Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think I am stuck with showing closedness of the range of a given operator. Given a sequence $(X_n)$ of closed subspaces of a Banach space $X$. Define $Y=(\oplus_n X_n)_{\ell_2}$ and set $T\colon Y\to X$ by $T(x_n)_{n=1}^\infty = \sum_{n=1}^\infty \frac{x_n}{n}$. Is the range of $T$ closed?

share|improve this question
add comment

1 Answer

The range is not necessarily closed. For example, if $X=(\oplus_{n \in \mathbb{N}} X_n)_{\ell_2}$=Y:

if $T(Y)$ is closed, $T(Y)$ is a Banach space. $T$ is a continuous bijective map from $X$ onto $T(Y)$, so is an homeomorphism (open map theorem) . But $T^{-1}$ is not continuous, because $T^{-1}(x_n)=nx_n$, for $x_n \in X_n$. So $T(Y)$ is not closed.

share|improve this answer
    
If $X_1=X_2$, then $T$ is not injective. How do you identify $X$ with $Y$? Even when $X$ is a Hilbert space, this need not be injective. –  Ville Jun 26 '12 at 15:30
    
@Ville: francis-jamet is just giving an example, and it is one in which $X_1\neq X_2$. In some cases your map will have closed range. E.g., if $X_n=X$ for all $n$, then $T$ is onto. –  Jonas Meyer Aug 1 '12 at 6:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.