limsup.

**Showcase_22** · March 11th 2009, 09:59 AM

Let $(a_n; n \geq 1), \ (b_n;n \geq 1)$ and $(\lambda_n; n \geq 1)$ be three sequences of real numbers.

Suppose that $\lambda_n \in [0,1]$ for every n, and let $c_n=\lambda_n a_n+(1-\lambda_n)b_n$ . Assuming that limsup a_n and limsup b_n are finite, prove the following inequality:

$limsup \ c_n \leq max (limsup\ a_n, limsup\ b_n)$

This is what I tried:

$(1-\lambda_n)b_n > (1-\lambda_n)a_n$
$\Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_na_n+(1-\lambda_n)a_n=a_n$

$\Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ a_n$

Assume $b_n>a_n$

$\lambda_nb_n> \lambda_na_n$

$\Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_nb_n+(1-\lambda_n)b_n=b_n$

$\Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ b_n$

Therefore $limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq max(limsup \ a_n,limsup \ b_n)$

This is only true if $a_n>b_n$ eventually or $b_n>a_n$ eventually. Is there a way to adapt this proof for sequences which constantly oscillate between $a_n>b_n$ or $b_n>a_n$ ? (ie. $a_n=5 \ (n \ even), \ 2 \ (n \ odd).$

What can you say about the value of $limsup_{n \rightarrow \infty} c_n$ as compared to $limsup_{n \rightarrow \infty} a_n$ and $limsup_{n \rightarrow \infty}b_n$ ?

Justify your answer.

I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

Does anyone have any ideas?

**bkarpuz** · March 11th 2009, 11:15 AM

Originally Posted by Showcase_22

This is what I tried:

$(1-\lambda_n)b_n > (1-\lambda_n)a_n$
$\Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_na_n+(1-\lambda_n)a_n=a_n$

$\Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ a_n$

Assume $b_n>a_n$

$\lambda_nb_n> \lambda_na_n$

$\Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_nb_n+(1-\lambda_n)b_n=b_n$

$\Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ b_n$

Therefore $limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq max(limsup \ a_n,limsup \ b_n)$

This is only true if $a_n>b_n$ eventually or $b_n>a_n$ eventually. Is there a way to adapt this proof for sequences which constantly oscillate between $a_n>b_n$ or $b_n>a_n$ ? (ie. $a_n=5 \ (n \ even), \ 2 \ (n \ odd).$

I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

Does anyone have any ideas?

I hope this helps.
Set $f_{n}(\lambda):=\lambda a_{n}+(1-\lambda)b_{n}$ for $\lambda\in[0,1]$ , then $f_{n}$ is nondecreasing if $a_{n}-b_{n}\geq 0$ and decreasing if $a_{n}-b_{n}<0$ (check the derivative of $f_{n}$ ).
Now, considering the nondecreasing/decreasing nature of $f_{n}$ , define
$c_{n}^{\ast}:=\begin{cases} f_{n}(1)=a_{n},& a_{n}\geq b_{n}\\ f_{n}(0)=b_{n},& a_{n}<b_{n}\end{cases}$
for $\mathbb{N}$ .
Then, we have $c_{n}=f_{n}(\lambda_{n})\leq c_{n}^{\ast}=\max\{a_{n},b_{n}\}$ for all $n\in\mathbb{N}$ .

**Showcase_22** · March 11th 2009, 11:41 AM

Does what you have imply:

$limsup \ c_n=limsup f_n(\lambda) \leq limsup \c*_n=max \{limsup \ a_n, limsup \ b_n \}$ ?

Would the addition of this step then make it a complete proof?

**bkarpuz** · March 11th 2009, 11:54 AM

Originally Posted by Showcase_22

Does what you have imply:

$\limsup c_n=\limsup f_n(\lambda) \leq \limsup c^{*}_n=\max \{\limsup a_n, \limsup b_n \}$ ?

Would the addition of this step then make it a complete proof?

Sure!
I wonder from where you got this problem?

