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As for the second question, this is really just me wondering on definitions. For a function $f:\mathbb{R}\to\mathbb{R}$ define

$$\limsup_{x\to+\infty}f(x):=\lim_{x\to+\infty}\sup_{z\geq x}f(x).$$

Just like the sequence definition. If you want to instead send the argument to $-\infty$ it seems it should look like

$$\limsup_{y\to-\infty}f(y):=\lim_{y\to-\infty}\sup_{w\leq y}f(w).$$

However, I can't seem to find anywhere that does this. Has anyone encountered this before?

Regarding the nested $\limsup$'s, with the above definition and $f:\mathbb{R}^2\to\mathbb{R}$, would it follow that

$$\limsup_{x\to+\infty}\limsup_{y\to-\infty}f(x,y)=\limsup_{y\to-\infty}\limsup_{x\to+\infty}f(x,y)?$$

The only way I can think to do this is breaking it into cases, whether the $\limsup$'s are finite or not, then using $\varepsilon$'s and $\delta's$'s and the like, but when we write out, for the left side, the limit then the supremum then the next limit then the next supremum I get tangled up. In particular (finite case), for a fixed $y$ call inner supremum $L_y$. Then for all $\varepsilon>0$ there exists some $z_{y,\varepsilon}$ so that $|f(y,z_{y,\varepsilon})-L_y|<\varepsilon$ and $z_{y,\varepsilon}\leq y$. But at this point I don't see to send $y$ to $-\infty$, so to speak.

Thanks in advance.

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  • $\begingroup$ Please clarify, is there a typo in the second $\lim \sup$ in your question? $\endgroup$ – vadim123 May 9 '13 at 13:17
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    $\begingroup$ Note $\limsup_{x\to+\infty}f(x)=\inf_{x\in \Bbb R}\sup_{z\geq x}f(x)$. Now it's easier to deal with nested birds. $\endgroup$ – user59671 May 9 '13 at 13:27
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    $\begingroup$ Alternatively, you can define $\limsup_{x\to -\infty} f(x) = \limsup_{x\to\infty} f(-x)$. $\endgroup$ – Thomas Andrews May 9 '13 at 13:28
  • $\begingroup$ vadim123: I do not see one. CutieKrait: Ahh, that does make it a little easier to look at. Thanks. Thomas Andrews: Yeah, that makes life a lot easier. I'll use this. Thanks! $\endgroup$ – danzibr May 9 '13 at 14:22
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You can't swap $\limsup$ any more than you can swap limits. If $\lim_{x\to \infty} f(x)$ exists then $\lim_{x\to\infty} f(x)=\limsup_{x\to\infty} f(x)$, and the same is true for $x\to -\infty$.

But even if $\lim_{x\to\infty} \lim_{y\to\infty} f(x,y)$ and $\lim_{y\to\infty} \lim_{x\to\infty} f(x,y)$ both exist, they need not be equal.

For example, if $f(x,y)=\left(\frac{x^2}{1+x^2}\right)^{|y|}$ then $$\lim_{x\to\infty} \lim_{y\to\infty} f(x,y) = 0$$ while $$\lim_{y\to\infty} \lim_{x\to\infty} f(x,y) = 1$$

We don't need to deal with the case of $x\to -\infty$ any differently, because we can easily show that $\limsup_{x\to -\infty} f(x) = \limsup_{x\to\infty} f(-x)$.

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