. Because the modulus is 1, we have: $$ {\left| f(x) \right|}^2 = f(x) f^*(x) = 1 $$ Taking a Fourier transform on both sides and rearranging a bit, and we end up with: $$ \int_{-\infty}^{\infty} dq'\, \hat{f}(q') \, \hat{f^*}(q' - q) = {\left(2\pi\right)}^2 \delta(q) \qquad (1) $$ So this thing has a $\delta$-function correlation. Eq. (1) leads to the same result we could get via Parseval's theorem: $$ \int_{-\infty}^{\infty} dq\, {\left|\hat{f}(q)\right|}^2 = \infty $$ By integrating $q$ from $-\infty$ to $+\infty$ in Eq. (1), we obtain $$ \left|\int_{-\infty}^{\infty} dq\, \hat{f}(q)\right| = 2\pi\, . $$ This is a special case of a more general result which immediately follows from the fact that $f$ has unit modulus: $$ \left|\int_{-\infty}^{\infty} dq\, \hat{f}(q)\, e^{i q x}\right| = 2\pi\, , $$ a relationship which must hold for all $x$.
While these are interesting to me, they aren't quite what I had in mind. If I know, say, the real part of $\hat{f}(q)$, is there some way of calculating the imaginary part from that? Or, if I have its amplitude, is there some way of calculating the phase, or vice versa?
Edited to add more observations:
We can always divide the phase function into an odd and an even part: $$ \phi(x) = \phi_e(x) + \phi_o(x)\, , $$ where, of course $$ \phi_e(x) = \frac{1}{2}\left(\phi(x) + \phi(-x)\right) $$ and $$ \phi_o(x) = \frac{1}{2}\left(\phi(x) - \phi(-x)\right)\, . $$ Then $$ f(x) = f_1(x) \times f_2(x) = e^{i \phi_e(x)}\times e^{i \phi_o(x)}\, , $$ and $$ \hat{f}(q) = \frac{1}{2\pi}\left(\hat{f}_1 * \hat{f}_2\right)(q)\, . $$ Since $f_1(x)$ is even, so too must be $\hat{f}_1(q)$. Furthermore, $f_2(x)$ satisfies $f_2(-x) = f_2^*(x)$, which means that $\hat{f}_2(q)$ must be real. Thus $\hat{f}(q)$ is the convolution of an even function and a real function. Does that narrow things down, or can any function be expressed that way?
