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Let $(\kappa_t)_{t\ge0}$ be a semigroup of linear, conservative, contractive and nonnegative operators on $C_0(E)$ with $$(\kappa_tf)(x)\xrightarrow{t\to0}f(x)\tag1\;\;\;\text{for all }x\in\mathbb R$$ for all $f\in C_0(\mathbb R)$. Let $(\mathcal D(A),A)$ denote the generator of $(\kappa_t)_{t\ge0}$. Suppose we know that $$\kappa_tf-f=\int_0^t\kappa_s Af\:{\rm d}s\;\;\;\text{for all }f\in\mathcal D(A)\tag2.$$

How can we conclude that $(\kappa_t)_{t\ge0}$ is strongly continuous on $\mathcal D(A)$?

Clearly, $(2)$ again yields $(1)$, but I don't see why the convergence is uniform with respect to $x$.

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  • $\begingroup$ $C_{0}(\mathbb{R})$ denotes the space of continuous functions vanishing at infinity, right? $\endgroup$
    – sharpe
    Nov 24, 2018 at 10:27
  • $\begingroup$ @sharpe Yes, that's correct. $\endgroup$
    – 0xbadf00d
    Nov 24, 2018 at 11:26
  • $\begingroup$ $(2)$ gives $$\|\kappa_t f-f\|_{\infty} \leq t \|Af\|_{\infty}$$ which immediately yields the desired convergence. $\endgroup$
    – saz
    Nov 30, 2018 at 14:00
  • $\begingroup$ @saz Is it possible to obtain the strong continuity without $(2)$? math.stackexchange.com/questions/3096864/… $\endgroup$
    – 0xbadf00d
    Feb 2, 2019 at 17:39

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