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

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Sep 29, 2018 at 8:37	comment	added	Jochen Glueck		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, $f \in W^{1,1}(I)$.
Sep 20, 2018 at 20:41	comment	added	Pietro Majer		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?
Sep 20, 2018 at 20:35	history	edited	Pietro Majer	CC BY-SA 4.0	added 8 characters in body
Sep 19, 2018 at 21:57	history	edited	Pietro Majer	CC BY-SA 4.0	added 4 characters in body
Sep 19, 2018 at 21:51	vote	accept	Saj_Eda
Sep 19, 2018 at 21:30	history	edited	Pietro Majer	CC BY-SA 4.0	u->t
Sep 19, 2018 at 20:35	history	edited	Pietro Majer	CC BY-SA 4.0	added 4 characters in body
Sep 19, 2018 at 20:27	history	answered	Pietro Majer	CC BY-SA 4.0