<p><strong>For Q 1</strong>:</p>
<p>Let me give you a preference relation on <span class="math-container">$\mathbb{R}^2_+$</span></p>
<p><span class="math-container">$(x_1, y_1) \succsim (x_2, y_2)$</span> if and only if
<span class="math-container">$(x_1^2 + y_1^2 &gt; x_2^2 + y_2^2)$</span> or <span class="math-container">$(x_1^2 + y_1^2 = x_2^2 + y_2^2 \ \wedge x_1 \geq x_2)$</span></p>
<p>This gives the following strict preference relation:
<span class="math-container">$(x_1, y_1) \succ (x_2, y_2)$</span> if and only if
<span class="math-container">$(x_1^2 + y_1^2 &gt; x_2^2 + y_2^2)$</span> or <span class="math-container">$(x_1^2 + y_1^2 = x_2^2 + y_2^2 \ \wedge x_1 &gt; x_2)$</span></p>
<p>and indifference relation:
<span class="math-container">$(x_1, y_1) \sim (x_2, y_2)$</span> if and only if <span class="math-container">$(x_1=x_2 \wedge y_1 = y_2)$</span></p>
<p>Clearly, this preference is strictly monotone, 1's definition holds and 2nd does not hold.</p>
<p>Now you can try and construct one on <span class="math-container">$\mathbb{R}^2$</span> (there are many)</p>
<p><strong>For Q 2</strong>:</p>
<p>Let me again give you a utility function defined on <span class="math-container">$\mathbb{R}^2_+$</span> that satisfy monotonicity, Varian's definition, but does not satisfy definition 2 of convexity. You can try and find an example for <span class="math-container">$\mathbb{R}^2$</span> yourself (there are many examples).</p>
<p><span class="math-container">\begin{eqnarray*} u(x,y) = \begin{cases} x+ y &amp; \text{if } x + y &lt; 2 \\ 1 + x &amp; \text{if } x + y = 2 \text{ and } x \geq 1 \\ 4 - x &amp; \text{if } x + y = 2 \text{ and } x &lt; 1 \\ x + y + 2 &amp; \text{if } x + y &gt; 2 \end{cases} \end{eqnarray*}</span></p>

CC BY-SA 4.0
2022-06-18T23:26:28.477Z
I tried constructing examples along the lines of these two, but I couldn't. Maybe they have to be constructed differently.
Kur_Kush
CC BY-SA 4.0
2022-06-18T23:50:04.047Z
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
CC BY-SA 4.0
2022-06-24T04:19:44.577Z
Can I please see the counterexample for the $X = \mathbb{R}^2$ case?
Kur_Kush
CC BY-SA 4.0
2022-06-16T16:21:26.670Z
2022-06-18T01:56:18.300Z
2022-06-18T01:56:18.300Z
Amit
<p>Varian's Intermediate Microeconomics describes convexity as <span class="math-container">$$\text{Given } x, y \in X: x \sim y \implies \forall t \in [0,1], tx + (1-t)y \succeq x,y$$</span></p>
<p>The other definition I read everywhere is: <span class="math-container">$$\text{Given } x, y \in X: x \succeq y \implies \forall t \in [0,1], tx + (1-t) y \succeq y$$</span></p>
<p>Consider <span class="math-container">$X = \mathbb{R}^{2}$</span>. The first definition does not imply the second when we have, for example, <span class="math-container">$U(x_1,x_2) = \begin{cases} 0 \text{ if } x_1 + x_2 = 1 \\ 2 \text{ if } x_1 + x_2 &gt; 1 \\ 1 \text{ if } x_1 + x_2 &lt; 1 \end{cases}$</span>.</p>
<p><strong>Assume <span class="math-container">$X = \mathbb{R}^2$</span> and preferences are complete, transitive and strictly monotone.</strong> (Strict monotonicity is defined as <span class="math-container">$y \geq x \ (y \neq x) \implies y \succ x$</span> where <span class="math-container">$(y_1, y_2) = y \geq x = (x_1, x_2)$</span> means <span class="math-container">$y_1 \geq x \land y_2 \geq x_2$</span>.) <strong>Following are two questions that I thought of and was unable to prove/disprove :</strong></p>
<ol>
<li>Can we have a preference relation that satisfies the three assumptions and the first definition of convexity but not the second one?</li>
<li>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?</li>
</ol>

