# Linear Utility?

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.

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 $$(- 0.5 + 0.25) / 0.5 = - 0.5 > - 0.75 = (-1 + 0.25) / 1.$$

• 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 $X$. Nov 15 '19 at 22:26
• 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. Nov 16 '19 at 2:38
• 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? Nov 16 '19 at 14:48
• 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. Nov 16 '19 at 14:54