# growth bound for solution of an ordinary integro-differential equation

I am considering the following ordinary integro-differential equation $$A g = \sigma^2(y) + \int_\mathbb{R} \nu(y,dz) z^2$$ where $$A = b(y) \partial + \frac{1}{2} \sigma^2(y) \partial^2 + \int_\mathbb{R} \nu(y,dz) (e^{z \partial} - 1 - z \partial)$$ $\nu(y,\cdot)$ is a Levy measure for every $y$ and $$b(y) = - \frac{1}{2} \sigma^2(y) -\int_\mathbb{R} \nu(y,dz) (e^{z} - 1 - z)$$ I use the notation $e^{z \partial}$ to indicate the shift operator $e^{z \partial} f(y) = f(y+z)$. Those with a background in probability will recognize $A$ as the generator of a Levy-type process.

I would like to find conditions on $\sigma$ and $\nu$ such that the classical solution $g$ of the top equation is exponentially bounded. That is, there exists $C,b>0$ such that $$g(y) \leq C \exp ( b |y| ) .$$ If $\sigma(y)=a$ and $\nu(y,dz)=\mu(dz)$ then it is easy to check that the solution is $g(y) = Q y$ where $Q$ is a constant $$Q = \frac{a^2 + \int \mu(dz) z^2 }{a^2/2 + \int \mu(dz) (e^z - 1 - z)} .$$ So, I imagine that bounded coefficients would yield a solution that satisfies the exponential growth bound. But, I do not know how to show this. By "bounded" levy measure I mean $\sup_y \nu(y,B) \leq \overline{\nu}(B)$. I am happy to consider stronger conditions if needed.

I was hoping there might be a general result concerning pseudo-differential operators that might show this (the generator $A$ can be thought of as a $\Psi$DO the generates a Markov process). For example, maybe if the symbol of $A$ satisfies certain conditions then I could say something about the solution $g$. But, I am not an expert in this area, so any help would be greatly appreciated.