 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-326/4/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 326 / 4 / a / Solution by

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

For a convex functional

J

and

p \in \partial J (v)

, the generalized Bregman distance is

D_{J}^{p} (u, v) = J (u) - J (v) - ⟨ p, u - v ⟩ \geq 0.

The inequality is exactly the subgradient inequality.

Suppose

J

is strictly convex and

u \neq = v

, with both values finite. If the distance vanished, the supporting affine function at

v

would also meet

J

at

u

. At

w = (1 - t) v + t u

,

0 < t < 1

, the subgradient inequality would give

J (w) \geq J (v) + t ⟨ p, u - v ⟩ = (1 - t) J (v) + t J (u),

contradicting strict convexity. Hence zero distance implies

u = v

.

Without strict convexity, take

J (s) = ∣ s ∣

,

v = 1

, and

p = 1 \in \partial J (1)

. For every

u \geq 0

,

D_{J}^{1} (u, 1) = u - 1 - (u - 1) = 0,

including

u = 2 \neq = 1

. Thus zero Bregman distance need not identify a point. It identifies points on the same supporting face, as explained by zero Bregman distance and supporting faces.

 Read the full article

 New to topics? Read the docs here!