**Opalg** · March 11th 2009, 11:55 AM

Let $C = \limsup c_n$ , let $\varepsilon>0$ and let $N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}$ . By the definition of limsup, the set N is infinite. Let $N_1 = \{n\in N: a_n\geqslant c_n\}$ and let $N_2 = \{n\in N: b_n\geqslant c_n\}$ . Then $N\subseteq N_1\cup N_2$ , so at least one of the sets $N_1,\,N_2$ must be infinite. Say $N_1$ is infinite. That says that there are infinitely many values of n for which $a_n>C-\varepsilon$ , and so $\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon$ . Since this holds for all $\varepsilon>0$ , it follows that $\max\{ \limsup a_n,\,\limsup b_n\}\geqslant C$ .

**Showcase_22** · March 11th 2009, 12:04 PM

WOW!! Well firstly I got this problem from an analysis book. I just wondered what the answer was since someone pointed out my proof wasn't completely correct.

For the last part of the question I don't actually think anything definite can be stated, but I can't justify it. It would be better if it asked me to prove something else! =D

Also:

$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon $

Where does this part come from?

**bkarpuz** · March 11th 2009, 12:21 PM

Originally Posted by Opalg

Let $C = \limsup c_n$ , let $\varepsilon>0$ and let $N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}$ . By the definition of limsup, the set N is infinite. Let $N_1 = \{n\in N: a_n\geqslant c_n\}$ and let $N_2 = \{n\in N: b_n\geqslant c_n\}$ . Then $N\subseteq N_1\cup N_2$ , so at least one of the sets $N_1,\,N_2$ must be infinite. Say $N_1$ is infinite. That says that there are infinitely many values of n for which $a_n>C-\varepsilon$ , and so $\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon$ . Since this holds for all $\varepsilon>0$ , it follows that $\max\{ \limsup a_n,\,\limsup b_n\}\geqslant C$ .

Hi Opalg, having your answer read carefully, I cant see where the requirement $\lambda_{n}\in[0,1]$ is used.
Do you think it is extra?

Showcase_22,
$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon $
this is just a typo, it should be
$ \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon $

**Opalg** · March 11th 2009, 12:24 PM

Originally Posted by Showcase_22

For the last part of the question I don't actually think anything definite can be stated, but I can't justify it. It would be better if it asked me to prove something else! =D

I agree with that. I don't see that there is anything else that can be said about limsup(c_n)

Originally Posted by Showcase_22

$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon $

Where does this part come from?

It follows from the definition of limsup that if there are infinitely many values of n for which $a_n>C-\varepsilon$ , then $\limsup a_n>C-\varepsilon$ .

**Showcase_22** · March 11th 2009, 12:32 PM

Originally Posted by bkarpuz

Showcase_22,
$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon $
this is just a typo, it should be
$ \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon $

So is this a typo? Surely the maths you posted implies what Opalg originally put? (or am I wrong? :s)

I'll also try and find a way to justify the fact that limsup $c_n$ has no other special qualities apart from what has just been proved. It's a little like proving the impossible! 8-p

**Opalg** · March 11th 2009, 12:32 PM

Originally Posted by bkarpuz

Hi Opalg, having your answer read carefully, I cant see where the requirement $\lambda_{n}\in[0,1]$ is used.

The condition $\lambda_{n}\in[0,1]$ says that $c_n$ lies between $a_n$ and $b_n$ . That implies that at least one of $a_n$ and $b_n$ must be $\geqslant c_n$ . In my previous comment, that is the reason that $N\subseteq N_1\cup N_2$ .

Originally Posted by bkarpuz

Showcase_22,
$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon $
this is just a typo, it should be
$ \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon $

No it's not a typo, it needs to be what I stated.

**bkarpuz** · March 11th 2009, 12:37 PM

Originally Posted by Opalg

The condition $\lambda_{n}\in[0,1]$ says that $c_n$ lies between $a_n$ and $b_n$ . That implies that at least one of $a_n$ and $b_n$ must be $\geqslant c_n$ . In my previous comment, that is the reason that $N\subseteq N_1\cup N_2$ .

