I'm studying a proof of a large deviations principle and I'm having trouble with a part that is concerned with the semigroup property of a family of operators.
Assumptions
$(X_t)_{t\geq0}$ is a symmetric random walk on $\mathbb Z^d$. $\mathbb E_x$ designates the expectation given that $X_0=x$. Let $Q \in \mathbb Z^d$ be a finite connected subset. $\mathbb R^Q$ is the set of functions from $Q$ to $\mathbb R$. $\ell_t$ defined by $\ell_t(z)=\int_0^t \mathbb 1_{\{X_s=z\}}\mathrm ds=\int_0^t \delta_{X_s}(z)\mathrm ds$ for $z\in\mathbb Z^d$ and $t>0$ are the local times of the walk, measuring the amout of time that the random walk spends in state $z$ up to time $t$.
For an arbitrarily fixed $V\in\mathbb R^Q$ and $t>0$ define the operator $P_t^V$ on $\mathbb R^Q$ as $$P_t^V f(x)=\mathbb E_x \left[\mathrm e^{\int_0^tV(X_s)\mathrm ds}f(X_t) \mathbb 1_{\{\mathrm{supp}(\ell_t)\subset Q\}} \right]$$ for $x\in Q$ and $f\in\mathbb R^Q$.
Problem
I don't get a remark in the proof that claims it's easy to check that the family $(P_t^V)_{t\geq0}$ is a strongly continuous semigroup of linear, continuous operators on $\mathbb R^Q$. Unfortunately, I have no prior knowledge of semigroups and I don't quite get why this is true.
Mentioning continuous, I should say a word about the norm on $\mathbb R^Q$: Since $Q$ is finite, we can identify $\mathbb R^Q$ with the Hilbert space $\ell^2(Q)$ (of sequences in $\mathbb R$ indexed over $Q$) and consider the standard dot product on that space as well as the $2$-norm induced by that dot product.
However, I'm mainly concerned with the algebraic properties of $(P_t^V)_{t\geq0}$ because only those are relevant for the proof. That $P_0^V$ is the identity operator/matrix is easy to see even for me. But I'm heaving trouble to ascertain that $P_t^V \circ P_t^V=P_{t+s}^V$.
Let $x\in Q$ and $f\in\mathbb \ell^2(Q)$ be arbitrarily chosen. Then we get something like $$P_t^V\left(P_s^V f(x)\right)=\mathbb E_x \left[\mathrm e^{\int_0^tV(X_h)\mathrm dh}P_s^Vf(X_t) \mathbb 1_{\{\mathrm{supp}(\ell_t)\subset Q\}} \right]=\mathbb E_x \left[\mathrm e^{\int_0^tV(X_s)\mathrm ds}\mathbb E_{X_t} \left[\mathrm e^{\int_0^sV(X_h)\mathrm dh}f(X_s) \mathbb 1_{\{\mathrm{supp}(\ell_s)\subset Q\}} \right] \mathbb 1_{\{\mathrm{supp}(\ell_t)\subset Q\}} \right]$$ which doesn't really make a lot of sense to me...
Can someone help me properly deal with this operator and maybe even attach some meaning to it?
We will begin with $P^V_{s+t}f$. It is convenient to extend $f$ and $V$ to all of $\mathbb{R}^{\mathbb{Z}^d}$ by setting them to be zero outside of $Q$.
The first step is to explicitly split the random variable into the part that depends on information up to and including time $t$, and the part that depends on information after time $t$.
$$e^{\int_0^{t+s} V(X_u)\,du} f(X_{t+s}) =\underbrace{\vphantom{f}e^{\int_0^t V(X_u)\,du}}_{\mbox{past}}\,\,\,\underbrace{e^{\int_t^{t+s} V(X_u)\,du} f(X_{t+s})}_{\mbox{future}}$$
Taking the conditional expected value with respect to $\cal F_t$, we get $$\mathbb{E}\left(e^{\int_0^{t+s} V(X_u)\,du} f(X_{t+s})\mid {\cal F}_t\right) = e^{\int_0^t V(X_u)\,du}\ \mathbb{E}_{X_t}\left(e^{\int_0^s V(X_u)\,du} f(X_s)\right)=e^{\int_0^t V(X_u)\,du}\, (P^V_sf)(X_t) $$ Taking expectations on both sides with respect to $\mathbb{P}_x$ gives $$P^V_{t+s}f(x)=P^V_t\left(P^V_sf\right)(x) $$