Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y| \cap \overline D \ne \emptyset}\}.$$$$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y|} \cap \overline D \ne \emptyset\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$
(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.
Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.