Let $G \subset \mathbb{R}^2$ be a convex, closed set with $\emptyset \neq \partial G \neq G$. Then given a unit vector $v \in \mathbb{R}^2$ (direction) we say that a point $x_0 \in \partial G$ is illuminated by "light" of direction $v$ iff the line $L = \{x_0+tv|t \geq 0\}$ contains an interior point of $G$.
Let $I_v$ be the set of points in $\partial G$ which are illuminated by "light" of direction $v$. We say that a set $V$ of unit vectors illuminates $G$ iff $$ \bigcup_{v \in V} I_v = \partial G$$
Let $c(G)$ denote the cardnality $|I|$ of the smallest set $I$ (not necessary unique) of unit vectors that illuminates $G$.
Question:
Is there a (unbounded) convex set $G \subset \mathbb{R}^2$ with the properties above such that $c(G) = \infty$?
(One can prove that if $G$ is bounded, then $c(G) = 4$ iff $G$ is a parallelogram and $c(G) = 3$ otherwise)
If $G$ is unbounded then $c(G)\leq 2$.
1) If there is a whole line contained in $G$ it is not difficult to see that $G$ must be either a half plane or a stripe between two parallel lines, using the fact that a convex set is the intersection of half planes. In this case $c(G)\leq2$. So let us assume from now on there isn't a whole line inside $G$.
2) Yet there must be a half line entirely contained in $G$.
To see this take any point in $ G$, wlog $0$, and take a sequence $x_n\in G$ which goes to $\infty$ and extract a subsequence such that $x_n/|x_n|\to\nu\in \mathbb S^1$. By convexity the segment $\overline{0x_n}$ is contained in $G$ and by closure the limit, which is the half line $l_\nu$ in direction $\nu$, must be contained in $G.$
3) For every $x\in\partial G$, the half line $x+l_\nu$ is contained in $G$ (again by convexity interpolating with $l_\nu$).
4) The projection $G_{\nu^\perp}$ of $G$ on $\nu^\perp$ is an interval, with at most two extremes. From here we see that $3$ directions are sufficient for the worst case of $G_{\nu^\perp}$ having two extremes: $\nu$ and two other directions close to $\nu$, one from each "side".
5) Actually we can do better and use just $2$ directions: take $x\in\partial G$ whose projection on $\nu^\perp$ lies in the interior of the segment $G_{\nu^\perp}$. The directions from which $x$ is illuminated are an open set which contains $\nu$, and taking any two distinct directions in this set, one from each side of $\nu$, will do.