# Solving a Convex Optimization Problem Using KKT Multipliers

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

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

I asked the same question for a possible solution via using a programming language. Here I am searching an analytical solution and my questions are as follows:

Is it possible to use the Lagrangian multipliers approach and find some parameteric forms for $$f$$ anf $$g$$. These parameters can then be determined by imposing the KKT constraints. I have my own work but I cannot obtain the sets $$E_k$$ nicely.

According to my work $$g/f$$ must be constant on some Lebesgue measure positive set. However, If I make some examples I can see that this must not necessarily be the case. Why? did I do something wrong or something missing? (this part seems to be resolved after Michael's comments and corrections)

Here is my own 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

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

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

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

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

After here I am unable to specify the sets $$E_k$$ in terms of $$f_L$$, $$f_U$$, $$g_L$$, $$g_U$$ and Lagrangian parameters. The same goes to $$h_0$$ and $$h_1$$.

• Your Lagrangians are ill posed. Your Lagrange multipliers $\lambda_*$ should be functions, and you need to integrate them over $\mathbb{R}$ Aug 10, 2017 at 1:47
• I think so, because they should hold for every $y$, unlike the other constraints with $\mu_*$. I need to integrate them over $\mathbb{R}$ sounds also nice but where and how? Aug 10, 2017 at 1:54
• Just integrate that product you have there. Aug 10, 2017 at 1:59
• @MichaelGrant I included the integral there, though I didnt understand well. I think that the Lagrangian parameters (functions) will force this these inequalities to hold. Aug 10, 2017 at 2:14
• @MichaelGrant I think now it is clear why the programming approach was giving almost nowhere constant $g/f$. Becase $\lambda_*$ are not necessarily constant functions. Aug 10, 2017 at 2:23