Let $T$ be the operator defined on the Sobolev space $H^1((0,1))$ by $$T:H^1((0,1)) \longrightarrow \mathbb{R} \\ f \mapsto f(0).$$ This operator is clearly a finite rank operator and thus it is compact. Am I right?. Thank you.
2026-03-25 06:19:20.1774419560
Compactness of the trace operator in dimension 1
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Yes, you are correct. Any $\Bbb R$-linear map to $\Bbb R$ has an image of dimension at most $1$, which means that the operator is of finite rank, which means that the operator is compact.