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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?
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  • $\begingroup$ 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. $\endgroup$
    – Herr K.
    Jun 16 at 16:04
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    $\begingroup$ @HerrK. I have made the required changes. I hope that suffices. $\endgroup$
    – Kur_Kush
    Jun 16 at 18:34
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    $\begingroup$ @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. $\endgroup$
    – Amit
    Jun 16 at 18:40
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    $\begingroup$ If you are still interested in the edited out question: If preferences are complete, transitive, and continuous, then both definitions are equivalent. $\endgroup$ Jun 18 at 6:11
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    $\begingroup$ @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? $\endgroup$
    – Kur_Kush
    Jun 20 at 1:25

1 Answer 1

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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*}

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  • $\begingroup$ I tried constructing examples along the lines of these two, but I couldn't. Maybe they have to be constructed differently. $\endgroup$
    – Kur_Kush
    Jun 18 at 23:26
  • $\begingroup$ 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. $\endgroup$
    – Amit
    Jun 18 at 23:50
  • $\begingroup$ Can I please see the counterexample for the $X = \mathbb{R}^2$ case? $\endgroup$
    – Kur_Kush
    Jun 24 at 4:19

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