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If we have a scalar product $\left<\cdot, \cdot\right>:X\times X \to \mathbb{K}$, where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$, than one has for the norm $\lVert x \rVert := \sqrt{\left<x, x\right>}$ that $$ \langle x, \ y \rangle = \frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 \right) \ \forall x,y \in X$$ if $\mathbb{K}=\mathbb{R}$ and a similar formula if $\mathbb{K}=\mathbb{C}$.

By this, we can get our scalar product if we have given only the norm. But what if that norm is not induced by any scalar product and we use the above formula to define the map $\left<\cdot, \cdot\right>$?

Clearly, one would still have $\lVert x \rVert = \sqrt{\left<x, x\right>}\ \forall x \in X$. But has $\left<x, y\right>$ also some meaning if $x \neq y$?

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  • $\begingroup$ You can always define a map. However, it is not bilinear (or sesquilinear for $\mathbb{K}=\mathbb{C}$) if the norm does not come from an inner product. $\endgroup$ – user10354138 Nov 6 '18 at 16:05
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You can define such a map $\langle\cdot,\cdot\rangle:X\times X \to \mathbb K$ but it won't be an inner product so I don't see what it should be good for. A norm is induced by an inner product (and induces one) if and only if it satisfies the parallelogram relation $$2\Vert x\Vert+2\Vert y\Vert=\Vert x+y\Vert^2+\Vert x-y\Vert^2.$$ I'm not sure whether this gives you a satisfying answer but I don't really see what one could say about this additionally.

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  • $\begingroup$ I was just interested whether this map has some other nice properties. $\endgroup$ – Jakob B. Nov 6 '18 at 16:31
  • $\begingroup$ I see. Well, I don't think that there are any particularly interesting properties if it doesn't give you an inner product. $\endgroup$ – James Nov 7 '18 at 8:16

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