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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$\begingroup$ Does it mean anything? Looks to me like you're binding $h$ twice. Did you mean something like $\displaystyle \limsup_{h \to \infty}\ \sup_{x \in \mathbb R}\ A(x,h)$? $\endgroup$ – Ilmari Karonen Aug 28 '11 at 0:38
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$\begingroup$ Exactly, but I have no idea how to make it like this! $\endgroup$ – Joan Aug 28 '11 at 0:41
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$\begingroup$ You can right-click the expression in my comment and select "Show Source" to see how I did it. $\endgroup$ – Ilmari Karonen Aug 28 '11 at 0:44
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$\begingroup$ There are many different ways of defining the limit superior; one of them is the following: the limit superior is the supremum of the set of limit points. $\endgroup$ – Arturo Magidin Aug 28 '11 at 3:13
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$\begingroup$ This question seems to be related: math.stackexchange.com/questions/49699/… $\endgroup$ – Martin Sleziak Aug 28 '11 at 10:05
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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.
