 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/31j/a/solution

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

f

\partial f (x) = {g : f (y) \geq f (x) + g^{T} (y - x) for every y} .

For

f (x) = γ ∣ x ∣

\partial f (x) = ⎩ ⎨ ⎧ {γ}, [- γ, γ], {- γ}, x > 0, x = 0, x < 0.

The defining inequality shows directly that

0 \in \partial f (x)

exactly when

x

is a global minimizer.

f (λ x + (1 - λ) y) < λ f (x) + (1 - λ) f (y)

for distinct

x, y

and

0 < λ < 1

. Two distinct minimizers would make the midpoint have a strictly smaller value, so a minimizer is unique.

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