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Consider a preference relation $\succeq$ on $X\subseteq\mathbb R^2$. If $\succeq$ satisifies: $$ \begin{align} &1.\mbox{ }(a_1,a_2)\succeq (b_1,b_2)\implies(a_1+t,a_2+s)\succeq (b_1+t,b_2+s),\forall t,s\\ &2.\mbox{ }a_1\geq b_1 \mbox{ and } a_2\geq b_2 \implies (a_1,a_2)\succeq (b_1,b_2)\mbox{ (and the analogous for }\succ\mbox{)}\\ &3.\mbox{ Continuity } \end{align} $$ Then would there exists a linear representation for $\succeq$?


When $X=\mathbb R^2$, the old proof is copied here:

Step1: For each vector $(x,y)$, there is a unique $z\in \mathbb R$ such that $(x,y) \sim (z,z)$. WLOG assume $x \geq y$. Then to see this claim, first notice by A2 that $(x,x) \succeq (x,y) \succeq (y,x)$. Then traveling along the $45^\circ$ from $(y,y)$ to $(x,x)$, A3 ensures the existence of our $z$. (Strict) Monotonicity assures uniqueness in the obvious way. Let $u: (x,y) \mapsto z$ where $z$ is defined in this way.

Step2: Now let $(x,y) \sim (z,z)$ and $(x',y') \sim (z',z')$. Then by A1 and transitivity we have $(x+x',y+y') \sim (z+z',z+z')$. Additivity+transitivity implies linearity.


However, in our case of $X\subseteq \mathbb R^2$, for example let's set $X=[2,3]\times [2,3]$, then step 2 does not work anymore: because if $x,x'\in [2,3]$, then $x+x'\not\in [2,3]$.

Therefore, I hypothesize that the preference is not necessarily linear. It can be a power function like $u(x,y)=ax^b+cy^d$ where $a,b,c,d$ can be positive or negative. Also, $u$ must be analytic.

For 3+ dimensions, the $u$ must be separable.

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Assumptions $1$ to $3$ are sufficient to obtain a linear representation when $X$ is open and convex. We proceed in two steps.


Step $1$:

We will repeatedely use the following consequence of continuity and $A1$: If $x \sim x^{\prime}$, then $x \sim x + \lambda (x^{\prime} - x)$ for every $\lambda \in \mathbb{R}$ such that $x + \lambda (x^{\prime} - x) \in X$.

First, $x + 0.5 (x^{\prime} - x)$ is contained in $X$ as we assumed $X$ to be convex. Second, by completeness of the relation, $x \succeq x + 0.5 (x^{\prime} - x)$ or $x \preceq x + 0.5 (x^{\prime} - x)$. Assume the latter (the other case is treated similarly). By $A1$, $$x \preceq x + 0.5 (x^{\prime} - x) \quad \Leftrightarrow \quad x + 0.5 (x^{\prime} - x) \preceq x + 0.5 (x^{\prime} - x) + 0.5 (x^{\prime} - x).$$ The latter comparison is equivalent to $x + 0.5 (x^{\prime} - x) \preceq x^{\prime}$. This shows $$x \preceq x + 0.5 (x^{\prime} - x) \preceq x^{\prime} \sim x.$$

Repeating this argument, we conclude that for any $n, k \in\mathbb{N}$ with $k \leq 2^{n}$ the comparison $$ x \sim x + \frac{k}{2^{n}}(x^{\prime} - x) $$ holds. For any number $\lambda \in [0, 1]$ there is a sequence of fractions of the form $k / 2^{n}$ converging to $\lambda$. Since the relation is continuous, this establishes that $x \sim x + \lambda (x^{\prime} - x)$ for any $\lambda \in [0, 1]$.

