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Let $T(t)_{t \ge 0}$ be a strongly continuous semigroup on a Hilbert space $H.$

Then, one can consider the function

$f(t_1,t_2):= T(t_1)S T(t_2)x$ where $x$ is a fixed element of the Hilbert space and $S$ a bounded operator.

Obviously this function is continuous componentwise (by strong continuity of the semigroup). I am wondering however whether it is also continuous as a map $f:\mathbb R_{\ge 0}^2 \rightarrow H.$

Obviously, continuity holds if the semigroups are uniformly continuous, so the question is only interesting in the case of only strongly continuous semigroups.

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Strongly continuous semigroups are locally bounded (in $t$), hence if we have a sequence $(t_n)$ converging to $t_1$ and another one $(t_n^{\prime})$ converging to $t_2$ then the sequence $(T(t_n)S T(t_n^{\prime}))$ converges to $T(t_1)S T(t_2)$, since we have a product of two bounded, strongly convergent sequences and multiplication is continuous in strong operator topology on bounded subsets.

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  • $\begingroup$ @ I assume the continuity of multiplication on bounded sets follows somehow from the open mapping theorem or is it simpler? $\endgroup$
    – Sascha
    Commented Apr 20, 2018 at 14:53
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    $\begingroup$ @Sascha: Simpler. If $A_n \to A$, $B_n \to B$ strongly, write $A_n B_n -A B = (A_n B_n - A_n B) + (A_n B - AB)$, and for the first term, use the fact that the sequence $A_n$ is bounded. $\endgroup$
    – Nate Eldredge
    Commented Apr 20, 2018 at 15:28
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    $\begingroup$ Maybe it is worthwhile to point out the following two small remarks: (i) To see that @Nate Eldredge's argument implies joint strong continuity of operator multiplication on bounded sets, one actually has to take $(A_n)$ and $(B_n)$.to be nets (not only sequences). (ii) On the other hand, operator multiplication is jointly sequentially strongly continuous on the entire operator space (not only on sets which are a priori bounded) since a convergent operator sequence is automatically bounded due to the uniform boundedness theorem. $\endgroup$
    – Jochen Glueck
    Commented Apr 20, 2018 at 19:21

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