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 $\succsim$.


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:

  1. $u(y, x_{L}') \leq u(y,x_{L}) \leq u(y',x_{L}') \leq u(y',x_{L})$

and not

  1. $u(y, x_{L}') \leq u(y',x_{L}') < u(y,x_{L}) \leq u(y',x_{L})$

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.

  • $\begingroup$ Perhaps I misunderstand the problem, but it seems to me that you are trying to prove that all $u(x) with$ some basic property represent quasilinear preferences, whereas in the forward direction you only have to prove that there is at least one $u(x)$ that represents the quasilinear preference. I.e.: you only have to show that it is possible to have a function $u(x)$ that is consistent with your criteria. $\endgroup$
    – Giskard
    Sep 19, 2015 at 17:45
  • $\begingroup$ @denesp What you say is true. I must admit that the actual question I was facing gave a specific form for $\phi$, but I thought that posting and seeing the solution for a more general question might resolve the issue. However, I have decided to post the specific example for $\phi$ actually given in the question. The problem I face is still the same, but now with a particular form for $u(x)$. $\endgroup$
    – möbius
    Sep 19, 2015 at 18:00

1 Answer 1


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 $x_L(y') < x_{L}'$.
Then $$ u(y',x_{L}') < u(y,x_{L}) $$ cannot hold as $$ u(y,x_{L}) = x_{L} + \phi(y) = x_{L} - x_{L}(y) = 0 $$ and $$ u(y',x_{L}') = x_{L}' + \phi(y') = x_{L}' - x_{L}(y') > 0, $$ because of what we have shown earlier, $x_L(y') < x_{L}'$.


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