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Suppose $f(z)$ is analytic in the open unit disc and $|f(z)|<1$ there. Suppose further that $f(0) =0$ and $f'(0) = a \neq 0$. Show that there is a disc of positive radius $|z|<\rho$ such that for $z_1$ and $z_2$ in the disc, $$f(z_1)=f(z_2) \Longrightarrow z_1=z_2.\tag{1}$$ Find an estimate for $\rho$. Try to make the estimate as sharp as you can. Hint: $$f(z_2)-f(z_1) = \int_{z_1}^{z_2}f'(z)dz = a(z_2-z_1)+ \dotsb .$$

My answer so far: I can show (1) in a neighborhood of zero as follows: $z=0$ is an isolated zero of $f$ by the isolated zero theorem (note $f$ is non-constant since $f'(0)\neq 0$). Therefore there is a closed neighborhood $\overline{B_{\rho}(0)}$ of zero such that $f\neq 0$ on the punctured disc $\overline{B_{\rho}(0)} \setminus \{0\}$. Let $M = \max_{z\in \partial{B_{\rho}(0)}}|f(z)|$. Now for $w\in B_{M}(0)$, we have that $|f(z)-0|> |w-0|$ for $z\in \partial B_\rho(0)$, so $f(z)-0$ and $f(z)-0 + (w-0)=f(z)-w$ have the same number of zeros in $\overline{B_{\rho}(0)}$, meaning that $f$ is one-to-one there.

For the estimate, I'm having more trouble. I know by Schwarz's theorem that $|f(z)|<|z|$ in the disc, and $a<1$, unless $f(z)=\lambda z$ for $\lambda \in S^1$, in which case both are equalities. But how can I use this?

If I try to use the hint, I get (I think) $$f(z_2)-f(z_1) = a(z_2-z_1)+ \frac12 f''(0)(z_2^2 - z_1^2) + \dotsb$$ but I'm not sure what to do with this.

Eric Auld
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  • Try to apply the inverse function theorem. Although, I don't know if you will get an estimate for $\rho$. – Mustafa Said Dec 23 '13 at 22:38
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    Here's a hint, Eric: $f$ will be one-to-one until you hit a radius $\rho$ on which $f$ has a critical point. (The local behavior at a critical point tells you that the function will fail to be one-to-one on little disks centered at that point.) – Ted Shifrin Dec 23 '13 at 22:49
  • @TedShifrin Thanks for the hint. I can prove that $f$ will not be one-to-one on a neighborhood of a critical point by a Rouche's argument like the one above. But I'm having a harder time proving that if $f$ is analytic with $f'\neq 0$ in $B_{\rho}(0)$, then $f$ is one-to-one on $B_{\rho}(0)$. I can show that $f$ is locally one-to-one by the above argument...any hints for extending this? – Eric Auld Dec 23 '13 at 23:56
  • Yes, of course I'm wrong, as $f(z)=e^z$ illustrates. Let's try to use the Argument Principle. Consider $$g(w)=\frac1{2\pi i}\int_{|z|=R}\frac{f'(z)}{f(z)-w}dz.$$ Can we find the estimate on $|w|<\rho$ to ensure that $g=1$? – Ted Shifrin Dec 24 '13 at 00:14
  • @TedShifrin Isn't f(z) = $e^z$ is single valued if we limit ourselves to $0 \le \theta < 2\pi $? And f(0) $\ne$ 0 in this case. – Betty Mock Dec 24 '13 at 00:45
  • @BettyMock: You mean if you limit $|y|<\pi$, say. I was merely pointing out the error in my argument, not giving a counterexample. – Ted Shifrin Dec 24 '13 at 01:01

4 Answers4

1

Hint. Try to find a maximum disk $D(0,r)$ where the inverse of $f$ is definable as a holomorphic function.

Yiorgos S. Smyrlis
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1

I interpret the hint as the suggestion to apply the Mean Value Inequality to $f(z)-az$: $$|f(z_2)-f(z_1) - a(z_2-z_1)| \le |f'(\xi)-a| |z_2-z_1|$$ where $\xi$ is on the line segment between $z_1$ and $z_2$. So, you have injectivity in $|z|<\rho$ if you can show that $|f'-a|<|a|$ there.

One way to go from here is to use Cauchy's integral formula: $$ f'(z)-f'(0) = \frac{1}{2\pi i} \int_{|\zeta|=r} \left(\frac{1}{(\zeta-z)^2} - \frac{1}{\zeta^2}\right)f(\zeta)\,d\zeta $$ This way, the given bound for $|f|$ fits right, and the rest is algebra around $$ \frac{|z| \, |2\zeta-z|}{|\zeta-z|^2 |\zeta|^2} $$ Pushing $r\to 1^-$, you can get this bounded by
$$ \frac{\rho (2+\rho) }{(1-\rho)^2 } $$ As long as this thing is $<|a|$, injectivity holds.

The beginning of the solution can be worded differently. Write $f(z)=a(z+g(z))$. If you can find a region where $g$ is a strict contraction, then $z+g(z)$ is injective there (easy to see). The rest is about getting $|g'|<1$.

Post No Bulls
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Alternatively,try to find the minimum $ρ>0$ such that if $f(z_1)=f(z_2)$ then $z_1\neq z_2$.

This has logic because the $z_1,z_2$ that have the property $(1)$ lie in a disk centred at $0$. So all the others will be outside.

Now let $z_1,z_2$ such that $f(z_1)=f(z_2)$ .Then from the hint we have $a(z_2-z_1)+c_2(z_2^2-z_1^2)+...+c_n(z_2^n-z_1^n)+...=0$ $(*)$ with $$c_n=\frac {f^{(n_)}(0)}{n!}$$.

Now for these $z_2\neq z_1$ we can divide with $z_2-z_1$ the $(*)$.Until when you can divide with $z_2-z_1$? I mean which is the biggest $k$ such that if you divide with $(z_2-z_1)^k$ everything is OK?

Haha
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We have f(0) = 0 and f'(0) $\ne$ 0. Let c > 0 be such that f has no zeros other than 0 in |z| < c, and for some $z_1, z_2$ in this domain we have $f(z_1) = f(z_2)$ . We have

$2 \pi if(z_1) = \oint_{|z| = c} f(z)/(z- z_1)dz = 2 \pi if(z_2) = \oint_{|z| = c} f(z)/(z- z_2)dz.$ So these two integrals are equal.

This gives $\oint_{|z| = c} [f(z)/(z- z_1)] -[f(z)/(z- z_2)] dz = 0$ and
$\oint_{|z| = c} f(z)(z_1- z_2)/[(z-z_1)(z-z_2)] dz = 0.$

If f(z) $\ne$ 0 in this domain (except at zero) there are two cases:
1. f(z) does not have period 2$\pi$. Then the only term in this last integral that can be 0 is $(z_1- z_2)$ so we conclude $z_1= z_2$ in this domain.
2. f(z) is periodic in $2\pi$. Then f($2\pi$) = 0. Then c must be less than the smallest divisor of $2\pi$ in which f is periodic.

Suppose $f(z_0) = 0$ where $z_0 \ne$ 0 and |$z_0$| is the smallest modulus of the zeros of f (outside of 0). Then $f(z_0) = f(0)$, so we have a duplicate value.

Thus c must be less than the modulus of the nearest zero or singularity of f.

If there is a smaller bound on c this argument does not show it.

Betty Mock
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