Parallelogram law in Normed vector space without an inner product

Question

Let $V$ be any $\mathbb{K}$-vector space with norm $\|\cdot\|\,.$

I know that the Parallelogram law holds if the norm is induced by some inner product $\langle\cdot,\cdot\rangle$, i.e.

$$ \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\skp}[2]{\left\langle#1,#2\right\rangle}\begin{array}{rcl} \norm{a+b}^2+\norm{a-b}^2 &=& \skp{a+b}{a+b} + \skp{a-b}{a-b} \\ &=& \skp{a}{a+b}+\skp{b}{a+b} + \skp{a}{a-b}-\skp{b}{a-b} \\ &=& \skp{a}{a}+\skp{a}{b}+\skp{b}{a}+\skp{b}{b}+\skp{a}{a}-\skp{a}{b}-\skp{b}{a}+\skp{b}{b}\\ &=& 2\left(\skp{a}{a}+\skp{b}{b}\right) \\ &=& 2\left(\norm{a}^2+\norm{b}^2\right) \end{array}$$

However, does the Parallelogram law hold if the norm is not induced by some inner product? Do you have a proof or a counter example for this case?

Answer 1

The paralelogram law holds if and only if the norm is induced by an inner product (over characteristic $\ne 2$):

Supposing the paralelogramma law,
Let $\langle a,b\rangle:=\displaystyle\frac12\left(\|a+b\|^2 - \|a\|^2-\|b\|^2\right)$.

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For $\Bbb K=\Bbb C$, we can define a hermitian inner product:
Let $\langle a,b\rangle:=\displaystyle\frac14\left(\|a+b\|^2+i\|a+ib\|^2-\|a-b\|^2-i\|a-ib\|^2\right)$.

Answer 2

It's a non-trivial exercise to show that $$ (a,b)=\frac{1}{2}(\|a+b\|^{2}-\|a\|^{2}-\|b\|^{2}) $$ actually defines an inner-product when the parallelogram law holds for a norm $\|\cdot\|$ over the field of real numbers. The scalar linearity can only be shown for rational numbers directly, and then a continuity argument is required to extend to all real scalars. The full proof is not simple, only assuming that $\|\cdot\|$ is a norm.