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Let $X$ be a Banach space and $$\mathrm C_0(\mathbb R,X):=\{f\colon\mathbb R\to X\colon f \text{ is continuous and } \lim\limits_{|t|\to\infty}f(t)=0\}$$ normed by $$\|f\|:=\sup\limits_{t\in\mathbb R}\|f(t)\|, \quad f\in \mathrm C_0(\mathbb R,X).$$ Let $(U(t,s))_{t\ge s}$ denote an evolution family of bounded linear operators on $X$ satisfying: $$U(t,r)U(r,s)=U(t,s)$$ and $U(t,t)=\operatorname{Id}$, the identity on $X$, for all real numbers $s\le r\le t$. Let $(R(t))_{t\ge 0}$ denote the right shift semigroup: $$(R(t)f)(s)=f(t-s)$$ for all $s\in\mathbb R$, $t\ge 0$, and $f\in \mathrm C_0(\mathbb R,X)$.

Does the evolution family and the right shift semigroup commute? More precisely, it is true that for any $T\ge 0$ $$U(t,t-s)R(s)f=R(s)U(t,t-s)f$$ holds for all $t,s\in[0,T]$ and $f$ from some appropriate space (which?)?

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In general no; this happens only in very special cases.

To make this simpler, let's suppose $U$ is a semigroup (a time-homogeneous evolution family), so $U(s,t) =U(t-s)$. Then formally, this commutation would only occur if the generator $A$ of $U(t)$ were to commute with the generator of $R(t)$, which is $\frac{d}{dx}$. So we get counterexamples by taking $U(t)$ to be generated by any operator $A$ which does not commute with $\frac{d}{dx}$.

Perhaps the simplest counterexample is to let $A$ be multiplication by any nonnegative non-constant continuous function $h$, so $U(t)f = e^{-th} f$. Then you can easily see that $$\begin{align*}(U(t)R(s)f)(x) &= e^{-th(x)} f(x-s) \\ (R(s) U(t) f)(x) &= e^{-th(x-s)} f(x-s)\end{align*}$$ which are not the same.

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