When is local Pareto-efficiency equivalent to global Pareto-efficiency?

Question

Given a state-space $X \subseteq \mathbb{R}^m$ and $n$ utility functions, $u_1,\ldots,u_n: X\to \mathbb{R}$, define a state $x\in X$ as Locally Pareto-efficient if it has an open neighborhood $N(x)$ such that no state in $N(x)$ Pareto-dominates $x$.

What conditions on the functions $u_1,\ldots,u_n$ guarantee that any locally Pareto-efficient state is also (globally) Pareto-efficient?

The simplest case is when $n=1$, so there is only one utility function. In this case, a state $x$ is locally Pareto-efficient if and only if it is a local maximum of $u_1$. It is well-known that, if $X$ is a convex set and $u_1$ is a concave function, then every local maximum of $u_1$ is a global maximum.

Based on this, my guess is that, if $X$ is a convex set and $u_1,\ldots,u_n$ are concave functions, then any locally Pareto-efficient state is also (globally) Pareto-efficient.

Is this correct?

Answer 1

Yes. The proof is virtually the same. Let $x^*\in X$ be a Pareto dominated state. There must be some $x\in X$ such that $u_i(x)>u_i(x^*)$ for some agent $i$ and $u_j(x)\geq u_j(x^*)$ for every agent $j$. By concavity, we have for all $\alpha\in (0,1)$ that $u_i(\alpha x^*+(1-\alpha)x)>u_i(x^*)$ for the same agent $i$ and $u_j(\alpha x^*+(1-\alpha)x)\geq u_j(x^*)$ for every agent $j$. Since every neighborhood of $x^*$ contains a point of the form $\alpha x^*+(1-\alpha)x$, the state $x^*$ cannot be a local Pareto optimum.