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)$?

  • $\begingroup$ Almost never. The operator valued function is in general not measurable (only if the generator is bounded). $\endgroup$ Commented Jun 4, 2017 at 7:14

1 Answer 1


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).

  • $\begingroup$ Thanks. If $T(t)$ is a uniformly contiouous semigroup, then the identity is true. Right? $\endgroup$
    – Y Chen
    Commented Jun 6, 2017 at 0:31
  • $\begingroup$ 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

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