prove that flat shape maximizes a functional

Question

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

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

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, but I cannot prove that (Matlab hints towards it).

Answer 1

Write $A / (G(\omega) + A) = f(\omega)$ and note that $f(\omega) > 0$ for all $\omega$. Also write $J[f] = I[G]$. Multiplying $I[G]$ by the denominator gives $\left( \int_{-\kappa \pi}^{\kappa \pi} f(\omega) \ d \omega \right)^2 - \left| \int_{-\kappa \pi}^{\kappa \pi} f(\omega) \exp(-i \omega) \ d \omega \right|^2$. Since $\left| \int g(x) \ d x \right| \leq \int |g(x)| \ d x$, the second term is less than or equal to the first term. In particular $I[G] \geq 0$ for all $G$. Finding $f$ for which the two terms are equal, if such $f$ exists, would solve the optimization problem.

Trying the function $f(\omega) = a/(2 \kappa \pi)$ or equivalently $G(\omega) = A /f(\omega) -A $ gives $J[f] = a \left(1 - \left(\sin(\kappa \pi)/\kappa \pi \right)^2 \right)$. This gets closer to zero as $a \rightarrow 0$, or equivalently, $G(\omega) \rightarrow \infty$.

Note that as $\kappa \downarrow 0$, $\sin(\kappa \pi) \approx \kappa \pi$, so that $I[G] \rightarrow 0$ as $\kappa \downarrow 0$ if you choose $G$ constant.

Answer 2

The easiest way to prove this is using variational calculus. You have to put $$ \delta I(G(\omega))=0. $$ The calculation is quite straigthforward and provides the condition $$ \delta G(\omega)=0 $$ and so the extremum is for $G(\omega)=G=constant$. Finally, from the condition you have to set $$ \int_{-k\pi}^{k\pi}G(\omega)=2k\pi G=1. $$ This gives the value of the extremum $G=\frac{1}{2k\pi}$.

Expanded on OP request: The idea behind functional calculus (calculus of variations) is to consider a class of functionals, as in your case, that can be amenable to a generalized differentiation. You can find all the rules and the definition of a functional derivative here but for a more serious approach some lectures as the ones I pointed out in the comment area are needed. Your case is particularly simple as one is left in each term with the variation of $G(\omega)$ and this must be zero to find an extremum.

Update on OP request: Let us introduce the following functional $$ Z_m[G]=\int_{-k\pi}^{k\pi}\frac{A}{G(\omega)+A}e^{-im\omega}d\omega $$ The functional we are considering takes the form $$ I[G]=Z_0[G]-\frac{Z_1^*[G]Z_1[G]}{Z_0[G]}. $$ Now we have $$ \delta Z_m[G]=-\int_{k\pi}^{-k\pi}\frac{A}{(G(\omega)+A)^2}\delta G(\omega)e^{-im\omega}d\omega. $$ Chain rule applies also to functionals and we can evaluate $\delta I[G]$ immediately to give $$ \delta I[G]=\delta Z_0[G]-\frac{Z_1^*[G]Z_1[G]\delta Z_0[G]-Z_0[G]\delta(Z_1^*[G]Z_1[G])}{Z_0^2[G]} $$ and we see that the condition $\delta G(\omega)=0$ sets the variation to zero. This solution is consistent with the given constraint provided $G=\frac{1}{2k\pi}$. The application of the constarint a posteriori fixes the value of the constant.

Further clarification for OP: I will show that a functional that does not depend from at least a first derivative is a constant in one dimension. Let us consider the functional $$ S=\int_a^bL(q(t),q'(t),t)dt. $$ The condition for the extremum just gives $\delta S=0$ yielding Euler-Lagrange equation $$ \frac{d}{dt}\frac{\partial L}{\partial q'(t)}=\frac{\partial L}{\partial q(t)}. $$ Then, if there is no dependence on derivative we are left with $\frac{\partial L}{\partial q(t)}=0$ that implies immediately $L=L(t)$ and $q(t)=constant$.