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Suppose $f:\mathbb{D}\to \mathbb{C}$ is holomorphic and the diameter d:= $sup_{z,w\in \mathbb {D}} \vert{f(z)-f(w)}\vert $ of the image of $f$. We know that 2$\vert f'(0) \vert = d $ if and only if $f$ is linear.

If $f(z)=z+\textbf{a}z^2$ for some nonzero $\textbf{a}\in \mathbb{C}$, then $f$ is not linear. We know that $f'(0)=1$. Then, the diameter of $f$ >2 because 2$\vert f'(0) \vert \leq d$ always satisfy. However, if $\vert \textbf{a} \vert$ be very small, I couldn't find the points $z,w \in \mathbb{\bar{D}}$ satisfying d(the diameter of $f$) > 2.

Any help is appreciated..!

Thank you.

  • Take a look here: https://math.stackexchange.com/questions/111809/inequality-relating-diameter-of-the-image-of-a-holomorphic-function-on-the-unit – Robert Z Sep 26 '17 at 14:36
  • I know how to prove that. But, if f is nonlinear, then the diameter of f must be bigger than 2 by the problem's consequence. I want to find some points $z,w \in \mathbb {\bar{D}}$ satisfying the diameter of f be bigger than 2. –  Sep 26 '17 at 14:47
  • Let $f(t) = \cos(t)+a \cos(2t+b)$. The claim is that $\sup_t |f(t)| > 1$. If $b = \pm \pi/2$ it is obvious : if $a \cos(b) > 0$ then $f(0) > 1$, otherwise $f(\pi) < -1$. If $b = \pm \pi/2$ then look at $f(\epsilon), f(\epsilon+1)$ – reuns Sep 26 '17 at 14:48

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