Let $T(t)$ be a semigroup of bounded linear operators on a Banach space $X$. When does the following hold $$ \int_0^t T(s)x ds = \Big(\int_0^t T(s) ds\Big)x, \quad x \in X \, , $$ where $ t \in (0,1)$?
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$\begingroup$ Almost never. The operator valued function is in general not measurable (only if the generator is bounded). $\endgroup$– András BátkaiCommented Jun 4, 2017 at 7:14
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1 Answer
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Following ideas of John von Neumann,
J. von Neumann, Über einen Satz von Herrn M. H. Stone, Ann. Math. (2) 33, 567-573 (1932). ZBL0005.16402,
it can be shown that if a function satisfies the semigroup law and is measurable, then it is already continuous. Hence your identity is only true if the generator is bounded.
See also
Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036., Exercise I.1.7.(2).
answered Jun 4, 2017 at 7:39
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$\begingroup$ Thanks. If $T(t)$ is a uniformly contiouous semigroup, then the identity is true. Right? $\endgroup$– Y ChenCommented Jun 6, 2017 at 0:31
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$\begingroup$ Yes. But then, the identity of the right-hand side is also just a Riemann integral of a continuous function. In that case, the generator is a bounded operator, which is almost never satisfied in relevant applications. $\endgroup$ Commented Jun 6, 2017 at 6:27