Krdytkyu

Exercise :

Let $T:ell^infty to ell^infty$ be an operator such that :
$$T(x_1,x_2,dots) = bigg(frac{1}{2}x_2,frac{1}{3}x_3,dotsbigg)$$
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 ?

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 $z=(z_1,z_2,z_3,ldots)inell^infty$. Then,
$$T(z)=left(frac{z_2}{2},frac{z_3}{3},frac{z_4}{4},ldotsright)$$
so that
$$big|T(z)big|_{infty}=supleft{frac{|z_k|}{k},Big|,k=2,3,4,ldotsright}leq supleft{frac{|z|_infty}{k},Big|,k=2,3,4,ldotsright}=frac{|z|_infty}{2},.$$
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 $xinell^infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=left(sum_{k=1}^infty,frac{y_k}{k!},sum_{k=2}^infty,frac{2!y_k}{k!},sum_{k=3}^infty,frac{3!y_k}{k!},ldotsright)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.