<p>The following is slightly adapted from</p>
<blockquote>
<p>Mas-Colell, Andreu. &quot;Pareto optima and equilibria: the finite
dimensional case.&quot; <em>Advances in equilibrium theory</em>. Springer, Berlin,
Heidelberg, 1985. 25-42.</p>
</blockquote>
<p>Consider an exchange economy in which everyone has the consumption set <span class="math-container">$\mathbb{R}^l_+$</span>, the aggregate endowment is <span class="math-container">$\omega$</span>, and free disposal is possible. By abuse of notation, we can use <span class="math-container">$\leq$</span> both for the coordiantewise order of vectors and the usual order on the real line.</p>
<p>The feasible set is
<span class="math-container">$$X^*=\{(x_1,\ldots,x_N)\in \mathbb{R}_+^{lN}\mid x_1+\cdots+x_n\leq\omega\}.$$</span>
If every agent <span class="math-container">$i$</span> has a monotone, continuous, and concave utility function <span class="math-container">$u_i$</span> such that <span class="math-container">$u_i(0)=0$</span>, then the utility possibility set
<span class="math-container">$$U=\Big\{\big(u_1(x_1),\ldots,u_n(x_N)\big)\mid (x_1,\ldots,x_N)\in X^*\Big\}$$</span>
is convex. For notation, if <span class="math-container">$x=(x_1,\ldots,x_N)\in X^*$</span>, we write <span class="math-container">$u(x)$</span> for <span class="math-container">$\big(u_1(x_1),\ldots,u_N(x_N)\big)$</span>.</p>
<p>To see that <span class="math-container">$U$</span> is convex, note that first that <span class="math-container">$X^*$</span> is convex and for every allocation <span class="math-container">$x\in X^*$</span>, if <span class="math-container">$0\leq x'\leq x$</span>, then <span class="math-container">$x'\in X$</span>. Now, let <span class="math-container">$u,u'\in U$</span> and <span class="math-container">$\alpha\in [0,1]$</span>. There are <span class="math-container">$x,x'\in X^*$</span> such that <span class="math-container">$u(x)=u$</span> and <span class="math-container">$u(x')=u'$</span>. Since all utility functions are concave, we have <span class="math-container">$u(\alpha x+(1-\alpha)x')\geq\alpha u+(1-\alpha)u'$</span>. Since each utility function is monotone and has the value <span class="math-container">$0$</span> at <span class="math-container">$0$</span>, there exist a vector <span class="math-container">$x''$</span> satisfying <span class="math-container">$0\leq x''\leq\alpha x+(1-\alpha)x'$</span> such that
<span class="math-container">$u(x'')=\alpha u+(1-\alpha)u'$</span> and <span class="math-container">$x''\in X^*$</span>.</p>

CC BY-SA 4.0
2022-05-29T20:16:47.757Z
Thank you very much for the insightful answer! I notice that in this setting, the utility of each agent only depends on his consumption, not on any other's consumption. Is there any result about the more general case that the utility of each agent depends on the consumption of everyone?
RPG
CC BY-SA 4.0
2022-05-29T20:27:18.083Z
What is really relevant here are three things: 1. The feasible set is convex. 2. All utility functions are concave. 3. There is "free disposal in utility," if $u\in U$ and $u'\leq U$ (or $0\leq u'\leq u$ or something similar), then $u'\in U$. The fundamental problem is that the graph of a concave function is not concave, but its [hypograph](https://en.wikipedia.org/wiki/Hypograph_(mathematics)) is.
Michael Greinecker
CC BY-SA 4.0
2022-05-29T19:40:30.623Z
2022-05-29T19:40:30.623Z
Michael Greinecker
<p>Concavity and non-decreasing is not enough. Consider <span class="math-container">$X = [0,4]$</span>. <span class="math-container">$u_1(x) = x$</span> and <span class="math-container">$u_2(x) = \sqrt{x}$</span>. In this case, we get <span class="math-container">$U = \{(x,\sqrt{x})|x\in [0,4]\}$</span> which is not a convex set.</p>
<p>If all <span class="math-container">$u_i$</span>s are linear, we'll get the convexity of <span class="math-container">$U$</span>.</p>

CC BY-SA 4.0
2022-05-29T17:59:46.450Z
2022-05-29T17:59:46.450Z
Amit
<p>given a convex consumption set <span class="math-container">$X$</span>, and utility function of <span class="math-container">$N$</span> agents: <span class="math-container">$u_1,\cdots,u_N$</span>, under what condition, the utility set <span class="math-container">$U=\{u\in\mathbb{R}^N:\exists x\in X s.t. u_i(x)=u_i\forall i\}$</span> is convex?</p>
<p>Is it enough to let <span class="math-container">$u_i$</span> be concave and non-decreasing function? Thank you in advance!</p>

