# Is the range of this operator closed?

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?

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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.
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