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.



Strongly continuous semigroup: continuous or continuous componentwise?
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.


