Question about linear operator in $l_1$ space

Question

Let be $a\in{l^\infty}$ and the linear operator $T_a\colon l^1\longrightarrow{l^1}$ such that$$(\forall{x\in{l_1}})(\forall{k\in\mathbb{N}}):T_ax(k)=a(k)x(k)$$

Prove that:

$T_a$ is an isomorfism if and only if $\inf\{|a(k)|\}>0$.

Thanks.

Answer 1

If $\inf\bigl\{|a(k)|\,|\,k\in\mathbb{N}\bigr\}>0$, then $T_a$ is an isomorphism, since $T_{1/a}$ is a continuous inverse, where, of course, $(1/a)(k)=\frac1{a(k)}$.

If $\inf\bigl\{|a(k)|\,|\,k\in\mathbb{N}\bigr\}=0$, then $T_a$ is not an isomorphism. This is clear if $a(k)=0$ for some natural $k$, since then $T_a$ is not even injective. Otherwise, it is still true that $T_{1/a}$ is the inverse of $T_a$. However, it is not continuous. For each $n\in\mathbb N$, pick $k\in\mathbb N$ such that $\bigl|a(k)\bigr|<\frac1n$. Let $x_n\in l^1$ be such that $x_n(k)=1$ and $x_n(k')=0$ is $k'\neq k$. Then $\|x_n\|_1=1$, but $\|T_{1/a}(x_n)\|_1\geqslant n$.