# 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?

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.

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).