The claim we are asked to prove in Petr Mandl's boook (An analytical treatment of one-dimensional Markov chains) in exercise 3 of chapter 3 (pg 49) is
3. Under the assumptions of Theorem 3, $\mathscr{D}(A)$ is dense in $C_I$ if $(41)$ is true; in the opposite case $\{F\in\mathscr{D}(A):AF\in C'\}$ is dense in $C'$.
Condition (41) and theorem 3 are:
We already know that $\mathcal{D}(D_m D_p^+)$ is dense. If we are to prove that $\mathcal{D}(A)$ is also dense and if $\mathcal{D}(A) = \{ F \in D_m D_p^+ \mid \phi_i(F) = 0\}$ and if there is a $G\in \mathcal{D}(D_m D_p^+) \setminus \mathcal{D}(A) $ then we can find a sequence of $G_n \in \mathcal{D}(A) $ such that $G_n \to G$, but $\phi(G- G_n) = \phi(G)$ and so $F_n = G- G_n$ is such that $F_n \to 0$ and $\phi(F_n) \neq 0$.
But since we are trying to prove the other way arround, that is, that $$ \overline{\mathcal{D}(D_m D_p^+)} = \overline{\mathcal{D}(A)} $$
If we find $F_n \to 0$ such that $\phi(F_n) \to a >0$ we can for every $G\in \mathcal{D}(D_m D_p^+) \setminus \mathcal{D}(A) $ set $G_n = G - \frac{\phi(G)}{a} F_n$ and conclude that $G_n \to G$ and $G_n \in \mathcal{D}(A) $
Any idea on how to find such an $F_n$