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

Question

$\succsim$ 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.

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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?

Answer 1

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.