If $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ and $|a_1|\geq\sum\limits_{n=2}^\infty n|a_n|R^{n-1}$, then $f$ is injective on $D(0,R)$

Question

Let $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ and $R$ its radius of convergence. If $f$ is not constant on $D(0,R)$ and $|a_1|\geq\sum\limits_{n=2}^\infty n|a_n|R^{n-1}$, show that $f$ is injective on $D(0,R)$.

My attempt:

$$|a_1|=\left| \sum\limits_{n=1}^\infty na_nz^{n-1} - \sum\limits_{n=2}^\infty na_nz^{n-1}\right|\leq |f'(z)| + \left|\sum\limits_{n=2}^\infty na_nz^{n-1}\right|\leq |f'(z)| + \sum\limits_{n=2}^\infty n|a_n||z|^{n-1}\stackrel{(*)}{\leq} |f'(z)|+\sum\limits_{n=2}^\infty n|a_n|R^{n-1}$$

The inequality $(*)$ is strict if at least one of the $a_n$, $n\geq 2$ is not $0$. We can assume that, because otherwise, $f(z)=a_0+a_1z$, which is injective.

Hence, $|f'(z)|> |a_1|-\sum\limits_{n=2}^\infty n|a_n|R^{n-1}\geq 0$. In particular, $f'(z)\neq 0 \ \forall z\in D(0,R)$.

Now, I tried to use the Inverse Function Theorem (local version), but I don't think this suffices to show global injectivity in $D(0,R)$.

Answer 1

I'm just going to take $R=1$ (the only change would be that the very last bound in the last line would have an $R^{n-1}$) and call $D=D(0,1)$:

Suppose not. Let $z,w\in D,$ with $z\neq w$, but $f(z)=f(w).$ Then, $$\sum\limits_{n=0}^\infty a_n(z^n-w^n)=0.$$ Let us write this as $$a_1(z-w)+g(z)-g(w)=0,$$ where $g(z)=\sum\limits_{n=2}^\infty a_nz^n.$ Note that for any smooth path $\gamma:[0,1]\rightarrow D$ from $w$ to $z$, we have that $$g(z)-g(w)=\int_{\gamma} g'(s)ds.$$ By the maximum modulus principle, $z_0\in \partial \gamma([0,1])$ exists so that $|g'(s)|\leq |g'(z_0)|$ for any $s\in\gamma([0,1]).$ Since $D$ is open and connected, the integral is independent of path (for any path in $D$ starting at $z$ and ending at $w$). Since $D$ is convex, we can take a straight path, and hence \begin{align*}|g(z)-g(w)|&=\left|\int_{\gamma} g'(s)ds\right|\leq\int_{\gamma}|g'(s)||ds|\leq |g'(z_0)|\int_{\gamma} |ds|\\ &=|g'(z_0)|\int_0^1|\gamma'(t)| dt=|g'(z_0)||z-w|, \end{align*} since $\gamma$ is a straight line. Since $-a_1(z-w)=g(z)-g(w),$ we have $$|a_1||z-w|=|g(z)-g(w)|\leq |g'(z_0)||z-w|,$$ or $$|a_1|\leq |g'(z_0)|=\left|\sum\limits_{n=2}^\infty na_n z_0^{n-1}\right|\leq \sum\limits_{n=2}^\infty \left|na_n z_0^{n-1}\right|\leq \sum\limits_{n=2}^\infty n|a_n| |z_0|^{n-1}< \sum\limits_{n=2}^\infty n|a_n|,$$ a contractiction.

Answer 2

Since the inequality is equivalent to $|Ra_1|\geq\sum\limits_{n=2}^\infty n|a_n|R^n$, we can use the substitution $z \to Rz$ and assume wlog $R=1$; since $\alpha f$ is injective iff $f$ is injective for any constant $\alpha \ne 0$ and since $a_1 \ne 0$ (as otherwise $f=0$ contradicts the fact that $f$ is not constant), we can use $\frac{f}{a_1}$ so assume $a_1=1$

Then $f'(z)=1+\sum_{k \ge 2}ka_kz^{k-1}$ and:

$|\Re \sum_{k \ge 2}ka_kz^{k-1}| \le |\sum_{k \ge 2}ka_kz^{k-1}| \le \sum_{k \ge 2}k|a_k||z|^{k-1} \le \sum_{k \ge 2}k|a_k| \le |a_1|=1$.

This shows that $\Re f'(z) \ge 0$ on the open unit disc. By the minimum property for harmonic functions and noting that $f'(0)=1$, we must have strict inequality so $\Re f'(z)>0$ and then Noshiro-Warschawski (which is just an easy integration on segments) shows that $f$ is injective. Done!

(edit: for reference if $\Re f'>0$ in a convex domain $U$ and $z \ne w$, then $|f(z)-f(w)|=|\int_{[wz]}f'(y)dy|=|z-w||\int_0^1f'(tz+(1-t)w)dt|$ and $\int_0^1\Re f'(tz+(1-t)w)dt >0$ shows that $|\int_0^1f'(tz+(1-t)w)dt| \ne 0$, hence $f(z) \ne f(w)$)