Let $X$ be a Banach space, $T(t)$ be a strongly continuous semigroup on $X$, and $f\in L^1(0,\tau;X)$. It has been implied that the integral $$v(t)=\int_0^t T(ts)f(s)ds,\quad t\in [0,\tau]$$ is not always an element of $W^{1,1}(0,\tau;X)$. That seems odd to me. Can anyone think of an example?

$\begingroup$ I changed the title so as to yield some information about the mathematical contents, but this can maybe still be improved... $\endgroup$ – YCor Sep 19 '18 at 22:07
In other words, if $A$ is the infinitesimal generator of $T$, the mild solution of the abstract inhomogeneous Cauchy problem $$\begin{cases}\dot v =A v +f\\v(0)=0\end{cases}$$ needs not to be $W^{1,1}_{loc}(\mathbb{R}_+, X)$. For instance an $f\in C^0(\mathbb{R}_+,X)$ of the form $f(t):=T(t)x$ for some $x=x(\theta)\in X$, gives $v(t)=tT(t)x=tf(t)$ which has no reason to be in $W^{1,1}_{loc}(\mathbb{R}_+, X)$.
To justify the preceding claim it is sufficient to show that $f(t)$ itself, the mild solution to the homogeneous Cauchy problem $\dot f=Af$ with $f(0)=x$, may fail to be $W^{1,1}$ (even at any point). Consider e.g.
$X:=L^1(\mathbb{S}^1)$, $1$periodic one variable $L^1_{loc}$ functions;
$T :\mathbb{R}_+\times X\to X$ the left translation semigroup $T(t)x:=x(\cdot+t)$, whose infinitesimal generator is $A:=\partial_\theta$, with domain $D(A)=W^{1,1}(\mathbb{S}^1)$;
Then, for $x\in X$, $f(t):=T(t)x=x(\cdot+t)$ is, of course, the mild solution to $\dot f(t)=\partial_\theta f(t)$ with initial data $f(0)=x$, and defines a continuous path $f:\mathbb{R}\ni t \mapsto x(\cdot+t)\in X$. Saying, for some open interval $I$, that $f\in W^{1,1}(I;X)$ means there is $h\in L^1(I;X)\sim L^1(I\times \mathbb{S}^1)$ such that $f(t')f(t)=\int_t^{t'} h(s,\cdot)ds$ in $X$, that is $x(\theta+t')x(\theta+t)=\int_t^{t'} h(s,\theta)ds$ a.e., whence $x\in W^{1,1}(\mathbb{S}^1)$.

1$\begingroup$ For the semigroup of translations in the example, if $f \in W^{1,1}$ then $x\in W^{1,1}(\mathbb{S}^1)=D(A)$. Therefore $f\in C^1$! For solutions of an abstract Cauchy problem, $\dot f=Af$, $f(0)=x$, is this always true? $\endgroup$ – Pietro Majer Sep 20 '18 at 20:41

1$\begingroup$ No, this is not always true: Let $X = \ell^1$ and let $A(y_n) = (ny_n)$ whenever $(ny_n) \in \ell^1$. Choose $x = (1/n^2)$ and $I = (0,1)$. Note that $x \not\in D(A)$. Moreover, $\dot{f}(t) = (e^{nt}/n)$ and hence $\ \dot{f}(t) \ = \log(1e^{t})$ for all $t \in (0,1)$. Hence, we have $\int_0^1 \\dot{f}(t)\ \, dt = \int_0^1 e^{t}\log(1e^{t})/e^{t} \, dt = \int_1^{1/e} \log(1s)/s \, ds < \infty$, so $\dot{f}$ is Bochner integrable over $(0,1)$. Thus, $f \in W^{1,1}(I)$. $\endgroup$ – Jochen Glueck Sep 29 '18 at 8:37