Can we reconstruct a function $f$ by knowing its scalar product with its own shift, $\langle T_xf,f \rangle$?

Question

Assume $f:\mathbb{R} \to \mathbb{C}$ is a function, let's say square integrable. Assume we know the values $$ a(x) = \langle T_xf,f\rangle = \int f(y-x)\overline{f(y)} \, dy, $$

for $x \in \mathbb{R}$, i.e. we know all the scalar products of $f$ with its own shift $T_x$. Is there a way to construct one function $f$ such that $\langle T_xf,f\rangle = a(x)$? Of course, $f$ is not unique in this case.

Answer 1

Note that $$ \langle T_af,f\rangle = \langle\widehat{T_af},\widehat f\rangle = \langle M_a\widehat f,\widehat f\rangle = (|\widehat f|^2)^\vee(a), $$ where $(M_ag)(x) = e^{2\pi iax}g(x)$ is modulation and $f^\vee$ (resp. $\widehat f$) is the (inverse) Fourier transform of $f$.

So, you're asking basically if you can reconstruct $f$ from $|f|^2$ and the answer is of course no.