So, context.
Theorem. Let $D\subseteq\mathbb R^d$ be bounded and satisfy the Poincaré cone condition and let $\varphi:\partial D\to\mathbb R$ be $C^0$ boundary conditions. Let $(B_t)_{t\geq0}$ be a standard Brownian motion and $\tau:=\inf\{t\geq0:B_t\in\partial D\}$. Then a $C^2$ solution to the PDE $$\Delta u=0$$ in the interior of $D$ is $$u(x)=\mathbb E\left[\varphi(x+B_\tau)\right].$$ And we have $u\vert_{\partial D}=\varphi$.
This is a pretty well-known theorem. I'm trying to learn it right now. I'm hitting a stumbling block with regards to the Poincaré cone condition. First, I don't understand its statement in certain forms. One such statement I don't understand is from a set of lecture notes from Cambridge (by the legendary Dexter Chua) is
Definition. We say a domain $D$ satisfies the Poincaré cone condition if for any $x\in\partial D$, there is an open cone $C$ based at $X$ [sic] such that $$C\cap D\cap B(x,\delta)=\varnothing$$ for some $\delta\geq0$.
What is an "open" cone? Is it just any cone that contains $x$ minus its boundary? Do we require that $x\notin D$ or $x\notin C$, otherwise it trivially fails for all $x$? Where is this requirement coming from?
I have confusions with other (seemingly isomorphic) definitions of the Poincaré cone condition but I won't belabor the point here since I'm sure solving my confusion with this one will solve my confusion with others.
My next question is: What is the geometric intuition of this condition, in the context of the theorem above? I literally have no idea what this is giving me except a little bit of space in the proof of the earlier stated theorem. I'm not sure why this would be necessary for the proof, i.e. why is the proof necessarily impossible without this condition. Nor can I "see" what this property implies of my domain in the "visually intuitive" settings of 2- or 3-space.
An open cone (for this example, based at $0$) is any set $C \subseteq \mathbb{R}^n$ of the form $C = \bigcup_{t \in \mathbb{R}^+} tD$, where $D \subseteq S^{n-1}$ is a non-empty open set (easiest to imagine this if $D$ is a disc). There are a few key properties of cones which become essential in the proof you reference. One elementary observation is that the open cone takes up a non-zero fraction of the ball $B^{n-1}$. Therefore, since the distribution of Brownian motion is preserved under orthogonal transformations, there is a non-zero probability that Brownian motion hits the cone. Another useful property is that cones are invariant under scaling, which makes calculating probabilities related to the comparison of hitting times easier, given the scaling invariance and strong Markov property of Brownian motion. Lastly, the condition is fairly weak, so it is possible for the theorem to be applied to a large class of domains which are likely to occur naturally (for example, $U = D^{n-1}$).