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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

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    $\begingroup$ 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
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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.

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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.

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  • $\begingroup$ 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

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