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Let a function $F: \textbf{R}^2_{++} \rightarrow \textbf{R}$ satisfy the relation: if $F(x,y)=F(x',y')$ then $F(x,y)=F(x+x',y+y')$.

It is easy to prove that under the additional assumption of continuity $F$ must be of the form $F(x,y)=g(x/y)$, where $g(z)=F(z,1)$. Are there any other solution if the continuity assumption is replaced with strict monotonicity with respect to the first argument?

Thanks.

Mikhail
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1 Answers1

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Here is my attempt to give a counterexample. Please check.

Define $F$ by

$F(x,y)=x/y$ if $x<y$,

$F(x,y)=2x/y$ if $x>y$,

$F(x,y)=z$ if $x=y$, $x\in z\mathbb{Q_{++}}$, where $z\mathbb{Q_{++}}$ is given by $z\mathbb{Q_{++}}=\{x\in\mathbb{R_{++}}:x=zq, q\in \mathbb{Q_{++}} \}$, $z\in [1,2]$, and $\mathbb{Q_{++}}$ is the set of positive rational numbers.

Mikhail
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