 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/19g/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Take

f_{n} (t) = sin (2 πn t)

. Its domain norm is

1

, whereas its image norm is

1 + 2 πn

. Hence $T$ is unbounded, so it has no finite operator norm and is not a compact operator.

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