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](https://math.stackexchange.com/questions/2306202/convex-optimization-problem-outputs-a-unique-solution-if-f-l-leq-q-o-but-not-i) 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.](https://math.stackexchange.com/questions/2388437/solving-a-convex-optimization-problem-using-kkt-multipliers)



Uniqueness of a Solution for a Convex Optimization Problem
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.


