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I want to prove the following:

Suppose $\mathcal{H}_1$ and $\mathcal{H_2}$ are Hilbert spaces and let $T: \mathcal{D} \rightarrow \mathcal{H}_2$ be a closed operator, where $\mathcal{D} \subset \mathcal{H}_1$ denotes its domain. For any relatively compact subset $C$ of $\mathcal{D}$ we have that $T(C)$ is closed.

Any input is really appreciated.

Thanks, Giacomo

  • 1
    If $\mathcal H_1=\mathcal H_2=\mathbb R$, $T(x)=x$, $C=(0,1)$, $T(C)$ is not closed.2012-05-20
  • 1
    Thank you for the counterexample. It seems as one really needs compactness and not just relative compactness then.2012-05-20

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