Exercise 7, § 1

Let $E$ be a real or complex normed space, of dimension $\geqslant 2$, having the following property: there exists a real number $\gamma$ distinct from 0 and from $\pm 1$, such that the relation $\|x + y\| = \|x - y\|$ implies $\|x + \gamma y\| = \|x - \gamma y\|$.

(a) Now prove that if $\|x + y\| = \|x - y\|$, the convex mapping $\varphi : \xi \mapsto \|x + \xi y\|$ from $\mathbf{R}$ into $\mathbf{R}$ is constant in no interval. (Reasoning by contradiction, let $[\alpha, \beta]$ be the largest interval in which $\varphi$ is constant; show that there exists $\delta > \beta$ sufficiently close to $\beta$ and such that $\varphi(\delta) = \varphi(\beta)$, by remarking that if $\|u + v\| = \|u - v\|$, one has $\|u + \gamma^n v\| = \|u - \gamma^n v\|$ for every rational integer $n$.)

(b) Now prove that if $\|x + y\| = \|x - y\|$, one has $\|x + \xi y\| = \|x - \xi y\|$ for every real number $\xi$. (With the notations of a), remark that $\varphi$ admits a relative minimum at the point $\xi = 0$, using the fact that $\varphi(\gamma^n) = \varphi(-\gamma^n)$ for every rational integer $n$; deduce that one has identically $\varphi(\xi) = \varphi(-\xi)$ by remarking that, in the contrary case, one would have $\varphi(\lambda) = \varphi(\mu)$ for two numbers $\lambda, \mu$ such that $\lambda + \mu \neq 0$ and that in this case $\varphi$ would admit a relative minimum at the point $\frac{1}{2}(\lambda + \mu)$.)

(c) Deduce from b) that the norm on $E$ is prehilbertian. (Show first that if $\|x\| = \|y\|$, one has $\|\alpha x + \beta y\| = \|\beta x + \alpha y\|$ for every pair of real numbers $\alpha, \beta$ and that the equality $\|x + y\| = \|x - y\|$ implies $\|\alpha x + \beta y\| = \|\alpha x - \beta y\|$ for every pair of real numbers $\alpha, \beta$. Show then that, if $\|x\| = \|y\| = 1$ and $\|x + y\| = \|x - y\|$, one has the relation $\|(\alpha^2 - \beta^2)x + 2\alpha\beta y\| = \alpha^2 + \beta^2$ using the preceding results, and deduce the conclusion.)