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In order to define a dot product, can we apply the polarization identity to any norm ?

One exercise I had was based on the norm 1 : $$||u||_1=\sum{|u_i|}$$ I wonder if I could have used (wasn't asked...) the polarization identity : $$<u,v>=\frac{1}{2}(||u+v||_1^2-||u||_1^2-||v||_1^2)$$ in order to define a "dot product 1" : $<u,v>_1$ ?

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It doesn't pass the test of linearity.

To be an iner product space.

$\langle a\mathbf{u,v}\rangle = a\langle\bf{ u,v}\rangle$

let $\mathbf u = (1,1), \mathbf v = (1,-1)$

$\|\mathbf u\|_1 = 2, \|\mathbf v\|_1 = 2, \|\mathbf{u+v}\|_1 = 2,\frac 12(2^2 - 2^2 - 2^2) = -2$

$\|\mathbf 2u\|_1 = 4, \|\mathbf v\|_1 = 2, \|\mathbf{u+v}\|_1 = 4,\frac 12(4^2 - 4^2 - 2^2) = -2$

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  • $\begingroup$ Obvious, thanks ! So the identity polarization holds only for the euclidian norm, isn't it ? But does it really mean there is no dot product associated with the norm 1 ? $\endgroup$ – Andrew Jan 18 '18 at 18:44
  • $\begingroup$ I had to look it up. $\langle \mathbf{u,v}\rangle = \frac 14 (\|\mathbf{u+v}\|_1^2 - \|\mathbf{u-v}\|_1^2)$ $\endgroup$ – Doug M Jan 18 '18 at 18:46
  • $\begingroup$ Then, with your vectors $u=(1,1)$ and $v=(1,-1)$, you get $<u,v>=0$ if I made no mistake. I thought the different formulas for the identity polarization were equivalent ? $\endgroup$ – Andrew Jan 18 '18 at 19:02

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