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

CC BY-SA 4.0
2023-12-12T00:42:01.230Z
2023-12-12T00:42:01.230Z
Michael Greinecker
<p>Given a state-space <span class="math-container">$X \subseteq \mathbb{R}^m$</span> and <span class="math-container">$n$</span> utility functions, <span class="math-container">$u_1,\ldots,u_n: X\to \mathbb{R}$</span>, define a state <span class="math-container">$x\in X$</span> as <em>Locally Pareto-efficient</em> if it has an open neighborhood <span class="math-container">$N(x)$</span> such that no state in <span class="math-container">$N(x)$</span> Pareto-dominates <span class="math-container">$x$</span>.</p>
<p><strong>What conditions on the functions <span class="math-container">$u_1,\ldots,u_n$</span> guarantee that any locally Pareto-efficient state is also (globally) Pareto-efficient?</strong></p>
<p>The simplest case is when <span class="math-container">$n=1$</span>, so there is only one utility function. In this case, a state <span class="math-container">$x$</span> is locally Pareto-efficient if and only if it is a local maximum of <span class="math-container">$u_1$</span>. It is well-known that, if <span class="math-container">$X$</span> is a convex set and <span class="math-container">$u_1$</span> is a concave function, then every local maximum of <span class="math-container">$u_1$</span> is a global maximum.</p>
<p>Based on this, my guess is that, if <span class="math-container">$X$</span> is a convex set and <span class="math-container">$u_1,\ldots,u_n$</span> are concave functions, then any locally Pareto-efficient state is also (globally) Pareto-efficient.</p>
<p>Is this correct?</p>

<mathematical-economics><pareto-efficiency>
When is local Pareto-efficiency equivalent to global Pareto-efficiency?
Erel Segal-Halevi
# When is local Pareto-efficiency equivalent to global Pareto-efficiency?

(Asked by: username_0 on 2023-12-11)

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?

**Tags:** mathematical-economics, pareto-efficiency

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## 1 answers

### Answer by username_1 on 2023-12-12. Score: 3 | Accepted answer

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.


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