Spectrum of the unilateral shift

Shift identities: $(Sx{)}_n=x_{n-1}$ for $n\ge 1$ and $(Sx{)}_0=0$, so $\mathrm{\parallel }Sx\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }S\mathrm{\parallel }=1$; moreover $S-\lambda$ is injective for every $\lambda$: from $(S-\lambda )x=0$ the recursion $x_{n-1}=\lambda x_n$ gives $x=0$ for $\lambda \mathrm{\ne }0$ and $Sx=0$ gives $x=0$ for $\lambda =0$, since $S$ is injective.

Spectral containment: if $\mathrm{\mid }\lambda \mathrm{\mid }>1$ then $\mathrm{\parallel }S\mathrm{/}\lambda \mathrm{\parallel }<1$ and $S-\lambda =-\lambda (1-S\mathrm{/}\lambda )$ is invertible by the Neumann series; if $\mathrm{\mid }\lambda \mathrm{\mid }<1$ then $\mathrm{\parallel }(S-\lambda )x\mathrm{\parallel }\ge (1-\mathrm{\mid }\lambda \mathrm{\mid })\mathrm{\parallel }x\mathrm{\parallel }$ by [L2], so $S-\lambda$ is bounded below.

Non-density in the open disc: for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ the vector $c:=({\bar{\lambda }}^n{)}_{n\ge 0}$ lies in ${\mathrm{\ell }}^2$ and annihilates the range: for every $x\in {\mathrm{\ell }}^2$, $⟨(S-\lambda )x,c⟩={\sum }_n{x}_n(\overline{c_{n+1}}-\lambda \overline{c_n})={\sum }_n{x}_n({\lambda }^{n+1}-\lambda {\lambda }^n)=0$. So $\mathrm{ran}(S-\lambda )$ is not dense for $\mathrm{\mid }\lambda \mathrm{\mid }<1$, and ${\sigma }_{cp}(S)\supseteq D$.

Approximate eigenvectors on the circle: for $\mathrm{\mid }\lambda \mathrm{\mid }=1$ and $N\ge 1$ put $v_N:=N^{-1\mathrm{/}2}{\sum }_{k<N}{\lambda }^{-k}{e}_k$, a unit vector; since $S{v}_N=N^{-1\mathrm{/}2}{\sum }_{k<N}{\lambda }^{-k}{e}_{k+1}$, the interior terms cancel in $(S-\lambda )v_N$ and only the two boundary terms survive, $(S-\lambda )v_N=N^{-1\mathrm{/}2}({\lambda }^{-(N-1)}{e}_N-\lambda e_0),$ so $\mathrm{\parallel }(S-\lambda )v_N\mathrm{\parallel }=\sqrt{2}{N}^{-1\mathrm{/}2}\le 2N^{-1\mathrm{/}2}\to 0$ and $S-\lambda$ is not bounded below.

Density on the unit circle: if $c\perp \mathrm{ran}(S-\lambda )$ then the same computation gives $c_{n+1}=\bar{\lambda }{c}_n$ for all $n$, that is, $c_n={\bar{\lambda }}^n{c}_0$. For $\mathrm{\mid }\lambda \mathrm{\mid }=1$ this makes $\mathrm{\mid }{c}_n\mathrm{\mid }=\mathrm{\mid }{c}_0\mathrm{\mid }$ for every $n$, so $c\in {\mathrm{\ell }}^2$ forces $c_0=0$ and the orthogonal complement is $\{0\}$: the range is dense there. For $\mathrm{\mid }\lambda \mathrm{\mid }<1$, on the other hand, $\mathrm{\mid }{c}_n\mathrm{\mid }=\mathrm{\mid }\lambda {\mathrm{\mid }}^n\mathrm{\mid }{c}_0\mathrm{\mid }$ decays, the vector $c=({\bar{\lambda }}^n{)}_{n\ge 0}$ is a nonzero element of ${\mathrm{\ell }}^2$ orthogonal to the range, and the range is not dense — the non-density already computed in [step 2.1].

Point spectrum empty and residual spectrum: injectivity is [step 1.1], so ${\sigma }_p(S)=\mathrm{\varnothing }$; for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ the range is non-dense by [step 2.1], so $\lambda \in {\sigma }_r(S)$ and also $\lambda \in {\sigma }_{cp}(S)$; for $\mathrm{\mid }\lambda \mathrm{\mid }>1$ the operator is invertible by [step 1.2], so those points are outside every spectral set.

Combining: $\sigma (S)=\bar{D}$ because $\mathrm{\mid }\lambda \mathrm{\mid }>1$ gives invertibility [step 1.2] and every $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$ lies in ${\sigma }_c(S)$ or ${\sigma }_r(S)$ by [step 3.2] and [step 2.2]; ${\sigma }_{ap}(S)=\mathbb{T}$ by [step 1.2] (bounded below inside the disc), [step 2.2] (on the circle) and [step 1.2] again (invertible, hence bounded below, outside); ${\sigma }_c(S)=\mathbb{T}$ because those points are injective with dense range [step 1.1, step 3.1] and non-surjective (else invertible); ${\sigma }_r(S)={\sigma }_{cp}(S)=D$ by [step 2.1], [step 3.1] and [step 3.2].