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Disclaimer: Please bear with me, the question isn't as complicated as it looks like, but I wasn't able to find any simplification for which no counterexample comes to my find.

Let $(E,\mathcal E,\lambda),(E',\mathcal E',\lambda')$ be measure spaces, $I$ be a finite nonempty set, $p,q_i$ be positive probability densities on $(E,\mathcal E,\lambda)$, $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$ measurable with $\lambda'\circ\varphi_i^{-1}=q_i\lambda$, $w_i:E\to[0,1]$ be $\mathcal E$-measurable with $\sum_{i\in I}w_i=1$ and $r:(E'\times I)^2\to[0,\infty)$ be $\left(\mathcal E'\otimes2^I\right)^{\otimes2}$-measurable with $\sum_{j\in I}\int\lambda'({\rm d}y')r((x',i),(y',j))=1$ for all $(x',i)\in E\times I$. Note that $$\tilde p(x',i):=\left(w_i\frac p{q_i}\circ\varphi_i\right)(x')\;\;\;\text{for }(x',i)\in E'\times I$$ is a probability density on $(E'\times I,\mathcal E'\otimes 2^I,\lambda'\otimes\zeta)$, where $\zeta$ denotes the counting measure on $(I,2^I)$. For simplicity, let $$\sigma((x,i),(y,j)):=r\left(\left(\varphi_i^{-1},i\right),\left(\varphi_j^{-1},j\right)\right)\;\;\;\text{for }(x,i),(y,j)\in E\times I.\tag1$$

I want to choose $(w_i)_{i\in I}$ and $r$ (everything else being fixed) minimizing the quantity $$\eta_{w,\:r}:=\sup_{\tilde g\in L^2\left(\tilde p\left(\lambda'\otimes\zeta\right)\right)}\Psi_{w,\:r}(\tilde g)\tag2,$$ where $$\Psi_{w,\:r}(\tilde g):=\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)\left|\tilde g\left(\varphi_i^{-1}(x),i\right)-\tilde g\left(\varphi_i^{-1}(y),j\right)\right|^2.$$$$\Psi_{w,\:r}(\tilde g):={\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)\left|\tilde g\left(\varphi_i^{-1}(x),i\right)-\tilde g\left(\varphi_i^{-1}(y),j\right)\right|^2}.$$ Are we able to reduce this problem? For example, are we able to show that if $(\hat w_i)_{i\in I}$, $\hat r$ is another choice for $(w_i)_{i\in I}$, $r$ and \begin{equation}\begin{split}&{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)}\\&\;\;\;\;\le{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(\hat w_i(x)p(x)q_j(y)\hat\sigma((x,i),(y,j)),\hat w_j(y)p(y)q_i(x)\hat\sigma((y,j),(x,i))\right)}\end{split}\tag3,\end{equation} ($\hat\sigma$ being defined for $\hat r$ as in $(1)$), then $\eta_{w,\:r}\le\eta_{\hat w,\:\hat r}$?

Disclaimer: Please bear with me, the question isn't as complicated as it looks like, but I wasn't able to find any simplification for which no counterexample comes to my find.

Let $(E,\mathcal E,\lambda),(E',\mathcal E',\lambda')$ be measure spaces, $I$ be a finite nonempty set, $p,q_i$ be positive probability densities on $(E,\mathcal E,\lambda)$, $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$ measurable with $\lambda'\circ\varphi_i^{-1}=q_i\lambda$, $w_i:E\to[0,1]$ be $\mathcal E$-measurable with $\sum_{i\in I}w_i=1$ and $r:(E'\times I)^2\to[0,\infty)$ be $\left(\mathcal E'\otimes2^I\right)^{\otimes2}$-measurable with $\sum_{j\in I}\int\lambda'({\rm d}y')r((x',i),(y',j))=1$ for all $(x',i)\in E\times I$. Note that $$\tilde p(x',i):=\left(w_i\frac p{q_i}\circ\varphi_i\right)(x')\;\;\;\text{for }(x',i)\in E'\times I$$ is a probability density on $(E'\times I,\mathcal E'\otimes 2^I,\lambda'\otimes\zeta)$, where $\zeta$ denotes the counting measure on $(I,2^I)$. For simplicity, let $$\sigma((x,i),(y,j)):=r\left(\left(\varphi_i^{-1},i\right),\left(\varphi_j^{-1},j\right)\right)\;\;\;\text{for }(x,i),(y,j)\in E\times I.\tag1$$

I want to choose $(w_i)_{i\in I}$ and $r$ (everything else being fixed) minimizing the quantity $$\eta_{w,\:r}:=\sup_{\tilde g\in L^2\left(\tilde p\left(\lambda'\otimes\zeta\right)\right)}\Psi_{w,\:r}(\tilde g)\tag2,$$ where $$\Psi_{w,\:r}(\tilde g):=\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)\left|\tilde g\left(\varphi_i^{-1}(x),i\right)-\tilde g\left(\varphi_i^{-1}(y),j\right)\right|^2.$$ Are we able to reduce this problem? For example, are we able to show that if $(\hat w_i)_{i\in I}$, $\hat r$ is another choice for $(w_i)_{i\in I}$, $r$ and \begin{equation}\begin{split}&{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)}\\&\;\;\;\;\le{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(\hat w_i(x)p(x)q_j(y)\hat\sigma((x,i),(y,j)),\hat w_j(y)p(y)q_i(x)\hat\sigma((y,j),(x,i))\right)}\end{split}\tag3,\end{equation} ($\hat\sigma$ being defined for $\hat r$ as in $(1)$), then $\eta_{w,\:r}\le\eta_{\hat w,\:\hat r}$?

Disclaimer: Please bear with me, the question isn't as complicated as it looks like, but I wasn't able to find any simplification for which no counterexample comes to my find.

