If $I:(l^{1},\|.\|_{2})\rightarrow (l^{1},\|.\|_{1})$(where $l^{1}$ is the vector space of all sequences $\{x_{n}\}$ in $\mathbb{C}$ such that $\sum |x_{n}|<\infty , \|x\|_{1}=\sum_{n} |x_{n}|,\|x\|_{2}=(\sum_{n} |x_{n}|^{2})^{\frac{1}{2}}$) is the identity map, then
$A.$ Both $I$ and $I^{-1}$ are continuous.
$B.$ $I$ is continuous but $I^{-1}$ is NOT continuous.
$C.$ $I^{-1}$ is continuous but $I$ is NOT continuous.
$D.$ Neither $I$ nor $I^{-1}$ is continuous.
According to me as $(l^{1},\|.\|_{1})$ will gives weaker topology than $ (l^{1},\|.\|_{2})$ so $I$ is continuous but not $I^{-1}.$ Am i right? Please give me some details for this question. Thanks in advance.
Let $x_n$ denote the sequence with $1's$ in the first $n$ places, and $0$'s elsewhere. If $I$ were continuous, then there would exists a constant $M > 0$ such that $$ \|Ix_{n}\| \le M\|x_{n}\|,\;\;\;\\ \|x_{n}\|_{\ell^1} \le M \|x_{n}\|_{\ell^2},\\ n \le M \sqrt{n}. $$ This cannot hold for all $n$, which proves that $I$ is not continuous.
On the other hand, consider the natural map $\mathscr{I} : \ell^1 \rightarrow \ell^2$ that takes a sequence in $\ell^1$ to the same sequence in $\ell^2$. This map is between Banach spaces. And you can show that this map is closed because if $\{ x_n \} \subset \ell^1$ converges to $x\in\ell^1$ in the norm of $\ell^1$, and if $\{\mathscr{I}x_n\}\subset\ell^2$ converges to $y\in \ell^2$ in the norm of $\ell^2$, then $x=y$ follows because $\{ (x_n)_m \}$ is a sequence in $\mathbb{C}$ that must converge to $x_m$, and must also converge to $y_m$. Hence $x=y$, which proves that $I^{-1}$ is closed. By the Closed Graph Theorem, $\mathscr{I}$ is a bounded operator. Hence $I^{-1}$ is continuous.