I have the following convex optimization problem ($q_0$ and $q_1$ are unkown): $$\begin{array}{ll} \text{maximize} & \displaystyle\int_{\Omega} q_1^u{q_0}^{1-u}\mathrm{d}\mu\\ \text{subject to} & \displaystyle\int_{\Omega} q_0 = 1,\quad \displaystyle\int_{\Omega} q_1 =1 \\ & f_l \leq {q_0} \leq f_u\\ & g_l \leq q_1 \leq g_u\end{array}$$ where $u\in(0,1) $ and $$\int_{\Omega}f_l \mathrm{d}\mu\leq 1,\quad\int_{\Omega}g_l \mathrm{d}\mu\leq 1$$

$$\int_{\Omega}f_u \mathrm{d}\mu\geq 1,\quad\int_{\Omega}g_u \mathrm{d}\mu\geq 1$$ Here, $q_0,q_1$ are density functions on $\Omega$, and $f_l,f_u,g_l,g_u$ are some known positive functions on $\Omega$.

I try to solve this optimization problem and I get exactly the same solution for every $u$, if I set $f_u=\infty$ and $g_u=\infty$, i.e., if there are only lower bounds for $q_0$ and $q_1$. If I also put $g_u$ and $f_u$ as some functions which are bounded above by $B<\infty$, the solution is not independent of $u$, I get different solutions for every $u$. Can one show this using math?

P.S: In the examples I had, $g_u$ and $f_u$ were also integrable over $\Omega$.

Addendum: I tried with $f_l=0$ and $g_l=0$ and for this case, namely with only upper bounding functions, I also get results independent of $u$, namely the same result for every $u$.

  • $\begingroup$ What is the variable you're optimizing over?? $\endgroup$
    – user856
    Jun 1, 2017 at 22:04
  • $\begingroup$ @Rahul please see the edit. thanks. $\endgroup$ Jun 1, 2017 at 22:08
  • $\begingroup$ what is $\Omega$? what is $\mu$? I think the answer changes a lot if $\mu$ is the Lebesgue measure or if $\mu$ is, say, a Delta dirac. Also, are the functions $f_l$, $g_l$, $f_u$, $g_u$ continuous? As it is, your problem is a bit too vague . $\endgroup$
    – Gio67
    Jun 10, 2017 at 6:54
  • $\begingroup$ but you are right, it is a very interesting question! $\endgroup$
    – Gio67
    Jun 10, 2017 at 7:00
  • $\begingroup$ @Gio67Let $\Omega=\mathbb{R}$ for example and in this case let $\mu$ be the Lebesgue measure. Again in this case $f_l,g_l,f_u,g_u$ are continuous functions. I didnt define these functions exactly, because one could also choose $\Omega$ for example a discrete set and $\mu$ would be the counting measure. Or one could also choose $\Omega$ as some interval of real numbers. I dont think that the result changes as long as suitable $\mu$ for the related $\Omega$ is chosen. Lets say $\Omega=\mathbb{R}$. $\endgroup$ Jun 10, 2017 at 13:53

1 Answer 1


I am not really answering your question, so don't worry about the bounty. Just some random thoughts. Don't have time to do more. I am using the Lebesgue measure.

If $\int f_{u}=1$ and $\int g_{u}=1$ then since the functions $s\in (0,\infty)\mapsto s^{u}$ and $s\in(0,\infty)\mapsto s^{1-u}$ are increasing, you have that $q_{0}^{u}q_{1}^{u}\leq f_{u}^{u}g_{u}^{1-u}$ and so $f_{u}$ and $g_{u}$ are the unique solution.

If $\int f_{u}>1$ and $\int g_{u}=1$, then the solution $q_{u}$ cannot always coincide with $f_{u}$ since $\int q_{0}=1$ and so $q_{0}<f_{u}$ in some set $E\subseteq\Omega$ with positive measure. If $\int f_{l}=1$, then since $f_{l}\leq q_{0}$ and $1=\int f_{l}=\int q_{0}$, you have $\int(q_{0}% -f_{l})=0$ and so $q_{0}-f_{l}=0$ for $\mu$ a.e. $x\in\Omega$. Thus $f_{l}$ and $g_{u}$ are the unique solution.

If $\int f_{l}<1$, assume that there exists a solution $q_{0}$ and assume that for some $n$ the set $G_{n}=\{x\in G:f_{l}(x)+\frac{1}{n}<q_{0}(x)<f_{u}% (x)-\frac{1}{n}\}$ has positive measure. Take $\varphi=0$ outside $G_{n}$, $|\varphi|\leq\frac{1}{2n}$ in $G_{n}$, and $\int_{G_{n}}\varphi=0$. Setting $F(q)=\int q^{u}g_{u}^{1-u}$, you have% $$ g(t):=F(q_{0}+t\varphi)\geq g(0)=F(q_{u}) $$ for every $t$ small and so $$ 0=\frac{dg}{dt}(0)=\int uq_{0}^{u-1}g_{u}^{1-u}\varphi=u\int_{G_{n}}% q_{0}^{u-1}g_{u}^{1-u}\varphi. $$

