Prove by means of proposition and proposition that under the assumptions of theorem the subspace $F\colon=\dom L\cap L_\infty(\mu)$ is a core for $L$. 2. Conclude that $\ker L+L(F)$ is dense and for all $f=Lg\in L(F)$ we have: $\norm{A_tf}_\infty\leq2\norm g_\infty/t$.
Since both $\dom L$ and $L_\infty(\mu)$ are $P_t$ invariant it suffices by proposition to prove that $F$ is dense: By proposition we conclude that for all $f\in L_\infty(\mu)\cap L_1(\mu)$: $u(t)\colon=\int_0^t P_sf\,ds$ is both in $L_\infty(\mu)$ and in $\dom L$. Since $u(t)/t$ converges in $L_1(\mu)$ to $f$ the $L_1(\mu)$-closure of $F$ contains the dense subspace $L_\infty(\mu)\cap L_1(\mu)$ of $L_1(\mu)$.
2. As $F$ is a core $L(F)$ is dense in $\im L=L(\dom L)$ and thus $\ker L+L(F)$ is dense in $L_1(\mu)$. Finally for $f=Lg$ and $g\in F$ it follows that $$ A_tf =\frac1t\int_0^t P_sLg\,ds =\frac1t\int_0^t \ttd sP_sg\,ds =\frac1t(P_sg-g) $$ and the $L_\infty(\mu)$-norm of this is bounded by $2\norm g_\infty/t$.