Find the spectrum of $Ax(t)=x(t+s)$ Let be $\mathcal{C}_\mathbb{R}$ the set of the continuous and bounded function in $\mathbb{R}$ with norm $ \|x\|=\underset{t\in\mathbb{R}}{sup}|x(t)|$ Consider the operator $A$ such that $Ax(t)=x(t+s)$ for some $s\in\mathbb{R}$. Show that the spectrum of A is the unit sphere $=\{\lambda\in\mathbb{C}: |\lambda|=1\} $.
Please, I don't know how to aboard this exersice. I find the norm of $A$, $\|A\|=1$ so $\sigma(A)\subseteq \bar{D}(0,1)$. Then I was trying to show that if $|\lambda|<1$ then $A-\lambda Id$ is invertible, so I began from show that it's injective but I only find that $|x(t+s)|<|x(t)|$ for all $ t\in\mathbb{R}$ and I don't know why this implies that $x=0$.
Or maybe this is not te correct way to make this exercise please could you give me some hints?
 A: $\|Ax\|=\|x\|$ and $A$ is surjective: Given $y$ note that $Ax=y$ where $x(t)=y(t-s)$. Since $\|A\| =1$ we see that $\sigma (A) \subseteq \{\lambda: |\lambda | \leq 1 \}$. Using the fact that $\|A^{-1}\| \leq 1$ we see that  $\sigma (A) \subseteq \{\lambda: |\lambda | \geq 1 \}$. To finish the proof note that $|\lambda|=1$ implies that thre exists a continuous function $x$ with $Ax=\lambda x$.
[ Suppose $s>0$. Let $x$ be any continuous function on $[0,s]$ with $x(0)=x(s)=0$, say $x(t)=s(t-s)$. Define $x$ on $[ns,(n+1)s]$ by $x(t)=\lambda ^{n} x(t-ns)$. You can check that this function is continuous, bounded and $Ax=\lambda x$].
A: This is definitely not enough to finish the exercise, but a couple things you could notice:

*

*$A$ is an isometry, i.e. $\|Ax\|=\|x\|$.


*if $|\lambda|<1$, then $$ \|(A-\lambda I)x\|=\|Ax-\lambda x\|\geq\|Ax\|-|\lambda|\,\|x\|=(1-|\lambda|)\|x\|,$$ so $A-\lambda I$ is bounded below which shows that it is injective and has closed range.


*when $|\lambda|=1$, you can show that $\lambda$ is an eigenvalue by constructing an eigenvalue explicitly. Not needed, by the multiplicity of each eigenvalue is infinite.
