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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!

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  • $\begingroup$ Should $X$ be the set of allocations? $\endgroup$ May 29, 2022 at 18:28
  • $\begingroup$ @MichaelGreinecker Yes, I am actually considering a resource allocation problem. $\endgroup$
    – RPG
    May 29, 2022 at 18:41
  • $\begingroup$ @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. $\endgroup$
    – Amit
    May 30, 2022 at 3:23

2 Answers 2

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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 $x''\in X^*$.

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  • $\begingroup$ 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? $\endgroup$
    – RPG
    May 29, 2022 at 20:16
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    $\begingroup$ 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 is. $\endgroup$ May 29, 2022 at 20:27
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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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