The exercise I try to solve states: "Let $\,f\,$ be analytic in $\,D:=\{z\in\mathbb{C}\;|\;|z|<1\}\,$ , and such that $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$

Prove that $\,f\,$ is $\,1-1\,$ in $\,D\,$.

My thoughts: The condition $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$

means the range of the analytic function $\,f'\,$ misses lots of points on the complex plane, so applying Picard's Theorem (or some extension of Liouville's) we get that $\,f'(z)=w=\,$ constant, from which it follows that $\,f\,$ is linear on $\,D\,$ and thus $\,1-1\,$ there.

Doubts: $\,\,(i)\,\,$ This exercise is meant to be from an introductory first course in complex functions, so Picard's theorem seems overkill here...yet I can't see how to avoid it.

$\,\,(ii)\,\,$ Even assuming we must use Picard's Theorem, the versions of it I know always talk of "entire functions", yet our function $\,f\,$ above is analytic only in the open unit disk. Is this a problem? Perhaps it is and thus something else must be used...?

Any help will be much appreciated.

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    $\begingroup$ Yes, it is a problem to use Picard's theorem here. The function $f(z) = z^2$ misses lots of points if you restrict it to a unit disk, but it certainly is not constant! You would have a case if $f$ had an essential singularity, but this is certainly not true here. $\endgroup$ – Erick Wong Jun 18 '12 at 3:06

There is a very similar answer here (feel free to ignore the question if it looks intimidating). The basic idea is to use the triangle inequality applied to $f(z)-z$ to show that $|f(x)-f(y)| > 0$ for any distinct $x,y$. More precisely your assumption allows us to show $|f(x)-f(y)| \ge \frac12 |x-y|$.

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    $\begingroup$ Thank you @Erick. Very nice link and very nice answer there. +1 $\endgroup$ – DonAntonio Jun 18 '12 at 3:40
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    $\begingroup$ Thanks for the link. I left the final step there to the reader's imagination. It could be the triangle inequality, but it could also be pure logic: $|g(z)-g(\zeta)|<|z-\zeta|$ implies $g(z)-g(\zeta)\ne z-\zeta$, which is the same as $f(z)\ne f(\zeta)$. $\endgroup$ – user31373 Jun 18 '12 at 23:30

Use the fact that the inverse of $f$ is holomorphic in every disk centered at some $w_0 \in f(D)$ since $$ \lim_{w \to w_0} \frac{f^{-1}(w) - f^{-1}(w_0)}{w-w_0} = \lim_{z \to z_0} \frac{z - z_0}{f(z) - f(z_0)} \to \frac{1}{f'(z_0)}, $$ and the condition $|f'(z)-1| < \frac 12$ ensures that $f'(z_0) \neq 0$. Therefore $f$ is injective. If you fill in the blanks this is all you need.

EDIT : Yeah... I thought I had it. But as comments noticed I don't think my approach is so straight forward. As requested, I'll fill in some blanks that I thought helped. Recall that $$ \frac 1{2\pi i} \int_C \frac{f'(z)}{f(z)} \, dz = Z_f - P_f $$ for a meromorphic function, $Z_f$ standing for the number of zeros of $f(z)$ in the interior of $C$ and $P_f$ the number of poles (assuming there are no zeros/poles on the contour). Since in our context the function is analytic, it is in particular holomorphic, hence has no poles over $D$, so that this integral counts the number of zeros. Let $z_0 \in D$, and in the above integral, define $N(w)$ by substituting $f(z)$ above by $f(z) - w$ :
$$ N(w) = \frac 1{2 \pi i} \int_{C_r} \frac{f'(z)}{f(z) - w} \, dz $$ where $C_r$ stands for a circle of sufficiently small radius around $z_0$ (small enough so that $N(f(z_0)) = 1$), so that the integral counts how many $z$ are such that $f(z) = w$ close enough to $z_0$. You can show that $N(w)$ is continuous, so that since $N(f(z_0)) = 1$, $N(w) = 1$ by continuity for all $w$ such that $|w-f(z_0)| < r$. This means $f$ is locally injective, and furthermore $f(D)$ is open since $|f'(z) - 1| < 1/2$ implies $f'(z) \neq 0$, hence $f(z)$ is not constant (this would be the only case where $f(D)$ would not be open).

Hope that helps,

  • $\begingroup$ Thanks a lot, @Patrick. Very nice trick. +1 $\endgroup$ – DonAntonio Jun 18 '12 at 3:43
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    $\begingroup$ Could you fill in some of the blanks? The function $f(z) = e^z$ satisfies $f'(z) \ne 0$ for all $z$, but it isn't injective at all. $\endgroup$ – Erick Wong Jun 18 '12 at 5:08
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    $\begingroup$ Note that $f'(z) \neq 0$ only gives local injectivity. $\endgroup$ – mrf Jun 18 '12 at 8:56
  • $\begingroup$ Another doubt after thinking over this stuff: can we freely talk of $\,f^{-1}\,$ to evaluate the limit in the above answer before we even know $\,f\,$ is $\,1-1\,$ ? I don't mean the set theoretical $\,f^{-1}\,$ but actually the inverse function, which requires injectivity, at least locally if I'm not wrong... $\endgroup$ – DonAntonio Jun 18 '12 at 16:13
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    $\begingroup$ @Erick : I didn't notice it was your answer! I +1'ed it by the way. $\endgroup$ – Patrick Da Silva Jun 19 '12 at 2:04

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