Let \[ f(z)=\sum_{n=0}^{\infty}a_nz^n \] be analytic in the unit disk. Assume \(a_1\ne0\) and \[ \sum_{n=2}^{\infty}n|a_n|\le |a_1|. \] Prove that \(f\) is injective.
Solution
1. If \(z,w\) lie in the unit disk and \(z\ne w\), then\[ \frac{f(z)-f(w)}{z-w} = a_1 + \sum_{n=2}^{\infty} a_n \sum_{j=0}^{n-1}z^{n-1-j}w^j. \]
For every \(n\ge1\), \[ z^n-w^n = (z-w)\sum_{j=0}^{n-1}z^{n-1-j}w^j. \] The coefficient hypothesis implies \[ \sum_{n=2}^{\infty}|a_n|<\infty. \] Thus, for fixed \(z,w\) in the unit disk, the relevant power series are absolutely convergent and the identity may be applied term by term to \(f(z)-f(w)\). The \(n=0\) terms cancel, giving the displayed formula.
2. Put\[ r\coloneqq\max\{|z|,|w|\}<1. \]Then the higher-order part in step 1 satisfies\[ \left| \sum_{n=2}^{\infty} a_n \sum_{j=0}^{n-1}z^{n-1-j}w^j \right| < |a_1|. \]
For \(n\ge2\), \[ \left| \sum_{j=0}^{n-1}z^{n-1-j}w^j \right| \le nr^{n-1} \le nr. \] Hence \[ \begin{aligned} \left| \sum_{n=2}^{\infty} a_n \sum_{j=0}^{n-1}z^{n-1-j}w^j \right| &\le r\sum_{n=2}^{\infty}n|a_n|\\ &\le r|a_1|\\ &< |a_1|, \end{aligned} \] because \(r<1\) and \(a_1\ne0\).
3. For distinct \(z,w\) in the unit disk,\[ f(z)\ne f(w). \]
By steps <1>1–<1>2 and the reverse triangle inequality, \[ \left| \frac{f(z)-f(w)}{z-w} \right| \ge |a_1| - \left| \sum_{n=2}^{\infty} a_n \sum_{j=0}^{n-1}z^{n-1-j}w^j \right| > 0. \] Therefore \((f(z)-f(w))/(z-w)\ne0\), and since \(z-w\ne0\), one has \(f(z)-f(w)\ne0\).
4. Thus \(f\) is injective on the unit disk.
Step 3 is exactly the injectivity condition.
5. Q.E.D.
Step 4 proves the assertion.