 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-2/20h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

g

satisfies condition (a), then

g_{i} \geq 0

for every

i

. Outside

B

, the rows of

P

and

P

agree, so

g_{i} \geq (P g)_{i} = (P g)_{i}

. On

B

the modified row is the identity row, so

(P g)_{i} = g_{i}

. Thus

g \geq P g .

This conclusion uses no boundary-value assumption on

g

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