<p>It's well known that a convex preference implies quasiconcave utility functions. Since quasiconcavity need not imply concavity, it's easy to find examples of a non-concave utility function representing a convex preference.</p>
<p>For example: <span class="math-container">$u(x,y)=(x+y)^3$</span>. The preference this function represents is convex (though not strictly so), as can be seen from its linear indifference curves. The function is quasiconcave, as evidenced by the convex upper contour sets. Lastly, the function is not concave, as betrayed by the exponent.</p>

CC BY-SA 4.0
2020-08-30T09:44:23.827Z
Thanks! A follow up question: Let's say the utility function is not even quasiconcave, is using Lagrangian to maximize the function possible?
megg
CC BY-SA 4.0
2020-08-30T17:14:30.407Z
@megg: What exactly do you mean by not quasiconcave? Strictly convex? Or neither quasiconcave nor quasiconvex? At any rate, it may be worth asking a separate question.
Herr K.
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2020-08-31T13:51:08.517Z
@Tomcat: Yes, you're right: a utility function is quasiconcave iff the preference it represents is convex. But regarding your reinterpretation of OP's question: $(x+y)^3$ is both convex and quasiconcave on $\mathbb R^2_+$. Likewise in $\mathbb R$, any convex function that is also monotonic, e.g. $\mathrm e^x$, is also quasiconcave (as well as quasiconvex).
Herr K.
CC BY-SA 4.0
2020-08-30T05:01:34.290Z
2020-08-30T05:01:34.290Z
Herr K.
<p>Example given by Herr K. is perfect. Let me give another example of a dis-continuous utility function which is quasi-concave, but not concave.</p>
<p>Consider <span class="math-container">$u:\mathbb{R}^2_+ \rightarrow \mathbb{R}$</span> defined as follows
<span class="math-container">$u(x, y) = \lfloor x\rfloor$</span>, where <span class="math-container">$\lfloor x \rfloor$</span> is the greatest integer less than or equal to <span class="math-container">$x$</span>.</p>
<p><span class="math-container">$u$</span> is quasi-concave because upper level sets will be of the type <span class="math-container">$[n,\infty) \times \mathbb{R}_+$</span>, where <span class="math-container">$n\in \mathbb{Z}_+$</span>, which is a convex set.</p>
<p>It is not concave because it is discontinuous at some interior points in the domain of <span class="math-container">$u$</span>.</p>

CC BY-SA 4.0
2022-05-05T13:22:11.477Z
2022-05-05T13:47:35.167Z
2022-05-05T13:47:35.167Z
Amit
<p>Can preference be convex when utility is not a concave function (e.g. <span class="math-container">$U=x_1^2 + x_2^2$</span>)?</p>

<microeconomics><utility><preferences>
Convex Preference but Convex Utility
megg
# Convex Preference but Convex Utility

(Asked by: username_0 on 2020-08-30)

Can preference be convex when utility is not a concave function (e.g. $U=x\_1^2 + x\_2^2$)?

**Tags:** microeconomics, utility, preferences


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## 2 answers

### Answer by username_1 on 2020-08-30. Score: 7

It's well known that a convex preference implies quasiconcave utility functions. Since quasiconcavity need not imply concavity, it's easy to find examples of a non-concave utility function representing a convex preference.

For example: $u(x,y)=(x+y)^3$. The preference this function represents is convex (though not strictly so), as can be seen from its linear indifference curves. The function is quasiconcave, as evidenced by the convex upper contour sets. Lastly, the function is not concave, as betrayed by the exponent.

> **Comments:**
> **username_0**: Thanks! A follow up question: Let's say the utility function is not even quasiconcave, is using Lagrangian to maximize the function possible?  
> **username_1**: @username_0: What exactly do you mean by not quasiconcave? Strictly convex? Or neither quasiconcave nor quasiconvex? At any rate, it may be worth asking a separate question.  
> **username_1**: @Tomcat: Yes, you're right: a utility function is quasiconcave iff the preference it represents is convex. But regarding your reinterpretation of OP's question: $(x+y)^3$ is both convex and quasiconcave on $\mathbb R^2_+$. Likewise in $\mathbb R$, any convex function that is also monotonic, e.g. $\mathrm e^x$, is also quasiconcave (as well as quasiconvex).  

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### Answer by username_2 on 2022-05-05. Score: 2

Example given by username_1. is perfect. Let me give another example of a dis-continuous utility function which is quasi-concave, but not concave.

Consider $u:\mathbb{R}^2\_+ \rightarrow \mathbb{R}$ defined as follows
$u(x, y) = \lfloor x\rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.

$u$ is quasi-concave because upper level sets will be of the type $[n,\infty) \times \mathbb{R}\_+$, where $n\in \mathbb{Z}\_+$, which is a convex set.

It is not concave because it is discontinuous at some interior points in the domain of $u$.


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