 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-71/1/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write the Bernstein polynomial in its partition-normalized Bernstein basis:

B_{n} (f, x) = k = 0 \sum n f (k / n) (k n) x^{k} (1 - x)^{n - k} .

For

0 \leq x \leq 1

, every weight is nonnegative, and the binomial theorem gives their sum as

(x + 1 - x)^{n} = 1

. Consequently

∣ B_{n} (f, x) ∣ \leq \sum_{k = 0}^{n} ∣ f (k / n) ∣ (k n) x^{k} (1 - x)^{n - k} \leq ∥ f ∥_{\infty} .

Taking the supremum norm proves $∥ B_{n} (f) ∥_{\infty} \leq ∥ f ∥_{\infty}$ . The construction yields degree at most

n

, possibly lower for special functions.

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