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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?

Kyan Cheung
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    Approximate $f$ by a smooth function with compact support. For the smooth function with compact support you can use the mean value theorem. – Severin Schraven May 23 '23 at 04:50

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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 $t\neq 0$.

Severin Schraven
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  • "standard fact" unfortunately not :/ I haven't done that much reading into numerical analysis and I have not heard of Meyers–Serrin until today. I've been treating this more as an introduction to the theoretical aspects of C0-semigroups, while skipping over the numeric aspects of the lectures. – Kyan Cheung May 23 '23 at 05:48
  • That has nothing to do with numerics (at least to me that is a standard PDE fact if anything, but then a lot of numerics is about PDEs, don't know what your understanding of numerics is). It's covered in pretty much any introductory book that deals with Sobolev spaces. – Severin Schraven May 23 '23 at 05:55
  • So TL;DR approximate by a continuous function, where we know the conclusion holds? – Akiva Weinberger Oct 12 '23 at 03:47
  • Not quite, differentiable (or smooth as used here) function with compact support and show conclusion holds – Kyan Cheung Oct 12 '23 at 04:10
  • @AkivaWeinberger If you want to use continuous and compactly supported then that works too :) Then we just use uniform continuity. But yes, at the end of the day the idea that we restrict to a nice function space and then use density to extend to more general functions. – Severin Schraven Oct 12 '23 at 05:09
  • @SeverinSchraven I did mean to say differentiable. D'oh! (After all, the point is that $\phi(s+t)-\phi(s)\approx\phi'(s)t$.) But I had completely missed that compact support is necessary. Incidentally, do we need $C^\infty$, or is $C^1$ enough? – Akiva Weinberger Oct 12 '23 at 06:08
  • @AkivaWeinberger Surely $C^1$ is enough. I only took smooth functions out of habit. As pointed out above, compact supported and continuous would be fine too as such functions would be uniformly continuous and so the difference between the shifted version would be small in $L^\infty$. However, we cannot drop compact support as we are integration over a set of infinite measure. – Severin Schraven Oct 12 '23 at 16:21