Parallelogram law in Normed vector space without an inner product

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?

• It's a common exercise that if the parallelogram law holds, the norm is induced by an inner product. Known as the polarization identity. – Daniel Fischer Jan 16 '14 at 23:16

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 $\langle a,b\rangle:=\displaystyle\frac14\left(\|a+b\|^2+i\|a+ib\|^2-\|a-b\|^2-i\|a-ib\|^2\right)$.

• Precisely that "The paralelogram law holds if and only if the norm is induced by an inner product." @user127001 – Pedro Tamaroff Jan 16 '14 at 23:36
• 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. – Berci Jan 16 '14 at 23:36
• 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. – Disintegrating By Parts Jan 17 '14 at 6:59

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.

• 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. – DrHAL Sep 4 '16 at 4:12
• @DrHAL Axler's Linear Algebra Done Right, 3rd edition, Question 21. – 1729_SR Aug 15 '20 at 20:56