Consider the following case then: $\max_{y} \int y(x) w(x) dx $ subject to $\int y(x)^2dx=1$, which has the trivial solution $y(x)=w(x)$. However, following your approach yields $\delta y(x)=0$ and a flat shape appears. The problem is that the functional derivative is not what you write butinstead reads $$\delta Z_m[G]=\frac{A}{(G(\omega)+A)^2}e^{-imw}.$$ This complicates things greatly. Have I misunderstood the whole concept...?

Jon, Is this really correct...? I thought that the functional derivative didn't inlcude the integral sign and the variation of G(w)...? In fact, with your reasoning, would a flat shap always be the maximizer to any functional as the functional derivative will always be zero?

Jon, I dont get this...I took the functional derivative and obtained an equation system through a Lagrange multiplier. However, I get stuck since the functional derivative is horrible. Could you please share a few details how you obtained the condition you mentioned above?

Thanks, Do you foresee any possibilities to actually solve the system? I tried it out, but get stuck pretty fast....

Thanks. This was only a part of a bigger thing. Would you mind also taking a look at the following? $I(G(\omega)) = \int_{-\kappa\pi}^{\kappa\pi} \frac{A}{G(\omega)+A}d\omega-\frac{| \int_{-\kappa\pi}^{\kappa\pi} \frac{A}{G(\omega)+A}\exp(-i\omega)d\omega|^2}{ \int_{-\kappa\pi}^{\kappa\pi} \frac{A}{G(\omega)+A}d\omega}$, where $\kappa<1$, $A>0$, and $G(\omega)\geq 0$. Now I would like to minimize $I(G(\omega))$ under the constraint of unit area of $G(\omega)$, i.e., $\int_{-\kappa \pi}^{\kappa \pi} G(\omega)d\omega=1$. My hypothesis is that a flat $G(\omega)=1/2\kappa\pi$ is optimal.