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Possible Duplicate:
Norms Induced by Inner Products

I am trying to prove to that if a norm of a vector space satisfies the parallelogram law ($\| \vec x + \vec y \|^2 + \| \vec x - \vec y\|^2 = 2 \| \vec x \|^2 + 2 \| \vec y \|^2$), then it can be defined by an inner product.

I know that for an inner product to exist for a given norm, it must take the form $(\vec x, \vec y) = \frac{\| \vec x + \vec y \|^2 - \| \vec x \|^2 - \| \vec y \|^2}{2} = \frac{\| \vec x + \vec y \|^2 - \| \vec x -\vec y \|^2}{4}$. (In fact, the two expressions are equal because we are assuming the parallelogram law holds.)

However, from here I am stuck trying to prove that the product defined above satisfies $(\vec x+\vec y, \vec z) = (\vec x,\vec z) + (\vec y,\vec z)$ and $(c\vec x, \vec y) = c(\vec x, \vec y)$.

Can someone give me a hint on where to begin? So far, I've only been taking the definition of $(\vec x,\vec y)$ and blindly applying the parallelogram law. This has not been working very well.

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marked as duplicate by t.b., Andrés E. Caicedo, Ross Millikan, joriki, kahen Jun 7 '11 at 14:52

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    $\begingroup$ I added a quite detailed solution to the question linked to above. $\endgroup$ – t.b. Jun 7 '11 at 10:13
  • $\begingroup$ Thanks! I'm sorry I didn't find that when I searched for this question. $\endgroup$ – Alan C Jun 8 '11 at 0:14
  • $\begingroup$ Alan: No problem, I hope the solution there helps. If not, please leave comments and ask, I'd love to improve that text so that it is maximally helpful. $\endgroup$ – t.b. Jun 8 '11 at 0:15