<p>If I'm reading right, you define a preference <span class="math-container">$\succsim$</span> on <span class="math-container">$\mathbb{R}^N$</span> to be <strong>linear</strong> if, for all <span class="math-container">$x,y$</span> and all <span class="math-container">$\alpha \in \mathbb{R}^N$</span>,
<span class="math-container">$$
x \succsim y \implies x + \alpha \succsim y + \alpha.
$$</span>
and <span class="math-container">$\succsim$</span> satisfies <strong>Condition 1</strong> if, only for <span class="math-container">$\alpha \ge 0$</span>, we have:
<span class="math-container">$$
x \succsim y \implies x + \alpha \succsim y + \alpha.
$$</span>
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).</p>
<blockquote>
<p><strong>Proof</strong>: Suppose <span class="math-container">$\succsim$</span> (and <span class="math-container">$\succ$</span>) satisfy condition 1, and suppose <span class="math-container">$x \succsim y$</span>.  Fix <span class="math-container">$\alpha \in \mathbb{R}^N$</span> arbitrarily.  We want to show that <span class="math-container">$x + \alpha \succsim y+ \alpha$</span>. Define:
<span class="math-container">$$
\alpha^+ = \big(\max\{\alpha_1, 0\}, \ldots, \max\{\alpha_N, 0\}\big),
$$</span>
and
<span class="math-container">$$
\alpha^- = \big(\min\{\alpha_1, 0\}, \ldots, \min\{\alpha_N, 0\}\big),
$$</span>
and let <span class="math-container">$x' = x + \alpha^-$</span>, <span class="math-container">$y' = y + \alpha^-$</span>.  By condition 1 applied to <span class="math-container">$\succ$</span>, and the assumption that <span class="math-container">$x \succsim y$</span>, we have that <span class="math-container">$x' \succsim y'$</span>.  But then by condition 1 for <span class="math-container">$\succsim$</span>, we have that <span class="math-container">$x' + \alpha^+ \succsim y'+ \alpha^+$</span>.  However, <span class="math-container">$x'+ \alpha^+ = x + \alpha$</span>, and <span class="math-container">$y' + \alpha^+ = y + \alpha$</span>. If instead <span class="math-container">$x \succ y$</span> an analogous argument holds. QED</p>
</blockquote>
<p>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 <span class="math-container">$U(x) = \langle x ,\lambda \rangle$</span>.  You also need either continuity or a stronger monotonicity assumption. For a counterexample, consider <span class="math-container">$N=1$</span> and a preference represented by any discontinuous solution to the <a href="https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation" rel="nofollow noreferrer">Cauchy functional equation</a>.</p>

