Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a way to show that $\liminf \limits _{k\rightarrow \infty} f_k = \lim \limits _{k\rightarrow \infty} f_k$. The only way I can think of is by showing $\liminf \limits _{k\rightarrow \infty} f_k = \limsup \limits _{k\rightarrow \infty} f_k$. Is there another way?

Edit: Sorry, I should have mentioned that you can assume that $\{f_k\}_{k=0}^\infty$ where $f_k:E \rightarrow R_e$ is a (lebesque) measurable function

share|improve this question
1  
Are you asking how to show this if the limit exists, because in general a limit may not exist, whereas the $\liminf$ always exists (in an extended sense). –  copper.hat Oct 17 '12 at 0:18
    
No not existence but equality. To restate the question, under what conditions would the limit inferior be equal to the limit? –  rioneye Oct 17 '12 at 0:27
    
I gave necessary and sufficient conditions below. –  copper.hat Oct 17 '12 at 1:18
add comment

2 Answers

up vote 1 down vote accepted

The following always holds: $\inf_{k\geq n} f_k \leq f_n \leq \sup_{k\geq n} f_k$. Note that the lower bound in non-decreasing and the upper bound is non-increasing.

Suppose $\alpha = \liminf_k f_k = \limsup_k f_k$, and let $\epsilon>0$. Then there exists a $N$ such that for $n>N$, we have $\alpha -\inf_{k\geq n} f_k < \epsilon$ and $\sup_{k\geq n} f_k -\alpha < \epsilon$. Combining this with the above inequality yields $-\epsilon < f_k - \alpha< \epsilon$ from which it follows that $\lim_k f_k = \alpha$.

Now suppose $\alpha = \lim_k f_k$. Let $\epsilon >0$, then there exists a $N$ such that $-\frac{\epsilon}{2}+\alpha < f_k< \frac{\epsilon}{2}+\alpha$. It follows from this that $-\epsilon + \alpha \leq \inf_{k\geq n} f_k \leq \sup_{k\geq n} f_k < \epsilon+\alpha$, and hence $\liminf_k f_k = \limsup_k f_k = \alpha$.

Hence the limit exists iff the $\liminf$ and $\limsup$ are equal.

share|improve this answer
add comment

What you have is incorrect. For instance, consider $$f_k = \begin{cases} 0 & \text{if }k \text{ is even}\\1 &\text{if }k \text{ is odd} \end{cases}$$

$\liminf f_k = 0$, $\limsup f_k = 1$ while $\lim f_k$ does not exist.

share|improve this answer
    
But isn't it true that if $\liminf \limits _{k\rightarrow \infty} f_k = \limsup \limits _{k\rightarrow \infty} f_k$ is true then $\liminf \limits _{k\rightarrow \infty} f_k = \lim \limits _{k\rightarrow \infty} f_k = \limsup \limits _{k\rightarrow \infty} f_k$? I am just curious to know if there is another method to prove the same result. –  rioneye Oct 17 '12 at 0:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.