Exercise :
Let $T:\ell^\infty \to \ell^\infty$ be an operator such that : $$T(x_1,x_2,\dots) = \bigg(\frac{1}{2}x_2,\frac{1}{3}x_3,\dots\bigg)$$ Show that for all $y \in \ell^\infty$, the equation $$x = y + Tx$$ has a unique solution.
Attempt :
I have proved that $T$ is a linear operator. Now, $\ell^\infty$ is the space defined as : $$\ell^\infty = \{x =(x_n) : \|x\|< \infty\} \quad \|x\| :=\sup|x_n|$$ From the definition of the norm over $\ell^\infty$, we can observe that
$$\|T(x_1,x_2,\dots)\|<\|(x_1,x_2,\dots)\|$$
This means that there exists an $M<1$, such that :
$$\|T(x_1,x_2,\dots)\|\leq M\|(x_1,x_2,\dots)\|$$
Thus, $T$ is a bounded linear operator $T \in B(\ell^\infty)$ with $\|T\| \leq M <1$.
Now, it is
$$x = y + Tx \Leftrightarrow x-Tx = y \Leftrightarrow(1-T)x=y$$
where $1$ is the identity operator.
But $\ell^\infty$ is a Banach space and since $\|T\| <1$, then it is :
$$(1-T)^{-1}=\sum_{n=0}^\infty T^n \Leftrightarrow (1-T)^{-1}y=\sum_{n=0}^\infty T^ny$$
Thus $x = \sum_{n=0}^\infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y \in \ell^\infty$.
Question : Is my approach correct and rigorous enough ?