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 ?
Best Answer
You are correct and the proof looks sufficiently rigorous. However, $\|T\|_\text{op}$ can be computed exactly. That is, $\|T\|_\text{op}=\dfrac12$. To show this, let $z=(z_1,z_2,z_3,\ldots)\in\ell^\infty$. Then, $$T(z)=\left(\frac{z_2}{2},\frac{z_3}{3},\frac{z_4}{4},\ldots\right)$$ so that $$\big\|T(z)\big\|_{\infty}=\sup\left\{\frac{|z_k|}{k}\,\Big|\,k=2,3,4,\ldots\right\}\leq \sup\left\{\frac{\|z\|_\infty}{k}\,\Big|\,k=2,3,4,\ldots\right\}=\frac{\|z\|_\infty}{2}\,.$$ Note that the equality holds for $z=(0,1,0,0,0,\ldots)$. This implies $\|T\|_{\text{op}}= \dfrac{1}{2}$.
You can write an explicit solution $x\in\ell^\infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=\left(\sum_{k=1}^\infty\,\frac{y_k}{k!},\sum_{k=2}^\infty\,\frac{2!y_k}{k!},\sum_{k=3}^\infty\,\frac{3!y_k}{k!},\ldots\right)$$ Nonetheless, you did sufficient and good work. I was just making additional comments.