Regarding unital positive operators Let $\Omega$ be a domain in $\mathbb{C}^n$. Let $\mathbb{D}$ denote the open unit disc in $\mathbb{C}$. Let $C_b(\Omega)$ and $C_b(\mathbb{D})$ denote the space of all bounded continuous complex valued functions on $\Omega$ and $\mathbb{D}$ respectively.
Let $T:C_b(\Omega)\longrightarrow C_b(\mathbb{D})$ be a positive linear operator which is unital.
Suppose $f\in C_b(\Omega)$ be such that $f(z)\neq 0$ for any $z\in \Omega$. Will it imply that $Tf(y)\neq 0$ for every $y\in\mathbb{D}$?
If not then under what additional conditions will $T$ satisfy this property?
 A: No, this already fails for linear functionals. For instance, let $\zeta \in \beta \mathbb{D} \setminus \mathbb{D}$, where $\beta \mathbb{D}$ is the Stone-Cech compactification. Then $f \mapsto f(\zeta)$ is a positive unital map (even multiplicative) from $C_b(\mathbb{D})$ into $\mathbb{C}$, and any function in $C_b(\mathbb{D})$ which is nonzero on $\mathbb{D}$ but vanishes on the boundary will then be a counterexample. (You can easily modify this to go into $C_b(\mathbb{D})$ by identifying $\mathbb{C}$ with the constant functions.)
I don't think there will be any good condition which guarantees the result you want.
A: Nik Weaver has already explained in his answer that this does not hold, in general. On the positive side, here is a sufficient condition for the implication to be true:
Proposition. Let $\Omega_1, \Omega_2$ be topological spaces and let $T: C_b(\Omega_1) \to C_b(\Omega_2)$ be a positive linear operator such that $T1 = 1$. Suppose in addition that $T$ has the following continuity property:
$(*)$ If a sequence $(g_n) \subseteq C_b(\Omega_1)$ is bounded in supremum norm and converges pointwise to $g \in C_b(\Omega_1)$, then $(Tg_n)$ converges pointwise to $Tg$.
Then $Tf$ has no zeros whenever $0 \le f \in C_b(\Omega_1)$ has no zeros.
Proof. Assume that $0 \le f \in C_b(\Omega_1)$ has no zeros. Then $(nf) \land 1$ converges pointwise to $1$ as $n \to \infty$. Hence, $T\big((nf) \land 1\big)$ converges pointwise to $T1 = 1$ as $n \to \infty$. But we have
$$
 nTf \ge T\big((nf) \land 1\big)
$$
for each $n$, so $Tf$ cannot be $0$ at any point of $\Omega_2$. qed
Remark 1. In the statement (and proof) of the proposition, the function $1$ can be replaced with any other function $0 \le h \in C_b(\Omega_1)$ that does not have any zeros.
Remark 2. The continuity condition $(*)$ is much more common than one might, maybe, expect at first glance: it is satisfied for all transition operators that are given by measurable transition kernels (a class of operators which occurs frequently in stochastic analysis).
