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Let be $H$ a Hilbert space and $T$:$H\rightarrow{H}$ a continuous linear operator. Can we conclude that T(H) is closed?

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closed as off-topic by Jason, Davide Giraudo, Xam, Ethan Bolker, Siong Thye Goh Nov 11 '17 at 20:08

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  • $\begingroup$ So, don't we need to use that T is continuous? $\endgroup$ – bemath Nov 11 '17 at 16:49
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No. Hint: If there exists $c>0$ such that $||Tx||\ge c||x||$ then it does follow that TH is closed. So your counterexample cannot be "bounded below"...

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  • $\begingroup$ So, T(H) is not closed? $\endgroup$ – bemath Nov 11 '17 at 16:56
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$T$ is a bounded, injective linear map between Banach spaces with closed range if and only if $T$ is bounded below (i.e., $\exists$ $c>0$ such that for all $x$, $\|Tx\| \geq c\|x\|$).

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$T(H)$ is not necessarily closed in $H.$ For example let $\{e_n:n\in \Bbb N\}$ be a Hilbert-space basis for $H$ and let $T(e_n)=n^{-1}e_n.$

Then $\sum_{n\in \Bbb N}r_n e_n\in T(H)$ iff $\sum_{n\in \Bbb N}n^2r_n^2<\infty.$

For $k\in \Bbb N$ let $y(k)=\sum_{n=1}^k n^{-1}e_n.$ Each $y(k)\in T(H),$ and $(y(k))_{k\in \Bbb N}$ is a Cauchy sequence in $H$ converging to $y=\sum_{n\in \Bbb N}n^{-1}e_n.$ But $y\not \in T(H).$

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