Let $H$ be a Hilbert space and let $T \in \mathcal{B}(H)$. Define what it means for $T$ to be bounded below. Prove that, if $L T=I$ for some $L \in \mathcal{B}(H)$, then $T$ is bounded below.

Prove that an operator $T \in \mathcal{B}(H)$ is invertible if and only if both $T$ and $T^{*}$ are bounded below.

Let $H$ be the sequence space $\ell^{2}$. Define the operators $S, R$ on $H$ by setting

$$S(\xi)=\left(0, \xi_{1}, \xi_{2}, \xi_{3}, \ldots\right), \quad R(\xi)=\left(\xi_{2}, \xi_{3}, \xi_{4}, \ldots\right),$$

for all $\xi=\left(\xi_{1}, \xi_{2}, \xi_{3}, \ldots\right) \in \ell^{2}$. Check that $R S=I$ but $S R \neq I$. Let $D=\{\lambda \in \mathbb{C}:|\lambda|<$ $1\}$. For each $\lambda \in D$, explain why $I-\lambda R$ is invertible, and define

$$R(\lambda)=(I-\lambda R)^{-1} R$$

Show that, for all $\lambda \in D$, we have $R(\lambda)(S-\lambda I)=I$, but $(S-\lambda I) R(\lambda) \neq I$. Deduce that, for all $\lambda \in D$, the operator $S-\lambda I$ is bounded below, but is not invertible. Deduce also that $\operatorname{Sp} S=\{\lambda \in \mathbb{C}:|\lambda| \leqslant 1\}$.

Let $\lambda \in \mathbb{C}$ with $|\lambda|=1$, and for $n=1,2, \ldots$, define the element $x_{n}$ of $\ell^{2}$ by

$$x_{n}=n^{-1 / 2}\left(\lambda^{-1}, \lambda^{-2}, \ldots, \lambda^{-n}, 0,0, \ldots\right) .$$

Prove that $\left\|x_{n}\right\|=1$ but that $(S-\lambda I) x_{n} \rightarrow 0$ as $n \rightarrow \infty$. Deduce that, for $|\lambda|=1, S-\lambda I$ is not bounded below.