Consider the Hilbert space $\mathcal{H}=l^2(\mathbb{Z})$ and define the left shift operator $\mathcal{L}:\mathcal{H} \rightarrow \mathcal{H}$ by
$$
\mathcal L (a_n) = (b_n) \qquad \text{ where } \qquad b_n=a_{n+1}
,$$
so $\mathcal L$ is the right-shift operator.
My first issue is with notation, what does $\mathcal L -zI$ mean for $z \in \mathbb{C}$? Does this mean $$\mathcal L-I((a_n)) = (a_{n+1}-a_n)?$$
Moreover, I'm supposed to show $\mathcal L-zI$ is not invertible for $|z| =1$ by showing that $$(a_n) = \frac{1}{|n|+1}$$ is in $\ell^2(\mathbb{Z})$, yet not in the image of $\mathcal L$ for $z=1$, and then adjust for other values of $z$. Though, I don't quite understand how this shows that $\mathcal L -zI$ isn't surjective.. was this perhaps a typo? Should I show $(a_n)$ isn't in the image of $\mathcal L-I$ and then adjust for values of $z$? It doesn't seem like it would make sense because surely it's in the image of $\mathcal L$, as the left shift operator is $\mathcal L$'s adjoint and inverse, with both being norm-preserving, so this wouldn't make sense.
I don't quite see how to begin with this either way, so any hints would be wonderful. Indeed, this is homework. (The next part involves the construction of an inverse to $\mathcal L -z I$ using the power series for $(1-x)^{-1}$.)
Best Answer
Answer. $S=\{z\in\mathbb C: \lvert\,z\rvert=1\}$.
Explanation. The space $\ell^2(\mathbb Z)$ is isometric to $L^2(\mathbb T)$, where $\mathbb T$ is the unit circle (equivalently, the domain of $2\pi-$periodic functions), and this is done by the transfomation $$ \varPhi\big((a_k)_{k\in\mathbb Z}\big)=\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx}, $$ as $\|(a_k)_{k\in\mathbb Z}\|_{\ell^2}=\big\|\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx}\big\|_{L^2}$. The shift $S$ in $L^2(\mathbb T)$ becomes a multiplication operator, by $\mathrm{e}^{ix}$, and hence, its the spectrum coincides with its essential range.