Does continous extendability of a derivative to the boundary of a half-space imply differentiability? (This is a simplified version of this question.)
Let $\mathbb H^n:=\mathbb R^{n-1}\times[0,\infty)$. Let $\Omega\subset\mathbb H^n$ be open with respect to $\mathbb H^n$. Let $\Omega^o$ denote the interior of $\Omega$ with respect to $\mathbb R^n$.
Suppose $f:\Omega\to\mathbb R$ and $g:\Omega\to\mathbb R^n$ are continuous. Further, assume that $f$ is differentiable on $\Omega^o$ and
$$\nabla (f|_{\Omega^o})=g|_{\Omega^o}$$
Does this imply that $f$ is differentiable on $\Omega$ with derivative $g$? I. e., is it true that for all $x\in\Omega$ the following holds?
$$f(y)=f(x)+\langle g(x),y-x\rangle+o(\|y-x\|)$$

Fix $x\in \Omega$ and $\epsilon > 0$. We want to show that there exists some $\delta > 0$ such that for $y\in\Omega$ with $\|y-x\| < \delta$ it follows 
$$ |f(y) - f(x) - \langle g(x), y-x \rangle| < \epsilon \|y -x\|.$$


*

*Notice that for $x$ there exists some $\eta > 0$ such that for every $y\in\mathbb R^n$ with $\|y-x\|\le\eta$ and $y_n > 0$ it follows $y\in\Omega^o$.
Let $C = \bar U_\eta(x)$ be the closed $\eta$-neighborhood of $x$ in $\Omega$.
As $g$ is uniformly continuous on the compact set $C$, there exists a $\tau > 0$ such that 
$$ \| g(v) - g(u) \| < \frac\epsilon2 $$
holds for all $u, v \in C$ with $\|v - u\| < \tau$.

*For $\delta = \min(\frac\tau3, \eta)$ let $U = U_\delta(x)$ be the $\delta$-neighborhood of $x$ in $\Omega$. Notice that
$$ U\cap \Omega^o = \{ u\in U \mid u_n > 0 \}$$
is convex.
Thus, for every $u, v\in U \cap \Omega^o$, as $g$ is continuous, we have
$$ |f(v) - f(u) - \langle g(u), v - u \rangle | \le \max_{z\in U} \| g(z) - g(u) \| \| v - u \| \le \frac\epsilon2 \|v-u\|. $$

*Now let $u \to x$ and $v\to y\in U$... and we are done. 
