<p>This is not true.</p>

<p>Let <span class="math-container">$n=1$</span> and define <span class="math-container">$u(x)=\min{\{x,0\}}$</span>. Let <span class="math-container">$\succsim$</span> be the preference relation represented by <span class="math-container">$u$</span>. This preference relation is continuous and convex. We also have <span class="math-container">$x\sim y$</span> implies <span class="math-container">$x+a\sim y+a$</span> for any <span class="math-container">$a\geq0$</span> and <span class="math-container">$x,y\in\mathbb R$</span>. But let <span class="math-container">$x=0$</span>, <span class="math-container">$y=1$</span>, and <span class="math-container">$a=-1$</span>. Then <span class="math-container">$x\sim y$</span>, but <span class="math-container">$y+a=0\succ -1=x+a$</span>, thus <span class="math-container">$x+a\nsim y+a$</span> and <span class="math-container">$\succsim$</span> is not linear.</p>

CC BY-SA 4.0
2020-03-26T01:43:56.497Z
This is a good answer! However, in economics, most if not utility functions are strictly increasing. It would be very helpful if you could give me a hint on how to obtain a counterexample that is also strictly increasing in both elements.
High GPA
CC BY-SA 4.0
2020-03-26T08:49:22.080Z
@HighGPA: Well, that's a whole new thing, and I guess linearity would then indeed follow. Try posting this as a new question.
VARulle
CC BY-SA 4.0
2020-03-27T21:54:40.323Z
I believe so. I posted that altered question on the Math forum. Do you believe that I could post it on the Econ forum again? https://math.stackexchange.com/questions/3591086/forall-a0-fx-fy-leftrightarrow-fxa-fya-implies-forall-b-f/3592171#3592171
High GPA
CC BY-SA 4.0
2020-03-27T22:49:22.607Z
@HighGPA: If you frame it as a question on monotone, continuous, convex preferences on $\mathbb{R}^n$, then it belongs to economic theory, I'd say.
VARulle
CC BY-SA 4.0
2020-03-29T02:46:43.940Z
Hi, I post a new question and it seems like that I find a proof. Do you mind to check it?https://economics.stackexchange.com/questions/35670/a-geq-0-x-succsim-y-implies-xa-succsim-ya-so-the-preference-is-linear
High GPA
CC BY-SA 4.0
2020-03-23T14:32:00.027Z
2020-03-23T14:32:00.027Z
VARulle
<p><span class="math-container">$\succsim$</span> is a countinuous and convex weak order.</p>

<p><span class="math-container">$x,y,a$</span> are vectors in <span class="math-container">$\mathbb R^n$</span></p>

<p>We say <span class="math-container">$a\geq0$</span> if all directions of the vector <span class="math-container">$a$</span> is greater or equal to zero.</p>

<p>We want to prove (or disprove by counterexample) that: </p>

<p>Suppose <span class="math-container">$x\sim y$</span> implies <span class="math-container">$x+a\sim y+a$</span> for any <span class="math-container">$a\geq0$</span> and <span class="math-container">$x,y\in\mathbb R^n$</span>, </p>

<p>Then the preference is linear. </p>

<hr>

<p>One definition of linear preference is that <span class="math-container">$x\sim y$</span> implies <span class="math-container">$x+a\sim y+a$</span> for any <span class="math-container">$x,y,a$</span>.</p>

CC BY-SA 4.0
2020-03-21T14:48:42.850Z
Should be $x\sim y$ in the title I guess.
VARulle
CC BY-SA 4.0
2020-03-22T00:08:00.980Z
@VARulle Thank you for pointing that out!
High GPA
<microeconomics><game-theory><preferences>
$a\geq 0$, $x\sim y$ implies $x+a\sim y+a$ so the preference is linear?
High GPA
# $a\geq 0$, $x\sim y$ implies $x+a\sim y+a$ so the preference is linear?

(Asked by: username_0 on 2020-03-21)

$\succsim$ is a countinuous and convex weak order.

$x,y,a$ are vectors in $\mathbb R^n$

We say $a\geq0$ if all directions of the vector $a$ is greater or equal to zero.

We want to prove (or disprove by counterexample) that:

Suppose $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$,

Then the preference is linear.

---

One definition of linear preference is that $x\sim y$ implies $x+a\sim y+a$ for any $x,y,a$.

**Tags:** microeconomics, game-theory, preferences

> **Comments:**
> **username_1**: Should be $x\sim y$ in the title I guess.  
> **username_0**: @username_1 Thank you for pointing that out!  

---

## 1 answers

### Answer by username_1 on 2020-03-23. Score: 2 | Accepted answer

This is not true.

Let $n=1$ and define $u(x)=\min{\{x,0\}}$. Let $\succsim$ be the preference relation represented by $u$. This preference relation is continuous and convex. We also have $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R$. But let $x=0$, $y=1$, and $a=-1$. Then $x\sim y$, but $y+a=0\succ -1=x+a$, thus $x+a\nsim y+a$ and $\succsim$ is not linear.

> **Comments:**
> **username_0**: This is a good answer! However, in economics, most if not utility functions are strictly increasing. It would be very helpful if you could give me a hint on how to obtain a counterexample that is also strictly increasing in both elements.  
> **username_1**: @username_0: Well, that's a whole new thing, and I guess linearity would then indeed follow. Try posting this as a new question.  
> **username_0**: I believe so. I posted that altered question on the Math forum. Do you believe that I could post it on the Econ forum again? https://math.stackexchange.com/questions/3591086/forall-a0-fx-fy-leftrightarrow-fxa-fya-implies-forall-b-f/3592171#3592171  
> **username_1**: @username_0: If you frame it as a question on monotone, continuous, convex preferences on $\mathbb{R}^n$, then it belongs to economic theory, I'd say.  
> **username_0**: Hi, I post a new question and it seems like that I find a proof. Do you mind to check it?https://economics.stackexchange.com/questions/35670/a-geq-0-x-succsim-y-implies-xa-succsim-ya-so-the-preference-is-linear  

---