CC BY-SA 4.0
2022-06-16T16:04:47.087Z
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.
Herr K.
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2022-06-16T16:46:54.673Z
In the second question, what is the forth assumption: utility representation or continuity of preference?
Amit
CC BY-SA 4.0
2022-06-16T18:34:21.320Z
@HerrK. I have made the required changes. I hope that suffices.
Kur_Kush
CC BY-SA 4.0
2022-06-16T18:35:59.870Z
@Amit The fourth assumption is the existence of utility representation. I read that utility representation exists iff preferences are complete, transitive and continuous. Since the first two assumptions are already there, I added the fourth assumption (continuity of preferences) for 2.
Kur_Kush
CC BY-SA 4.0
2022-06-16T18:40:23.843Z
@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
CC BY-SA 4.0
2022-06-16T19:07:37.737Z
@Amit Okay, that's right. I have made the required changes.
Kur_Kush
CC BY-SA 4.0
2022-06-17T02:57:52.457Z
@Kur_Kush Double implication in the first definition is not very clear. Do you mean: [1] $(\forall x, y)(\forall t)(x\sim y \iff tx+(1-t)y\succsim x, y)$ or [2] $(\forall x, y)(x\sim y \iff (\forall t) tx+(1-t)y\succsim x, y)$ ?
Amit
CC BY-SA 4.0
2022-06-17T09:59:26.437Z
@Amit Thanks for pointing it out; it's $\implies$ and not $\iff$.
Kur_Kush
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2022-06-18T02:09:23.470Z
@Kur_Kush I have updated the answer with the example of a utility function with the desired properties.
Amit
CC BY-SA 4.0
2022-06-18T06:11:21.747Z
If you are still interested in the edited out question: If preferences are complete, transitive, and continuous, then both definitions are equivalent.
Michael Greinecker
CC BY-SA 4.0
2022-06-18T23:14:13.127Z
@MichaelGreinecker Yes, please! I initially thought that might be possible, but I couldn't come up with a proof and gave up.
Kur_Kush
CC BY-SA 4.0
2022-06-18T23:35:00.200Z
@Kur_Kush You can try to prove it. Let me tell you one way to approach it: You can proceed by contradiction. Suppose $x\succsim y$, and there exists $t_1$ such that $y \succ t_1x +(1-t_1)y$. Then by continuity, you can find $t_2<t_1$, such that $y \sim t_2x +(1-t_2)y$. This is a contradiction.
Amit
CC BY-SA 4.0
2022-06-19T02:28:56.413Z
@Amit How is $y \sim t_2 x + (1-t_2) y$ by continuity?
Kur_Kush
CC BY-SA 4.0
2022-06-19T02:36:03.347Z
@Kur_Kush The argument to show that such a $t_2$ exists is standard. I have already given you the entire proof. You just need do connect the pieces. Take that as an exercise.
Amit
CC BY-SA 4.0
2022-06-20T01:25:24.990Z
@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?
Kur_Kush
CC BY-SA 4.0
2022-06-20T01:29:47.930Z
@Kur_Kush Of course you can do that without using the utility function. That is the exercise.
Amit
CC BY-SA 4.0
2022-06-24T04:17:19.037Z
Update: I was able to show without using utility functions (or Debreu and IVT) that the two definitions are equivalent when continuity (rather than monotonicity) is assumed.
Kur_Kush
<microeconomics><mathematical-economics><consumer-theory><preferences><choice-theory>
Convexity of preferences (dissimilar definitions)
Kur_Kush
# Convexity of preferences (dissimilar definitions)

(Asked by: username_0 on 2022-06-16)

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?

**Tags:** microeconomics, mathematical-economics, consumer-theory, preferences, choice-theory
> **Comments:**
> **username_1**: 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.  
> **username_2**: In the second question, what is the forth assumption: utility representation or continuity of preference?  
> **username_0**: @HerrK. I have made the required changes. I hope that suffices.  
> **username_0**: @Amit The fourth assumption is the existence of utility representation. I read that utility representation exists iff preferences are complete, transitive and continuous. Since the first two assumptions are already there, I added the fourth assumption (continuity of preferences) for 2.  
> **username_2**: @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.  
> **username_0**: @Amit Okay, that's right. I have made the required changes.  
> **username_2**: @Kur_Kush Double implication in the first definition is not very clear. Do you mean: [1] $(\forall x, y)(\forall t)(x\sim y \iff tx+(1-t)y\succsim x, y)$ or [2] $(\forall x, y)(x\sim y \iff (\forall t) tx+(1-t)y\succsim x, y)$ ?  
> **username_0**: @Amit Thanks for pointing it out; it's $\implies$ and not $\iff$.  
> **username_2**: @Kur_Kush I have updated the answer with the example of a utility function with the desired properties.  
> **username_3**: If you are still interested in the edited out question: If preferences are complete, transitive, and continuous, then both definitions are equivalent.  
> **username_0**: @MichaelGreinecker Yes, please! I initially thought that might be possible, but I couldn't come up with a proof and gave up.  
> **username_2**: @Kur_Kush You can try to prove it. Let me tell you one way to approach it: You can proceed by contradiction. Suppose $x\succsim y$, and there exists $t_1$ such that $y \succ t_1x +(1-t_1)y$. Then by continuity, you can find $t_2<t_1$, such that $y \sim t_2x +(1-t_2)y$. This is a contradiction.  
> **username_0**: @Amit How is $y \sim t_2 x + (1-t_2) y$ by continuity?  
> **username_2**: @Kur_Kush The argument to show that such a $t_2$ exists is standard. I have already given you the entire proof. You just need do connect the pieces. Take that as an exercise.  
> **username_0**: @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?  
> **username_2**: @Kur_Kush Of course you can do that without using the utility function. That is the exercise.  
> **username_0**: Update: I was able to show without using utility functions (or Debreu and IVT) that the two definitions are equivalent when continuity (rather than monotonicity) is assumed.  

---

## 1 answers

### Answer by username_2 on 2022-06-16. Score: 3

**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\*}

> **Comments:**
> **username_0**: I tried constructing examples along the lines of these two, but I couldn't. Maybe they have to be constructed differently.  
> **username_2**: 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.  
> **username_0**: Can I please see the counterexample for the $X = \mathbb{R}^2$ case?  

---

