I am working on a problem from an old complex analysis qual, and have run across the following problem:

Let $f(z)$ be holomorphic on $D(0,1)$, such that $|f(z)|\leq 1$ for all $|z|<1$. If $f(0) = f'(0) = 0$, prove $|f''(0)|\leq 2$.

After working through it for awhile, I realized we definitely needs Schwarz' Lemma for the proof. I'm just unsure of how to handle the second derivative. I think working with the cases of $f(z) = 1$ and $f(z) < 1$ is the way to start.


You can apply the maximum principle to $\dfrac{f(z)}{z^2}$.

You can also apply the Cauchy integral formula

$$f''(0) = \frac{2}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{f(\zeta)}{\zeta^3}\,d\zeta,$$

use the standard estimate, and let $r\uparrow 1$.

  • $\begingroup$ How does the maximum principle apply for $\frac{f(z)}{z^2}$? $\endgroup$ – Keith Jul 14 '17 at 5:07
  • 2
    $\begingroup$ @Keith Since $f(0) = f'(0) = 0$, $g(z) = \frac{f(z)}{z^2}$ has a removable singularity at $0$. Thus, considering it removed, $g$ is a holomorphic function on the disk, and we can apply the maximum principle to $g$ (on the disk $D(0,r)$ for $0 < r < 1$, yielding $\lvert g(z)\rvert \leqslant 1/r$ for $\lvert z\rvert \leqslant r$, then let $r\to 1$ to conclude $\lvert g(z)\rvert \leqslant 1$ for all $z\in D(0,1)$). $\endgroup$ – Daniel Fischer Jul 14 '17 at 11:53

Hint: Apply the Schwarz lemma to $f(z)/z$.


It seems as according to Harald Hanche-Olsen

We can define g(z) = f(z)/z

Use product rule to take the derivative and apply Schwarz lemma

to conclude the third coefficient in the taylor series expansion is

less than one.


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