# closedness of image of closed, unbounded operator

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

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If $\mathcal H_1=\mathcal H_2=\mathbb R$, $T(x)=x$, $C=(0,1)$, $T(C)$ is not closed. –  Davide Giraudo May 20 '12 at 17:17
Thank you for the counterexample. It seems as one really needs compactness and not just relative compactness then. –  giacomo_C May 20 '12 at 17:23

Quoting Davide Giraudo's comment (so that this has an answer):

If $\mathcal H_1=\mathcal H_2=\mathbb R$, $Tx=x$, $C=(0,1)$, then $C$ is relatively compact but $TC$ is not closed.

(If you upvote this then the question will no longer be bumped by Community.)

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@tomasz: Oops! I had intended to do so but apparently I forgot to check the box. It's fixed now. Thanks! –  Nate Eldredge Aug 21 '12 at 0:28
Take a sequence $(y_n)$ in $T(C)$ such that $y_n\to y$. We want to show that $y\in T(C)$. Writing $y_n=Tx_n$, the sequence $(x_n)$ is in a compact set $C$, hence it has some convergent subsequence $x_{n_k}\to x$. We have $Tx_{n_k}\to y$; now apply closedness of the operator.
$C$ is relatively compact, not compact. And we are not sure that $x\in C$, except I am missing something. –  Davide Giraudo May 20 '12 at 17:16
$x\in C$ follows from closedness, does it not? But yes, relative compactness seems to be an issue... –  Eepzy May 20 '12 at 17:21