= Solution
The inference is most useful when the contributing normalized amplitudes are large. Positivity gives a precise illustration: let $f_h=F(h)/F(0)$ and regard $\rho(x)/F(0)$ as a probability density over a period. From $\langle\cos^2(2\pi hx)\rangle\geq\langle\cos(2\pi hx)\rangle^2$,
$$
f_{2h}\geq2f_h^2-1.
$$
Thus $|f_h|>1/\sqrt2$ forces $F(2h)>0$. If $F(h)$ is weak, this bound allows either sign; a small $F(2h)$ also contributes too little to the density sketch to impose a persuasive sign constraint. Large unnormalized intensity alone should not be confused with a large amplitude relative to $F(0)$.
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