I have this part:

enter image description here

Where $$\underline{F(+\infty)}=\liminf_{x\rightarrow +\infty} F(x), \overline{F(-\infty)}=\limsup_{x\rightarrow-\infty} F(x)$$

My question is: How does property $(3.12)$ follow from the given information?

Thank you in advance.

  • $\begingroup$ I don't think your question is clear. The author chose to define $c_\epsilon$ and $d_\epsilon$ in order to make an interesting mathematical point. What is your question about property 3.12? Your wording isn't clear. $\endgroup$
    – jdods
    Mar 9, 2015 at 13:54
  • $\begingroup$ I don't understand why we have (3.12) $\endgroup$
    – Vrouvrou
    Mar 9, 2015 at 13:56
  • $\begingroup$ If you like, you could edit your question to state: "How does property 3.12 follow from the given information?" $\endgroup$
    – jdods
    Mar 9, 2015 at 14:00
  • 1
    $\begingroup$ Important question is what is your definition of $\limsup$. (Maybe you could include your definition in the post.) There are several equivalent definitions of limit superior. The one mentioned here might be what you need. This post and this post also compare two definitions of limit superior. $\endgroup$ Mar 9, 2015 at 14:27
  • $\begingroup$ but this is the definition for a sequence not for a function ! $\endgroup$
    – Vrouvrou
    Mar 9, 2015 at 15:10

1 Answer 1


Hint: If for some $\epsilon>0$ and for any $K>0$, $F(t)<c_\epsilon$ for some $t>K$, how could the $\liminf$ be larger than $c_\epsilon$?

More details:

Assume $\liminf_{x\rightarrow\infty} F(x)=\infty$ and there exists $\epsilon$ such that for any $K>0$, $F(t)<1/\epsilon=c_\epsilon$ for some $t>K$.

This means that no matter how far out you look towards positive infinity, there will always be an $x$ value such that $F(x)<1/\epsilon$, which is finite. Therefore, the limit infemum cannot possibly be infinite as there is always a larger $x$ value which gives a function output lower than $1/\epsilon$.

Similar arguments will work in the remaining 3 cases.

When you have trouble understanding the mathematics, especially complicated notation like this problem, it's a good idea to look at a more concrete example.

Let $F(x)=x$, so that $\liminf_{x\rightarrow+\infty} F(x)=+\infty$. Now fix $\epsilon$, say $\epsilon=1/10$. Therefore $c_\epsilon=2$. Now we can let $K$ be any number larger than $10$, thus for any $x>K$, $F(x)>c_\epsilon$.

Try to find the opposite of property 3.12 here. Can you find an $\epsilon$ such that $F(x)$ will always be smaller than $1/\epsilon$ if it is diverging to infinity?

  • $\begingroup$ I don't understand you $\endgroup$
    – Vrouvrou
    Mar 9, 2015 at 14:06
  • $\begingroup$ I've "negated" property 3.12. The definition of $\liminf$ requires the function to "go above" $\liminf$ eventually. $c_\epsilon$ is either an arbitrarily large number (if the $\liminf$ is infinity) or a number arbitrarily close, but below the $\liminf$ if it is finite. $\endgroup$
    – jdods
    Mar 9, 2015 at 14:10
  • $\begingroup$ $\liminf_{x\rightarrow+\infty} F(x)=\lim_{x\rightarrow +\infty} (\inf_{y\geq x} F(y))$ why you say there is always $x$ such that $F(x)<1/\varepsilon$ ? $\endgroup$
    – Vrouvrou
    Mar 9, 2015 at 14:19
  • $\begingroup$ I'm doing a proof by contradiction. I assume that property 3.12 does not hold and arrive at a contradiction, therefore property 3.12 must hold. $\endgroup$
    – jdods
    Mar 9, 2015 at 14:20
  • $\begingroup$ when $\liminf_{x\to +\infty} F(x)=x_0<\infty$ and $F(t)< x_0-\varepsilon$ where is the contradiction please $\endgroup$
    – Vrouvrou
    Mar 9, 2015 at 15:09

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