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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 6 / 5 / b / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 6 5 b
Created 2026-10-03 Updated 2026-10-06  0 By others on same topic  0 Discussions Create my own version
Every finitely supported sequence v lies in Y: if vn​=0 for n>N, set xn​=∑k=nN​vk​ for n≤N and xn​=0 afterwards. Then x is a bounded sequence and x−Sx=v. Consequently L(v)=0. The finitely supported sequences are dense in the space of sequences converging to zero c0​ for the supremum norm, by truncation, so continuity of the bounded linear functional L gives L∣c0​​=0. For x in the real convergent sequence space, with ordinary limit ℓ, we have x−ℓ1∈c0​. Therefore L(x)=ℓL(1)=n→∞lim​xn​​.

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