I've been self-studying $C_0$-semigroups through Innsbruck's online lectures and one of the exercises is to show that the left-shift operator
$$(S(t)f)(s)=f(t+s)$$
forms a strongly continuous semigroup over $L^p(\mathbb{R})$ space. (for $1\leq p<\infty$)
It feels like it has an easy proof, but honestly I'm quite lost as
I've been thinking of a following proof:
Our goal is to show that for any $\varepsilon>0$, there exists a $h>0$ such that $$\int^\infty_{-\infty}|f(t+h)-f(t)|^p\,dt<\varepsilon$$ provided that $\int^\infty_{-\infty}|f(t)|^p\,dt$ is finite.
Consider an interval $I=[a,b]$ such that $\int_{\mathbb{R}\setminus[a,b]}|f(t)|^p\,dt<\delta$. Then, $$\int_{\mathbb{R}\setminus[a-h, b]}|f(t+h)-f(t)|^p\,dt\leq \int_{\mathbb{R}\setminus I}|2f(t)|^p\,dt\leq 2^p\delta$$ We then consider the integral in the bound $[a-h, b]$. As $f(t)\in L^p$, it must be measureable.
Now consider intervals of width $\alpha$, which cover the reals. Consider each interval's preimage from $f$, intersected with $[a-h, b]$, and denote it $S$. $S$ is measureable, so it is composed of countable unions of intervals. As a result, it is possible to construct a countable cover of $[a-h, b]$ (denoted $C$, where in each interval $S$, $x_1, x_2\in S\implies|f(x_1)-f(x_2)|<\alpha$.
Consider $M=\{|S|:S\in C\}$. $M$ is bounded below by $0$, and $\sum M=b-a+h=l$, so there exists a supremum of $M$. At the same time, $[l-\varepsilon, l]\cap M$ must be finite, so there exists a maximal element of $M$. We can repeat this process to arrange the nonzero elements of $M$ by size.
There are a finite number of intervals $S=[c,d]$ with width $w\geq h$. For each, we know that for $t\in [c, d-h]$, we have $|f(t+h)-f(t)|^p\leq \alpha^p$. Denote these intervals $S'$ and their union $C'$. As $h$ decreases, $C'$ will gradually encompass all of $I$, minus a set with measure $0$.
Hence, we can write our integral as \begin{align*} \int^\infty_{-\infty}|f(t+h)-f(t)|^p\,dt&=\int_{\mathbb{R}\setminus[a-h, b]}|f(t+h)-f(t)|^p\,dt+\int_{C'}|f(t+h)-f(t)|^p\,dt+\int_{[a-h, b]\setminus C'}|f(t+h)-f(t)|^p\,dt \\ &\leq2^p\delta+(b-a+h)\alpha^p+\int_{[a-h, b]\setminus C'}|f(t+h)-f(t)|^p\,dt \end{align*} By making $\alpha$ and $h$ arbitrarily small, we can then make $(b-a+h)\alpha^p$ and $[a-h, b]\setminus C'$ arbitrarily small.
The main issue I see with this proof is that there isn't a straightforward way of bounding $\int_{[a-h, b]\setminus C'}|f(t+h)-f(t)|^p\,dt$. It would be possible if $f$ is bounded, but it isn't.
People have said that proving this is easy but I really can't see how one can prove it. Can anyone explain?
Let $1\leq p <\infty$. Let $f\in L^p(\mathbb{R})$ and $\varepsilon>0$ arbitrary, but fixed. Pick a function $\varphi\in C_c^\infty(\mathbb{R})$ such that $\Vert f-\varphi\Vert_{L^p(\mathbb{R})}\leq \varepsilon/3$ (given the question you are asking, I assume that this is a standard fact to you). Then we have
$$ \Vert S(t)(f)-f \Vert_{L^p(\mathbb{R})} \leq \Vert S(t)(f-\varphi) \Vert_{L^p(\mathbb{R})} + \Vert S(t) \varphi - \varphi\Vert_{L^p(\mathbb{R})} + \Vert f-\varphi \Vert_{L^p(\mathbb{R})}. $$
Translation does not change the $L^p$-norm of a function, so we get
$$ \Vert S(t)(f)-f \Vert_{L^p(\mathbb{R})} \leq 2 \Vert f -\varphi \Vert_{L^p(\mathbb{R})} +\Vert S(t)(\varphi) -\varphi \Vert_{L^p(\mathbb{R})} \leq 2\varepsilon/3 + \Vert S(t)(\varphi) -\varphi \Vert_{L^p(\mathbb{R})}. $$
However, we know that
$$\vert S(t)(\varphi)(s)-\varphi(s) \vert = \vert\varphi(s+t)-\varphi(s) \vert \leq \Vert \varphi'\Vert_{L^\infty(\mathbb{R})} \cdot \vert t \vert. $$
Thus, we get
$$ \Vert S(t) \varphi - \varphi \Vert_{L^p(\mathbb{R})} \leq \Vert \varphi'\Vert_{L^\infty(\mathbb{R})} \vert \text{supp}(\varphi) \vert^{1/p} \cdot \vert t \vert. $$
Thus, choosing $\vert t \vert$ sufficiently small (with respect to the choice of $\varphi$), we get
$$ \Vert S(t)(f) -f\Vert_{L^p(\mathbb{R})} \leq \varepsilon. $$
As usual, evil things happen for $p=\infty$. The semigroup is not continuous, as we have
$$ \Vert S(t)1_{[0,1]} - 1_{[0,1]} \Vert_{L^\infty(\mathbb{R})} = 1 $$
for all $t\neq 0$.