To get indifference for arbitrary $\lambda \geq 0$, find $n\in \mathbb{N}$ such that $n \leq \lambda \leq n + 1$. Note that $x \sim x^{\prime}$ iff $x^{\prime} \sim x^{\prime} + (x^{\prime} - x)$ iff $x^{\prime} + (x^{\prime} - x) \sim x^{\prime} + 2(x^{\prime} - x)$, etc. Thus, $x \sim x + n (x^{\prime} - x ) \sim x + (n+1) (x^{\prime} - x )$. Since $\lambda$ may be written as a convex combination of $n$ and $n+1$, earlier arguments now imply that $x\sim x + n (x^{\prime} - x ) \sim x + \lambda (x^{\prime} - x )$. Lastly, when $\lambda \leq 0$, go through the same steps but consider $x - n (x^{\prime} -x)$, etc.


Step $2$:

Let $x$ be an arbitrary element of $X$. By continuity and monotonicity we may find a point $x^{\prime}$ unequal to $x$ such that $x\sim x^{\prime}$. The argument is as follows: For $\varepsilon$ sufficiently small, the points $x - \varepsilon (1, 1)$ and $x + \varepsilon (1, 1)$ are contained in $X$ as $X$ is open. By monotonicity, $$x - \varepsilon (1, 1) \prec x \prec x + \varepsilon (1, 1).$$ Then for $\varepsilon^{\prime}$ sufficiently small, $$x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) \prec x \prec x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$$ by continuity (and for small $\varepsilon^{\prime}$ these points are once again contained in $X$). Now consider the line segment between $x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) $ and $ x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$ and use continuity and the fact that $X$ is convex.

Now suppose $y$ and $y^{\prime}$ are two other distinct points in $X$ such that $y \sim y^{\prime}$. Without loss, let's label the points such that $x_{1} < x_{1}^{\prime}$ and $y_{1} < y_{1}^{\prime}$. We will show that $x^{\prime} - x$ and $y^{\prime} - y$ are parallel. If they are not, then we may assume that $(x^{\prime}_{2} - x_{2}) / (x_{1}^{\prime} - x_{1}) > (y^{\prime}_{2} - y_{2}) / (y_{1}^{\prime} - y_{1}) $ is satisfied (the argument is analogous for the other case).

By $A1$, $$x \sim x + \lambda_{x} (x^{\prime} - x) \Rightarrow y \sim y + \lambda_{x} (x^{\prime} - x)$$ for all $\lambda \in [0, 1]$. Also, $y \sim y + \lambda_{y} (y^{\prime} - y)$ for all $\lambda_{y} \in [0, 1]$ since $y\sim y^{\prime}$. All of these comparisons are well-defined as $X$ is convex. Choosing $\varepsilon > 0$ sufficiently small, these comparisons hold, in particular, for $\lambda_{x} = \varepsilon / (x_{1}^{\prime} - x_{1})$ and $\lambda_{y} = \varepsilon / (y_{1}^{\prime} - y_{1})$. (Choosing $\varepsilon$ sufficiently small is necessary to guarantee $\lambda_{x}, \lambda_{y} \in (0, 1)$.) Now we note that \begin{align*} \lambda_{x} (x^{\prime} - x) &= \varepsilon \begin{pmatrix} 1 \\ \frac{x_{2}^{\prime} - x_{2}}{x_{1}^{\prime}- x_{1}} \end{pmatrix}, \\ \text{and}\quad \lambda_{y} (y^{\prime} - y) &= \varepsilon \begin{pmatrix} 1 \\ \frac{y_{2}^{\prime} - y_{2}}{y_{1}^{\prime}- y_{1}} \end{pmatrix}. \end{align*} From our assumptions we then conclude that $\lambda_{x} (x^{\prime} - x) > \lambda_{y} (y^{\prime} - y)$. Given monotonicity, this contradicts the fact that $y\sim y + \lambda_{x} (x^{\prime} - x)$ and $y \sim y + \lambda_{y} (y^{\prime} - y)$.

