Prove that the phase retrieval problem is non-convex The  phase retrieval problem consists of recovering phase information from given intensity measurements, as shown in the image below from Deep phase retrieval: analyzing over-parameterization in phase retrieval.



I don't understand why it is a non-convex optimization problem. Can someone please help me on how to prove it? Or is there any paper or book about this topic?
 A: Consider the simple example where $n = 1$, $m = 1$, $a_1 = 1$, $y_1 = 1$, i.e. we are estimating a $1$-dimensional complex variable $z \in \mathbb{C}$ from one measurement $1 = |z|^2$.
Then, $f(z) = (1-|z|^2)^2$, which satisfies $f(1)=f(-1) = 0$, but $f(0) = 1$. Hence, $$f\left(\dfrac{-1+1}{2}\right) = f(0) = 1 > 0 = \dfrac{0+0}{2} = \dfrac{f(-1)+f(1)}{2},$$ and thus, $f$ is not convex.
A: It seems a bit strange to "blame" the non-convexity on the "square modulus operation"  here. The least-squares phase retrieval problem can be casted even without the square on the modulus, yet it would still be non-convex. The issue here is that both the maps $\mathbf{z} \mapsto y_r - |\mathbf{a}_r ^* \mathbf{z} |$ and $\mathbf{z} \mapsto y_r - |\mathbf{a}_r ^* \mathbf{z} |^2$, despite being concave, are / can be negative for certain values of $\mathbf{z}$ (that will depend on $\mathbf{a}_r$); thus the non-convexity arises when they are composed with a convex non-monotonic function (i.e. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$).
