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I need to show that $-\infty \leq \liminf_{k \to \infty}f_k \leq \limsup_{k \to \infty}f_k \leq \infty$ , where $f_k$ is a sequence of functions from $\mathbb{R}^n$ to $\mathbb{R}$. This seems inherently true, but for some reason, I am having difficulty with the details.

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what definition of liminf and limsup are you using? Also did you intend to write $x\rightarrow\infty$? as opposed to $k\rightarrow\infty?$. Or are you taking the infimum over $k$ first and then taking a limit in $x$? –  Alex R. Nov 6 '12 at 1:15
    
No I did intend for it to be the limit as k approaches infinity. –  Angelo Christophell Nov 6 '12 at 15:23
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I assume that what you want to show is that $(\liminf_{k\to\infty}f_k)(x)\le(\limsup_{k\to\infty}f_k)(x)$ holds for all $x$ and that the function $\liminf f_k$ is defined by $(\liminf_{k\to\infty}f_k)(x)=\liminf_{k\to\infty}(f_k(x))$, and likewise for $\limsup f_k$.

Let $n\in\mathbb N$. Now by definition, $\inf_{k\ge n}f_k(x)$ is a lower bound for the set $\{f_k(x)\mid k\ge n\}$ and $\sup_{k\ge n}f_k(x)$ is an upper bound for it. Since $n\ge n$, $f_n(x)$ is an element of this set and we get $$\inf_{k\ge n}f_k(x)\le f_n(x)\le\sup_{k\ge n}f_k(x).$$ Thus $\inf_{k\ge n}f_k(x)\le\sup_{k\ge n}f_k(x)$ holds for every $n$, and we have that $$\lim_{n\to\infty}\inf_{k\ge n}f_k(x)\le\lim_{n\to\infty}\sup_{k\ge n}f_k(x).$$

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I cannot understand why you are focusing on a sequence of functions. Would anything change if $\{f_k\}_k$ were a sequence of numbers?

Anyway, it all depends on the definition of $\liminf$ and $\limsup$. Usually, given a sequence $\{p_k\}_k$ of real numbers, you consider the set $E$ of those points $p \in \mathbb{R}$ such that a subsequence of $\{p_k\}_k$ converges to $p$. By definition, $\liminf_{k \to +\infty} p_k = \inf E$ and $\limsup_{k \to +\infty} p_k = \sup E$. Since $\inf E \leq \sup E$, the conclusion should be clear.

If you want to use an equivalent definition for $\liminf$ and $\limsup$, then you might need to work a little bit.

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