Let $X = \{ u \in C(\mathbb{R}) : \lim_{t \rightarrow \pm \infty} u(t) = 0 \}$ with the supremum norm on X.
Given the operators $S(t): X \rightarrow X, u \mapsto S(t)u := u(\bullet -t)$, so S(t) is the rightshift operator.
I want to show that $(S(t))_{t \in \mathbb{R}}$ is a group that is not uniformly continuous.
Uniform continuity on a group is defined as:
$ \Vert S(t) - Id \Vert_{L(X)}$ as $t \rightarrow 0$, where L(X) is the norm defined as $\sup_{u \in X} \frac{\Vert S(t)u \Vert_\infty}{\Vert u \Vert_\infty}$
Just by the definition I get:
$ \Vert S(t) - Id \Vert_{L(X)} = \sup_{u \in X} \frac{ \sup_{x \in \mathbb{R}} | u(x-t) - u(x) |}{\sup_{x} |u(x)|}$
Now I guess it might be possible to choose some $u_0 \in X$ for which one can show that this is larger than some $\epsilon > 0$, but to me it seems as if this shouldn't be possible, since $u_0$ is supposed to be continuous. Can someone give some help please?
Let $u$ be an element of $X$ whose supremum norm is $1$ . Define $u_n(x):= u(nx)$. Then for all $t$, $$ \left\lVert S(t)-\operatorname{Id}\right\rVert_X\geqslant \sup_{n\geqslant 1}\sup_{x\in\mathbb R} \lvert u_n(x-t)-u_n(x)\rvert=\sup_{n\geqslant 1} \sup_{x\in\mathbb R}\lvert u(nx-nt)-u(nx)\rvert=\sup_{n\geqslant 1} \sup_{x'\in\mathbb R}\lvert u(x'-nt)-u(x')\rvert $$ hence for each integer $N$, $$ \left\lVert S(t)-\operatorname{Id}\right\rVert_X\geqslant \sup_{n\geqslant N} \sup_{x \in\mathbb R}\lvert u(x -nt)-u( x)\rvert \geqslant \sup_{n\geqslant N} \sup_{x \lt 0}\lvert u(x -nt)-u( x)\rvert\geqslant \sup_{n\geqslant N}\left( \sup_{x \lt 0}\lvert u( x)\rvert -\sup_{x \lt 0}\lvert u( x-nt)\rvert\right). $$ which gives, in view of $\sup_{n\geqslant N}\sup_{x \lt 0}\lvert u( x-nt)\rvert\leqslant \sup_{s\leqslant -Nt}\lvert u( s)\rvert$ that $$ \left\lVert S(t)-\operatorname{Id}\right\rVert_X\geqslant\sup_{x \lt 0}\lvert u( x)\rvert -\sup_{s\leqslant -Nt}\lvert u( s)\rvert $$ for $t\gt 0$ and $N\geqslant 1$. Letting $N$ going to infinity shows that $$ \left\lVert S(t)-\operatorname{Id}\right\rVert_X\geqslant\sup_{x \lt 0}\lvert u( x)\rvert. $$