# Lyapunov-type function in a non locally-compact space and boundedness of the average

Set-up and question. Let $$\mathcal{X}$$ be a complete separable metric space which is not locally-compact. Let $$V: \mathcal{X} \to [0; +\infty]$$ be a function and $$(X_t)_{t\geq 0}$$ a Markov process in $$\mathcal{X}$$ such that $$P\{ V(X _t) < \infty \text{ for all } t \geq 0 \} = 1$$. Let $$g: \mathcal{X} \to \mathbb{R}$$ be such that $$V(X _t) - \int ^t _0 g(X _s) ds \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \quad \quad \quad \quad \quad \quad \quad (1)$$ is a martingale, where $$g(x) \leq c - V(x), \quad \text{ for all } x \in \{z \in \mathcal{X}: V(z) < \infty \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ for some $$c >0$$. Is it possible to deduce that $$\limsup\limits _{t \to \infty} \mathbb{E} V(X _t) < \infty? \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

My thoughts. If $$V$$ was 'norm-like' in the sense that $$\{x \in \mathcal {X} : V (x) \leq r \}$$ was compact for $$r >0$$ and $$\mathcal{X}$$ was locally-compact, we would get ergodicity from (1) and (2) and thus (3) would hold. I think that without $$V$$ being norm-like and local compactness of $$\mathcal{X}$$ ergodicity cannot follow from (1) and (2), however a weaker property like (3) should hold. I have an idea how to prove (3) if additionally $$(X_t)_{t\geq 0}$$ is assumed to be continuous a.s. (finding an ergodic birth-death process which would stochastically dominate $$(V(X_t))_{t\geq 0}$$), however without continuity assumption I am currently out of ideas. I did not find any similar result in the literature, but any related reference would be kindly appreciated.

Similar posts. The set-up is similar to this, however the assumptions and the questions are different.

Edit Edited the condition on $$g$$. The initial version was not well formulated and resulted in a triviality as was pointed out by Iosif Pinelis.

The conditions $$g\ge0$$, (2), and $$0\le V(X_t)<\infty$$ almost surely (a.s.) imply $$V(X_t)\le c$$ a.s. (Indeed, on the event $$\{V(X_t)<\infty\ \forall t\ge0\}$$ -- which is of probability $$1$$, we have $$0\le g(X_t)\le c-V(X_t)$$ and hence $$0\le c-V(X_t)$$ and $$V(X_t)\le c$$.) So, $$\limsup_{t\to\infty}EV(X_t)\le c<\infty$$ if $$c<\infty$$.
Added in response to the edit of the OP: Even when the condition $$g\ge0$$ is removed, the conclusion (3) holds. Indeed, letting $$h(t):=Eg(X_t)$$ and $$v(t):=EV(X_t)$$, and using the martingale condition, we have $$$$v(t)-\int_0^t h(s)\,ds=E\Big(V(X_t) - \int^t_0 g(X_s)\, ds\Big)=EV(X_0),$$$$ whence $$v'=h$$. Also, by (2), $$$$h(t)=Eg(X_t)\le c-EV(X_t)=c-v(t),$$$$ whence $$v'(t)+v(t)\le c$$, or, for $$u(t):=v(t)e^t$$,
$$$$u'(t)\le ce^t$$$$ and hence $$u(t)\le u(0)+ce^t$$, that is, $$$$EV(X_t)=v(t)\le u(0)e^{-t}+c,$$$$ so that $$$$\limsup_{t\to\infty}EV(X_t)\le c<\infty,$$$$ as desired.