0

If $Q\subset \mathbb{C}$ is a square in the complex plane and $f:Q\rightarrow \mathbb{C}$ is a function which is continuous on $Q$ and holomorphic in its interior which satisfies $|f(z)|=1$ for every $z\in \partial Q$, then can $f$ can be extended to a holomorphic function on a neighborhood of $Q$?

Thanks you for any insight.

user54631
  • 707

1 Answers1

1

Generally, piecewise analytic boundary does not allow for holomorphic extension because singularities develop at the joints. This was discussed. However, you have a square, which has the nice property of tiling the plane by reflection.

So, you can actually do this using three reflections around every corner. For simplicity, I take the square $Q=\{x+iy : 0<x<1, 0<y<1\}$ and consider the corner at $0$. Extend $f$ by letting

  • $f(x+iy) = 1/\overline{f(x-iy)}$ when $x>0$, $y<0$
  • $f(x+iy) = 1/\overline{f(-x+iy)}$ when $x<0$, $y>0$
  • $f(x+iy) = f(-x-iy)$ when $x<0$, $y<0$

Of course the division creates poles in places, but there is a punctured neighborhood of $0$ in which the extended map is holomorphic. And since it's bounded there, the singularity at $0$ is removable. Same for other corners.

Community
  • 1