TITLE: What are the properties of the fourier transform of a phase-only function?
QUESTION [9 upvotes]: Given a function of the form:
$$
f(x) = e^{i\phi(x)} | \phi(x)\in\Re
$$
What are the properties of its Fourier transform?
For instance, purely real functions have Fourier transforms with symmetric properties. Can anything similar be said about Fourier transforms of phase-only functions?

REPLY [4 votes]: A "phase-only function" $f(x)$ as you call it can be equivalently expressed as a function for which $|f(x)|=1$ for all $x$. Or in other words for all $x\in\mathbb R$,
$$
f(x)\cdot \overline{f(x)}=1.\qquad(1)
$$
Let $g(y)$ denote the Fourier transform of $f(x)$, which is actually a tempered distribution for this class of functions (we will ignore this distinction for the rest of this paragraph). Since Fourier transforms interchange multiplication with convolution, intuitively $(1)$ should transform into
$$
\delta(y)=g(y)*\overline{g(-y)},
$$
which can be interpreted (by definition of convolution) as saying that $g$ is orthogonal to all of its non-zero translates. Indeed, intuitively one has that
$$
g(y)*\overline{g(-y)}=\int_{\mathbb R}g(t)\overline{g(t-y)}\ dt=\langle g(t),g(t-y)\rangle.
$$
For this to "equal" the delta "function" means that, for $y\not=0$, the function $g(t)$ is orthogonal to its translate by $y$.
Now that we have the intuitive picture, we can proceed to formalize it. Let $\tau_z$ denote the operator that shifts a tempered distribution by $z$. (See this post by Terry Tao for a precise definition of what this means, along with a good outline of formal properties for working with tempered distributions.)

Claim. The Fourier transform of a "phase-only function" is a tempered distribution $g$ with the property that for all $z\in\mathbb R\setminus \{0\}$, there exists a sequence of Schwarz functions $g_n$ (possibly depending on $z$) converging to $g$ in the sense of tempered distributions, with the property that $$\lim_{n\to\infty}\langle \tau_z g,g_n\rangle=0.$$ 


Explicit examples.

In the case of a linear phase $f(x)=e^{i(ax+b)}$, the Fourier transform is a multiple of a dirac spike, which obviously satisfies the condition (since any non-zero translate has the spike somewhere else, meaning that it is orthogonal to the first spike).
In the case of a quadratic phase $f(x)=e^{-ix^2/2}$, the Fourier transform is a constant times $e^{iy^2/2}$, and we can see that it is orthogonal to its translate by $z\in\mathbb R\setminus \{0\}$ since we can take a sequence of functions $g_n$ by restricting $e^{iy^2/2}$ to the interval $[-\frac{2\pi n}{|z|},\frac{2\pi n}{|z|}]$ and obtaining
$$
\langle e^{i(y+z)^2/2},g_n\rangle=\int_{-2\pi n/|z|}^{2\pi n/|z|}e^{i(y+z)^2/2}e^{-iy^2/2}\ dy=0,
$$
since the integrand is a constant multiple of $e^{iyz}$, which has period $2\pi/|z|$.