# Defining scalar product from norm

1. Dec 11, 2008

### mma

Euclidean norm is defined usually as|v|2= g(v,v), where g is a nondegenerate, positive definite, symmetric bilinear form. But how can make it backwards? What properties must norm have that g(v,w) = (|v+w|2 - |v|2 - |w|2)/2 be a positive definite, symmetric bilinear form?

2. Dec 11, 2008

### lurflurf

The parrallelogram identity
(1/2)(|x+y|^2+|x-y|^2)=|x|^2+|y|^2
Watch out
(1/2)(|v+w|^2 - |v|^2 - |w|^2)
is not the form of an inner product over other fields or with different conjugations
for example the complex inner product is
<v|w>=(1/2)(|v+w|^2 - |v|^2 - |w|^2)+(i/2)(|v+iw|^2 - |v|^2 - |w|^2)
where i*i=-1

3. Dec 11, 2008

### mma

Cool! Thanks! Now I see that this is a theorem of von Neumann and Jordan. Is the proof very long? I haven't found it anywhere, only remarks that it is complicated enough. Do you know if it is available somewhere online?

4. Dec 12, 2008

### lurflurf

If you have access tohttp://www.jstor.org (perhaps from a school of library) you can access these

A Characterization of Inner Product Spaces
Neil Falkner
The American Mathematical Monthly, Vol. 100, No. 3 (Mar., 1993), pp. 246-249 (article consists of 4 pages)

On Inner Products in Linear, Metric Spaces
P. Jordan and J. V. Neumann
The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723 (article consists of 5 pages)

or you could track down hardcopy at a univercity library.

The proof is a few pages, you may be able to do it yourself. Here is a start
4<x|y>+4<z|y>=2|x+y|^2+2|z+y|-2|x|^2-4|y|^2-2|z|^2
=|x+2y+z|^2+|x+y|^2-4|y|^2
=4<x+z|y>

Homogeneity is harder, it is obvious that
<r x|y>=r <x|y>
when r is rational
when r is real more is needed (like Cauchy-Swartz)

5. Dec 12, 2008

### mma

Thank you!

6. Dec 13, 2008

### mma

Sorry, but this seems false (subststitute for example y=z=0, x<>0).

I made a derivation, perhaps isn't too elegant, but I hope that good:

$$4<x|y>+4<z|y>$$ applying the defiition of scalar product:

$$=2|x+y|^2+2|z+y|^2 -2|x|^2-4|y|^2-2|z|^2$$ applyig the parallelogram law to (x+y) and (z+y):

$$= |x+2y+z|^2 + |x-z|^2 -2|x|^2-4|y|^2-2|z|^2$$ applyig the parallelogram law to (x+y+z) and (y):

$$= 2|x+y+z|^2 - |x+z|^2 + 2|y|^2 + |x-z|^2 -2|x|^2-4|y|^2-2|z|^2$$ applyig the parallelogram law to x and z:

$$= 2|x+y+z|^2 - |x+z|^2 + 2|y|^2 + 2|x|^2 + 2|y|^2 - |x+z|^2 -2|x|^2-4|y|^2-2|z|^2$$

$$=2|x+y+z|^2-2|x+z|^2-2|y|^2$$ applying the defiition of scalar product:

$$=4<x+z|y>$$