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The following functional arises in an information theoretic problem that I work on currently.

$I(G(\omega)) = \int\limits_{-\kappa\pi}^{\kappa\pi}d\omega \frac{A}{G(\omega)+A}-\frac{| \int\limits_{-\kappa\pi}^{\kappa\pi}d\omega \frac{A}{G(\omega)+A}\exp(-i\omega)|^2}{ \int\limits_{-\kappa\pi}^{\kappa\pi}d\omega \frac{A}{G(\omega)+A}}$,

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., $\int\limits_{-\kappa \pi}^{\kappa \pi}d\omega\, G(\omega)=1$.

My hypothesis is that a flat $G(\omega)=\frac{1}{2}\kappa\pi$ is optimal, but I cannot prove that (Matlab hints towards it).

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Fred, I think your question will be moved to soon. – Yrogirg Aug 7 '12 at 14:08
As this seems to have been reduced to pure math at this point I have asked the math mods if they want it. In any case, welcome to Physics.SE, Fred. – dmckee Aug 7 '12 at 14:34
Crossposted to – Qmechanic Sep 15 '14 at 0:23

This hypothesis is not right. The first integral alone $\int {A\over G(\omega) +A } d\omega$ has a local minimum at the constant function (and a global minimum too), so that it's first variation is zero, but the numerator of the second term doesn't have a vanishing variational derivative, so it isn't extremal for a constant, so there are easy counterexamples close to a constant function.

To see that a constant locally minimizes the first integral, you can expand in a Taylor series in G

$$ \int 1 - {G\over A} + {G^2\over 2A^2} d\omega $$

The first two terms are constant (since you have a constraint on the total integral), and the last term is positive definite quadratic form, so the constant is a local minimum. You can show it's a global minimum too using convexity arguments (the second variation of the function is everywhere positive definite).

Knowing this, the first variation of the numerator near the constant function is all you need to check. This variation is proportional to

$$2\mathrm{Re} (\int e^{i\omega'-i\omega} \delta G(\omega) d\omega'd\omega) $$

which is not zero for general $\delta G$. If you want an explicit counterexample, use a little positive bump for $\delta G$ anywhere that $\cos(\omega)$ is negative, or a little negative bump where $\cos(\omega)$ is positive (the real part can be taken inside the integral, so you can see what a $\delta G$ perturbation does explicitly).

The solution to your extremization problem can be given as a nonlinear integral equation using the method of Lagrange multipliers. Add the term

$$ \int \lambda G $$

to your functional, and minimize the sum. The equation doesn't simplify very much, and should be solved numerically, and for this purpose, you can just make a grid and minimize the values by steepest decent, starting from the constant. You can also solve the integral equation from the calculus of variations by iterations, starting from a constant, and this is roughly equivalent.

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