If $\Omega$ is convex and $|f'(z)-1|<1$ then $f$ is injective 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

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