

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Let $$f(z) = z - \sum^\infty_{n=2} a_nz^n.$$ What is the largest ball around $0$ where $f$ is injective?

If we restrict to the case where $a_n \geq 0,$ it seems the radius should be given exactly by the minimum positive zero of $f'(x).$ Is there an easy complex analysis proof of this fact in this special case?

share|cite|improve this question
up vote 7 down vote accepted

Here is a supporting evidence for the conjecture made in the end. Let $f$ be a plynomial (or an entire function). We have $f'=1-P$; where $P$ is a power series with positive coefficients. I claim that the zero of $f'$ which is closest to the origin is positive. Indeed this zero is the closest singularity to the origin of the function $$1/f'=1/(1-P)=1+P+P^2+P^3+... .$$ But this is a power series with positive coefficients, therefore the closest singularity to the origin must be on the positive ray by Pringsheim's theorem.

However, this does not prove the conjecture, even for polynomials, because there can be other obstacles to injectivity, different from a zero of $f'$. One has to prove that these other obstacles cannot occur for such series.

EDIT: In fact the conjecture is true! Let $f(z)=z-Q(z)$, $Q$ has positive coefficients and double zero at zero. Suppose without loss of generality that the zero of $f'$ of the smallest modulus is outside the disc $D$ of radius $r$ with center at the origin.

We wish to prove that $f$ is injective in this disc. Consider $P=Q'$. Then $P(0)=0$ and $P$ has positive coefficients. So $P$ is increasing on $(0,r)$ and cannot take the value $1$. Thus $|P(z)|$ is less than $k$ in $D$, with some $k\in(0,1)$, because coefficients are positive.

Now we prove the statement by contradiction. Suppose $f(z_1)=f(z_2)$ with $z_j$ in the unit disc. Integrating along the straight line segment connecting $z_1$ and $z_2$, we obtain $$0=f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(\zeta)d\zeta=z_2-z_1-\int_{z_1}^{z_2}P(\zeta)d\zeta.$$ Estimating the integral, we obtain $$|z_1-z_2|=\left|\int_{z_1}^{z_2}P(\zeta)d\zeta\right|\leq k|z_1-z_2|,$$ which is a contradiction. This proves your conjecture.

share|cite|improve this answer
Ralph: my last edit shows that your conjecture is true (and simple, as you suggested:-) Congratulations: nice result! I will use it as an exercise in my Complex Variables course. – Alexandre Eremenko Mar 24 '13 at 21:40
It is some analogue of an algebraic theorem I proved, I wanted to know if it had a simple proof in complex variables. You proof exhibits this fact. In fact it should be that for holomorphic maps $F:\mathbb{C}^n\rightarrow \mathbb{C}^n$ of the form $$F(z)= \text{id}(z) + O(\|z\|^2)$$ that the above holds (for appropriate statements of what the domain could be, some kind of polydisks.) Thank you for your time. – J. E. Pascoe Mar 24 '13 at 22:26
For a complete, edited proof, see, I am tired of struggling with Math Jack! – Alexandre Eremenko Mar 24 '13 at 22:26
The crucial thing which makes this true is that coefficients, except the first one, are negative. Just O(|z|^2) is certainly not enough. Can you state the precise conjecture for several variables? – Alexandre Eremenko Mar 24 '13 at 22:30
And what is the "algebraic theorem" you proved? – Alexandre Eremenko Mar 24 '13 at 22:31

de Branges's theorem / Bieberbach's conjecture says that if $f$ is injective on the unit disc then $|a_n| \leq n$. Then if $f$ is injective on the ball of radius $r$ than by rescaling $|a_n| \leq n /r^{n_1}$, so we have the upper bound

$r \leq \sup_{n\geq 2} \left( \frac{a_n}{n}\right)^{\frac{-1}{n-1} } $

For all $a_n$ nonnegative, your bound is at least as strong as this one, so this cannot be used to construct counterexamples.

share|cite|improve this answer
I believe the general claim is that you consider $$f(z)=z-\sum_{n\geq 2} a_nz^n.$$ Create some standardized version of the function $$\hat{f}(z)=z-\sum_{n\geq 2} |a_n|z^n,$$ and denote the first positive zero of $\hat{f}'(x)$ as $r_1.$ Then, $$r_1 \leq r \leq \text{above given bound from the Bieberbach conjecture}.$$ – J. E. Pascoe Mar 24 '13 at 4:54
Furthermore, should be sharp if all $a_n \geq 0.$ – J. E. Pascoe Mar 24 '13 at 4:55
This gives an estimate for the radius of injectivity in terms of the coefficients, however this does not address the conjecture made in the end: that the radius of injectivity is EQUAL to the smallest positive zero of $f'$. This seems to be an interesting conjecture... – Alexandre Eremenko Mar 24 '13 at 14:37
I don't disagree. I just said what I could figure out. – Will Sawin Mar 24 '13 at 14:44
The "context" is very helpful Will! – J. E. Pascoe Mar 24 '13 at 18:23

The question can be answered in a general way by is the following way.

You take a function on some polydisk $$f(z) = id(z) - \sum_{|I|\geq2} a_Iz^I.$$ Where $a_I\geq 0.$ As in the paper we're going to evaluate on the commuting tuples of matrices. (This satisfies the conditions on the domains used in the paper) The derivative of f (on commuting tuples of matrices) satisfies the following inequality $$\|Df_i(z)[h]\|\geq Df_i(\|z\|)[\|h\|]$$ (where we're abusing $\|x\| = (\|x_1\|,\ldots,\|x_n\|).)$ Thus, if the derivative is singular, which is equivalent to noninjectivity by the theorem, at some matrix tuple $z$ in some direction $h$, we have that $$0\geq Df_i(\|z\|)[\|h\|]$$ Thus, $$\langle Df(\|z\|)[\|h\|],\|h\| \rangle \leq 0.$$ So $Df[\|z\|]$ has a nonpositive eigenvalue. Thus, on the line segment between $0$ and $\|z\|$ there is a point $w$ such that $Df[w]$ is singular and positive semidefinite. (The derivative at zero is the identity and derivative is real symmetric by assumption) Thus $f$ is not injective on the domain.

share|cite|improve this answer

Your Answer


