Setting
-------


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.


Question
--------


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..



$L^2$-valued integral as parameter integral
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 $\Phi: (0,1) \to L^2(0,1)$ is integrable and $\Phi(t)(x)$ is $L^2\big((0,1)\times (0,1)\big)$, then we can also regard the $L^2(0,1)$-valued integral $\int\_0^1 \Phi(t) \,\mathrm{d}t$ as the parameter integral $x\mapsto \int\_0^1 \Phi(t)(x) \,\mathrm{d}t$.


