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When studying analysis, I met with the problem of lim sup at the infinity. The limit superior of $f$ at $y$ is defined as:

$$\begin{aligned} \underset{x\to y}{\overline\lim} f(x) &= \inf_{\delta > 0}\ \sup f \restriction (X\cap B(\delta,y)-\{y\}) \\ &=\lim_{\delta \to 0^+}\; \sup f \restriction (X\cap B(\delta,y)-\{y\}). \end{aligned}$$

Then, I wonder whether $\displaystyle \underset{x\to \infty}{\overline\lim} f(x)$ should be defined as:

$$\begin{aligned} \underset{x\to \infty}{\overline\lim} f(x) &= \underset{k\to \infty}{\overline\lim}\; \sup_{j\ge k} f \restriction (X\cap \{x\ge j\}) \\ &= \inf_{k\to \infty}\; \sup_{j\ge k} f \restriction (X\cap \{x\ge j\}). \end{aligned}$$

I just wonder whether I made up right mathematical expression for limit superior of $f$ at infinity.

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What is $X$? Is it an arbitrary metric space? In that case, what does $\{x \ge j\}$ mean? –  Nate Eldredge Jan 11 '13 at 2:41
    
@NateEldredge We have not talked about the concept of metric space. In our context, $X$ is an unbounded subset of $\Bbb R ^m$. <br/>In word, I think the lim sup f at infinity(=L, say) should be: $\forall \varepsilon \gt 0, \exists j \gt 0,$ s.t. $\forall x\gt j,\ |\sup f(x) -L|\lt \varepsilon$. I have trouble in converting this into expression. –  Scorpio19891119 Jan 11 '13 at 3:00
    
If $X \subset \mathbb R^m$ for $m > 1$, how do you define $x \ge y$ for $x,y \in X$? –  Ilmari Karonen Jan 11 '13 at 5:15

1 Answer 1

You're thinking along the right lines, but unfortunately it appears you haven't covered the theory that lets you talk meaningfully about lim sup in the more general setting of topological spaces.

There are two things that go wrong in trying to apply the metric space definition to the problem:

  • $\infty$ is not a real number, and
  • even if it were, the $\delta$-ball around $\infty$ wouldn't make much sense.

The second problem is most easily fixed by moving the discussion to the realm of topological spaces, so since we have to do that, let's use this opportunity to introduce the extended reals, $\overline{\mathbb R} = [-\infty,\infty]$, i.e. the real line with a point added "infinitely far to the left" and another "infinitely far to the right". What are the open sets of this concoction? Well if $U \subseteq \mathbb R$ is open, then it's also open in the extended reals. And if $U \subseteq \overline{\mathbb R}$ contains $\infty$, then it's open when it's bounded below and when $U \cap \mathbb R_+$ is open in $\mathbb R$.

Next we need the topological definition of limit superior that only makes reference to the open sets. Following Wikipedia's article on limit superior and inferior in slightly less generality, we consider a function $f: E \to \mathbb R$, where $E \subseteq \overline{\mathbb R}$, and a limit point $a$ of $E$.

$\displaystyle\limsup_{x\to a} f(x) = \inf \{ \sup \{ f(x) : x \in E \cap U - \{a\} \} : U\ \text{open}, a \in U, E \cap U - \{a\} \neq \emptyset \}.$

And indeed one finds that everything works and when one unwinds the definition to the concrete situation you're in, one finds (for the case $E = (b,\infty)$) that one gets what you've written below (here I'm making some assumptions about what you mean with your notation, but it seems reasonable enough).

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