In turn, given $\psi$ with $|\psi|\leq\frac{1}{4n(1+\mu(G_{0}))}$ and $\psi=0$ outside $G_{n}$, we have that $\varphi:=\psi-\int_{G_{n}}\psi$ satisfies $|\varphi|\leq\frac{1}{2n}$ in $G_{n}$, and $\int_{G_{n}}\varphi=0$ and so by Fubini's theorem$$ 0=\int_{G_{n}}q_{0}^{u-1}g_{u}^{1-u}\left( \psi-\int_{G_{n}}\psi \,d\mu\right) \,d\mu=\int_{G_{n}}\psi\left( q_{0}^{u-1}g_{u}^{1-u} -\int_{G_{n}}q_{0}^{u-1}g_{u}^{1-u}\,d\mu\right) \,d\mu. $$ Since this is true for every $\psi$ with $|\psi|\leq\frac{1}{4n(1+\mu(G_{n}% ))}$, it should imply that $q_{0}^{u-1}g_{u}^{1-u}-\int_{G_{n}}q_{0} ^{u-1}g_{u}^{1-u}\,d\mu=0$ for $\mu$ a.e. $x\in G_{n}$. So $q_{0}^{u-1} g_{u}^{1-u}$ is constant in $G_{n}$, which implies that $q_{0}=c_{n}g_{u}$ in $G_{n}$ for some $c_{n}>0$. But $G_{n}\subseteq G_{n+1}$ and so $c_{n} =c_{n+1}$. Thus $q_{0}=cg_{u}$ in $\bigcup G_{n}=\{x\in G:f_{l}(x)<q_{0} (x)<f_{u}(x)\}$.

This shows that if $\int g_{u}=1$ then either there is no solution $q_{0}$ or there is and it has the form$$ q_{0}(x)=\left\{ \begin{array} [c]{ll} f_{u}(x) & x\in E_{1},\\ cg_{u}(x) & x\in E_{2},\\ f_{l}(x) & x\in E_{3}. \end{array} \right. $$ I guess that taking simple choices of functions, such as, $f_{l}=g_{l}=0$, $g_{u}=1$, $f_{u}=2$ and $\Omega=[0,1]$, one could try to see if there is uniqueness or not.

I don't have time to do the interesting case $\int g_{u}>1$, $\int f_{u}>1$, $\int g_{l}<1$,$\int f_{l}<1$.

Edit The solution always exist. provided $$ \left( s,t\right) \mapsto s^{u}t^{1-u}=f(s,t) $$ is concave. So if $$ \ell=\sup\int q_{1}^{u}q_{0}^{1-u}$$ and you construct a sequence $(q_{1,n},q_{0,n})$ such that $$ \ell=\lim_{n\rightarrow\infty}\int q_{1,n}^{u}q_{0,n}^{1-u}$$ then if $g_{u}$ and $f_{u}$ are bounded then you can find a subsequence $(q_{1,n_{k}},q_{0,n_{k}})\overset{\ast}{\rightharpoonup}(q_{1},q_{0})$ in $L^{\infty}$. Since $f$ is concave $$ \ell=\limsup_{k\rightarrow\infty}\int q_{1,n_{k}}^{u}q_{0,n_{k}}^{1-u}\leq\int q_{1}^{u}q_{0}^{1-u}. $$ Again by weak star convergence, if $\Omega$ has finite measure $1=\int q_{1,n_{k}}\rightarrow\int q_{1}$ and $1=\int q_{0,n_{k}}\rightarrow\int q_{0}$. The bounds should also be satisfied. $\int_{E}f_{l}\leq\int_{E}% q_{0,k}\rightarrow\int_{E}q_{0}$ so $\int_{E}f_{l}\leq\int_{E}q_{0}$ for every measurable set $E$ so $f_{l}\leq q_{0}$ and the same is true for the other constraints. So you do have a maximum.

  • $\begingroup$ Thank you for the answer. Actually, the equality cases are not the most important. The most important is "the interesting case" you defined at the end. For this case, there is definitely a solution. Please have a look at this: math.stackexchange.com/questions/2262762/… $\endgroup$ Jun 11, 2017 at 21:19
  • $\begingroup$ there is a huge difference between the dicrete case and the infinite dimensional one. In the discrete case you can use Bolzano's theorem to conclude that bounded sequences have a subsequence which converges strongly. In the infinite dimensional case, if you take a sequence $\{(q_{0,n},q_{1,n})\}$ satisfying your bounds, all you can say is that a subsequence converges weakly star in $L^\infty(\Omega)$, but then you would need your functional to be upper semicontinuous and it is not. $\endgroup$
    – Gio67
    Jun 11, 2017 at 21:24
  • $\begingroup$ Take a look at this en.wikipedia.org/wiki/… it explains it a little bit in the infinite dimensional case. This is why I asked what measure and what was $\Omega$. It makes a huge difference. $\endgroup$
    – Gio67
    Jun 11, 2017 at 21:26
  • $\begingroup$ I agree about the difference. I will look at it a bit deeper. But I just choose say $f_l=0.5N(-1,1)$, $f_u=2N(-1,1)$, $g_l=0.5N(1,1)$, $g_u=2N(1,1)$. and solve the problem and I always get a solution. The function is concave and the constraints are linear.. $\endgroup$ Jun 11, 2017 at 21:30
  • $\begingroup$ I take it back what I wrote before. In general the product of two concave functions is not concave but in your case it is, so the solution always exists. You are right. I edited my answer $\endgroup$
    – Gio67
    Jun 11, 2017 at 22:03

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