## Limit hulls?

In the course of trying to explain $\liminf$ and $\limsup$ “visually” I was led to this variation on the notion of convex hull. Given a set $A$ in the plane, the convex hull is constructed by considering all the half-planes $H$ that contain $A$ and taking their intersection. The requirement $A\subset H$ can be relaxed for instance by saying that $A\setminus H$ has measure zero. But what happens if we instead ask that $A\setminus H$ be bounded? Suppose for instance that $A$ is the first quadrant. Then any half-plane of the form $H_t=\{v: v\cdot (1,1)>t\}$, for $t>0$, is a candidate and we see that their intersection is empty. But it isn’t empty if we pass to the projective plane. In that case we would get an arc of the line at infinity, which could maybe be called the “limit hull” of $A$. It seems the case, but I haven’t sat down to check, that whenever $A$ is unbounded, its limit hull is non-empty.

The connection with $\liminf$ was as follow. Let $y=f(x)$ be the graph of your favorite function. First define the infimum of $f$ “visually” by saying that “we draw a horizontal line under the graph and then raise it as far as we can until it bumps into the graph of $f$. The $\liminf_{x\rightarrow+\infty} f(x)$ can be defined analogously by drawing a horizontal line under the graph and continue raising it as long as the set of $x$‘s where $f$ is under the line forms a bounded set.