Bourbaki

Exercise 6, § 1

Topological Vector Spaces, V, p. 62

Bourbaki marks this one as harder than the rest.

a) Let E be a real or complex normed space, of dimension ≥2\geq 2. Show that for every x≠0x \neq 0 in E and every real number α>0\alpha > 0, there exists an element yy in E such that ∥y∥=α\|y\| = \alpha and ∥x+y∥2=∥x∥2+∥y∥2\|x + y\|^2 = \|x\|^2 + \|y\|^2. b) Suppose that if the vectors x,yx, y in E satisfy the relation ∥x+y∥2=∥x∥2+∥y∥2\|x + y\|^2 = \|x\|^2 + \|y\|^2, then we also have ∥x−y∥2=∥x∥2+∥y∥2\|x - y\|^2 = \|x\|^2 + \|y\|^2. Show that the norm on E is prehilbertian. (Using a), reduce to the case in exerc. 4 : restricting to the case where E is 2-bidimensional, we prove that if ∥x1∥=∥x2∥=1,y=12(x1−x2)\|x_1\| = \|x_2\| = 1, y = \frac{1}{2}(x_1 - x_2) and if z∈Ez \in E is such that ∥y∥2+∥z∥2=∥y+z∥2=1\|y\|^2 + \|z\|^2 = \|y + z\|^2 = 1, then z=12(x1+x2)z = \frac{1}{2}(x_1 + x_2) or z=−12(x1+x2)z = -\frac{1}{2}(x_1 + x_2). c) Suppose that, for every vector x≠0x \neq 0 in E, the set H of all vectors yy satisfying ∥x−y∥2=∥x∥2+∥y∥2\|x - y\|^2 = \|x\|^2 + \|y\|^2 is stable under addition. Show that the conclusion of b) holds. (Reduce to the case where E is real and of dimension 2. Using a) and the compactness of the unit ball in E, show that H is a closed set containing at least two distinct half-lines with origin 0 ; prove that these two half lines are opposite to each other by observing that, if not, the convex set which they generate would be contained in H and will contain either xx or −x-x.)

Solution

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