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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!
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 $x''\in X^*$.
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 $U$.
