and $(t,x)\in(0,T)\times\mathcal{H}$. Then for $s-t \to 0^+$ $$\begin{gathered}
\label{gconv1} \frac{1}{s-t} \left ( g(s,x_{t,x}(s)) - g(t,x) \right
) \leq g_t(t,x)
\\
+ \frac{1}{s-t} \int_t^s {\left\langle}D g (t,x), b(t,x,u(r)) {\right\rangle}{\mathrm{d}}r +
o(1)\end{gathered}$$ where $o(1)$ is uniform in $u(\cdot) \in \mathcal{U}[t,T]$

To prove the statement when $x \ne 0$ we use the fact that, in this case (see [@LiYong] page 241, equation (3.11)), $$\|x_{t,x}(s)\|  \leq \|x\| + \int_t^s {\left\langle}\frac{x}{\|x\|},
b(t,x,u(r)) {\right\rangle}{\mathrm{d}}r + o(s-t)$$ So we have $$\begin{gathered}
\label{eq:prooflemmag}
g(s,x_{t,x}(s))-g(t,x) = \eta(s) g_0(\|x_{t,x}(s)\|) -\eta(t)g_0(\|x\|)\\
\leq \eta(s) g_0\left ( \|x\| + \int_t^s {\left\langle}\frac{x}{\|x\|}, b(t,x,u(r)) {\right\rangle}{\mathrm{d}}r + o(s-t) \right ) - \eta(t)g_0(\|x\|) \\
\leq \eta'(t) g_0(\|x\|) (s-t) + \eta(t) g_0'(\|x\|) \left (
\int_t^s {\left\langle}\frac{x}{\|x\|}, b(t,x,u(r)) {\right\rangle}{\mathrm{d}}r \right ) + o(s-t) \\
= g_t(t,x) (s-t) + \int_t^s {\left\langle}D g(t,x) , b(t,x,u(r)) {\right\rangle}{\mathrm{d}}r +
o(s-t)\end{gathered}$$ where $o(s-t)$ is uniform in $u(\cdot)$. When $x=0$, using the fact that $g'_0(0)=0$, we get $$g(s,x_{t,x}(s))-g(t,x)=g_t(t,x) (s-t) + o(s-t+\|x_{t,x}(s)\|)$$ and (\[gconv1\]) follows upon noticing that $\|x_{t,x}(s)\|\le
C(s-t)$ for some $C$ independent of $u(\cdot) \in \mathcal{U}[t,T]$.

\[thexistence\] Let Hypotheses \[hpD2onb\] and \[hpD3onLandh\] hold. Then the value function $V$ (defined in (\[deterministicvaluefunction\])) is a viscosity solution of the HJB equation (\[deterministicHJB\]).

The proof is quite standard and can be obtained with small changes (due to 