# manipulating limsup

• Mar 5th 2009, 11:06 AM
claire511
manipulating limsup
I have to answer a question about limsups but im a bit confused...

Suppose an, bn, cn and dn are sequences of real numbers with dn in [0,1] for every n and cn=dnan+(1-dn)bn. Assume limsupan and limsupbn are finite, prove:

limsupcn <= max(limsupan,limsupbn)

I think ive done this part. But then the question asks 'What can you say about the value of limsupcn as compared to limsupan and limsupbn? Justify your answer'.

Well, clearly from the previous proof, limsupcn<=max(limsupan,limsupbn) but what else can i say? There must be a stronger condition that i'm missing but i don't know what it is! i attempted bounding it below to show limsupcn=>min(limsupan,limsupbn) but i don't think this is true.

any help would be appreciated. thankyou
• Aug 20th 2011, 04:28 AM
dgomes
Re: manipulating limsup
$\textbf{Proof.} \text{ We have that:}$

$\text{i) } d_{n} \in \left [ 0, 1 \right ] \ , \ \forall \ n \in \mathbb{N}$
$\text{ii) } c_{n} = d_{n}a_{n} + (1 - d_{n})b_{n}$
$\text{iii) } \limsup \left \{ a_{n} \right \} < \infty \text{ and } \limsup \left \{ b_{n} \right \} < \infty$

$\text{Note that, } \forall \ n \in \mathbb{N} \ , \ d_{n}a_{n} + (1 - d_{n})b_{n} \text{ is a convex combination between } a_{n} \text{ and } b_{n} \text{. Hence:}$

$(d_{n}a_{n} + (1 - d_{n})b_{n} \leq a_{n}) \text{ or } (d_{n}a_{n} + (1 - d_{n})b_{n} \leq b_{n}) \ , \ \forall \ n \in \mathbb{N}$

$\Rightarrow (c_{n} \leq a_{n}) \text{ or } (c_{n} \leq b_{n}) \ , \ \forall \ n \in \mathbb{N}$

$\Rightarrow (\limsup \left \{ c_{n} \right \} \leq \limsup \left \{ a_{n} \right \} ) \text{ or } (\limsup \left \{ c_{n} \right \} \leq \limsup \left \{ b_{n} \right \} )$

$\Rightarrow \limsup \left \{ c_{n} \right \} \leq \max \left \{ \limsup \left \{ a_{n} \right \}, \limsup \left \{ b_{n} \right \} \right \}$
$\text{Q.E.D.}$