[Fall 2002 Problem #8] Suppose \(f\) and \(g\) are holomorphic mappings of the unit disc \(D\) into an open domain \(\Omega\), \(f\) is one-to-one, \(g(D)\subseteq f(D)\), and \(f(0) = g(0)\). Show that \(g(\abs{z} < r) \subset f(\abs{z} < r)\) for each \(0 < r < 1\). (The first part of the problem asks for a statement of the Schwarz Lemma.)
Erratum: the source omits the hypothesis \(g(D)\subseteq f(D)\). Without it the conclusion is false. For example, with \[ \Omega=D,\qquad f(z)=\frac z2,\qquad g(z)=z, \] all printed hypotheses hold, but \[ g(D_r)=D_r \] is not contained in \[ f(D_r)=D_{r/2}. \] The corrected statement adds the subordination hypothesis \(g(D)\subseteq f(D)\).
Solution
Write \[ D_r=\{z\in\CC:\abs{z}<r\}. \]
1. Schwarz’s lemma states: if \(\varphi:\DD\to\DD\) is holomorphic and\(\varphi(0)=0\), then\[ \abs{\varphi(z)}\leq\abs{z} \]for every \(z\in\DD\).
This is Schwarz’s lemma; the source’s first part asks for its statement.
2. The inverse map\[ f^{-1}:f(D)\longrightarrow D \]is holomorphic.
Since \(f\) is one-to-one and holomorphic, it is a conformal bijection from \(D\) onto the domain \(f(D)\). In particular its derivative does not vanish, so the holomorphic inverse function theorem gives a holomorphic local inverse at every point of \(f(D)\). These local inverses agree because \(f\) is globally one-to-one, and therefore form the holomorphic inverse \(f^{-1}\).
3. Define\[ \varphi=f^{-1}\circ g. \]Then \(\varphi:\DD\to\DD\) is holomorphic and satisfies \(\varphi(0)=0\).
The hypothesis \(g(D)\subseteq f(D)\) makes the composition well-defined. Step 2 shows that it is holomorphic. Moreover, \[ \varphi(0) = f^{-1}(g(0)) = f^{-1}(f(0)) = 0. \]
4. For every \(0<r<1\),\[ \varphi(D_r)\subseteq D_r. \]
If \(z\in D_r\), then \(\abs{z}<r\). By step 1 applied to the function in step 3, \[ \abs{\varphi(z)} \leq \abs{z} < r. \] Thus \(\varphi(z)\in D_r\).
5. For every \(0<r<1\),\[ \boxed{ g(D_r)\subseteq f(D_r). } \]
By the definition of \(\varphi\), \[ g=f\circ\varphi. \] Therefore step 4 gives \[ g(D_r) = f(\varphi(D_r)) \subseteq f(D_r). \]
6. Q.E.D.
Step 5 is the required inclusion.