CC BY-SA 4.0
2022-05-29T18:28:12.943Z
Should $X$ be the set of allocations?
Michael Greinecker
CC BY-SA 4.0
2022-05-29T18:41:19.770Z
@MichaelGreinecker Yes, I am actually considering a resource allocation problem.
RPG
CC BY-SA 4.0
2022-05-30T03:23:32.050Z
@RPG I think you can edit your question and specify that $X$ is set of allocations. It is not the same thing as consumption set. Also, redefine the utility possibility set accordingly.
Amit
<utility>
What conditions imply convex utility set?
RPG
# What conditions imply convex utility set?

(Asked by: username_0 on 2022-05-29)

given a convex consumption set $X$, and utility function of $N$ agents: $u\_1,\cdots,u\_N$, under what condition, the utility set $U=\{u\in\mathbb{R}^N:\exists x\in X s.t. u\_i(x)=u\_i\forall i\}$ is convex?

Is it enough to let $u\_i$ be concave and non-decreasing function? Thank you in advance!

**Tags:** utility

> **Comments:**
> **username_1**: Should $X$ be the set of allocations?  
> **username_0**: @MichaelGreinecker Yes, I am actually considering a resource allocation problem.  
> **username_2**: @username_0 I think you can edit your question and specify that $X$ is set of allocations. It is not the same thing as consumption set. Also, redefine the utility possibility set accordingly.  

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## 2 answers

### Answer by username_1 on 2022-05-29. Score: 3

The following is slightly adapted from

> Mas-Colell, Andreu. "Pareto optima and equilibria: the finite
> dimensional case." *Advances in equilibrium theory*. Springer, Berlin,
> Heidelberg, 1985. 25-42.

Consider an exchange economy in which everyone has the consumption set $\mathbb{R}^l\_+$, the aggregate endowment is $\omega$, and free disposal is possible. By abuse of notation, we can use $\leq$ both for the coordiantewise order of vectors and the usual order on the real line.

The feasible set is
$$X^\*=\{(x\_1,\ldots,x\_N)\in \mathbb{R}\_+^{lN}\mid x\_1+\cdots+x\_n\leq\omega\}.$$
If every agent $i$ has a monotone, continuous, and concave utility function $u\_i$ such that $u\_i(0)=0$, then the utility possibility set
$$U=\Big\{\big(u\_1(x\_1),\ldots,u\_n(x\_N)\big)\mid (x\_1,\ldots,x\_N)\in X^\*\Big\}$$
is convex. For notation, if $x=(x\_1,\ldots,x\_N)\in X^\*$, we write $u(x)$ for $\big(u\_1(x\_1),\ldots,u\_N(x\_N)\big)$.

To see that $U$ is convex, note that first that $X^\*$ is convex and for every allocation $x\in X^\*$, if $0\leq x'\leq x$, then $x'\in X$. Now, let $u,u'\in U$ and $\alpha\in [0,1]$. There are $x,x'\in X^\*$ such that $u(x)=u$ and $u(x')=u'$. Since all utility functions are concave, we have $u(\alpha x+(1-\alpha)x')\geq\alpha u+(1-\alpha)u'$. Since each utility function is monotone and has the value $0$ at $0$, there exist a vector $x''$ satisfying $0\leq x''\leq\alpha x+(1-\alpha)x'$ such that
$u(x'')=\alpha u+(1-\alpha)u'$ and $x''\in X^\*$.

> **Comments:**
> **username_0**: Thank you very much for the insightful answer! I notice that in this setting, the utility of each agent only depends on his consumption, not on any other's consumption. Is there any result about the more general case that the utility of each agent depends on the consumption of everyone?  
> **username_1**: What is really relevant here are three things: 1. The feasible set is convex. 2. All utility functions are concave. 3. There is "free disposal in utility," if $u\in U$ and $u'\leq U$ (or $0\leq u'\leq u$ or something similar), then $u'\in U$. The fundamental problem is that the graph of a concave function is not concave, but its [hypograph](https://en.wikipedia.org/wiki/Hypograph_(mathematics)) is.  

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### Answer by username_2 on 2022-05-29. Score: 2

Concavity and non-decreasing is not enough. Consider $X = [0,4]$. $u\_1(x) = x$ and $u\_2(x) = \sqrt{x}$. In this case, we get $U = \{(x,\sqrt{x})|x\in [0,4]\}$ which is not a convex set.

If all $u\_i$s are linear, we'll get the convexity of $U$.


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