# $a\geq 0$, $x\sim y$ implies $x+a\sim y+a$ so the preference is linear?

$$\succsim$$ 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 $$x,y,a$$.

• Should be $x\sim y$ in the title I guess. Mar 21 '20 at 14:48
• @VARulle Thank you for pointing that out! Mar 22 '20 at 0:08

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.
• @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. Mar 27 '20 at 22:49