I would like your help to show that the system below may have no solutions.
Let $\mathcal{Y}\equiv \{0,1\}$. Let $\mathcal{V}\equiv \mathbb{R}$, not finite. Let $w: \mathcal{V}\rightarrow \mathbb{R}$ denote a probability density function on $\mathcal{V}$. Let $q: \mathcal{Y}\rightarrow [0,1]$ denote a probability mass function on $\mathcal{Y}$. Let $u: \mathcal{Y}\times \mathcal{V}\rightarrow \mathbb{R}$.
Consider the system of equations/inequalities below. The unknown is the mixed joint density $f: \mathcal{Y}\times \mathcal{V}\rightarrow \mathbb{R}$. The functions $w$, $q$, and $u$ and the sets $\mathcal{V},\mathcal{Y}$ are known. $$ (*) \quad \begin{cases} &(1) \quad \sum_{y\in \mathcal{Y}}f(y,v) =w(v) \quad \forall v \in \mathcal{V},\\ &(2) \quad \int_{v\in \mathcal{V}} f(y,v) dv=q(y)\quad \forall y\in \mathcal{Y},\\ &(3) \quad\int_{v\in \mathcal{V}} f(1,v) *(u(1,v)-u(0,v)) \text{ }dv \geq 0,\\ &(4) \quad\int_{v\in \mathcal{V}} f(0,v) *(u(0,v)-u(1,v))\text{ }dv \geq 0.\\ \end{cases} $$
Question: Show that $(*)$ may have no solutions.
Note 1: The case with $\mathcal{V}$ finite (hence, $w$ and $f$ are probability mass functions) has been studied here. The answer shows that the system may not have a solution. I want to generalise the claim (or prove the opposite) when $\mathcal{V}$ is not finite.
Note 2: The case with $u(1,v)\equiv v$ and $u(0,v)\equiv 0$ has been studied here. The answer shows that the system may not have a solution. I want to generalise the claim (or prove the opposite) for any $u$.