I have the following convex optimization problem: $$\begin{array}{ll} \text{maximize}_{{f,g}} & \displaystyle\int_{\Omega} g^u{f}^{1-u}\mathrm{d}\mu\\ \text{subject to} & \displaystyle\int_{\Omega} f \mathrm{d}\mu= 1,\quad \displaystyle\int_{\Omega} g\mathrm{d}\mu =1 \\ & f_L \leq {f} \leq f_U\\ & g_L \leq g \leq g_U\end{array}$$ where $u\in(0,1) $ and $$\int_{\Omega}f_L \mathrm{d}\mu< 1,\quad\int_{\Omega}g_L \mathrm{d}\mu< 1$$

$$\int_{\Omega}f_U \mathrm{d}\mu> 1,\quad\int_{\Omega}g_U \mathrm{d}\mu> 1$$ Here, $f$ and $g$ are distinct density functions and $f_L,f_U,g_L,g_U$ are some known positive functions on $\Omega$.

Claim: The solution is unique and it is the same for every $u\in(0,1)$, if $f_U=\infty$ and $g_U=\infty$, i.e., if there are only lower bounds, or $f_L=0$ and $g_L=0$, i.e., if there are only upper bounds. Else, the solution is also unique but it is not the same for all $u$.

Question: Are these claims true, especially the last one?

$\mu$ can be the Lebesgue measure, although I believe the same holds for the counting measure and the discrete sets. The set $\Omega$ can be $\mathbb{R}$ or an interval of real numbers.

I am especially interested in the last ''not'' case, and for this case if necessary $g_U$ and $f_U$ can be assumed to be integrable over $\Omega$.

I had previously asked this question at math.stackexchange but with no answers.

Addendum: For $f_U=\infty$ and $g_U=\infty$ I know that $$\frac{g}{f}(y)=\begin{cases}c_1\quad\mbox{if}\quad g/f<c_1\\ h(y)\quad\mbox{if}\quad c_1\leq g/f \leq c_2\\ c_2\quad\mbox{if}\quad g/f>c_2\\\end{cases}$$ where $c_1$ and $c_2$ are some constants and $h$ is a function of the bounding functions for example $g_L/f_L$. I think the same is true for the case $f_L=0$ and $g_L=0$.

I have some work on the solution of the problem with KKT multipliers, which may help solving this problem too.

  • $\begingroup$ @YoavKallus exactly!! I will add this information to the question. Thx for reminding. $\endgroup$ Commented Jul 8, 2017 at 14:49
  • $\begingroup$ For each fixed $u \in (0,1)$ there can be infinitely many solutions depending on $f_{u}, g_{u}, f_{\ell}, g_{\ell}$. Consider the scenario when $0\leq f_{\ell}=g_{\ell}<f_{u}=g_{u}$ a.e.. In this case the extremizers are $q_{0}=q_{1}=h$ where $h$ is any integrable function with $f_{\ell} \leq h \leq f_{u}$ and $\int_{\Omega}h d\mu=1$. Indeed, by Holder's inequality we always have $$ \int_{\Omega} q_{1}^{u}q_{0}^{1-u} d\mu \leq \left( \int_{\Omega} q_{1}\right)^{u}\left(\int_{\Omega} q_{0} d\mu \right)^{1-u}=1 $$ and equality is attained if and only if $q_{1}(x) = \lambda\, q_{0}(x)$. $\endgroup$ Commented Jul 8, 2017 at 18:16
  • $\begingroup$ @PaataIvanisvili $q_0$ and $q_1$ are distinct. Sorry I forgot it in the question. Editing now. $\endgroup$ Commented Jul 8, 2017 at 18:25
  • $\begingroup$ In this case the maximizer $(q_{1}, q_{0})$ does not exist when $0\leq f_{\ell} = g_{\ell} < f_{u}=g_{u}$ a.e.. So the question about uniqueness does not make sense. $\endgroup$ Commented Jul 8, 2017 at 18:58
  • $\begingroup$ There is no meaning of the question if one gets a result $q_0=q_1$. One chooses normally the functions $f_l$, $g_l$, $f_u$ and $g_u$ such that $q_0$ is distinct from $q_1$. Namely $f_u=g_u$ and $f_l=g_l$ are not allowed. Basilcally, this case is out of consideration. $\endgroup$ Commented Jul 8, 2017 at 19:15

