# Convolution with semigroup: does this belong to the Sobolev space $W^{1,1}$?

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 $$v(t)=\int_0^t T(t-s)f(s)ds,\quad t\in [0,\tau]$$ is not always an element of $W^{1,1}(0,\tau;X)$. That seems odd to me. Can anyone think of an example?

• I changed the title so as to yield some information about the mathematical contents, but this can maybe still be improved... – 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 $$x\in W^{1,1}(\mathbb{S}^1)$$.
• 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? – Pietro Majer Sep 20 '18 at 20:41
• 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 $\| \dot{f}(t) \| = -\log(1-e^{-t})$ for all $t \in (0,1)$. Hence, we have $\int_0^1 \|\dot{f}(t)\| \, dt = \int_0^1 -e^{-t}\log(1-e^{-t})/e^{-t} \, dt = \int_1^{1/e} \log(1-s)/s \, ds < \infty$, so $\dot{f}$ is Bochner integrable over $(0,1)$. Thus, $f \in W^{1,1}(I)$. – Jochen Glueck Sep 29 '18 at 8:37