# Show that a system with integral equations may have no solutions

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 $$u$$.

• I haven't dived very deep into the question, but if you have a counterexample with $\mathcal{V}$ finite, can you not just divide $\mathbb{R}$ into finitely many pieces and treat each piece as a single element? Sep 26, 2022 at 14:12
• Ideally yes, but I'm not sure how to formally aggregate things up as I need to go from a PMF to a PDF...
– TEX
Sep 26, 2022 at 14:42
• The same thing I did in the other thread still works here. vectors are now the functions, and the inner product is $r\cdot s = \int_{\cal V} r(v)s(v)\,dv$. There is some question of whether $c(v) = 1$ defines an allowable vector, but it was only used for a shorthand anyway, so that question can be bypassed. The condition on $z$ as the end which leads to no possible solution has to be stated here as $u(1,v) \ge u(0,v)$ for all $v$ and $\int_{\cal V}(u(1,v)-u(0,v)\, dv > 0$. Sep 26, 2022 at 16:53
• @PaulSinclair I guess this is a trivial question, but I wonder if the procedure you worked out here math.stackexchange.com/questions/4536086/… applies also to the case $\mathcal{V}=\mathbb{R}^K$ for $K>1$. In the comment above, you hinted that it should work for $K=1$. If this is not trivial, please let me know, and I will open a bounty.
– TEX
Dec 8, 2022 at 20:09