# prove that flat shape maximizes a functional

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).

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I do the straightforward compution $$\frac{d}{d\epsilon} I(G(\omega)+\epsilon h(\omega)))$$ where $h(\omega)$ is an arbitrary test function, and then I plug in $\epsilon=0$. This, however, gives me an expression that is not that easy to penetrate really. Am I on the right track? – Tommaso Jul 3 '12 at 22:25

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.

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It seems to me that you forgot the normalizing condition $\int G = 1$. – Benoît Kloeckner Jun 30 '12 at 11:08

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 $q(t)=constant$.

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Thanks, Can you please provide a few details as I dont know how to handle the case when there is not a single integral in var. Calc. Can I simply use the chain rule? Also, how about the following generalization: Let $b_k =\int G(\omega)\exp(-i\pi k)d\omega$ for $b=0,\ldots,L$. Define $\bar{B}=Toeplitz([b_0,\ldots,b_{L-1}])$ and $\bar{b}=[b_1,\ldots,b_L]$. Now I would like to minimize $I(G(\omega))=b_0-\bar{b}\bar{B}^{-1}\bar{b}_0^*$, which reduces to the original problem if $L=1$. Is the minimizer still flat for any $L$? – user24799 Jun 30 '12 at 6:32
Pierre, you can extend this to any number of dimensions. Variation calculus has no limitations on this side. You have to apply it to each component and the conclusions are consistent. – Jon Jun 30 '12 at 13:57
Jon, can you please hint me on thr direction I should take to resolve this problem. I do understand that the method should be var calc. But I do not know how it should be applied in this case. – user24799 Jul 1 '12 at 6:06
Is it write it correctly? I see a $\overline{b}_0^*$ multiplying Toeplitz matrix but should it be $\overline{b}^*$? Finally, it appears matrix $\overline{B}$ has all non-null elements. Is it so? As you may know, variation calculus applies to matrices as well. You can check math.uni-leipzig.de/~miersemann/variabook.pdf. – Jon Jul 1 '12 at 9:02
No, it should be $\bar{b}^*$ of course. Can ypu hint me how you obtained thr solution for $L=1$. Var calc is not my field and I only knpw the basic examples, like the braichostrone etc. My problem seems touger as I need to deal with multiplications of integrals. The pdf you sent doesn't seem to contains such examples. – user24799 Jul 1 '12 at 12:23

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