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

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?

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.