I need help with this problem.
Problem. Consider the Banach space $\ell^1$. Let $I$ be the identity operator and $S:\ell^1 \to \ell^1$ be the shift operator, i.e. $S((x))_k=x_{k+1}$. Now define the family of operators $T_\alpha :=I+\alpha S$ for $\alpha\in\mathbb{R}.$ Determine all values of $\alpha\in\mathbb{R}$ such that the inverse of $T_\alpha$ is bounded.
The notion of inverse that is used here is as an operator $T_\alpha^{-1}:\mathcal{R}(T_\alpha)\to \ell^1$. Where $\mathcal{R} $ denotes the range. For the inverse operator to exist only injectivity is needed.
Attempt.
The previous problem asked for values of $\alpha$ such that the inverse exist. By some straightforward calculations I found that for $|\alpha|\leq 1$ the inverse exists. After doing this and that I found the inverse, namely: \begin{align} T^{-1}_\alpha y=\left(d(1),\ d(2),\ ......\right) \end{align} where $$d(j) = \sum_{k\geq j}(-\alpha)^{k-j}y(k)$$
Now I split cases.
Case 1 $\alpha=1$ (case $\alpha=-1$ is analoguos). Take $x_n=\frac{1}{n}(1,-1,...,1,-1,0,0,0)$ with $n$ nonzero elements. Clearly $\Vert x_n\Vert =1$. We have for this $x$: \begin{align} d_n(j)=\frac{1}{n}\sum_{k=j}^n(-1)^{k-j}(-1)^{k-1}=\frac{n+1-j}{n}(-1)^{j+1} \end{align} So: \begin{align} \Vert T_\alpha^{-1} x_n\Vert = \sum_{j=1}^n \frac{n+1-j}{n}=\frac{n+1}{2}\to \infty \end{align} Hence $T_\alpha^{-1}$ is not bounded for $\alpha=1$. Something similar can be done for the case $\alpha=-1$.
Case 2 $|\alpha|<1$. I can do something similar as above, but that is too constructive and lost insight. So I thought to do it in an easier way. I thought about the following theorem that says that the existence of some $b>0$ such that for all $x\in\ell^1$: $\Vert T_\alpha x \Vert\geq b\Vert x\Vert$ implies the boundedness of $T^{-1}_\alpha$.
I wanted to use it like this. Let $x\neq 0$. One has by the reverse triangle inequality: $$\frac{\Vert T_\alpha x \Vert}{\Vert x\Vert } \geq \frac{1}{\Vert x\Vert }\bigg| \Vert Ix\Vert-|\alpha|\Vert Sx\Vert\bigg|$$ One knows that $\Vert S\Vert =1$, but that did not brought me anywhere.
Question. Is my solution for $\alpha=1$ correct? Is there easier ways to do this exercise, expecially for the case $|\alpha|<1$?
For $\alpha=0$ you have $T_\alpha=I$, so from now on let us assume that $\alpha\ne0$.
When $|\alpha|<1$, we can calculate the inverse of $T_\alpha$ explicitly via the Neumann series: namely, $$ T_\alpha^{-1}=\sum_{n=0}^\infty (-1)^n\alpha^nS^n. $$ So
Note that $$T_\alpha=I+\alpha S=\alpha\left(\frac1\alpha\,I+S\right).$$ Seeing it like this, we know that $T_\alpha$ is invertible precisely when $-1/\alpha\not\in\sigma(S)$.
We have $\|S\|=1$; by the general result that $|\lambda|>\|S\|$ implies that $\lambda\not\in\sigma(S)$, we immediately get that $|1/\alpha|>1$ implies $T_\alpha$ invertible. This is another way to see that $T_\alpha$ is invertible when $|\alpha|<1$.
For any $\lambda$ with $|\lambda|<1$, if $x=(\lambda^n) $, then $Sx=\lambda x$, and so $\lambda\in\sigma (S) $. As the spectrum is always closed, we have $$\sigma (S)=\{\lambda:\ |\lambda|\leq1\}. $$ So