CC BY-SA 4.0
2022-08-18T13:25:22.773Z
Nice answer! I am surprised that non-satiation isn't needed. Let me double check.
High GPA
CC BY-SA 4.0
2022-08-18T13:38:49.553Z
$\alpha^-\leq 0$, 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: $[x\succsim y\implies x+a\succsim y+a]\iff[ x+a\prec y+a \implies x\prec y]$.
High GPA
CC BY-SA 4.0
2022-08-18T14:31:56.137Z
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)
Pete Caradonna
CC BY-SA 4.0
2022-08-18T15:10:32.457Z
I've made an edit to show that my answer both assumes that an analogue of condition 1 holds for strict preferences, but that it shows that under these hypotheses, both linearity and its strict analogue hold.  The inclusion of the strict analogue is pretty standard for these kind of invariance axioms in the literature; case in point they let you get what you were after without continuity or LNS.  That said if the intent behind your question was to look at whether CONT & LNS are substitutes for this condition, this may be less helpful.
Pete Caradonna
CC BY-SA 4.0
2022-08-19T09:10:52.167Z
There is a slight difference, though. Linearity implies the strict analogue of linearity, so it is natural to also assume the strict analogue. But, the Condition 1 does *not* implies the strict analogue of Condition 1 without additional assumptions (I guess).
High GPA
CC BY-SA 4.0
2022-08-19T09:13:18.533Z
Intuitively, linearity is a "two-directional" condition, while Condition 1 is a one-directional condition, which makes the problem a bit non-obvious. Fishburn had multiple papers on similar "one-directional" invariance conditions, in which if you assume the strict analogue the problem becomes trivial.
High GPA
CC BY-SA 4.0
2022-08-17T20:15:26.110Z
2022-08-18T15:15:40.020Z
2022-08-18T15:15:40.020Z
Pete Caradonna
<p><span class="math-container">$\succsim$</span> is a continuous and local non-satiate 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">$x\succsim y$</span> implies <span class="math-container">$x+a\succsim y+a$</span> for any <span class="math-container">$a\geq0$</span> and <span class="math-container">$x,y\in\mathbb R^n$</span> (condition 1),</p>
<p>Then the preference is linear.</p>
<hr />
<p>The <strong>definition of linear preference</strong> is that <span class="math-container">$x\succsim y$</span> implies <span class="math-container">$x+a\succsim y+a$</span> for any <span class="math-container">$x,y,a$</span>.</p>
<p>Proof by contradiction. Suppose <span class="math-container">$\succsim$</span> is not linear then there exists <span class="math-container">$x\succsim y$</span> but <span class="math-container">$x+a\prec y+a$</span>.</p>
<p>By non-satiation and continuity, there exists <span class="math-container">$x+\epsilon\succ y$</span> and <span class="math-container">$x+a+\epsilon \prec y+a$</span></p>
<p>Denote <span class="math-container">$x'=x+\epsilon$</span></p>
<p>Here if <span class="math-container">$a_i\geq 0$</span> or <span class="math-container">$a_i\leq 0$</span> for all index <span class="math-container">$i\in\{1,..,n\}$</span>  then the proof is done.</p>
<p>Now suppose that <span class="math-container">$a_i\geq0$</span> for some indexes but <span class="math-container">$a_j\leq 0$</span> for some other indexes.</p>
<p>Let <span class="math-container">$c_i=\min\{0,a_i\}$</span></p>
<p><span class="math-container">$v:=x'+c$</span> is a point such that <span class="math-container">$v\leq x'$</span> and <span class="math-container">$v\leq x'+a$</span></p>
<p><span class="math-container">$w:=y+c$</span> is a point such that <span class="math-container">$w\leq y$</span> and <span class="math-container">$w\leq y+a$</span></p>
<p>If <span class="math-container">$v\succsim w$</span>, then by condition (1) we must have <span class="math-container">$x'\succsim y$</span> and <span class="math-container">$x'+a\succsim y+a$</span>, contradition!</p>
<p>If <span class="math-container">$v\precsim w$</span>, then by condition (1) we must have <span class="math-container">$x'\precsim y$</span> and <span class="math-container">$x'+a\precsim y+a$</span>, contradition!</p>
<p>Is the proof sounds rigorous?</p>

CC BY-SA 4.0
2020-03-30T12:40:47.373Z
That's maybe not the most elegant way to write it down, but yes, I think it's correct.
VARulle
<microeconomics><game-theory><supply-and-demand><preferences><decision-theory>
$a\geq 0$, $x\succsim y$ implies $x+a\succsim y+a$ so the preference is linear?
High GPA
# $a\geq 0$, $x\succsim y$ implies $x+a\succsim y+a$ so the preference is linear?

(Asked by: username_0 on 2020-03-29)

$\succsim$ 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?

**Tags:** microeconomics, game-theory, supply-and-demand, preferences, decision-theory
> **Comments:**
> **username_1**: That's maybe not the most elegant way to write it down, but yes, I think it's correct.  

---

## 1 answers

### Answer by username_2 on 2022-08-17. Score: 2

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](https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation).

> **Comments:**
> **username_0**: Nice answer! I am surprised that non-satiation isn't needed. Let me double check.  
> **username_0**: $\alpha^-\leq 0$, 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: $[x\succsim y\implies x+a\succsim y+a]\iff[ x+a\prec y+a \implies x\prec y]$.  
> **username_2**: 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)  
> **username_2**: I've made an edit to show that my answer both assumes that an analogue of condition 1 holds for strict preferences, but that it shows that under these hypotheses, both linearity and its strict analogue hold.  The inclusion of the strict analogue is pretty standard for these kind of invariance axioms in the literature; case in point they let you get what you were after without continuity or LNS.  That said if the intent behind your question was to look at whether CONT & LNS are substitutes for this condition, this may be less helpful.  
> **username_0**: There is a slight difference, though. Linearity implies the strict analogue of linearity, so it is natural to also assume the strict analogue. But, the Condition 1 does *not* implies the strict analogue of Condition 1 without additional assumptions (I guess).  
> **username_0**: Intuitively, linearity is a "two-directional" condition, while Condition 1 is a one-directional condition, which makes the problem a bit non-obvious. Fishburn had multiple papers on similar "one-directional" invariance conditions, in which if you assume the strict analogue the problem becomes trivial.  

---

