Let us regard the Hilbert space $L^2(0,1)$ and the $C_0$-semigroup $(T(t))_{t\geq 0}$ defined by $$ T(t):\left\{ \begin{array}{rml} L^2(0,1) & \to & L^2(0,1), \\ [f]_{\sim} &\mapsto &\left[x \mapsto \begin{cases} f(x+t), & \text{if}\; x+t<1\\ 0, & \text{else} \end{cases} \right]_{\sim}. \end{array} \right. $$ It is easy to verify that this is indeed a $C_0$-semigroup. Therefore, the mapping $t \mapsto T(t)f$ is a continuous mapping from $L^2(0,1)$ to $L^2(0,1)$. Consequently the $L^2(0,1)$-valued integral $$ g := \int_0^1 T(t)f \,\mathrm{d}t $$ exists.


In order to get some information about the behavior of $g$ it would be nice to regard $g$ as a parameter integral. Hence I am interested in the following equality $$ g(x) = \Big(\int_0^1 T(t)f \,\mathrm{d}t\Big) (x)\stackrel{?}{=} \int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t . $$ Or with a different notation $$ g = \int_0^1 \big(x \mapsto \big(T(t)f\big)(x) \big)\,\mathrm{d}t \stackrel{?}{=} \Big(x\mapsto\int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t\Big) . $$ The evaluation mapping is neither continuous nor well-defined on $L^2$. So I think it is not trivial to justify this step.

It seems quite common to evaluate such $L^2(0,1)$-valued integrals by interpreting it as a parameter integral, so I guess that there is a theorem which justifies that. It would be really great if someone had a reference.

Solution for this special case

In this particular case I think I have a solution. I know that every convergent sequence in $L^2$ has a subsequence which converges even point-wise a.e.. Since $$ g_n := x\mapsto \sum_{i=1}^{n} \frac{1}{n} \Big(T\Big(\frac{i}{n}\Big)f\Big)(x) $$ converges to $g$ and every subsequence of $g_n(x)$ converges in $\mathbb{R}$ to the same limit for a.e. $x\in (0,1)$, the point-wise limit of $g_n$ has to coincide with $g$ a.e..

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    $\begingroup$ Very good question! I've also encountered this issue several times, and found it quite subtle. I have two remarks: (1) I have difficulties to follow your solution of the special case: why does $g_n$ converge to $g$ almost everywhere? More precisely, how do you know that the mapping $t \mapsto T(t)f(x)$ is Riemann integrable for almost every $x$? (2) Do you have any specific situations/applications in mind where you wish to apply this? I often found it quite helpful to drop the almost-everywhere-perspective and work with duality instead. $\endgroup$ – Jochen Glueck May 18 '18 at 21:01
  • $\begingroup$ @JochenGlueck (1) You have a good point I don't know if $t\mapsto T(t)f(x)$ is Riemann integrable. I know that it is $L^1$. Maybe someone can find a justification for this. (2) The specific situation is the setting I stated. I want to calculate the domain of the infinitesimal generator of this $C_0$-semigroup. $\endgroup$ – Nathanael Skrepek May 19 '18 at 21:03
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    $\begingroup$ Do you really have difficulties with the particular example? Or does it only exemplify a general problem? Note that already your definition of $T_t$ is formally not correct (because the elements of $L^2$ aren't functions but equivalent classes. You could define $T_t(f)$ by the formula for continuous functions, check continuity and extend by general abstract nonsense to all of $L^2$. $\endgroup$ – Jochen Wengenroth May 22 '18 at 14:08
  • $\begingroup$ @JochenWengenroth As I already mentioned in my question: It seems quite common to interpret $L^2$-valued integrals as parameter integrals eventhough they are a priori different things. I realized that when I stumbled on this example. I edited the definition of $T(t)$. I already have an answer to my question. I will post it soon. $\endgroup$ – Nathanael Skrepek May 23 '18 at 8:24

The trick is to show that both functions $\Big(\int_0^1 T(t)f \,\mathrm{d}t\Big) (x)$ and $\int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t$ induce the same element in the dual space. Let $h \in L^2(0,1)$ be arbitrary. Since the scalar product is continuous in both arguments, we have $$ \Big\langle h, \int_0^1 T(t)f \,\mathrm{d}t \Big\rangle = \int_0^1 \big\langle h, T(t)f \big\rangle \,\mathrm{d}t = \int_0^1 \int_0^1 h(x) \big(T(t)f\big)(x) \,\mathrm{d}x\,\mathrm{d}t $$ It is easy to check that $(t,x) \mapsto \big(T(t)f\big)(x)$ is an element of $L^2\big((0,1)\times (0,1)\big)$ which allows us to use Fubini $$ = \int_0^1 h(x)\int_0^1 \big(T(t)f\big)(x) \,\mathrm{d}t\,\mathrm{d}x = \Big\langle h, x \mapsto \int_0^1 \big(T(t)f\big)(x)\,\mathrm{d}t \Big\rangle . $$

Actually, if $f: (0,1) \to L^2(0,1)$ is integrable and $f(t)(x)$ is $L^2\big((0,1)\times (0,1)\big)$, then we can also regard the $L^2(0,1)$-valued integral $\int_0^1 f(t) \,\mathrm{d}t$ as the parameter integral $x\mapsto \int_0^1 f(t)(x) \,\mathrm{d}t$.


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