$A$ to be $\mathcal{S}(A) = \{ g \in G | A+g = A\}.$ Clearly, $0 \in \mathcal{S}(A) \le G$. Let $A,B \subseteq G$, finite and nonempty sets, with $S= \mathcal{S}(A+B)$. Then, Kneser’s theorem affirms that $|A+B| \ge |A+S| + |B+S| -|S|$.

Let $P$ be the period of $f$, which we know divides $M$. Define $A_i = \{0,\cdots,P-1\} \cap \{x  | f(x) = i\}$. We have thanks to **(P2)** that $|A_0| = |A_1| = P/2$. We will now take $G = \mathbb{Z}/P\mathbb{Z}$, with $A=B=A_0$. By the definition of the period, we know that $|S|=1$. We then have that $|A_0+A_0| \ge P-1$.

Set $Y = \{ y : y \in \{0,\cdots,P-1\} | g(y) = 1 \text{ and } g(y+1) = 0\}$. By **(P3)**, we know that $P-1 \ \in Y$. To solve , we thus need to find $a$ and $l$ such that $a+l \in Y \cap (A_0+A_0)$ (this equation is in $\mathbb{Z}/P\mathbb{Z}$, not merely $\mathbb{Z}$). Because of the cardinality bounds $|Y|\ge 1$ and $|A+A|\ge P-1$, this argument can only fail to provide a $\mu$ if both bounds are actually equalities. In that case we would thus have $A_0+A_0 = \{0,\cdots,P-2\}$, with a unique drop for $f$ along one period (this is 