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.

Possible Duplicate:
Norms Induced by Inner Products and the Parallelogram Law

enter image description here

So suppose we are given a norm on a vector space.

If the Parallelogram law holds does that automatically mean we have the inner product which we can find using the Polarisation identity? Or is showing the Parallelogram law holds not sufficient to show that there exists an associated inner product?

Also, given that the Parallelgram law fails, e.g. $\Vert(x_1,x_2)\Vert_1 = |x_1| + |x_2|$ in $\ell^1(2)$, is there any significance in considering the Polarisation identity?

share|improve this question

marked as duplicate by t.b., Adam Rubinson, Jonas Teuwen, Martin Sleziak, Matt N. May 27 '12 at 8:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
The validity of the Parallelogram Law is a necessary and sufficient condition for the existence of a scalar product inducing the given norm, and in the affirmative case the scalar product is uniquely determined by the Polarization identity. I have to search a reference perhaps Serge Lang Real and Functional Analysis. –  Giuseppe Tortorella May 26 '12 at 17:10
1  
Thanks t.b. . That link is exactly what I am looking for. It's a shame someone has already asked it and has got loads of points for the question (I quite like points lol). Edit: btw I voted to close. I encourage a few other people who are reading this to vote to close. –  Adam Rubinson May 27 '12 at 7:34
    
Or should I just delete this thread? –  Adam Rubinson May 27 '12 at 8:46
    
@Adam: Just leave it. There's nothing wrong with closed questions. They make finding the answers easier. –  t.b. May 27 '12 at 9:30
add comment

1 Answer 1

Well, you know the definition of angle in an IPS? You can use this to find the i.p. if you know the norm and angle between two vectors: $$x\cdot y=||x||\cdot||y||\cdot\cos\theta$$ with $\,\theta\,=$ the angle between vectors $\,x\,,\,y\,$

share|improve this answer
    
How do you find the angle between two vectors for a general NLS (one which has an associated IPS)? For example, in l^2, what is the angle between two vectors? The only way I can see to find out what the angle is is to use theta = cos^-1[<x,y>/(||x||||y||)], that is, to find theta you must use the fact that you already know what the inner product is. –  Adam Rubinson May 27 '12 at 7:21
    
Here is where the "if you know the angle" part comes in. –  DonAntonio May 27 '12 at 11:32
1  
I don't get what you are trying to say. Suppose we know the vector space and the norm. The definition of the "angle" is only available to us if the inner product actually exists, which is not always the case. For example,how you would you go about finding the angle of the 1-norm space as in the OP? Anyway, the (somewhat nice) conclusion of all this is that, given a NLS, the Parallelogram Law determines IF there even exists such an inner product. If there does, then the Polarisation Identity tells us EXACTLY what the inner product (we want) is! Proving all this is a different (but doable) task. –  Adam Rubinson May 27 '12 at 11:55
    
If you're given enough info to deduce the angle, or if you're directly given the angle. That's all. Otherwise you'll have to go another way. Of course, IF you're given the parallelogram identity then the polarisation identity tells you how to find the i.p. –  DonAntonio May 27 '12 at 12:00
1  
"Enough info to deduce the angle". Info like what? –  Adam Rubinson Jun 3 '12 at 0:11
show 2 more comments

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