Three-point test for being holomorphic

This is a marvelous exercise in complex analysis; I heard it from Steffen Rohde but don’t remember the original source.

Let {D=\{z\in \mathbb C\colon |z|<1\}}. Suppose that a function {f\colon D\rightarrow D} satisfies the following property: for every three points {z_1,z_2,z_3\in D} there exists a holomorphic function {g\colon D\rightarrow D} such that {f(z_k)=g(z_k)} for {k=1,2,3}. Prove that {f} is holomorphic.

No solution here, just some remarks.

  • The domain does not matter, because holomorphicity is a local property.
  • The codomain matters: {D} cannot be replaced by {\mathbb C}. Indeed, for any function {f\colon D\rightarrow\mathbb C} and any finite set {z_1,\dots, z_n\in D} there is a holomorphic function that agrees with {f} at {z_1, \dots, z_n} — namely, an interpolating polynomial.
  • Two points {z_1,z_2} would not be enough. For example, {f(z)=\mathrm{Re}\,z} passes the two-point test but is not holomorphic.

Perhaps the last item is not immediately obvious. Given two points {z_1,z_2\in D}, let {x_k=\mathrm{Re}\,z_k}. The hyperbolic distance {\rho} between {z_1} and {z_2} is the infimum of {\displaystyle \int_\gamma \frac{1}{1-|z|^2}} taken over all curves {\gamma} connecting {z_1} to {z_2}. Projecting {\gamma} onto the real axis, we obtain a parametrized curve {\tilde \gamma} connecting {x_1} to {x_2}.

Projection does not increase the hyperbolic distance.
Projection does not increase the hyperbolic distance.


\displaystyle  \int_{\tilde \gamma} \frac{1}{1-|z|^2} =    \int_{\tilde \gamma} \frac{1}{1-|\mathrm{Re}\,z|^2}\le  \int_{\gamma} \frac{1}{1-|\mathrm{Re}\,z|^2}\le \int_\gamma \frac{1}{1-|z|^2}

it follows that {\rho(x_1,x_2)\le \rho(z_1,z_2)}. That is, {f} is a nonexpanding map in the hyperbolic metric of the disk.

We can assume that {x_1\le x_2}. There is a Möbius map {\phi} such that {\phi(z_1)=x_1}; moreover, we can arrange that {\phi(z_2)} is a real number greater than {x_1}, by applying a hyperbolic rotation about {x_1}. Since {\phi} is a hyperbolic isometry, {\rho(x_1,\phi(z_2))\ge \rho(x_1,x_2)}, which implies {\phi(z_2)\ge x_2}. Let {\lambda(z)=x_1+(z-x_1)\dfrac{x_2-x_1}{\phi(z_2)-x_1}}; this is a Euclidean homothety such that {\lambda(x_1)=x_1} and {\lambda(\phi(z_2))= x_2}. By convexity of {D}, {\lambda(D)\subset D}. The map {g=\lambda\circ \phi} achieves {g(z_k)=x_k} for {k=1,2}.

The preceding can be immediately generalized: {f} 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.

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