compute the spectrum of an operator Let $C_{0}(\mathbb R)$ be the space of complex continuous valued functions on $\mathbb R$ which vanish at $\infty$ equipped with the sup-norm. Let $S:C_{0}(\mathbb R) \to C_{0}(\mathbb R)$ given by $(Sf)(t)=f(t+s)$ for a fixed $s\in \mathbb R$. 
 Prove that the spectrum of $\sigma(S)$ is given by the set $\sigma(S)=\{z\in \mathbb C ; ||z||=1\}$.
I proved that if $|\lambda|\ne 1$ then $\lambda \in p(T)$. I want to prove that if $|\lambda|=1$ then $\lambda \in \sigma(T)$
Since if $|\lambda|=1$ I proved that $\lambda I-T$ it's injective. It remains to prove that $\lambda I-T$ it's not surjective.
Thanks for the tips! But it remains to prove that if $||z||=1$ then $z\in \sigma(T)$. And I don't know how
 A: This won't be a complete answer, but is just to indicate an idea of why the statement might be true:
The space $\mathcal C_0$ is dense in $L^2$, and so the spectrum of $S$ on $\mathcal C_0$ should be closely related to its spectrum as an operator on $L^2$.
On the other hand, $L^2$ is morally the direct sum of the the various one-dimensional spaces spanned by the functions $e^{ixy}$; more precisely, the theory
of the Fourier transform exhibits $L^2$ as the direct integral of these lines.
On the the line spanned by $e^{ixy}$, the operator $S$ acts as multiplictation by $e^{isy}$.  As $y$ varies over all elements of $\mathbb R$, the eigenvalues
$e^{isy}$ vary over all elements of absolute value $1$ in $\mathbb C$.  Thus,
at least morally, the spectrum of $S$ on $L^2$ is equal to the set of elements of $\mathbb C$ of absolute value $1$, and so (again, at least morally), the
same should be true of $S$ acting on $C_0$.
A: If $0\le|\lambda|<1$, you can explicitly compute
$$
(\lambda I - T) ^{-1} = -\sum_{i=0}^{\infty} \lambda^{i}T^{-i-1}.
$$
You can verify this is, indeed, an inverse and it's continuous.
