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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?

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    $\begingroup$ It's a common exercise that if the parallelogram law holds, the norm is induced by an inner product. Known as the polarization identity. $\endgroup$
    – Daniel Fischer
    Jan 16, 2014 at 23:16

2 Answers 2

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

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

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  • $\begingroup$ Precisely that "The paralelogram law holds if and only if the norm is induced by an inner product." @user127001 $\endgroup$
    – Pedro ♦
    Jan 16, 2014 at 23:36
  • $\begingroup$ I have updated my answer. Check that the given definitions are indeed inner products and that $\langle a,a\rangle=\|a\|^2$ in both cases. $\endgroup$
    – Berci
    Jan 16, 2014 at 23:36
  • $\begingroup$ I'm not aware of norm making sense for a general field because norm requires absolute value. For real numbers, it is not obvious that a norm which satisfies the parallelogram law must be generated by an inner-product. It is not easy to show that the expression for the inner-product that you have given is bilinear over the reals, for example. A continuity argument is required to prove such a thing. The same thing is true for the case of complex scalars, too. $\endgroup$
    – Disintegrating By Parts
    Jan 17, 2014 at 6:59
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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.

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  • $\begingroup$ Do you know where I can find this kind of proof (full one)? I've been trying to show that every norm in $\mathbb{R}$ comes from an inner product. $\endgroup$
    – DrHAL
    Sep 4, 2016 at 4:12
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    $\begingroup$ @DrHAL Axler's Linear Algebra Done Right, 3rd edition, Question 21. $\endgroup$
    – EE18
    Aug 15, 2020 at 20:56

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