Exercise 4, § 1

a) Let E be a real or complex normed space, of dimension $\geq 2$, having the following property : the relation $\|x\| = \|y\|$ implies the inequality

Show that the norm on E is prehilbertian. (Reduce to the case when E is real and of dimension 2, by means of exerc. 1 and 2 of V, p. 60. In this case, let A be the unit ball of E, and let $f$ be the positive, separating, symmetric bilinear form such that $A \subset B_f$ and such that the volume of $B_f$ with respect to a given basis is the smallest possible. Let $x_1, x_2$ be the two points constructed in exerc. 3, c). Show that the points of intersection of the circle $f(z, z) = 1$ and the bisectors of the two vectors $x_1, x_2$ also belong to A, and conclude, by iteration, that $A = B_f$.) b) Let E be a real or complex normed space, of dimension $\geq 2$, having the following property : the relation $\|x + y\| = \|x - y\|$ implies $\|x + y\|^2 = \|x\|^2 + \|y\|^2$. Show that the norm on E is prehilbertian (reduce to a)). c) Prove the analogue of a) when we assume that the relation $\|x\| = \|y\|$ implies the inequality (use exerc. 3, d)).