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The statement is: Let $\Omega$ be an open convex subset of $\mathbb{C}$, and $f\in\mathcal{H}(\Omega)$ with $|f'(z)-1|<1$, for all $z\in\Omega$. Prove that $f$ is injective.

I want to know if my proof is correct.

Suppose there exist $\alpha,\beta\in\Omega$ with $\alpha\ne\beta$ and $f(\alpha)=f(\beta)$. Because $\Omega$ is convex, we can integrate over the segment $\alpha\beta$, which is in $\Omega$. Then we have $$ \left|\int_{\alpha}^{\beta} (f'(z)-1) \,dz\right| = \left|\left. f(z)-z\right]_{\alpha}^{\beta}\right|= |f(\alpha)-\alpha - f(\beta)+\beta| = |\beta -\alpha|. $$ By the other hand we have $$ \left|\int_{\alpha}^{\beta}( f'(z)-1) \,dz\right|\leq \int_{\alpha}^{\beta}|f'(z)-1||dz| < \int_{\alpha}^{\beta} |dz| = |\beta - \alpha| $$ but this is not possible.

Is this correct? Thanks

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  • $\begingroup$ Fine........... $\endgroup$ Jul 2 '19 at 17:43
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    $\begingroup$ @DavidC.Ullrich: Your edit makes my answer obsolete :) $\endgroup$
    – Martin R
    Jul 2 '19 at 17:44
  • $\begingroup$ @MartinR Sorry - the answer hadn't appeared here when I made the edit. $\endgroup$ Jul 2 '19 at 17:46
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    $\begingroup$ @MartinR I was upvoting your anwser when suddenly disapeared :(... thanks for the anwser $\endgroup$
    – JoseSquare
    Jul 2 '19 at 17:48
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community wiki answer to push it from unanswered queue: Yes, your proof is correct.

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