This question is closely related to Mas-colell, Whinston, Green: Microeconomic Theory, Question 3.C.5b
Let $\succsim$ be a strictly monotone, continuous, and rational preference relation on $(-\infty, \infty)\times \mathbb{R}^{L-1}_{+}$. Furthermore, suppose $\succsim$ is quasilinear in good $L$. Let us write $x \in X$ as $x=(y,x_{L})$ where $y \in \mathbb{R}^{L-1}_{+}$.
It can be shown that if $v(y,x_{L})$ is a utility function that represents $\succsim$, then there is a unique $x_{L}(y) \in \mathbb{R}$ such that $v(y,x_{L}(y))=0$.
Let $\phi(y)=-x_{L}(y)$ and show the utility function of the form $u(x) = x_{L} + \phi(y)$ represents $\succsim$.
Proof:
To show that $u(x)$ represents $\succsim$ I need to show that for every $x,x' \in X$, $x'\succsim x \iff u(x') \geq u(x)$.
However, I have found an example that may not be satisfied: suppose that $x$ and $x'$ are such that $x=(y,x_{L})$ and $x'=(y',x'_{L})$, with $x\precsim x'$, $(y,z_{L}) \precsim (y',z_{L})$ $\forall z_{L}$ and $(z,x'_{L}) \precsim (z,x_{L})$ $\forall z$.
These preferences imply:
$(y,x_{L}') \precsim (y,x_{L}) \precsim (y',x_{L}') \precsim (y',x_{L})$
To show that $u(x)$ is consistent with this I need to show that this preference implies:
- $u(y, x_{L}') \leq u(y,x_{L}) \leq u(y',x_{L}') \leq u(y',x_{L})$
and not
- $u(y, x_{L}') \leq u(y',x_{L}') < u(y,x_{L}) \leq u(y',x_{L})$
However I can find no reason why 2. presents any sort of contradiction! I have been thinking about this for two days, and am sure by now I am making some logical errors. Either the example I have come up with is not valid, or it must imply some sort of contradiction! Any comments will help. And please hints only, no full solutions.