 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/22h/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Suppose first that

∥ x_{n} - x_{\infty} ∥ \to 0

. The reverse triangle inequality gives

∥ x_{n} ∥ - ∥ x_{\infty} ∥ \leq ∥ x_{n} - x_{\infty} ∥ ⟶ 0,

so (i) implies (ii).

It also implies (iii). Given

ϵ > 0

, choose

N

so that

∥ x_{n} - x_{\infty} ∥^{2} < \frac{ϵ}{8} (n \geq N) .

I_{0}

so that the basis tail of

x_{\infty}

beyond

I_{0}

is less than

ϵ /8

. For

n \geq N

and

I \geq I_{0}

\sum_{i \geq I} ∣ ⟨ x_{n}, e_{i} ⟩ ∣^{2} \leq 2∥ x_{n} - x_{\infty} ∥^{2} + 2 \sum_{i \geq I} ∣ ⟨ x_{\infty}, e_{i} ⟩ ∣^{2} < \frac{ϵ}{2} .

There are only finitely many

n < N

, so increasing

I

makes each of their tails less than

ϵ

. Thus one

I

works for every

n

, proving (iii).

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