# Uniqueness of a Solution for a Convex Optimization Problem

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 $\Omega$.

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.

• @YoavKallus exactly!! I will add this information to the question. Thx for reminding. Commented Jul 8, 2017 at 14:49
• 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 $q_{1}(x) = \lambda\, q_{0}(x)$. Commented Jul 8, 2017 at 18:16
• @PaataIvanisvili $q_0$ and $q_1$ are distinct. Sorry I forgot it in the question. Editing now. Commented Jul 8, 2017 at 18:25
• 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. Commented Jul 8, 2017 at 18:58
• 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. Commented Jul 8, 2017 at 19:15

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 $$f=f_L, \quad f=f_u, \quad f_L and for $$L_1$$ we have $$g=g_L, \quad g=g_u, \quad g_L

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

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.