Is a linear combination of Markov generator a Markov generator?

Question

Let $X$ be a compact metric set and $\mathcal{C}(X)$ be the set on continuous real functions over $X$ endowed with the supremum norm $\|\cdot\|_{\infty}$. Let $\Omega_1$ and $\Omega_2$ be two Markov generators on $\mathcal{C}(X)$ (using the definition of Liggett) with domains $\mathcal{D}_1$ and $\mathcal{D}_2$ such that $\mathcal{D}_1\cap\mathcal{D}_2$ is dense in $\mathcal{C}(X)$.

Question: Is any positive linear combination of $\Omega_1$ and $\Omega_2$ ($\lambda\Omega_1+\mu\Omega_2$ where $\lambda,\mu>0$) a Markov generator?

If this is not true in general, is any of these two hypotheses required:

• $(\Omega_1+\Omega_2)$ is also a Markov generator
• $\Omega_1$ and $\Omega_2$ satisfy both the maximum principle (Proposition 2.2 of Liggett)

Answer 1

The answer to your question is positive under certain conditions. For example, by the "product formula" of H. Trotter, (https://www.ams.org/journals/proc/1959-010-04/S0002-9939-1959-0108732-6/S0002-9939-1959-0108732-6.pdf), if $\mathcal D_1\subset\mathcal D_2$, then the set of $\mu\ge 0$ such that $\Omega_1+\mu\Omega_2$ is a Markov generator is open in $[0,\infty)$ and contains $0$. Trotter provides other sufficient conditions, and several enlightening examples. [As I recall such things are discussed on the book of Ethier & Kurtz, but I do not have my copy at hand.]

Answer 2

I might be wrong, but it looks like the answer is negative.

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Let $X$ be the unit circle, that we identify with $[-\pi, \pi]$ with the endpoints glued together. Write $a_+ = \max\{a, 0\}$ and $a_- = \max\{-a, 0\}$. Consider a deterministic process $X_1(t)$ that moves to the right with velocity $1 + 2 (\sin x)_+^{1/2}$. Its generator is $$ \Omega_1 f(x) = (1 + 2 (\sin x)_+^{1/2}) f'(x) , $$ with domain $C^1(X)$.

Edit: In other words, $X_1(t)$ satisfies the differential equation $$ X_1'(t) = (1 + 2 (\sin X_1(t))_+^{1/2}) . $$ This is an autonomous first-order ODE, with a unique solution $$X_1(t) = F(F^{-1}(X_1(0)) + t),$$ where $F$ is the indefinite integral of $1 / (1 + 2 (\sin x)_+^{1/2})$. Clearly, $X_1(t)$ depends continuously on $t$ and the initial value $X_1(0)$, and hence $\Omega_1$ indeed generates a $C_0$ semigroup on $C(X)$. (This certainly follows from some general theorems, too.)

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Now consider a similar deterministic process $X_2(t)$ that moves to the left with velocity $-1 - 2 (\sin x)_-^{1/2}$. It is generated by $$ \Omega_2 f(x) = -(1 + 2 (\sin x)_-^{1/2}) f'(x) , $$ again with domain $C^1(X)$. The average of these two operators is $$ \Omega = \tfrac12 (\Omega_1 + \Omega_2) f(x) = ((\sin x)_+^{1/2} - (\sin x)_-^{1/2}) f'(x) $$ for $f$ in $C^1(X)$. However, this operator does not generate a $C_0$ semigroup on $C(X)$.

Indeed: suppose, contrary to our claim, that it does. Then the corresponding Markov process $X(t)$ necessarily satisfies the differential equation $$ X'(t) = ((\sin X(t))_+^{1/2} - (\sin X(t))_-^{1/2}) $$ as long as $X'(t) \in (-\pi, 0) \cup (0, \pi)$. When the starting point goes to $0^+$ or $0^-$, the solutions converge to two (non-unique) solutions with the starting point $0$ (they behave as $\pm \tfrac14 t^2$ as $t \to 0^+$), and so it is easy to see that $e^{t \Omega} f$ can fail to be continuous at $0$, contrary to our assumption.