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

  • $\begingroup$ Fine........... $\endgroup$ Jul 2 '19 at 17:43
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @MartinR I was upvoting your anwser when suddenly disapeared :(... thanks for the anwser $\endgroup$
    – JoseSquare
    Jul 2 '19 at 17:48

community wiki answer to push it from unanswered queue: Yes, your proof is correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.