This is a marvelous exercise in complex analysis; I heard it from Steffen Rohde but don’t remember the original source.
Let . Suppose that a function satisfies the following property: for every three points there exists a holomorphic function such that for . Prove that is holomorphic.
No solution here, just some remarks.
- The domain does not matter, because holomorphicity is a local property.
- The codomain matters: cannot be replaced by . Indeed, for any function and any finite set there is a holomorphic function that agrees with at — namely, an interpolating polynomial.
- Two points would not be enough. For example, passes the two-point test but is not holomorphic.
Perhaps the last item is not immediately obvious. Given two points , let . The hyperbolic distance between and is the infimum of taken over all curves connecting to . Projecting onto the real axis, we obtain a parametrized curve connecting to .
it follows that . That is, is a nonexpanding map in the hyperbolic metric of the disk.
We can assume that . There is a Möbius map such that ; moreover, we can arrange that is a real number greater than , by applying a hyperbolic rotation about . Since is a hyperbolic isometry, , which implies . Let ; this is a Euclidean homothety such that and . By convexity of , . The map achieves for .
The preceding can be immediately generalized: passes the two-point test if and only if it is a nonexpanding map in the hyperbolic metric. Such maps need not be differentiable even in the real-variable sense.
However, the three-point test is a different story.