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.

Let $E$ be Banach space over $\mathbb{R}$. Let $u$ and $v$ be such that $||u||=||v||=1$ and $||2u+v||=||u-2v||=3$.

How do we show that there exists a linear functional $f$ defined on all of $E$ such that $||f||=1$, $f(u)=1$ and $f(v)=1$?

share|improve this question
    
Please do not write in caps. Most of the time it is considered to be very rude to do so. Also, try to show us what you have tried before, so we can help you solve the problem, instead of giving you the solution on a silver platter. This way you will learn a lot more! –  sxd Aug 23 '12 at 23:15
    
I think you need to show first that $u$ and $v$ are linearly independent, so that they span a 2-dimensional subspace. –  timur Aug 24 '12 at 0:06
1  
I defined the functional: g defined in H=span{u,v} such that, g(au+bv)=a+b. Note that g(u)=g(v)=1 end g(2u/3+v/3)=1. But I can't show that ||g||=1. –  Valdinês Júnior Aug 24 '12 at 0:13
    
@DimitriSurinx I find it ridiculous to claim that typing in all caps (in this case) is rude. –  Quinn Culver Aug 24 '12 at 0:31

1 Answer 1

up vote 2 down vote accepted

Note first that $u$ and $v$ are linearly independent. Now consider the linear functional $g$ on the linear span $V$ of $u$ and $v$ such that $g(u) = 1$ and $g(v) = 1$. Thus $g(au + bv) = a+b$. What is the norm of this? Draw a picture of the unit ball of $V$, using the fact that it must be symmetric and convex...

share|improve this answer
    
I managed to solve. Thanks! –  Valdinês Júnior Aug 24 '12 at 3:50
    
@ValdinêsJúnior: Then upvote and accept the answer! –  timur Aug 24 '12 at 14:03

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.