1 Answer 1


I tried to work on the problem and I think I am able to resolve some points. Here is my work:

Consider the Lagrangians: $$L_0=\int_{\mathbb{R}} g^u{f}^{1-u}\mathrm{d}\mu+\int_{\mathbb{R}}\left(\lambda_0(f-f_L)+\lambda_{00}(f_U-f)\right)\mathrm{d}\mu+\mu_0\left(\int_{\mathbb{R}} f\mathrm{d}\mu-1\right)$$

$$L_1=\int_{\mathbb{R}} g^u{f}^{1-u}\mathrm{d}\mu+\int_{\mathbb{R}}\left(\lambda_1(g-g_L)+\lambda_{11}(g_U-g)\right)\mathrm{d}\mu+\mu_1\left(\int_{\mathbb{R}} g\mathrm{d}\mu-1\right)$$

Taking the Gateux derivatives of the Lagrangians, at the direction of funtions $\psi_0$ and $\psi_1$, respectively, leads to

$$\frac{\partial L_0}{\partial f}=\int\left((1-u)\left(\frac{g}{f}\right)^u+\lambda_0-\lambda_{00}+\mu_0\right)\psi_0\mathrm{d}\mu=0\quad\quad(1)$$

$$\frac{\partial L_1}{\partial g}=\int\left(u\left(\frac{g}{f}\right)^{u-1}+\lambda_1-\lambda_{11}+\mu_1\right)\psi_1\mathrm{d}\mu=0\quad\quad\quad\,\,\,(2)$$

Here according to Gateux derivative, $\psi_0$ and $\psi_1$ are arbitrary functions. I take them as integrable functions with $\int \psi_0 \mathrm{d}\mu=1$ and $\int \psi_1 \mathrm{d}\mu=1$

There are actually $3$ cases for each Lagrangian $L_0$ and $L_1$. For $L_0$ we have $$f=f_L, \quad f=f_u, \quad f_L<f<f_U$$ and for $L_1$ we have $$g=g_L, \quad g=g_u, \quad g_L<g<g_U$$

The conditions above $\partial L_0/\partial f=0$ and $\partial L_1/\partial g=0$ make sense only for the conditions $f_L<f<f_U$ and $g_L<g<g_U$.

Hence, I can write the maximizing functions as

$$f(y)=\begin{cases}f_L\quad y\in E_0\\ f_U\quad y\in E_1\\h_0\quad y\in E_2\\\end{cases}\quad g(y)=\begin{cases}g_L\quad y\in E_3\\ g_U\quad y\in E_4\\h_1\quad y\in E_5\\\end{cases}$$

Based on this result I have following conclusions:

$1.$ For the general case, namely if $g_L\neq \infty$ and $g_U\neq \infty$, the sets $E_k$ and functions $h_0$ and $h_1$ are necessarily dependent on $u$, although there may be special cases. This is because there is no simplification which can show that $E_k$ and functions $h_0$ and $h_1$ may be written independent of $u$.

$2.$ if $g_L= \infty$ and $g_U= \infty$, the results are independent of $u$ and I can confirm this with experiments. In order to show this analytically, first, I should be able to show that $g/f$ is constant in (1) and (2).

I am trying to explain this: $\lambda_{00}$ and $\lambda_{11}$ are zero, $\lambda_{0}$ and $\lambda_{1}$ are some positive functions. $\mu_0$ and $\mu_1$ must be negative constants, if not the equations (1) and (2) above wont be $0$. My problem is that $(1−u)(g/f)^u+\lambda_0$ can also be a constant function which is equal to $-\mu_0$, if $\lambda_0$ is carefully chosen. Then, the integral equation will hold while $g/f$ is not a constant. However, this is never the case. Am I wrong?

Please feel free to comment and post a new anwer based on mine. Because still there are missing points.


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