Consider the following cake eating problem: \begin{align} &v(x_0) = \max_{\{u(t)\}_{t \geq 0}}\int_0^\infty{e^{-rt}\ln(u(t))dt}\\ \text{s.t.} \quad &\dot x(t) = -u(t)\\ &u(t) \in [0,x(t)]\\ &x(0) = x_0 > 0\\ &x(t) \in X = [0,x_0]. \end{align} The value function $v(x) = (\ln(rx)-1)/r$ solves the Hamilton-Jacobi-Bellman equation \begin{align} rv(x) = \max_{u \geq 0}\{\ln(u) - v'(x)u\} \end{align} and yields linear feedback control $\mu(x) = rx$. Note that the set of feasible states $X = [0,x_0]$ includes zero (clearly one can eat the entire cake). For $\mu(x) = rx$ being optimal the following transverality condition needs to hold: \begin{align} \lim_{t \to \infty}e^{-rt}v(x) = 0 \quad \forall x \in X. \end{align}
Question: Since $v(0)$ is not defined, I suspect that the transversality condition is violated at $x = 0$. Is $\mu(x) = rx$ optimal? How do I prove it?
Workaround: Since the payoff function $\ln(u)$ is not defined for $u = 0$ we may restrict the set of control functions such that $(u(t),x(t)) > (0,0)$ for all $t \geq 0$.
Define the set of admissable control paths \begin{align} U = \left\{u : [0,\infty) \to \mathbb R_{++} ~\bigg|~ x(t) = x_0 - \int_0^t{u(s)ds} > 0 ~ \forall t \geq 0\right\}. \end{align} Then $x(t) \in (0,x_0]$ for all $t \geq 0$ and $u(\cdot) \in U$ and the transversality is satisfied on the space of admissable states. Then we may say that $u(t) = rx(t)$ is optimal for all admissable $u(\cdot) \in U$.