This question is about transversality condition in a stochastic control problem.
Transversality condition
Consider an infinite horizon stochastic control problem
$$ \sup_{u(\cdot) \in \mathcal{A}} E^x\Big[\int_0^{\infty} f(X^u_t, u_t) dt\Big] $$
where the state $X$ follows a controlled SDE
$$ dX^u = \mu(X, u) dt + \sigma(X, u) dB, \; X_0 = x. $$
The class of admissible controls is defined by (suppressing appropriate measurability assumptions)
$$ \mathcal{A} = \{ u(\cdot) | E[\int_0 ^{\infty} |f(X^u_t, u_t)| dt ] < \infty \}. $$
The variational inequality is
$$ \max_u f -\rho V + \mu V_x + \frac{1}{2} \sigma^2 V_xx = 0. $$
If a function $V$ solves the HJB equation with corresponding $u^*$, then we have the following verification theorem-type statement for sufficiency: If the following two conditions hold for all $x$
$$ 1. \;\; \liminf_{t \rightarrow \infty} e^{- \rho t} E^x[V(X_t^{u^*})] \geq 0 , \;\; \mbox{transversality condition}\\ 2. \;\; E^x[\int_0^{\infty} |V_x(X_t^{u^*}) \sigma(X_t^{u^*}, u^*)|^2] < \infty, \;\;\mbox{martingale condition} $$
then $u^*$ is an optimal control.
This can be argued as follows: for all $t > 0, x$ and $u(\cdot) \in \mathcal{A}$,
\begin{align*} V(x) &= e^{-\rho t} V(X^{u^*}_t) + \int_0^t e^{-\rho s} [\rho V(X^{u^*}_s) - \mu(X^{u^*}_s, u^*_s) V_x(X^{u^*}_s)) - \frac{1}{2} \sigma^2(X^{u^*}_s, u^*_s) V_{xx} (X^{u^*}_s] ds - \int_0 ^t V_x(X^{u^*}_s) \sigma dB \\ &\geq e^{-\rho t} V(X^{u^*}_t) + \int_0^t e^{-\rho t} f(X^{u}_s, u_s) ds - \int_0 ^t V_x(X^{u^*}_s) \sigma(X^{u^*}_s, u_s) dB \\ \end{align*}
Taking expectation on both sides gives \begin{align*} V(x) &\geq e^{-\rho t} E^x [ V(X^{u^*}_t)] + E^x [ \int_0^t e^{-\rho t} f(X^{u}_s, u_s) ds] - E^x [\int_0^t V_x(X^{u^*}_s) \sigma(X^{u^*}_s, u_s) dB] \\ \end{align*}
Taking the limit as $t \rightarrow \infty$,
(i) $e^{-\rho t} E^{x} [ V(X^{u^*}_t)]$ vanishes by the transversality condition.
(ii) $E^{x} [\int_0^t V_x \sigma dB] = 0$ for all $t$ by the martingale condition.
(iii) $E^{x} [ \int_0^t e^{-\rho t} f(X^{u}_s, u_s) ds]$ converges to $E^{x} [ \int_0^{\infty} e^{-\rho t} f(X^{u}_s, u^*_s) ds]$ by definition of admissibility and the dominated convergence theorem. Therefore $u^*$ is optimal.
A specific problem
$$ \max_{c, \psi} E^x[\int_0^{\infty} e^{-\rho s} (-e^{\gamma c_s}) ds ] $$
where the state evolves according to
$dX = (rX - c + \psi(\mu - r)) dt + \psi \sigma dB$.
Solving the HJB equation gives feedback/Markov control
\begin{align*} c(x) &= rx - \frac{1}{\gamma r}(r - \rho - \frac{1}{2} \frac{(\mu-r)^2}{\sigma^2}) \\ &\equiv rx - D, \end{align*}
$$ \psi(x) = \frac{\mu - r}{\sigma^2 r \gamma} \equiv E. $$
Now these solutions satisfy the above conditions for verification:
\begin{align*} dX^{u^*} &= (D + E(\mu -r)) dt + E \sigma dB \\ &\equiv F dt + G dB. \end{align*}
So $E[V(X^{u^*}_t)] = - e^{(F + \frac{1}{2} G^2)t}$ and the transversality condition holds under some restrictions on the parameters $(r, \mu,\sigma, \rho, \gamma)$. Similarly, $E[V_x(X^{u^*}_t)] = -e^{(r \gamma F + \frac{1}{2} r^2 \gamma^2 G^2)t}$ and the martingale condition holds under some restrictions on the parameters.
Question
In general one would like to interpret the transversality condition (in the form, say, $\lim e^{- \rho t} E^x [V(X^{u^*}_t)] = 0$) as "leaving no money on the table". The general argument seems to hold regardless of the sign of $f$. But something seems strange when the instantaneous utility $f$ is negative. In the specific problem above, $0$ in this case should be interpreted formally as infinity, and the transversality condition would say that $e^{- \rho t} E^x [V(X^{u^*}_t)]$ is infinity in the limit, which seems strange. Anyone have a interpretation/comments on this?