Let \(U = \{z \in \CC \mid \operatorname{Im}(z) \leq \frac{\pi}{2}\}\) and \(f\) be an entire function satisfying \(f(U) \subset U\), \(f(-1) = 0\), \(f(0) = 1\). Prove that \(f(z) = z + 1\).
Solution
Set \(g(z)=f(z)-1\) and \(H=\{z:\operatorname{Im}z<\pi/2\}\). Then \(g(-1)=-1\) and \(g(0)=0\).
1. The restriction \(g:H\to H\) is a holomorphic self-map.
Subtracting the real number one leaves imaginary parts unchanged, so \(g(U)\subset U\). Both \(-1\) and zero belong to \(H\), and their distinct images show that \(g\) is nonconstant there. By the open mapping theorem, \(g(H)\) is open in \(\CC\). It is contained in the closed half-plane \(U\), so it cannot contain a point of the boundary line: any neighborhood of such a point meets \(\CC\setminus U\). Hence \(g(H)\subset H\).
2. A disk normalization makes \(g\) the identity on \(H\).
Define \[ \psi(z)=\frac{z}{z-i\pi},\qquad \psi^{-1}(w)=\frac{i\pi w}{w-1}. \] The pole \(i\pi\) is outside \(H\). Direct calculation gives \[ 1-\abs{\psi(z)}^2 =\frac{\pi^2-2\pi\operatorname{Im}z}{\abs{z-i\pi}^2}>0 \quad(z\in H), \] and \[ \frac\pi2-\operatorname{Im}\frac{i\pi w}{w-1} =\frac{\pi(1-\abs{w}^2)}{2\abs{1-w}^2}>0 \quad(\abs{w}<1). \] Substitution verifies the inverse formulas. Thus \(\psi\) is a biholomorphism from \(H\) to the unit disk \(D\).
The holomorphic disk self-map \(G=\psi\circ g\circ\psi^{-1}\) fixes zero and \(\beta=\psi(-1)=1/(1+i\pi)\ne0\). The Schwarz lemma gives \(\abs{G(w)}\leq\abs{w}\). The quotient \(G(w)/w\) extends holomorphically through zero, has modulus at most one, and equals one at \(\beta\). The maximum modulus principle therefore makes that quotient identically one. It follows that \(G(w)=w\) on \(D\) and \(g(z)=z\) on \(H\).
3. The equality holds on the entire plane.
The entire function \(f(z)-z-1\) vanishes on the nonempty open set \(H\). The identity theorem on the connected plane makes it identically zero. Hence \(f(z)=z+1\) for every \(z\in\CC\). This translation indeed preserves \(U\) and has the two prescribed interpolation values.
4. Q.E.D.
Step 3 gives the required formula on the entire plane.