Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space, $X:\Omega\times[0,\infty)\times E\to E$ be $(\mathcal A\otimes\mathcal B([0,\infty))\otimes\mathcal E,\mathcal E)$-measurable and $$X^x_t:=X(\;\cdot\;,t,x)\;\;\;\text{for }(t,x)\in[0,\infty)\times E.$$ It's easy to see that $$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$ is a Markov kernel and $$(\kappa_tf)(x):=\int\kappa_t(x,{\rm d}y)f(y)=\operatorname E\left[f(X^x_t)\right]\tag1$$ for all $\mathcal E$-measurable $f:E\to\mathbb R$ with $(\kappa|f|)(x)<\infty$ for all $(t,x)\in[0,\infty)$.
(1): If $X(\omega,t,\;\cdot\;)$ is Fréchet differentiable, are we able to show that $\kappa_tf$ is Fréchet differentiable for all Fréchet differentiable $f:E\to\mathbb R$?
(2): What can we conclude (and how is this related to the former assumption) if $$E\ni x\mapsto C^0([0,\infty),L^p(\operatorname P,E))\;,\;\;\;x\mapsto X(\;\cdot\;,\;\cdot\;,x)\tag2$$ is Fréchet differentiable?
