Let $\phi \in \mathcal ( [0,1]^2)$ symetric , can we find a solution to the following minimisation problem?
$$ \inf \left\{ F[\nu] : \nu \in L^2 , \nu \geq 0, \int _0 ^1 \nu=1\right\}$$
with $$ F[\nu] := \frac{1}{2} \int_0^1 \nu^2(t) ~dt + \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~\nu(s) ~\nu(t) ~ds~dt$$
And if it exist such a $\nu$ what optimality conditions it satisfies ?
If I made no mistakes by making the Frechet derivative equals to zero we should have for any $h \in L^2$, $\int _0^1 h=0$ :
$$ \int_0^1 \nu(t)~h(t) ~dt = - \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~[\nu(s)~h(t)+\nu(t)~h(s)] ~ds~dt$$
After this point I can't see how to conclude exploring probably the symmetry of $\phi$. Could someone give me some advice or the solution if you see how to solve it easily?
Many thanks.
Here one idea of solution. Any corrections, additional steps, alternative solutions and suggestion are welcome.
Admitting that there is a solution for the optimisation problem, fact that still must be proved, we have:
$$ \int_0^1 \nu(t)~h(t) ~dt = - \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~[\nu(s)~h(t)+\nu(t)~h(s)] ~ds~dt$$
where $\phi$ is symmetric so it implies that
$$ \int_0^1 \nu(t)~h(t) ~dt = - \int_0^1\int_0^1 \phi(s,t) ~\nu(s)~h(t) ~ds~dt$$
therefore $\forall h \in L^2, ~ \int h =0$
$$ \int_0^1 \left( \nu(t) +\int_0^1 \phi(s,t) ~\nu(s) ~ds \right) ~h(t) ~dt = 0$$
then
$$ \nu(t) +\int_0^1 \phi(s,t) ~\nu(s) ~ds =0$$
so
$$ \int_0^1 \Phi (s) ~\nu(s) ~ds = 1$$
where $\Phi(s) := -\int_0^1 \phi(s,t) ~dt$.