No it's not a typo, it needs to be what I stated.

If its not a typo, then you should have written
$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon. $
In any case, there is a typo.

**Opalg** · March 11th 2009, 12:51 PM

Originally Posted by bkarpuz

If its not a typo, then you should have written
$ \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon. $
In any case, there is a typo.

Humph! Okay, I'll admit that.

**Laurent** · March 12th 2009, 03:34 AM

To bkarpuz, about your 1st post: I feel like you only prove $c_n\leq \max(a_n,b_n)$ in a complicated way, or maybe I missed something? Then you can deduce $\limsup_n c_n\leq \limsup_n \max(a_n,b_n)$ .

It is a true fact that $\limsup_n \max(a_n,b_n)=\max(\limsup_n a_n,\limsup_n b_n)$ , but not a trivial one. Actually this is a particular case of the initial problem, and the answer Opalg gave to it gives also a proof of this fact (it proves the inequality " $\leq$ ", and " $\geq$ " is almost trivial).

Here's a variant of Opalg's proof, maybe more complicated, but maybe more intuitive: there exists a subsequence $(c_{\phi(n)})_{n\geq 0}$ which converges toward $\limsup_n c_n$ . For any $n$ , either $c_{\phi(n)}\leq a_{\phi(n)}$ or $c_{\phi(n)}\leq b_{\phi(n)}$ , so that one case happens for infinitely many $n$ . For instance, there are infinitely many $n$ such that $c_{\phi(n)}\leq a_{\phi(n)}$ . Then we can pick a subsequence $(a_{\phi(\psi(n))})_{n\geq 0}$ of $(a_{\phi(n)})_{n\geq 0}$ which converges toward a limit $\ell\in\mathbb{R}\cup\{\pm\infty\}$ (for instance, you can take $\ell=\limsup_n a_{\phi(n)}$ ). From $c_{\phi(\psi(n))}\leq a_{\phi(\psi(n))}$ , we deduce (taking the limsups)

$\limsup_n c_n\leq \ell\leq \limsup_n a_n\leq \max(\limsup_n a_n,\limsup_n b_n)$ .

(we have $\ell\leq \limsup_n a_n$ because the limsup is the maximum of the limits of subsequences, a fact I already used before. )

Originally Posted by Showcase_22

I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

Does anyone have any ideas?

Well, you could say

$\min(\liminf_n a_n,\liminf_n b_n)\leq \liminf_n c_n\leq\limsup_n c_n \leq \max(\limsup_n a_n, \limsup_n b_n)$ ,

but you're right, you can't give any lower bound with limsups.

**bkarpuz** · March 12th 2009, 04:06 AM

Originally Posted by Laurent

To bkarpuz, about your 1st post: I feel like you only prove $c_n\leq \max(a_n,b_n)$ in a complicated way, or maybe I missed something? Then you can deduce $\limsup_n c_n\leq \limsup_n \max(a_n,b_n)$ .

It is a true fact that $\limsup_n \max(a_n,b_n)=\max(\limsup_n a_n,\limsup_n b_n)$ , but not a trivial one. Actually this is a particular case of the initial problem, and the answer Opalg gave to it gives also a proof of this fact (it proves the inequality " $\leq$ ", and " $\geq$ " is almost trivial).

I now think that proving

$c_{n}\leq\max\{a_{n},b_{n}\}$

was silly (because of the convexity), okay.

I am nowadays working on $\max$ and $\min$ values of $\lambda(\lambda-1)\cdots(\lambda-k)$ , and I think I considered that term in the form $a\lambda(\lambda-b)$ at first sight, and gave a proof similar (but complicated) to those in my works.
But I think that

$\limsup_{n\to\infty}\max\{a_{n},b_{n}\}=\max\bigg\ {\limsup_{n\to\infty}a_{n},\limsup_{n\to\infty}b_{ n}\bigg\}$

is trivial.

Thanks for the comments.

**ravenclaw14** · November 15th 2009, 11:10 AM

cool

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