In short:
If a norm $\Vert\cdot\Vert$ on a real vector space satisfies the parallelogram law, and $\langle x,y\rangle := \frac14\left(\Vert x+y\Vert^2 - \Vert x-y\Vert^2\right)$, then how can we show $ \langle\lambda x,y\rangle = \langle x,\lambda y\rangle $ for $\lambda\in \Bbb R$?
This is a request for clarification on this specific answer given by zodiac. The post is many years old, so I thought it better to open a new question rather than leave a comment.
The problem is to show that if $\Vert\cdot\Vert$ is a norm on a real vector space that satisfies the parallelogram law, $\Vert x+y\Vert^2 + \Vert x-y\Vert^2 = 2\Vert x\Vert^2 + 2\Vert y\Vert^2$, then $\Vert\cdot\Vert$ is induced by an inner product. We proceed by defining $$\langle x,y\rangle := \frac14\left(\Vert x+y\Vert^2 - \Vert x-y\Vert^2\right)$$ (from the polarization identity), and show that this is indeed an inner product, and that it induces $\Vert\cdot\Vert$. The hardest part is arguably to show scalar linearity, i.e. that $$ \langle\lambda x,y\rangle = \lambda \langle x,y\rangle. $$ The proofs I've seen show it first for $\lambda \in \Bbb Q$ and then appeal to some form of continuity of the purported inner product to extend to real scalars. However, the (or a) central step of the linked answer is instead to show $$ \langle\lambda x,y\rangle = \langle x,\lambda y\rangle. $$
How can we prove the above identity?
Of course, the proof should hopefully not use continuity, which is what made this specific answer particularly interesting. We may assume that all other axioms of inner products are already proven, including additive linearity. zodiac also gives the hint of using $\langle x+y,z\rangle = \langle x,z\rangle+ \langle y,z\rangle$ and $\langle -x,y\rangle =-\langle x,y\rangle =\langle x,-y\rangle$, both of which may be taken as given.
(I made an attempt where I considered the difference, regrouped, applied the parallelogram law in two places, and neatly ended up right where I started...)