Let $(E,\mathcal E,\lambda),(E',\mathcal E',\lambda')$ be measure spaces, $I$ be a finite nonempty set, $p,q_i$ be positive probability densities on $(E,\mathcal E,\lambda)$, $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$ measurable with $\lambda'\circ\varphi_i^{-1}=q_i\lambda$, $w_i:E\to[0,1]$ be $\mathcal E$-measurable with $\sum_{i\in I}w_i=1$ and $r:(E'\times I)^2\to[0,\infty)$ be $\left(\mathcal E'\otimes2^I\right)^{\otimes2}$-measurable with $\sum_{j\in I}\int\lambda'({\rm d}y')r((x',i),(y',j))=1$ for all $(x',i)\in E\times I$. Note that $$\tilde p(x',i):=\left(w_i\frac p{q_i}\circ\varphi_i\right)(x')\;\;\;\text{for }(x',i)\in E'\times I$$ is a probability density on $(E'\times I,\mathcal E'\otimes 2^I,\lambda'\otimes\zeta)$, where $\zeta$ denotes the counting measure on $(I,2^I)$. For simplicity, let $$\sigma((x,i),(y,j)):=r\left(\left(\varphi_i^{-1},i\right),\left(\varphi_j^{-1},j\right)\right)\;\;\;\text{for }(x,i),(y,j)\in E\times I.\tag1$$

I want to choose $(w_i)_{i\in I}$ and $r$ (everything else being fixed) minimizing the quantity $$\eta_{w,\:r}:=\sup_{\tilde g\in L^2\left(\tilde p\left(\lambda'\otimes\zeta\right)\right)}\Psi_{w,\:r}(\tilde g)\tag2,$$ where $$\Psi_{w,\:r}(\tilde g):={\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)\left|\tilde g\left(\varphi_i^{-1}(x),i\right)-\tilde g\left(\varphi_i^{-1}(y),j\right)\right|^2}.$$ Are we able to reduce this problem? For example, are we able to show that if $(\hat w_i)_{i\in I}$, $\hat r$ is another choice for $(w_i)_{i\in I}$, $r$ and \begin{equation}\begin{split}&{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)}\\&\;\;\;\;\le{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(\hat w_i(x)p(x)q_j(y)\hat\sigma((x,i),(y,j)),\hat w_j(y)p(y)q_i(x)\hat\sigma((y,j),(x,i))\right)}\end{split}\tag3,\end{equation} ($\hat\sigma$ being defined for $\hat r$ as in $(1)$), then $\eta_{w,\:r}\le\eta_{\hat w,\:\hat r}$?

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Can we conclude $\sup_g\int f_1g\le\sup_g\int f_2g$ from $\int f_1\le\int f_2$ in this situation?

Disclaimer: Please bear with me, the question isn't as complicated as it looks like, but I wasn't able to find any simplification for which no counterexample comes to my find.

Let $(E,\mathcal E,\lambda),(E',\mathcal E',\lambda')$ be measure spaces, $I$ be a finite nonempty set, $p,q_i$ be positive probability densities on $(E,\mathcal E,\lambda)$, $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$ measurable with $\lambda'\circ\varphi_i^{-1}=q_i\lambda$, $w_i:E\to[0,1]$ be $\mathcal E$-measurable with $\sum_{i\in I}w_i=1$ and $r:(E'\times I)^2\to[0,\infty)$ be $\left(\mathcal E'\otimes2^I\right)^{\otimes2}$-measurable with $\sum_{j\in I}\int\lambda'({\rm d}y')r((x',i),(y',j))=1$ for all $(x',i)\in E\times I$. Note that $$\tilde p(x',i):=\left(w_i\frac p{q_i}\circ\varphi_i\right)(x')\;\;\;\text{for }(x',i)\in E'\times I$$ is a probability density on $(E'\times I,\mathcal E'\otimes 2^I,\lambda'\otimes\zeta)$, where $\zeta$ denotes the counting measure on $(I,2^I)$. For simplicity, let $$\sigma((x,i),(y,j)):=r\left(\left(\varphi_i^{-1},i\right),\left(\varphi_j^{-1},j\right)\right)\;\;\;\text{for }(x,i),(y,j)\in E\times I.\tag1$$

I want to choose $(w_i)_{i\in I}$ and $r$ (everything else being fixed) minimizing the quantity $$\eta_{w,\:r}:=\sup_{\tilde g\in L^2\left(\tilde p\left(\lambda'\otimes\zeta\right)\right)}\Psi_{w,\:r}(\tilde g)\tag2,$$ where $$\Psi_{w,\:r}(\tilde g):=\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)\left|\tilde g\left(\varphi_i^{-1}(x),i\right)-\tilde g\left(\varphi_i^{-1}(y),j\right)\right|^2.$$ Are we able to reduce this problem? For example, are we able to show that if $(\hat w_i)_{i\in I}$, $\hat r$ is another choice for $(w_i)_{i\in I}$, $r$ and \begin{equation}\begin{split}&{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(w_i(x)p(x)q_j(y)\sigma((x,i),(y,j)),w_j(y)p(y)q_i(x)\sigma((y,j),(x,i))\right)}\\&\;\;\;\;\le{\sum_{i\in I}\int\lambda({\rm d}x)\sum_{j\in I}\int\lambda({\rm d}y)\min\left(\hat w_i(x)p(x)q_j(y)\hat\sigma((x,i),(y,j)),\hat w_j(y)p(y)q_i(x)\hat\sigma((y,j),(x,i))\right)}\end{split}\tag3,\end{equation} ($\hat\sigma$ being defined for $\hat r$ as in $(1)$), then $\eta_{w,\:r}\le\eta_{\hat w,\:\hat r}$?