# Values of $\alpha$ such that the inverse of a linear operator is bounded

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 $|\alpha| >1$ let $x_n=(-\frac 1 {\alpha})^{n}$. Then $\{x_n\} \in l^{1}$ and $(I+\alpha S )\{x_n\}=0$. Hence $T_\alpha$ is not one-to-one and it does not have an inverse. Jan 22 '18 at 7:31
• @KaviRamaMurthy yes!! That is also the example that I had. Jan 22 '18 at 10:29

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

$$T_\alpha$$ is invertible for all $$\alpha$$ with $$|\alpha|<1$$.

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

$$T_\alpha$$ is not invertible if $$|\alpha|\geq1$$.

• Thank you for your answer. It gave me new insights since it uses the concept of eigenvalues (is that so?) and spectrum which we did not cover yet. I think it is possible to do it without, but this answer is fine. Moreover the notion of invertible you use is as a function from $\ell^1$ to $\ell^1$ right? In class we use inverse as a function from the range to the domain. Only injectivity is needed for the latter. Jan 22 '18 at 10:27
• For example for $\alpha=-1$ the inverse exists I think. Take some $y\in\ell^1$ then $Ty=0$ implies $y_{j+1}=y_1$ since $y \in\ell^1$ we must have $y_1=0$ and therefore $y=0$. Hence injective so invertible. Jan 22 '18 at 10:27
• And for $|\alpha|>1$ there is no much worries since $T_\alpha$ is then not Injective. Jan 22 '18 at 10:50
• Sorry, I see that I misread your definition, and I was using the other shift. I'll edit soon. Jan 22 '18 at 11:12
• but the notion of inverse function that is used here is $T^{-1}_\alpha:\mathcal{R}(T_\alpha)\to\ell^1$. For this function to exist only injective $T_\alpha$ is needed. I must indeed be a little bit careful when talking about invertibility. Do you get it? Jan 22 '18 at 11:12