# The lack of uniform continuity of shift operator on $L^2$

When studying $c_0-$semigroups, I came accross a statement that if we define shift operator $(S(t)f)(x) = f(x+t)$ for $t>0$ on $L^2(\mathbb{R})$, then $S(t)$ forms a $c_0-$semigroup (that's easy) and the semigroup is not uniformly continuous since $||S(t) - I|| = 2$ for every $t>0$.

Sadly, I struggle to verify this identity. Clearly $||S(t) - I|| \leq2$ but I can't prove the identity. The best I could get was $||S(t) - I||\geq\sqrt{3}$ by taking $f(x) = \frac{1}{\sqrt{2t}}(\chi_{(0,t)}-\chi_{(-t,0)})$.

Consider the Rademacher functions $r_n(x) = \operatorname{sign}\sin 2^n x$, restricted to $[0, 2\pi]$ interval. Translating by $\pi/2^n$ changes $r_n$ almost into $-r_n$, except for two pieces of size $\pi/2^n$ on the boundary. Hence, the $L^2$ norm of $S(\pi/2^n)r_n-r_n$ is almost twice the norm of $r_n$.
This extends to general translation amounts $t$ by scaling.
• Thanks a lot - this works. To be honest, though, this is the first time I've seen Rademacher functions. The lecture notes I follow contains this remark (I should've mentioned it in the original question):"Fix $t > 0$ and consider $f(x)\in L^2(\mathbb{R})$ with unit norm and support very close to $x = 0$". Any idea what could the author mean? – user1321324 May 7 '16 at 16:13