I've got a contractive self-adjoint linear integral operator $\kappa$ of the form $$(\kappa g)(x):=g(x)+\int\lambda({\rm d}y)k(x,y)(g(y)-g(x))\;\;\;\text{for }g\in L^2(\mu),$$ where $k$ depends on the choice of a family of "weights" $(w_i)_{i\in I}$ (see definitions below). We may assume that the spectrum $\sigma(\kappa)$ is contained in $[-1,1)$.

> I want to determine a choice of $(w_i)_{i\in I}$ minimizing $\langle(1-\kappa)^{-1}f_0,f_0\rangle_{L^2(\mu)}$, where $f_0:=f-\int f\:{\rm d}\mu$ and $f$ is defined below. How can we solve this problem?

Maybe it's useful to consider the spectral measure $E:\mathcal B(\mathbb R)\to\mathfrak L(H)$ associated with $\kappa$ and write $$(1-\kappa)^{-1}=\int_{[-1,\:1)}\frac1{1-\lambda}\:E({\rm d}\lambda)\tag1,$$ but it's not clear to me how $E$ depends on $(w_i)_{i\in I}$.

*Definitions*:

 - $I$ is a finite set;
 - $p,q_i$ are probability densities on a measure space $(E,\mathcal E,\lambda)$;
 - $f\in\mathcal L^2(\lambda)$ with $\{p=0\}\subseteq\{f=0\}$;
 - $w_i:E\to\mathbb R$ is $\mathcal E$-measurable with $\{q_i=0\}\subseteq\{w_ip=0\}$ for $i\in I$ with $\{pf\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}$;
 - $\sigma_{ij}:E^2\to\mathbb R$ be $\mathcal E^{\otimes2}$-measurable with $\sigma_{ij}(x,y)=\sigma_{ji}(y,x)$ for all $(i,x),(j,y)\in I\times E$;
 - $\tau_{ij}(x,y):=w_i(x)q_j(y)\sigma_{ij}(x,y)$ for $(i,x),(j,y)\in I\times E$ satisfies $\sum_{j\in I}\int\lambda({\rm d}y)\tau_{ij}(x,y)=1$ for all $(i,x)\in I\times E$ and $$\alpha_{ij}:=\left.\begin{cases}\displaystyle1\wedge\frac{w_j(y)p(y)q_i(x)}{w_i(x)p(x)q_j(y)}&\text{, if }w_i(x)p(x)q_j(y)>0\\1&\text{, otherwise}\end{cases}\right\}$$ for $(i,x),(j,y)\in I\times E$
 - $k:=\sum_{i\in I}\sum_{j\in I}\alpha_{ij}\tau_{ij}$.