<p>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.</p>

<p>On to your question:
<br> 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 $x_L(y) = x_{L}$.
<br> (I am not crazy about the notation, feel free to edit my post as long as it remains consistent with the question.)
<br>
 Then we also have $0 \leq v(y',x_{L}')$, and because of monotonicity this means
$x_L(y') &lt; x_{L}'$.
<br> 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, $x_L(y') &lt; x_{L}'$.</p>

CC BY-SA 3.0
2015-09-20T10:25:55.257Z
2015-09-20T10:32:11.950Z
2015-09-20T10:32:11.950Z
Giskard
<p>This question is closely related to <em>Mas-colell, Whinston, Green: Microeconomic Theory</em>, <strong>Question 3.C.5b</strong></p>

<blockquote>
  <p>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 $y \in  \mathbb{R}^{L-1}_{+}$. </p>
  
  <p>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  $v(y,x_{L}(y))=0$. </p>
  
  <p>Let $\phi(y)=-x_{L}(y)$ and show the utility function of the form
  $u(x) = x_{L} + \phi(y)$ represents $\succsim$.</p>
</blockquote>

<p><strong>Proof:</strong></p>

<p>To show that $u(x)$ represents $\succsim$ I need to show that for every $x,x' \in X$, $x'\succsim x \iff u(x') \geq u(x)$.</p>

<p>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 $(z,x'_{L}) \precsim (z,x_{L})$  $\forall z$.</p>

<p>These preferences imply:</p>

<p>$(y,x_{L}') \precsim (y,x_{L}) \precsim (y',x_{L}') \precsim (y',x_{L})$</p>

<p>To show that $u(x)$ is consistent with this I need to show that this preference implies:</p>

<ol>
<li>$u(y, x_{L}') \leq u(y,x_{L}) \leq u(y',x_{L}') \leq u(y',x_{L})$</li>
</ol>

<p>and not</p>

<ol start="2">
<li>$u(y, x_{L}') \leq u(y',x_{L}') &lt;  u(y,x_{L}) \leq u(y',x_{L})$</li>
</ol>

<p>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.  </p>

CC BY-SA 3.0
2015-09-19T17:45:56.163Z
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.
Giskard
CC BY-SA 3.0
2015-09-19T18:00:04.480Z
@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 $u(x)$.
möbius
<microeconomics><consumer-theory><choice-theory>
Prove that a continuous $\succsim$ is quasilinear
möbius
# Prove that a continuous $\succsim$ is quasilinear

(Asked by: username_0 on 2015-09-19)

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:

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

and not

2. $u(y, x\_{L}') \leq u(y',x\_{L}') < u(y,x\_{L}) \leq u(y',x\_{L})$

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.

**Tags:** microeconomics, consumer-theory, choice-theory

> **Comments:**
> **username_1**: 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.  
> **username_0**: @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 $u(x)$.  

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## 1 answers

### Answer by username_1 on 2015-09-20. Score: 4 | Accepted answer

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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