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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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    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.2012-06-18

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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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    Thank you @Erick. Very nice link and very nice answer there. +12012-06-18
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    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 $f(z)\ne f(\zeta)$.2012-06-18
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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,

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    Thanks a lot, @Patrick. Very nice trick. +12012-06-18
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    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.2012-06-18
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    Note that $f'(z) \neq 0$ only gives *local* injectivity.2012-06-18
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    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...2012-06-18
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    @DonAntonio : I was reading notes at the moment I answered on complex analysis and assumed too much things in my answer. I wasn't careful : sorry for that.2012-06-18
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    @Patrick , don't worry about that: we all miss here and there. Thanks for the new intent. Yet I don't follow: what do you mean when write "small enough so that $\,N(f(z_0))=1\,$"?2012-06-18
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    I'm still not sure how you go from local injectivity at the end to injectivity. Isn't $e^{2\pi z}$ also locally injective? But it's not injective on the unit disk ($z=\pm \tfrac12 i$).2012-06-19
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    @DonAntonio: A non-zero holomorphic function has isolated zeros (else it agrees with $0$ on a convergent sequence), so within a small disk around $z_0$, $f(z)=w$ has only one solution, making $N(f(z_0))=1$.2012-06-19
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    @Erick Wong : I'm saying in my edit that I got it wrong ; i.e. that my ideas don't prove the desired statement. So yes, local injectivity doesn't imply injectivity ; note that $e^{2\pi z}$ does not satisfy $|f'(z) - 1| < \frac 12$, so your example only destroys "locally injective on $D$ \quad \Rightarrow \quad injective", but the statement may still be true (as I hope it is).2012-06-19
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    @PatrickDaSilva The statement is already known to be true, using Leonid's argument linked to from my answer.2012-06-19
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    @Erick : I didn't notice it was your answer! I +1'ed it by the way.2012-06-19