Prove that a continuous $\succsim$ is quasilinear 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


$u(y, x_{L}') \leq u(y',x_{L}') &lt;  u(y,x_{L}) \leq u(y',x_{L})$


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.
First off it seems to me you are proceeding in a manner more complicated than necessary. (Perhaps this is intentional because you wish to face a harder exercise.) Since it is given that $v(x)$ represents $\succsim$ you only have to show
$$
\forall x,x' \in X: v(x') \geq v(x) \iff u(x') \geq u(x).
$$
I think this would be a lot simpler than what you are currently doing.

On to your question:
 The function $u(x)$ is very well defined, as
$$
u(y,x_{L}) =  x_{L} + \phi(y) = x_{L} - x_L(y) 
$$
and we know that $v(y,x_L(y)) = 0$ describes the indifference curves of $\succsim$. The conditions you describe would violate this assumption because from
$$
(y,x_{L}) \precsim (y',x_{L}')
$$
it follows that
$$
v(y,x_{L}) \leq v(y',x_{L}').
$$
We can assume $v(y,x_{L}) = 0$, this is an affine transformation of any $v(y,x_{L})$. So $x_L(y) = x_{L}$.
 (I am not crazy about the notation, feel free to edit my post as long as it remains consistent with the question.)

 Then we also have $0 \leq v(y',x_{L}')$, and because of monotonicity this means
$x_L(y') &lt; x_{L}'$.
 Then
$$
u(y',x_{L}') &lt; u(y,x_{L})
$$
cannot hold as
$$
u(y,x_{L}) = x_{L} + \phi(y) = x_{L} - x_{L}(y) = 0
$$
and
$$
u(y',x_{L}') = x_{L}' + \phi(y') = x_{L}' - x_{L}(y') &gt; 0,
$$
because of what we have shown earlier, $x_L(y') &lt; x_{L}'$.