Minimizing the expectation of a functional of probability distribution subject to an entropy constraint Consider a PDF $\pi(x)$ for $x\in[0,1]$, and the following functional
$$
 F(\pi) = \mathbb{E}_\pi |x-y|  $$
It is minimized by any point mass, so to avoid such degeneracy I'd like to lower-bound the entropy of $\pi$:
$$
 \mathbb{E}_\pi \left[ \ln \pi \right] \geq C
$$
Is there an analytical solution to $\min_\pi F(\pi)$ subject to such an entropy constraint? In case of multiple solutions, I'd like the one(s) closest to some given $\pi_0$ (in the $L_p$ sense for a convenient $p$).
In the absence of analytical solutions, numerical methods would be useful. The entropy constraint can be addressed by a Lagrange multiplier, but perhaps there is some elegant way to deal with its non-linearity.
Now, the functional I'd really like to minimize over $L_1$ is
$$
 F_\alpha(\pi) = \mathbb{E_\pi} \left[ |x-y| \cdot (\pi^\alpha (x) + \pi(y)^\alpha) \right]
$$
for $1 \leq \alpha \leq 2$. Perhaps, start with $\alpha=1$.
The entropy constraint isn't critical, but I will start with some $\pi_{init}$ and would like to prevent unnecessary entropy loss if possible.
 A: Here is the lower bound of $0.49$ for all $\alpha\ge 1$. Note that $\min_\pi F(\pi)$ is a non-decreasing function of $\alpha$, so it is enough to consider $\alpha=1$. Also, the truth is about $0.55$ for $\alpha=1$ and $0.62$ for $\alpha=2$, so, if you care about entropy, the uniform distribution that gives $\frac 23$ for all $\alpha$ is not too far from the truth. 
If you do it numerically and try just to consider functions constant on short intervals, then do not forget the self-action on each interval in the sum to be optimized. It looks small compared to the whole for spread distributions but without it everything will quickly converge to a singleton. I made this error when trying to optimize numerically and it took me some time to realize and fix it.
Now the bound. I'll just consider $G(\pi)=\frac{F(\pi)}{2}=\iint \pi(x)^2\pi(y)|x-y|\,dx\,dy$. Note that this expression is invariant with respect to the scaling $\pi\mapsto c\pi(c\cdot)$, so the support restriction is not important. What matters is only that $\int\pi=1$.
Write the Lebesgue decomposition $\pi(x)=\int_{t>0}\chi_{E_t}(x)\,dt$ where $E_t$ is a decreasing nested family of measurable sets. Then 
$$
G(\pi)=\iint_{s,t>0}2sQ(s,t)\,ds\,dt
$$
where 
$$
Q(s,t)=\iint_{E_s\times E_t}|x-y|\,dx\,dy\,.
$$
It is not hard to see that if $0<\ell<L$ are lengths of $E_s,E_t$ (not necessarily in this order), then the least value of $Q$ is given by 2 concentric intervals, in which case it is
$$
\frac{\ell L^2}{4}+\frac{\ell^3}{12}\ge \frac{\ell L(\ell+L)}{6}\,.
$$
The RHS is just made up to get a symmetric expression in $s,t$, which leads to a nice optimization problem but costs about $12\%$ in the precision of the bound. Also it is clear from here that the only real competitors are symmetric unimodal distributions.
Now, putting $f(t)=|E_t|$, we see that we need to estimate
$$
\frac 16\iint 2s f(s)f(t)[f(s)+f(t)]\,ds\,dt=\frac 13\iint (s+t) f(s)^2f(t)\,ds\,dt\\
=\frac 13\int (s+a)f(s)^2\,ds
$$
from below under the conditions $\int f(s)\,ds=1,\int sf(s)\,ds=a>0$.
This is an almost pure Hilbert space question except for the positivity constraint. We'll just define $g_b(s)=\frac {(1-bs)_+}{a+s}$ for $b>0$ and use Cauchy-Schwarz:
$$
\left[\int (s+a)f(s)^2\,ds\right]\left[\int (s+a)g_b(s)^2\,ds\right]\ge
\left[\int (s+a)f(s)g_b(s)\,ds\right]^2
$$
The integral on the right is 
$$
\int f(s)(1-bs)_+\ge \int f(s)(1-bs)=1-ab\,,
$$
so we get the bound 
$$
\int (s+a)f(s)^2\,ds\ge (1-ab)^2\left[\int (s+a)g_b(s)^2\,ds\right]^{-1}
$$
as long as $ab<1$. Choosing $ab=\frac 14$, we get the right hand side independent of $a$ and equal approximately $0.735$ (the best choice is the solution of some transcendental equation, but we have lost $12\%$ already, so who cares now) giving $G(\pi)\ge 0.245$, as promised.
Simple functions that give values of $F(\pi)$ close to the true minimum (on $[-0.5,0.5]$ instead of $[0,1]$) are (properly normalized) $(1-2x)^2$ and  $1-x$ for $\alpha=1$ and $\alpha=2$ respectively. Those are not exact optimizers but you may find it hard to beat them when doing numeric optimization. 
