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

| cite | improve this question | | | | |
  • 1
    $\begingroup$ 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). $\endgroup$ – copper.hat Oct 17 '12 at 0:18
  • $\begingroup$ No not existence but equality. To restate the question, under what conditions would the limit inferior be equal to the limit? $\endgroup$ – rioneye Oct 17 '12 at 0:27
  • $\begingroup$ I gave necessary and sufficient conditions below. $\endgroup$ – copper.hat Oct 17 '12 at 1:18

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.

| cite | improve this answer | | | | |

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.

| cite | improve this answer | | | | |
  • $\begingroup$ 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. $\endgroup$ – rioneye Oct 17 '12 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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