 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/31j/b/solution

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

f : C \to R

is strictly convex when, for distinct

x, y \in C

and

0 < t < 1

f (t x + (1 - t) y) < t f (x) + (1 - t) f (y) .

Suppose distinct points

x

and

y

were both minimizers, with minimum value

m

. Since

C

is a convex set, their midpoint belongs to

C

, while strict convexity gives

f (\frac{x + y}{2}) < \frac{f ( x ) + f ( y )}{2} = m,

a contradiction. Thus any minimizer is unique, which is the uniqueness of a minimizer of a strictly convex function.

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