Shift Operator not continuous

Question

Let $X=L^1(\mathbb{R}^n)$ and $T:\mathbb{R} \rightarrow L(X)$, such that $(T(\tau)f)(t):=f(t+\tau)$. The question is: Is $T$ continuous?

Well, my idea was the following:

$|\tau_1-\tau_2| \le \delta \rightarrow ||T(\tau_1)-T(\tau_2)|| = sup_{f,||f||=1} \int |f(t+\tau_1) - f(t+\tau_2)|$

But the problem is: We could take any function with compact support(for example $[0,1]$), such that on intervals of length $|I|=\frac{\delta}{2}$, $f$ would be +1 on $[0,\frac{\delta}{2}]$ and negative one on $[\frac{\delta}{2},\delta]$ and so on. So the integral could become arbitrarily large. So I guess the shift-operator $T$ is not continuous, is this right?

Answer 1

This operator semigroup is continuous in the following sense: For every $u\in L^1(\mathbb R)$ $$ \lim_{h\to 0}\|T(t+h)u-T(t)u\|_{L^1(\mathbb R)}=0. $$ This is due to the fact that continuous compactly supported functions are dense in $L^1(\mathbb R)$ - $T(t)$ is strongly continuous.

However, the same operator semigroup IS NOT continuous in the operator norm, i.e., $\|T(t+h)-T(t)\|\not\to 0$, as $h\to 0$, as $\|T(t+h)-T(t)\|=2$, for all $h>0$, let $u$ be a function supported in $[0,h]$, and thus $\|(T(t+h)-T(t))u\|=2\|u\|$.

Answer 2

It is continuous.

It is enough to prove that $T$ is continuous at $\tau=0$. Given $f\in L^1$ we have to prove $$ \lim_{\tau\to0}\int_{\mathbb{R}^n}|f(x+\tau)-f(x)|\,dx=0. $$

Prove it first if $f$ is continuous with compact support, using the uniform continuity of $f$ in that case.

Then argue by density, using that the space of continuous functions with compact support is dense in $L^1$.

The argument works for $L^p$ with $1\le p<\infty$.