In the 1977 paper "Cone Conditions and Properties of Sobolev Spaces" by Adams and Fournier, the authors say that an open set $\Omega\subset\mathbb{R}^n$ satisfies the cone condition if each $x\in\Omega$ is the vertex of a cone $C_x$ contained in $\Omega$ congruent to a fixed cone $C$.
They then go on to define the weak cone condition.
Let $x\in\Omega$. Let $R(x)$ consist of all points $y\in\Omega$ such that the line segment from $x$ to $y$ lies completey $\Omega$. Let $\Gamma(x)=\{y\in R(x) : |x-y|<1\}$. $\Omega$ satisfies the weak cone condition if there is a number $\delta>0$ such that $\mu_n(\Gamma(x))\geq \delta$. Here $\mu_n(\Gamma(x))$ denotes the Lebesgue measure of $\Gamma(x)$.
The authors note that there are many domains satisfying the weak cone condition, but not the "strong" cone condition. Does anyone know an example of such a domain? I've tried to construct one but have had no luck.
Here are a couple of my thoughts:
Any domain $\Omega$ failing the cone condition must not contain a smallest cone. That is, there cannot be an $x\in\Omega$ such that for any other $y\in\Omega$, the cone $C_x$ may be rigidly transformed to have its vertex coincide with $y$, and $C_x\subset C_y$. If such an $x$ exists, then we can take our fixed reference cone to be $C_x$ and then $\Omega$ satisfies the cone condition.
If $\Omega$ fails the cone condition, but satisfies the weak cone condition, then there should be a sequence $\Gamma(x_n)$ for which the diameter of the interior of $\Gamma(x_n)$ tends to 0 as $n\to\infty$. If the diameter of the interior of $\Gamma(x)$ was bounded below for all $x$, there there would be a smallest ball contained in $\Gamma(x)$ for all $x$, and thus a smallest cone.
A simplistic example, but you can build others like it, in 2 and higher dimensions.
The idea is to start with the domain above the graph of the absolute value function, then fix a countable dense of points on the unit circle and delete the (open) line segments passing through the origin and these points (Draw a picture!). Denseness ensures no cone will fit centered at 0, while the countable property (and polar coordinates) ensure that $R(0)$ has full measure.
The above set is not open, however we can fix that in the following way (noting that we won't affect the failure of the cone property): Call $A_k$ the portion of our "domain" that lives at distance $d$ from the origin, with $2^{-k-1}<d\leq 2^{-k}$ from the origin (so the portion of our domain in an annulus with inner radius $2^{-k-1}$ and outer radius $2^{-k}$). Let $(a_j)_{j\geq0}$ be our dense set of points from before and $(l_j)_j$ the corresponding line segment. From $A_k$ we'll add back in the portions of $(l_j)_{j\geq k}$ (so from the outermost annulus $A_0$ we add all of them; on $A_1$ we add all except $l_0$, on $A_2$ we add all except $l_0,l_1$, and so on).
The new set, call it $\Omega$ is the union of $A_k'$, where these last sets are the $A_k$ with the corresponding portions of line segments added. Since $A_k'$ is a half open, half closed annulus with a finite number of segments, the only possible issue (we're trying to fit a small ball around every point in $\Omega$) is at the boundary of these sets. Here however, the only problem is if we're on a line segment that is deleted from $A_{k+1}'$ but not from $A_k'$, but this lies in the complement of $\Omega$ so we're in good shape.
Hopefully the above paragraph is no too obfuscating; the basic idea is: In deleting the "dense" set of line segments, we broke the open property of our set. So instead we delete one line segment at a time (this operation does preserve openness) and then we just arrange this so that as we tend to 0, we recover the dense set of directions (this is why we do things in $A_k$).
Added: I just noticed that if you keep the "upper" endpoints of your line segments in each $A_k'$ instead of the whole segments, it's easier to check that the weak cone condition is verified throughout the whole domain.
Another, less complicated (though also less satisfying somewhat), example is to take the absolute value function and scale it horizontally and translate it, so that as you move to $\infty$ in the horizontal direction, you get narrower and narrower corners. Added: This last example only works as one for which both properties fail.