How to show that $\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0$?

Question

For every $t\in[0, \infty)$ consider the bounded linear operator $S(t):C_{ub}(\mathbb R)\rightarrow C_{ub}(\mathbb R)$ given by $$S(t)u(x)=u(x+t),$$ where $C_{ub}(\mathbb R)$ is the set of all bounded and uniformly continuous functions over $\mathbb R$. How can I show that $$\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0?$$ The norm on the above limit is the operator norm whereas the norm on the space $C_{ub}(\mathbb R)$ is given by $\|f\|_\infty=\sup_{x\in \mathbb R}|f(x)|$..I've already spent a lot time on time and got nowhere..Any help will be valuable..Thanks

Answer 1

For a fixed $t\neq 0$, we can find $u_t$, a piecewise linear function for which $u_t(0)=0$, $u_t(t)=1$ and $0\leqslant u_t\leqslant 1$.

Answer 2

A theoretical sledgehammer: Otherwise the semigroup $S(t)$ would be uniformly continuous so that its infinitesimal generator $A(f)=f'$ would be everywhere defined. The existence of continuous but non-differential functions is thus a reason for the failure of uniform convergence.