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Consider a smooth function $f(x) \colon \mathbb{R}^2_+ \to \mathbb{R}_+$ such that $f$ is concave and positively homogeneous of order one. Consider a linear transform $P$ given by matrix $$ P = \begin{pmatrix} p_1 & 0 \\\ 0 & p_2 \end{pmatrix}, $$ where $p_1, p_2 > 0$ and $p_1 \neq p_2$ if $p_1 = 1$ of $p_2 = 1$. I have a hypothesis that the system of equations $$ \left \{ \begin{array}{rcl} f(x) & = & 1 \\ f(Px) & = & 1 \end{array} \right. $$ can't have more that one solution under some general conditions on function $f(x)$. Geometrically it means that curves $f(x) = 1$ and $f(Px) = 1$ can't have more than one common point (such curves are called pseudolines). Help me please with any idea to find such conditions on function $f(x)$. I tried to use some version of the global inverse function theorem, but unsuccessfully.

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I don't think this is true. If the set ${f(x) = 1}$ is two line segments connecting (0,10) to (1,1) to (10,0), and you contract in one direction and stretch in the other, you should be able to get one of the new line segments to cross both the originals. This doesn't give a smooth function but you can smooth things out near the corner without destroying the counterexample. – Mike Hall Apr 14 '12 at 20:19
This is true, thank you. I'll correct my question. – Nimza Apr 15 '12 at 18:59

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