Let
- $(E,\mathcal E,\lambda)$ be a measure space and $$\lambda g:=\int g\:{\rm d}\mu$$ for all $\mu$-integrable $g$
- $f:E\to[0,\infty)^3$ be $\mathcal E$-measurable with $\lambda|f|^2<\infty$
- $J$ be a finite nonempty set
- $h_j:E\to[0,\infty)$ be $\mathcal E$-measurable for $j\in J$ with $$\beta:=\sup_{j\in J}\left\|h_j\right\|\infty<\infty$$
- $B:=\{f\ne0\}$ and $$B_j:=B\cap\{h_j>0\}\;\;\;\text{for }j\in J$$
- $r$ be a probability density on $(E,\mathcal E,\lambda)$ with $B\subseteq\{r>0\}$
Assume $\lambda(B_j)\in(0,\infty)$ for all $j\in J$. I'd like to bound $$\max_{j\in J}\underbrace{\int_{B_j}\lambda({\rm d}y)\frac{\displaystyle\left|h_j(y)f(y)-\frac{\lambda\left(h_jf\right)}{\lambda\left(B_j\right)}\right|^2}{r(y)}}_{=:\:\theta_j}\tag1$$ by $$\int_B\lambda({\rm d}y)\frac{\left|f(y)-\gamma\lambda f\right|^2}{r(y)}\tag2$$ for a suitably chosen constant $\gamma\ge0$ which should not depend on $\left(h_j\right)_{j\in J}$.
I've already asked a simplified version of this question which received a nice probabilistic answer. However, my attempt to mimic the proof failed. We may clearly use monotonicity of integration again to obtain $$\theta_j\le\int_B\frac{\left|h_jf\right|^2}r\:{\rm d}\lambda-\frac2{\lambda(B_j)}\left\langle\int_B\frac{h_jf}r\:{\rm d}\lambda,\lambda(h_jf)\right\rangle+\left|\frac{\lambda\left(h_jf\right)}{\lambda(B_j)}\right|^2\int_B\frac1r\:{\rm d}\lambda\tag3$$ for all $j\in J$. Now we would need to substract $(2)$ from the right-hand side of $(3)$ and show that the result is negative.
If it's easier to prove, feel free to assume $\beta\le1$ or even $h_j=1_{\{\:h_j\:>\:0\:\}}$ for all $j\in J$. Moreover, if necessary, assume $\lambda(B)\in(0,\infty)$ so that, for example, $\gamma={\lambda(B)}^{-1}$ is a candidate choice.