To show an analytic function is one-to-one on the unit disk

Question

Let $\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$ be analytic in the unit disk $D_1(0)$ with $f(0) = 0$ and $f'(0) = 1$. Prove that if $\displaystyle \sum_{n=2}^\infty n|a_n| \le 1$, then $f$ is one-to-one in $D_1(0)$.

I am able to show that $f$ has a unique zero in $D_1(0)$ and $f$ is locally one-to-one, but I cannot go any further.

We may write $\displaystyle f(z) = z + \sum_{n=2}^\infty a_nz^n$. As $\displaystyle \sum_{n=2}^\infty n|a_n| \le 1$, we have $\displaystyle \sum_{n=2}^\infty |a_n| < 1$. So $$ \left|\sum_{n=2}^\infty a_n z^n \right|_{C_1} < 1 $$ By Rouché's Theorem, $f$ and $z$ have the same number of zeros inside $C_1$. So 0 is the unique zero of $f$.

Next, we let $g(z) = f'(z) -1$. So $\displaystyle g(z) = \sum_{n=2}^\infty na_nz^{n-1}$. Again, we use the assumption that $\displaystyle \sum_{n=2}^\infty n|a_n| \le 1$ to conclude that $g$ maps the unit disk into itself. Moreover, $g(0) = 0$. Apply the Schwarz's Lemma, we have $$ |f'(z)-1|\le |z|, z \in D_1(0) $$ which means that $f'(z) \ge 1-|z| >0$ for all $z \in D_1(0)$, i.e., $f$ is locally injective.

Answer 1

Suppose $f(z_1) = f(z_2)$ for some $z_1$ and $z_2$. Writing $f(z) = z + \sum_{n\geq2}a_n z^n = z + h(z)$, we have $$ 0 = f(z_1) - f(z_2) = z_2 - z_1 + \int_{z_1}^{z_2}h'(z)\,dz, $$ so the maximum distance between them is $$ |z_1-z_2| \leq |h'(z_0)||z_1-z_2|, $$ where $z_0$ is some point inside the line segment between $z_1$ and $z_2$. But from the assumption that $\sum_{n\geq2}n|a_n|\leq1$ we have that $|h'(z)|<1$ for all $z$ inside the unit disk, so $|z_1-z_2|$ must be zero.

Answer 2

Hints:

1. Show that $\operatorname{Re}f'>0$ in $D$.
2. Check that $\displaystyle \frac{f(b)-f(a)}{b-a} = \int_0^1 f'(a+t(b-a))\,dt$ for any $a,b\in D$ with $a\ne b$.
3. Combine 1 and 2.