Inverse linear operator is not an open map for a unbounded linear operator

Let $l^1= \{x =(x(1),x(2),x(3),...,x(n),...) | \sum_{n=1}^{\infty} |x(n)| \lt \infty\}$ be the sequence space equipped with the norm $||x||=\sum_{n=1}^{\infty} |x(n)|.$ Consider the subspace $X=\{x \in l^1 :\sum_{n=1}^{\infty} n|x(n)| \lt \infty\}$ and the linear transformation $T:X \rightarrow l^1$ given by $(Tx)(n)=nx(n)\ \ \text{for} \ \ n \in \Bbb N $. I want to prove the following two statements : $\\$ $(1). \ \ T^{-1} \ \text{is not an open map} , (2). T \ \ \text{is unbounded linear operator}.$ My attempt for $(2)$ $$||T||= sup A , \text{where A=}\{ \sum_{n=1}^{\infty} n x(n) |\sum_{n=1}^{\infty} |x(n)|=1 \}.$$ If anyhow i am able to show that $A$ is unbounded then i am done. Now i am unable to think sequences $x(n)$ such that i can make $\sum_{n=1}^{\infty} n x(n) $ as big as possible .Please help me to think of such sequences. My attempt for $(1)$, I know $T$ is surjective $(T^{-1} y)(n) = \frac{y(n)}{n}$. $$||(T^{-1}y)(n)|| \leq ||y(n)||$$. So $T^{-1}$ is bounded linear opeartor. Here $T^{-1}$ is not surjective so i can't use open mapping theorem. In order to prove that $T^{-1} \ \text{is not an open map}$ , i should find some open set in domain of $T^{-1}$ which is not mapping to open set in $X$. Please help me how to think such open set. Any help will be appriciated. Thank you

Solution 1:

I think you meant $X=\{x \in l^1 :\sum_{n=1}^{\infty} n|x(n)| \lt \infty\}$.

You are making things too complicated. $T$ is not continuous because $\|T(e_n)\|=n \to \infty$ and $\|e_n||=1$ where $e_n$ has $1$ in the $n-$th place and $0$ elsewhere.

$T$ is a bijection from $X$ to $\ell^{1}$. $T^{-1}$ is open iff it maps open set sets to open sets iff $T^{-1} (U)$ is open for any open set $U$. But this is exactly continuity of $T$. Since $T$ is not continuous it follows that its inverse is not an open map.