Show that the range $R(T)$ is not closed in $l^2$ norm for $T:(x_1, . . . , x_n, . . .) → (x_1, . . . ,\frac{1}{n}x_n, . . .)$

Question

Here is the question and the answer:

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I would like to ask in the first and second line, how does $R(T)$ being closed in $l^2$ and the open mapping theorem imply the inverse mapping $S$ being bounded.

Answer 1

A closed subspace of a Banach space is Banach. You can think of $T$ as a linear continuous bijection from $\ell^{2}$ onto $R(T)$ and open Mapping Theorem can be applied bacause $\ell^{2}$ and $R(T)$ are Banach spaces.

Answer 2

The range of the operator $T$ is dense as $e_n=T(ne_n).$ But the range is not equal $\ell^2$ as $\{1/n\}_{n=1}^\infty\notin T(\ell^2)$, (as observed by @gettha290krm) Indeed if $(Tx)_n={1/ n},$ then $x_n=1$ for any $n.$