What I have done so far:

Since $f$ is holomorphic on $U$ and $\overline{D}(0,1)$, we have by cauchy's integral formula we have \begin{align*} f'(0)=\frac{1}{2\pi i}\int_{\partial D(0,1)}\frac{f(w)}{w^2}dw \end{align*} So we have \begin{align*} |f'(0)|&=|\frac{1}{2\pi i}\int_{\partial D(0,1)}\frac{f(w)}{w^2}dw|\\ &\leq \frac{1}{2\pi}\int_{\partial D(0,1)}|\frac{f(w)}{w^2}dw|, \; \text{by the triangle and the modulus of $i$ being 1}\\ &\leq \frac{1}{2\pi}\max_{z,w\in \overline{D}(0,1)}\frac{f(w)-f(z)}{(w-z)^2} ||\partial D(0,1)||=\frac{1}{2\pi} \max_{z,w\in \overline{D}(0,1)}\frac{f(w)-f(z)}{(w-z)^2} 2\pi\\ &=\max_{z,w\in \overline{D}(0,1)}\frac{f(w)-f(z)}{(w-z)^2} \end{align*}

Of course by continuity the difference quotient can't blow up, but I am concerned this isn't the best way to approach the problem. I also have not used compactness of the closed disc, although that may be just to justify taking the max rather than the sup of f in the statement.


Notice that $$f^\prime(0) = \frac{1}{2 \pi i} \int_{\partial D(0;1)} \frac{f(\xi)}{\xi^2} d\xi = -\frac{1}{2\pi i} \int_{\partial D(0;1)} \frac{f(-\xi)}{\xi^2} d\xi,$$ whence $$ 2 f^\prime(0) = \frac{1}{2 \pi i} \int_{\partial D(0;1) } \frac{f(\xi) - f(-\xi)}{\xi^2} d\xi.$$ Use that $|\xi|=1$ along $\partial D(0;1)$ and $$|f(\xi) - f(-\xi)| \leqq \max_{\eta,\zeta \in \overline{D(0;1)}} |f(\eta)-f(\zeta)|,$$ if $|\xi|=1$.

  • $\begingroup$ where does the negative sign come from on the last expression in the first line $\endgroup$ – operatorerror May 9 '16 at 23:05
  • 1
    $\begingroup$ Apply Cauchy's integral formula to the derivative of $z \mapsto f(-z)$ at $z = 0$. $\endgroup$ – user179514 May 9 '16 at 23:11

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