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 $h$.
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 $h$.
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.