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To make it clear, this is the definition from wikipedia;

Let $X,Y$ be topological spaces and $E\subset X$. Let $Y$ be an ordered set and $f:E\rightarrow Y$ be a function.

Then, $\limsup_{x\to a} f(x) \triangleq \inf \{\sup \{f(x)\in Y|x\in U\cap E \setminus \{a\}\}\in Y|U \text{ is open}, U\cap E \setminus \{a\} ≠ \emptyset, a\in U\} \\ \liminf_{x\to a} f(x) \triangleq \sup \{\inf \{f(x)\in Y|x\in U\cap E \setminus \{a\}\}\in Y|U \text{ is open}, U\cap E \setminus \{a\} ≠ \emptyset, a\in U\}$

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Let $E\subset \mathbb{R}$ and $f:E\rightarrow \mathbb{R}$ be a function and $a$ be a limit point of $E$.

Then, it can be shown;

$\limsup_{x\to a} = \lim_{\epsilon\to 0} \sup\{f(x)\in \overline{\mathbb{R}}|x\in B(x,\epsilon)\cap E \setminus \{a\}\} \\ \liminf_{x\to a} = \lim_{\epsilon\to 0} \inf\{f(x)\in \overline{\mathbb{R}}|x\in B(x,\epsilon)\cap E \setminus \{a\}\}$.

(where $\overline{\mathbb{R}} = \mathbb{R} \cup \{+\infty,-\infty\}$)

Also, if $E$ is unbounded;

$\limsup_{x\to\infty}= \lim_{\epsilon\to\infty} \sup\{f(x)\in \overline{\mathbb{R}}| \epsilon < x\in E\} \\ \liminf_{x\to\infty}= \lim_{\epsilon\to\infty} \inf\{f(x)\in \overline{\mathbb{R}}| \epsilon < x\in E\}$.

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With this definition, what are useful properties of limit inferior and superior of $f:E\rightarrow \mathbb{R}$ where $E\subset \mathbb{R}$? So i can try to prove those properties :) (i.e. superadditivity)

(I'm asking this question, since i know there are many useful properties of limit inferior and superior of a sequence, so i think it has those properties too)(Since it seems it's a generalization of that of a sequence)

Till now, i have only shown that

$\limsup_{x\to a} f(x) = \liminf_{x\to a} f(x)=A$ iff $\lim_{x\to a} f(x)=A$.

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1 Answer 1

up vote 0 down vote accepted

You can reduce the function case to the sequence case by noting that $$ \limsup_{x\to a} f(x) = \sup_{\substack{(x_n) \in (E\setminus\{a\})^{\mathbb N}\\ x_n \to a}} \limsup_{n \to \infty} f(x_n). $$

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Best! Thank you –  Katlus Dec 7 '12 at 9:30

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