How can I minimize this integral functional on a function space?

45 Views Asked by At

I want to find a choice for the functions $w_i:E\to[0,\infty)$ satsifying $\sum_{i\in I}w_i=1$ which minimizes the functional $$\Phi(w):=\int_0^\infty\int\mu({\rm d}(i,x))\operatorname E_{(i,\:x)}\left[\exp\left(-r_0\int_0^t\frac{u\left(i,U_s^{(i)}\right)}{w_i\left(U_s^{(i)}\right)p_i\left(U_s^{(i)}\right)}\:{\rm d}s\right)\frac{u\left(i,U_t^{(i)}\right)}{w_i\left(U_t^{(i)}\right)p_i\left(U_t^{(i)}\right)}K^{(i)}_t\right]\:{\rm d}t$$ or at least a choice, which makes this functional small. Here, $\mu$ is a probability measure with density $u$ with respect to $$\Lambda(B):=\sum_{i\in I}\lambda(\{x:(i,x)\in B\})$$ and the $p_i$ are probability densities with respect to $\lambda$. Furthermore, $r_0>0$, $\left(K^{(i)}_t\right)_{t\ge0}$ is a nonnegative process and the $\left(U^{(i)}_t\right)_{t\ge0}$ are independent Markov processes (on the probability space with probability measure $\operatorname P_{(i,\:x)}$, for every $x$) whose invariant distribution has density $w_ip_i$ with respect to $\lambda$.

I'm uncertain what I should do. For the actual minimization, it seems like I should consider $w$ as an element of a suitable function Banach space (e.g. $L^2$) and then apply the Lagrange multiplier theorem. But how exactly do I compute the derivative ${\rm D}\Phi(w)$ here?

If that's too complicated, maybe a pointwise minimization (e.g. only minimizing $\frac{u(i,\:x)}{w_ip_i}$ might be sensible, but I'm not sure.

Another thought is: Aren't we able to write the inner expectation as $$\operatorname E_{(i,\:x)}\left[\frac{u\left(i,U_t^{(i)}\right)}{w_i\left(U_t^{(i)}\right)p_i\left(U_t^{(i)}\right)}\right]\operatorname E_{(i,\:x)}\left[\exp\left(-r_0\int_0^t\frac{u\left(i,U_s^{(i)}\right)}{w_i\left(U_s^{(i)}\right)p_i\left(U_s^{(i)}\right)}\:{\rm d}s\right)K^{(i)}_t\right]\tag1$$ since $\left(U^{(i)}_s\right)_{s\in[0,\:t)}$ and $U^{(i)}_t$ are independent by the Markov property? (Please assume that $K^{(i)}_t$ is $\sigma\left(\left(U^{(i)}_s\right)_{s\in[0,\:t)}\right)$-measurable here.)

Unfortunately, I'm stuck with all of this ideas ...