characterization of Euclidean norm by the parallelogram identity can any one give me a prove to the following:
A norm is Euclidean iff satisfies the parallelogram identity, 
$\|v+w\|=\sqrt{2\|v\|^2+2\|w\|^2-\|v-w\|^2}$
 A: Suppose $\|\cdot\|$ is a euclidean norm, then we will show it satisfies the parallelogram identity.  Consider $\|v+w\|^2$, this is equal to
$\left<v+w,v+w\right>= \left<v,v\right> + \left<v,w\right> + \left<w,v\right> + \left<w,w\right>$ where $\left<\cdot,\cdot\right>$ is the usual inner product defined as the non-negative square root of the norm.
Similarly, consider $\|v-w\|^2$, this is equal to 
$\left<v+w,v+w\right>= \left<v,v\right> - \left<v,w\right> - \left<w,v\right> + \left<w,w\right>.$
Where I am using the fact that $\left<ax,y\right> = a\left<x,y\right>$ and $\left<x,by\right> = \bar b\left<x,y\right>$ for (complex) scalars $a,b$ and vectors $x,y$. 
Hence $\|v+w\|^2 + \|v-w\|^2 = 2\|v\|^2 + 2\|w\|^2$.
It remains to show that if a norm, $\|\cdot\|$, satisfies the parralelogram identity than it must be euclidean.  Apparently the key here is to show that the polarization identity, $\frac14\left(\|x+y\|^2 -\|x-y\|^2 +i\|x+iy\|^2-i\|x-iy\|^2\right)$ defines an inner product.  The details can be found at http://www.pcs.cnu.edu/~jgomez/files/norm.pdf.  The proof uses the parralelogram identity and shows that the polarization identity satisfies all the axioms of an inner product.
