# Convexity of preferences (dissimilar definitions)

Varian's Intermediate Microeconomics describes convexity as $$\text{Given } x, y \in X: x \sim y \implies \forall t \in [0,1], tx + (1-t)y \succeq x,y$$

The other definition I read everywhere is: $$\text{Given } x, y \in X: x \succeq y \implies \forall t \in [0,1], tx + (1-t) y \succeq y$$

Consider $$X = \mathbb{R}^{2}$$. The first definition does not imply the second when we have, for example, $$U(x_1,x_2) = \begin{cases} 0 \text{ if } x_1 + x_2 = 1 \\ 2 \text{ if } x_1 + x_2 > 1 \\ 1 \text{ if } x_1 + x_2 < 1 \end{cases}$$.

Assume $$X = \mathbb{R}^2$$ and preferences are complete, transitive and strictly monotone. (Strict monotonicity is defined as $$y \geq x \ (y \neq x) \implies y \succ x$$ where $$(y_1, y_2) = y \geq x = (x_1, x_2)$$ means $$y_1 \geq x \land y_2 \geq x_2$$.) Following are two questions that I thought of and was unable to prove/disprove :

1. Can we have a preference relation that satisfies the three assumptions and the first definition of convexity but not the second one?
2. Can we have a utility function that describes a preference relation which satisfies the three assumptions and the first definition of convexity but not the second one?
• Can you be more specific about how your utility function satisfies the first definition but not the second? Your notations are also a little inconsistent: In the definitions, $x,y$ are (possibly) vectors, whereas in the utility function, $x,y$ are scalars. Jun 16 at 16:04
• @HerrK. I have made the required changes. I hope that suffices. Jun 16 at 18:34
• @Kur_Kush If preferences are complete, transitive and continuous then the utility representation exists, but the converse is not true. Example utility function $U$ given in the question above represents the discontinuous preference.
– Amit
Jun 16 at 18:40
• If you are still interested in the edited out question: If preferences are complete, transitive, and continuous, then both definitions are equivalent. Jun 18 at 6:11
• @Amit By Debreu's theorem, there exists continuous utility function. We have $u(x) > u(z)$, so every value in the interval $[u(z), u(x)]$, one of which happens to be $z = u(y)$, will be attained by the respective $z$-coordinate of the line segment connecting $x$ and $y$. Any idea how to do this without using utility functions, that is, just with the preference relation and from the definitions of continuity? Jun 20 at 1:25

For Q 1:

Let me give you a preference relation on $$\mathbb{R}^2_+$$

$$(x_1, y_1) \succsim (x_2, y_2)$$ if and only if $$(x_1^2 + y_1^2 > x_2^2 + y_2^2)$$ or $$(x_1^2 + y_1^2 = x_2^2 + y_2^2 \ \wedge x_1 \geq x_2)$$

This gives the following strict preference relation: $$(x_1, y_1) \succ (x_2, y_2)$$ if and only if $$(x_1^2 + y_1^2 > x_2^2 + y_2^2)$$ or $$(x_1^2 + y_1^2 = x_2^2 + y_2^2 \ \wedge x_1 > x_2)$$

and indifference relation: $$(x_1, y_1) \sim (x_2, y_2)$$ if and only if $$(x_1=x_2 \wedge y_1 = y_2)$$

Clearly, this preference is strictly monotone, 1's definition holds and 2nd does not hold.

Now you can try and construct one on $$\mathbb{R}^2$$ (there are many)

For Q 2:

Let me again give you a utility function defined on $$\mathbb{R}^2_+$$ that satisfy monotonicity, Varian's definition, but does not satisfy definition 2 of convexity. You can try and find an example for $$\mathbb{R}^2$$ yourself (there are many examples).

$$\begin{eqnarray*} u(x,y) = \begin{cases} x+ y & \text{if } x + y < 2 \\ 1 + x & \text{if } x + y = 2 \text{ and } x \geq 1 \\ 4 - x & \text{if } x + y = 2 \text{ and } x < 1 \\ x + y + 2 & \text{if } x + y > 2 \end{cases} \end{eqnarray*}$$

• I tried constructing examples along the lines of these two, but I couldn't. Maybe they have to be constructed differently. Jun 18 at 23:26
• No you don't need a different construction. I mean of course you can also do a different construction but the above method (with some modifications) will also work.
– Amit
Jun 18 at 23:50
• Can I please see the counterexample for the $X = \mathbb{R}^2$ case? Jun 24 at 4:19