Bourbaki

Proposition 5

Tag 07XP, permanent. Cite it as [Bourbaki, Tag 07XP].

Topological Vector Spaces, chapter V, § 1. Prehilbertian spaces and hilbertian spaces

— Let EE be a prehilbertian space. Let dd be a real number >0> 0, δ\delta a real number such that 0⩽δ⩽d0 \leqslant \delta \leqslant d. Let BB and B′B' be subsets of EE defined by ∥x∥<d\| x \| < d,

∥x∥≤d+δ respectively, and let A be a convex set contained in B′−B. Then for every pair of points x,y of A, we have ∥x−y∥≤12dδ (fig. 1). \|x\| \leq d + \delta \text{ respectively, and let } A \text{ be a convex set contained in } B' - B. \text{ Then for every pair of points } x, y \text{ of } A, \text{ we have } \|x - y\| \leq \sqrt{12d\delta} \text{ (fig. 1).}
In fact, we have 12(x+y)∈A\frac{1}{2}(x + y) \in A, hence ∥12(x+y)∥≥d\left\| \frac{1}{2}(x + y) \right\| \geq d; hence from (14) we get the inequality
∥12(x−y)∥2=12(∥x∥2+∥y∥2)−∥12(x+y)∥2≤(d+δ)2−d2≤3dδ \left\| \frac{1}{2}(x - y) \right\|^2 = \frac{1}{2}(\|x\|^2 + \|y\|^2) - \left\| \frac{1}{2}(x + y) \right\|^2 \leq (d + \delta)^2 - d^2 \leq 3d\delta
from which the proposition follows.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed volume. Not translated.