# Continuous linear operator un a Hilbert space [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jason, Davide Giraudo, Xam, Ethan Bolker, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.

• So, don't we need to use that T is continuous? – bemath Nov 11 '17 at 16:49

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"...
$T$ 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 $x$, $\|Tx\| \geq c\|x\|$).
$T(H)$ 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 $y\not \in T(H).$