# 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.


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. – DisintegratingByParts 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