I have the following problem:
Let $E=C(\mathbb{R})=\{u\in\mathbb{R}^{\mathbb{R}}\mid \sup_{t\in\mathbb{R}}|u(t)|<\infty\,\land\, $u$\text{ is continuous}\}$ be equipped with the norm $\|u\|_\infty=\sup_{t\in\mathbb{R}}|u(t)|$. Define $T\colon E\to E$ by $(Tu)(t)=u(t+s)$ where $s$ is a real non-zero constant. Prove $T\in\mathcal{L}(E)$ and that $\|T\|=1$. Find $\rho(T)$ (the set of all $\lambda\in\mathbb{R}$ that make $T-\lambda I$ bijective).
For any $y\in \mathbb{R}$, define $g_y\colon\mathbb{R}\to\mathbb{R}$ by $t\mapsto t+y$. I'll adopt the notation $g_s^+=T$. The first part is easy: Let $u,v\in E$ and $\lambda \in \mathbb{R}$, then $g_s^+(\lambda u+v)=(\lambda u+v)\circ g_s=\lambda u\circ g_s + v\circ g_s=\lambda g_s^+(u)+g_s^+(v)$ and $\|g_s^{+} (u)\|=\|u\circ g_s\|=\sup_{t\in\mathbb{R}}|u(t+s)|=\|u\|$ and thus $g_s^+\in\mathcal{L}(E)$ and $\|g_s^+\|=1$.
I'm now trying to prove that, if $|\lambda|>1$, then $\lambda \in \rho(g_s^+)$ (this was a hint on the exercise) but I don't really know how to show that $g_s^+-\lambda I$ is bijective for such $\lambda$. I'm looking for a hint or a solution on how to continue.
You know that $\|T\|=1$. In fact you've shown more: that $T$ is an isometry, that is $\|Tu\|=\|u\|$ for all $u$. Because you have all translations available, you also know that $T$ is invertible. So, for $\lambda\ne0$,
$$\tag1 T-\lambda I = -\lambda^{-1}T^{-1}(T^{-1}-\lambda^{-1}I). $$ This shows that $\lambda\in\rho(T)$ if and only if $\lambda^{-1}\in\rho(T^{-1})$.
Now suppose that $|\lambda|>1$. Then $$ T-\lambda I=-\lambda\,\big(I-\frac{T}\lambda\big). $$ Because $\|T/\lambda\|=1/|\lambda|<1$, it follows that $I-T/\lambda$ is invertible (via the Neumann series). Thus $\{|\lambda|>1\}\subset\rho(T)$. And this does not depend on $s$: so you also get that $\{|\lambda|>1\}\subset\rho(T^{-1})$. By $(1)$, this means that $\{|\lambda|<1\}\subset\rho(T)$. So the only possible values in $\sigma(T)$ are $1,-1$. We have that $T-I$ is not invertible, because it is zero on all constant functions. And $T+I$ is zero for instance on $$ f=\sin \frac{\pi x}s. $$ Thus $\rho(T)=\{\lambda\in\mathbb R:\ |\lambda|\ne1\}$.