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I'm a little confused about the definition of limit supremum; what does it mean that the following limit is finite?

$\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\; A(x,h)$ where $A(x,h)$ is a function of $x$, and $h$.

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    This question seems to be related: http://math.stackexchange.com/questions/49699/about-the-notion-of-limsup-and-liminf/2011-08-28

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It means that there exists some finite $H$ and $C$ such that for every $h\ge H$ and every $x$ in $\mathbb R$, $A(x,h)\le C$.

Proof:

If $H$ and $C$ as above exist, then $\limsup\limits_{h\to+\infty}\,\sup\limits_{x\in\mathbb R}A(x,h)\le C$ hence it is finite.

To prove the other direction, first recall that, for any function $B$ and any finite $c$, $\limsup\limits_{h\to+\infty}\,B(h)\le c$ means that, for every c'>c, there exists a finite $h_0$ such that B(h)\le c' for every $h\ge h_0$.

Hence $\limsup\limits_{h\to+\infty}\,B(h)$ is finite if and only if there exists some finite $C$ and $H$ such that $B(h)\le C$ for every $h\ge H$.

Apply this to $B(h)=\sup\limits_{x\in\mathbb R}A(x,h)$, hence $B(h)\le C$ for every $h\ge H$. Now, $A(x,h)\le B(h)$ for every $x$ in $\mathbb R$ and $B(h)\le C$ for every $h\ge H$, hence $A(x,h)\le C$ for every $h\ge H$ and $x$ in $\mathbb R$.

Done.