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

$$\succsim$$ is a continuous and local non-satiate 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 $$x\succsim y$$ implies $$x+a\succsim y+a$$ for any $$a\geq0$$ and $$x,y\in\mathbb R^n$$ (condition 1),

Then the preference is linear.

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

Proof by contradiction. Suppose $$\succsim$$ is not linear then there exists $$x\succsim y$$ but $$x+a\prec y+a$$.

By non-satiation and continuity, there exists $$x+\epsilon\succ y$$ and $$x+a+\epsilon \prec y+a$$

Denote $$x'=x+\epsilon$$

Here if $$a_i\geq 0$$ or $$a_i\leq 0$$ for all index $$i\in\{1,..,n\}$$ then the proof is done.

Now suppose that $$a_i\geq0$$ for some indexes but $$a_j\leq 0$$ for some other indexes.

Let $$c_i=\min\{0,a_i\}$$

$$v:=x'+c$$ is a point such that $$v\leq x'$$ and $$v\leq x'+a$$

$$w:=y+c$$ is a point such that $$w\leq y$$ and $$w\leq y+a$$

If $$v\succsim w$$, then by condition (1) we must have $$x'\succsim y$$ and $$x'+a\succsim y+a$$, contradition!

If $$v\precsim w$$, then by condition (1) we must have $$x'\precsim y$$ and $$x'+a\precsim y+a$$, contradition!

Is the proof sounds rigorous?

• That's maybe not the most elegant way to write it down, but yes, I think it's correct. Mar 30, 2020 at 12:40

If I'm reading right, you define a preference $$\succsim$$ on $$\mathbb{R}^N$$ to be linear if, for all $$x,y$$ and all $$\alpha \in \mathbb{R}^N$$, $$x \succsim y \implies x + \alpha \succsim y + \alpha.$$ and $$\succsim$$ satisfies Condition 1 if, only for $$\alpha \ge 0$$, we have: $$x \succsim y \implies x + \alpha \succsim y + \alpha.$$ Suppose we both require that Condition 1 also hold for strict preference, and ask that linearity also hold for strict preferences. Then I claim these properties are equivalent, without any further assumptions (such as local nonsatiation).
Proof: Suppose $$\succsim$$ (and $$\succ$$) satisfy condition 1, and suppose $$x \succsim y$$. Fix $$\alpha \in \mathbb{R}^N$$ arbitrarily. We want to show that $$x + \alpha \succsim y+ \alpha$$. Define: $$\alpha^+ = \big(\max\{\alpha_1, 0\}, \ldots, \max\{\alpha_N, 0\}\big),$$ and $$\alpha^- = \big(\min\{\alpha_1, 0\}, \ldots, \min\{\alpha_N, 0\}\big),$$ and let $$x' = x + \alpha^-$$, $$y' = y + \alpha^-$$. By condition 1 applied to $$\succ$$, and the assumption that $$x \succsim y$$, we have that $$x' \succsim y'$$. But then by condition 1 for $$\succsim$$, we have that $$x' + \alpha^+ \succsim y'+ \alpha^+$$. However, $$x'+ \alpha^+ = x + \alpha$$, and $$y' + \alpha^+ = y + \alpha$$. If instead $$x \succ y$$ an analogous argument holds. QED
A cautionary note on terminology, condition 1 (and hence equivalently what you're calling linear preferences) do not suffice, even with local nonsatiation, to guarantee existence of a linear utility representation, i.e. of the form $$U(x) = \langle x ,\lambda \rangle$$. You also need either continuity or a stronger monotonicity assumption. For a counterexample, consider $$N=1$$ and a preference represented by any discontinuous solution to the Cauchy functional equation.
• $\alpha^-\leq 0$, so we don't have this: "$x\succsim y\implies x'\succsim y'$". Perhaps I missed some of your points. I think, let $a\geq 0$, we have: $[x\succsim y\implies x+a\succsim y+a]\iff[ x+a\prec y+a \implies x\prec y]$. Aug 18, 2022 at 13:38
• But $\vert \alpha^-\vert$ (i.e. the component-wise absolute values of the vector $\alpha^-$) is component-wise non-negative. Now, we know $x' + \vert \alpha^- \vert \succsim y' + \vert \alpha^- \vert$ must be true (this is just $x \succsim y$). Suppose then that $x' \not \succsim y'$. Then as $\succsim$ is complete, $y' \succ x'$. Then by condition 1, it follows that $y' + \vert \alpha^- \vert\succ x' + \vert \alpha^- \vert$ which we know to be false by hypothesis. Thus $x' \succsim y'$. (Implicitly I'm also assuming the strict analogue of Condition 1 holds, which seems natural) Aug 18, 2022 at 14:31