So far we have thus shown that there is a vector $r$ such that $z \sim z^{\prime}$ only if $r \cdot z = r \cdot z^{\prime}$. To prove the converse, consider our initial points $x, x^{\prime}$ which we know to be equivalent and satisfy $r \cdot x = r \cdot x^{\prime}$. Now, if $r \cdot z = r \cdot z^{\prime}$, then travelling from $z$ to $z^{\prime}$ involves travelling in a line parallel to $x - x^{\prime}$. Indifference $z \sim z^{\prime}$ then follows from $A1$ and continuity.



One way to get away from a linear representation is to drop $A1$. For instance, if you replaced it with $x \sim y \Leftrightarrow x \sim x + \lambda (x - y)$ for all $\lambda \in \mathbb{R}$ (such that $x + \lambda (x - y) \in X$), then all utility representations may fail to be linear. Loosely speaking, if you only impose that each indifference curve is a hyperplane in $\mathbb{R}^{2}$, then restricting $X$ appropriately allows you to arrange indifference curves in a way such that they are not parallel and do not intersect. Of course, not restricting $X$ wouldn't work since nonparallel hyperplanes always intersect somewhere in $\mathbb{R}^{2}$.

Here is one such example: Let $X = (0, 1] \times [-1, 0]$ and let $u(x_{1}, x_{2}) = x_{2} / x_{1}$. In other words, the utility assigned to the point $x$ is the slope of the line segment that connects it to the origin. This utility function is strictly increasing in both arguments (since $x_{2} \leq 0 < x_{1}$ on $X$) and continuous. You can easily check that $$ \frac{x_{2}}{x_{1}} = \frac{y_{2}}{y_{1}} \Leftrightarrow \frac{x_{2}}{x_{1} } = \frac{x_{2} + \lambda (y_{2} - x_{2})}{x_{1}+ \lambda (y_{1} - x_{1})} \quad \forall \lambda $$ is satisfied, showing that indifference curves are hyperplanes. These hyperplanes are not parallel (by construction) and would only intersect at $(0, 0)$, but we removed the origin from $X$.

To see that this example violates $A1$ and linearity, consider the points $(0.5, -0.5)$ and $(-1, 1)$. Clearly, $- 0.5 / 0.5 = - 1 / 1$, i.e. $(0.5, -0.5) \sim (-1, 1)$. However, $$(0.5, -0.5 + 0.25) \succ (1, -1 + 0.25)$$ since $$(- 0.5 + 0.25) / 0.5 = - 0.5 > - 0.75 = (-1 + 0.25) / 1.$$

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  • $\begingroup$ Sorry but your third paragraph using A1 does not seem right in my limited intuition. Your third paragraph basically says that $x\sim 0.5x+0.5x'$; I don't know how to prove this. I guess you might try to prove it from $0.5x+0.5x\sim 0.5x+0.5x'$; however, first we have to notice that $0.5x$ is not necessarily ind to $0.5x'$. Additionally, $0.5x$ might not be in $X$. $\endgroup$ – High GPA Nov 15 '19 at 22:26
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    $\begingroup$ Sorry for being a little vague there; I added a longer explanation. The basic intuition is as follows: Imagine travelling from $x$ to $x^{\prime}$ via a finite number of equidistant steps in the direction $x^{\prime} - x$. Since the endpoints of that path are equivalent, $A1$ and transitivity are sufficient to conclude that all points along the path are equivalent. By continuity, choosing the steps of such a path sufficiently small implies that any points inbetween $x$ and $x^{\prime}$ must be equivalent. $\endgroup$ – induction_is_a_laddah Nov 16 '19 at 2:38
  • $\begingroup$ Thank you so much for your detailed answer! You are a very good teacher to me! Last question: would semicontinuity be enough to generate the proof in your Step 1? $\endgroup$ – High GPA Nov 16 '19 at 14:48
  • $\begingroup$ By the way, A2 is "monotonicity" and I think you used in step 2; but I am not saying that you are wrong: I agree that A2 can be dropped without affecting the result. $\endgroup$ – High GPA Nov 16 '19 at 14:54

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