<p>Concave utility functions are quasiconcave while convex utility functions are quasiconvex.</p>
<p>If <span class="math-container">$U()$</span> is convex then <span class="math-container">$-U()$</span> is concave and represents the 'opposite' preferences, so if you believe the first half of the above statement you can easily show that the second part is true.</p>
<hr />
<p>A function is both concave and convex if and only if
<span class="math-container">$$
\alpha f(x) + (1 - \alpha) f(y) = f\left( \alpha x + (1 - \alpha) y \right)
$$</span>
for <span class="math-container">$\alpha \in [0,1]$</span> and all <span class="math-container">$x,y$</span>.</p>
<p>E.g.; <em>linear</em> functions have this property.</p>
<hr />
<p>Because of the above, linear utility functions (but not only them) will be both quasiconvex and quasiconcave.</p>

CC BY-SA 4.0
2023-11-29T07:28:15.770Z
I showed that max is a convex function. So it means that it is not quasiconcave and preference relation is not convex. Correct?
Paul R
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2023-11-29T07:37:27.507Z
What are quasiconvex/quasiconcave preferences.
Michael Greinecker
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2023-11-29T07:48:24.443Z
Not preference, utility function.
Paul R
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2023-11-29T07:52:55.100Z
@MichaelGreinecker Oops, I don't know what I was thinking. Thanks!
Giskard
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2023-11-29T08:07:14.027Z
@Giskard The property you mentioned is not true for $f(x,y)=(x+y)^2$. Consider $\alpha=\frac{1}{2}$, $(x',y')=(0,0)$ and $(x'',y'')=(2,2)$. $\alpha f(x',y')+(1-\alpha)f(x'',y'')=8 > 4 = f(\alpha (x',y')+(1-\alpha) (x'',y''))$.
Amit
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2023-11-29T09:02:44.043Z
@Amit I've been out of the game for too long :/ Thank you!
Giskard
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2023-11-29T09:18:24.520Z
@Giskard Welcome back to the game :)
Amit
CC BY-SA 4.0
2023-11-29T07:09:49.710Z
2023-12-06T07:59:27.837Z
2023-12-06T07:59:27.837Z
Giskard
<p>It is known that if a utility function is concave, then it is quasiconcave, and the preference relation is convex.</p>
<p>What can be said if a utility function is convex?</p>
<p>I've found on the Internet then in this case the function can be both quasiconcave and quasiconvex (how is this possible? linear function?) and is in this case preference relation concave?</p>
<p>I've managed to show that the utility function <span class="math-container">$u(x_1,x_2)=\operatorname{max}(x_1,x_2)$</span> is convex. What can be said about the preference relation?</p>

<microeconomics><utility>
If utility function is convex, what can be said about preference relation?
Paul R
# If utility function is convex, what can be said about preference relation?

(Asked by: username_0 on 2023-11-29)

It is known that if a utility function is concave, then it is quasiconcave, and the preference relation is convex.

What can be said if a utility function is convex?

I've found on the Internet then in this case the function can be both quasiconcave and quasiconvex (how is this possible? linear function?) and is in this case preference relation concave?

I've managed to show that the utility function $u(x\_1,x\_2)=\operatorname{max}(x\_1,x\_2)$ is convex. What can be said about the preference relation?

**Tags:** microeconomics, utility


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## 1 answers

### Answer by username_1 on 2023-11-29. Score: 3 | Accepted answer

Concave utility functions are quasiconcave while convex utility functions are quasiconvex.

If $U()$ is convex then $-U()$ is concave and represents the 'opposite' preferences, so if you believe the first half of the above statement you can easily show that the second part is true.

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A function is both concave and convex if and only if
$$
\alpha f(x) + (1 - \alpha) f(y) = f\left( \alpha x + (1 - \alpha) y \right)
$$
for $\alpha \in [0,1]$ and all $x,y$.

E.g.; *linear* functions have this property.

---

Because of the above, linear utility functions (but not only them) will be both quasiconvex and quasiconcave.

> **Comments:**
> **username_0**: I showed that max is a convex function. So it means that it is not quasiconcave and preference relation is not convex. Correct?  
> **username_2**: What are quasiconvex/quasiconcave preferences.  
> **username_0**: Not preference, utility function.  
> **username_1**: @MichaelGreinecker Oops, I don't know what I was thinking. Thanks!  
> **username_3**: @username_1 The property you mentioned is not true for $f(x,y)=(x+y)^2$. Consider $\alpha=\frac{1}{2}$, $(x',y')=(0,0)$ and $(x'',y'')=(2,2)$. $\alpha f(x',y')+(1-\alpha)f(x'',y'')=8 > 4 = f(\alpha (x',y')+(1-\alpha) (x'',y''))$.  
> **username_1**: @username_3 I've been out of the game for too long :/ Thank you!  
> **username_3**: @username_1 Welcome back to the game :)  

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