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I need to show the above $\forall x,y,v \in V$ , a normed vector space on $\Bbb R$. A hint was given that i should first show that $$s:V \times V \to \Bbb R ; \: \:\: s(u,v):=\frac1 4 (\|u+v\|^2-\|u-v\|^2) $$ is a scalar product in order to prove the above but i neither understand how to show that $s$ is a scalar product nor how to use this in the prove (symmetry and positive definity are apparent but the liniarity troubles me). Can someone give me some tipps or ideas on how to solve the problem?

Thanks in advance

Michael Hardy
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DeltaChief
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    This is odd: by definition $;\left|v\right|=\sqrt{\langle v,v\rangle};$ . Perhaps you're trying to ask how to prove that the norm is induced by an inner product iff the parallelogram law is fulfilled? – DonAntonio Jun 05 '16 at 18:28
  • yes that's what i'm supposed to prove, sorry, i thought that was the same, i'll edit it – DeltaChief Jun 05 '16 at 18:29
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    Ok, and by the hint you give I understand you're working on an real vector space, right? Otherwise things can be messier over the complex. – DonAntonio Jun 05 '16 at 18:31
  • ah yes, sorry i forgot to add that – DeltaChief Jun 05 '16 at 18:34
  • https://en.wikipedia.org/wiki/Parallelogram_law $\qquad$ – Michael Hardy Jun 05 '16 at 18:39
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    Note that proper notation is $V\times V$, not $V\textrm x V$. Also, $|u|$ rather than $||u||$. The latter difference can be seen even more clearly by contrasting $||u|| ||v||$ with $|u||v|$. Just google "latex symbols" and you can find things like this. $\qquad$ – Michael Hardy Jun 05 '16 at 18:41
  • i also thought about the parallelogram law, but still, if i have $s(au,v)$ which should be $|a|s(u,v)$ but by using the law its $|a|^2 2||u||^2 + 2||v||^2$ – DeltaChief Jun 05 '16 at 18:42
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    http://math.stackexchange.com/q/21792 – Jean Marie Jun 05 '16 at 18:53

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