<p>If you have a single good, so that your commodity space is <span class="math-container">$\mathbb{R}$</span>, then every increasing function is quasi-concave and even strictly quasi-concave. So any non-concave but increasing function from <span class="math-container">$\mathbb{R}$</span> to <span class="math-container">$\mathbb{R}$</span> will give you the desired counterexample.</p>

CC BY-SA 4.0
2022-02-17T07:48:34.007Z
2022-02-17T07:48:34.007Z
Michael Greinecker
<p>For examples of real-valued function defined on subsets of <span class="math-container">$\mathbb{R}^2$</span>, consider:</p>
<ul>
<li><p><strong>Cobb Douglas</strong>: <span class="math-container">$u:\mathbb{R}^2_{++}\rightarrow\mathbb{R}$</span>, where <span class="math-container">$\mathbb{R}^2_{++}=\{(x,y)\in\mathbb{R}^2|x&gt;0, y&gt;0\}$</span> defined as <span class="math-container">$u(x, y) = x^\alpha y^\beta$</span> where <span class="math-container">$\alpha &gt; 0, \beta &gt; 0$</span>, and <span class="math-container">$\alpha+\beta&gt; 1$</span> is strictly quasiconcave but not concave.</p>
</li>
<li><p>Transformation of <strong>Quasi-linear</strong>: <span class="math-container">$u:\mathbb{R}^2_{+}\rightarrow\mathbb{R}$</span>, where <span class="math-container">$\mathbb{R}^2_{+}=\{(x,y)\in\mathbb{R}^2|x\geq 0, y\geq 0\}$</span> defined as <span class="math-container">$u(x, y) = (\alpha x + \beta \sqrt{y})^{\gamma} $</span> where <span class="math-container">$\alpha &gt; 0, \beta &gt; 0$</span>, and <span class="math-container">$\gamma&gt; 1$</span> is strictly quasiconcave but not concave.</p>
</li>
</ul>
<p>In fact, every strictly concave function is strictly quasi-concave, and every (strictly) increasing transformation of a strictly quasi-concave function is also strictly quasi-concave. Therefore, every (strictly) increasing transformation of a strictly concave function is also strictly quasi-concave, but the converse is not true. In this way you can take any strictly concave function and consider an appropriate strictly increasing transformation of the function so that the transformation of the function is not strictly concave.</p>

CC BY-SA 4.0
2022-05-22T02:16:17.360Z
2022-05-22T05:08:25.330Z
2022-05-22T05:08:25.330Z
Amit
<p>If we have complete, transitive, continuous, and strictly convex preferences, then all the utility functions that represent them are strictly concave. I know that strictly convex preferences are represented by a strictly <strong>quasi-concave</strong> utility, but I do not know how to prove that the utility function can be quasi-concave but does not necessarily have to be concave.</p>

CC BY-SA 4.0
2022-02-17T04:41:34.463Z
I think your first sentence is wrong. In particular, strict convexity of preference implies strict *quasi*concavity of utility and not necessarily the concavity of it.
Herr K.
CC BY-SA 4.0
2022-02-17T05:23:48.690Z
Thanks! Maybe it was my poor english, but my question is how can I show that the strict convexity implies strict quasiconcavity and not necessarily the concavity of it?
Giani
<mathematical-economics>
What it is a utility function that it is quasi-concave but not concave?
Giani
# What it is a utility function that it is quasi-concave but not concave?

(Asked by: username_0 on 2022-02-17)

If we have complete, transitive, continuous, and strictly convex preferences, then all the utility functions that represent them are strictly concave. I know that strictly convex preferences are represented by a strictly **quasi-concave** utility, but I do not know how to prove that the utility function can be quasi-concave but does not necessarily have to be concave.

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: I think your first sentence is wrong. In particular, strict convexity of preference implies strict *quasi*concavity of utility and not necessarily the concavity of it.  
> **username_0**: Thanks! Maybe it was my poor english, but my question is how can I show that the strict convexity implies strict quasiconcavity and not necessarily the concavity of it?  

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## 2 answers

### Answer by username_2 on 2022-02-17. Score: 11

If you have a single good, so that your commodity space is $\mathbb{R}$, then every increasing function is quasi-concave and even strictly quasi-concave. So any non-concave but increasing function from $\mathbb{R}$ to $\mathbb{R}$ will give you the desired counterexample.


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### Answer by username_3 on 2022-05-22. Score: 1

For examples of real-valued function defined on subsets of $\mathbb{R}^2$, consider:

* **Cobb Douglas**: $u:\mathbb{R}^2\_{++}\rightarrow\mathbb{R}$, where $\mathbb{R}^2\_{++}=\{(x,y)\in\mathbb{R}^2|x>0, y>0\}$ defined as $u(x, y) = x^\alpha y^\beta$ where $\alpha > 0, \beta > 0$, and $\alpha+\beta> 1$ is strictly quasiconcave but not concave.
* Transformation of **Quasi-linear**: $u:\mathbb{R}^2\_{+}\rightarrow\mathbb{R}$, where $\mathbb{R}^2\_{+}=\{(x,y)\in\mathbb{R}^2|x\geq 0, y\geq 0\}$ defined as $u(x, y) = (\alpha x + \beta \sqrt{y})^{\gamma} $ where $\alpha > 0, \beta > 0$, and $\gamma> 1$ is strictly quasiconcave but not concave.

In fact, every strictly concave function is strictly quasi-concave, and every (strictly) increasing transformation of a strictly quasi-concave function is also strictly quasi-concave. Therefore, every (strictly) increasing transformation of a strictly concave function is also strictly quasi-concave, but the converse is not true. In this way you can take any strictly concave function and consider an appropriate strictly increasing transformation of the function so that the transformation of the function is not strictly concave.


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