Limit superior in $\mathbb{R}^{n}$

Question

Can we define the concept of limsup and liminf in $\mathbb{R}^n$, the Euclidean $n$ space? Like for real valued function with domain in $\mathbb{R^n}$?

Answer 1

As Anne says in the comments, to give an exact analogue to $\limsup$ and $\liminf$, you need a certain order structure (specifically a complete lattice). You could, in theory, define such an order on $\mathbb{R}^n$ for $n > 1$, but it wouldn’t likely “look like” a meaningful order.

However, if you’re willing to weaken things a bit, you can get something similar to $\limsup$ and $\liminf$ to $\mathbb{R}^n$.

Instead of generalizing $\liminf$ and $\limsup$ to $\mathbb{R}^n$, consider generalizing the interval with endpoints $(\liminf, \limsup)$. For a sequence $(x_j)$ in $\mathbb{R}^n$, we can easily analogize the limiting interval $[\liminf x_j, \limsup x_j]$ in $\mathbb{R}$ with a limiting bounding ball in $\mathbb{R}^n$. To be precise, if $B_r(y)$ denotes the closed ball in $\mathbb{R}^n$ with center $y$ and radius $r$, we could look at the collection $\mathcal{B}$ of all bounding balls $B_r(y)$ such that $(x_n)$ is eventually inside $B_r(y)$. Let $R$ denote the infimum of radii in $\mathcal{B}$.

Proposition: Suppose $B_{r_j}(y_j)$ is a sequence of balls in $\mathcal{B}$ such that $r_j \rightarrow R$. Then $y_j$ converges to some value $y_0$. Further, this limiting value $y_0$ is independent of the sequence of balls chosen.

This proof is annoying, but can be done with some work.

A consequence of this proposition is that there is a limiting ball! This limiting ball is exactly the $B_R(y_0)$, and our analogue of the interval with endpoints $\liminf$ and $\limsup$.

The choice of a ball here was arbitrary. You could instead boxes, or perhaps convex hulls of points.