 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/21h/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

x, y \in ℓ^{2}

satisfy

∥ x ∥_{2} = ∥ y ∥_{2} = 1

\frac{x + y}{2}_{2}^{2} + \frac{x - y}{2}_{2}^{2} = \frac{∥ x ∥ _{2}^{2} + ∥ y ∥ _{2}^{2}}{2} = 1.

∥ x - y ∥_{2} \geq ε

, then

\frac{x + y}{2}_{2}^{2} \leq 1 - \frac{ε ^{2}}{4},

and therefore

\frac{x + y}{2}_{2} \leq 1 - \frac{ε ^{2}}{4} = 1 - δ, δ = 1 - 1 - ε^{2} /4 > 0.

This depends only on

ε

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