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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-326/3/2/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 326 / 3 / 2 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Let

G = E □ F

. To prove convexity, take

u_{1}, u_{2}

with finite

G

values,

0 < t < 1

, and

ε > 0

. By the definition of infimum, choose

v_{i}

with

E (v_{i}) + F (u_{i} - v_{i}) \leq G (u_{i}) + ε .

Put

u_{t} = t u_{1} + (1 - t) u_{2}

and

v_{t} = t v_{1} + (1 - t) v_{2}

E

and

F

gives

G (u_{t}) \leq E (v_{t}) + F (u_{t} - v_{t}) \leq t [E (v_{1}) + F (u_{1} - v_{1})] + (1 - t) [E (v_{2}) + F (u_{2} - v_{2})] \leq tG (u_{1}) + (1 - t) G (u_{2}) + ε .

Let

ε ↓ 0

. If either endpoint value is infinite, the convexity inequality is automatic, and

t = 0, 1

gives equality. Thus the infimal convolution of two convex functions is convex under the stated properness assumption. The proof uses approximate minimizing splits, so no attainment of the inner infimum is assumed.

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