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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-9/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 1 / a / Solution by

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

The statement requires

K \neq = \emptyset

; an empty set is closed and convex but contains no minimizer. Assume this necessary condition and put

d = in f_{g \in K} ∥ g - f ∥_{p}

. Since

K

is closed and

f \in / K

,

d > 0

. Choose a minimizing sequence

g_{j} \in K

with

∥ g_{j} - f ∥_{p} \to d

.

(g_{j} + g_{k}) /2

lies in the convex set

K

. Applying the stated Clarkson inequality to

g_{j} - f

and

g_{k} - f

gives

∥ g_{j} - g_{k} ∥_{p}^{p} \leq 2^{p - 1} (∥ g_{j} - f ∥_{p}^{p} + ∥ g_{k} - f ∥_{p}^{p}) - ∥ g_{j} + g_{k} - 2 f ∥_{p}^{p} \leq 2^{p - 1} (∥ g_{j} - f ∥_{p}^{p} + ∥ g_{k} - f ∥_{p}^{p}) - 2^{p} d^{p} .

The right side tends to zero, so

g_{j}

is a Cauchy sequence. By completeness of Lp spaces, proved in Question 2(a), it converges to some

h

. Closedness puts

h

in

K

, and continuity of the norm gives

∥ h - f ∥_{p} = d

. This is the closest point theorem for a closed convex subset of Lp. The same inequality applied to two minimizers forces their difference to be zero, so the minimizer is also unique.

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