Suppose that $f: D \to D$ is a holomorphic function on the open unit disc and $f(0)=0$.

If we set $\displaystyle \omega=\exp\left(\frac{2\pi i}{n}\right)$ for $n \in \mathbb{N}$ then show that for any $z \in D$ we have:

$$|f(\omega z)+f(\omega^2z)+\cdots+ f(\omega^nz)| \leq n |z|^n$$

Moreover, if there exists $z_0 \in D\setminus \{0\}$ such that the inequality becomes equality, prove that there exists $a \in \partial D$ such that $f(z)=az^n$

My attempt:

Since $f$ is a holomorphic function on the disc, it has a Taylor series around $z=0$:

$$f(z) = A_1z+A_2z^2+A_3z^3+ \cdots + A_nz^n+\cdots$$

It is easy to see that for $g(z) = f(\omega z)+f(\omega^2z)+\cdots+ f(\omega^nz)$ is an analytic function and $z=0$ is a zero of multiplicity $n$. (This is easily seen by adding the Taylor series for $f(\omega^i z)$ and using the fact that $1+\omega+\cdots+\omega^{n-1}=0$)

So, the Taylor series for $g(z)$ around $z=0$ is something like this:

$$g(z)= n \cdot \frac{f^{(n)}(0)}{n!} z^n + \cdots$$

Let's define the function:

$$h(z) = \begin{cases} \displaystyle \frac{g(z)}{z^n}, & \mbox{if } z \neq 0 \\ \displaystyle \frac{f^{(n)}(0)}{(n-1)!}, & \mbox{if } z = 0 \end{cases}$$

$h(z)$ is a holomorphic function. By using the Maximum principle, the maximum of $h(z)$ is attained at some point $m \in \mathbb{C}$ on the border, i.e. $|m|=1$, therefore:

$$|h(z)| \leq |g(m)| \leq |f(\omega z)+f(\omega^2z)+\cdots+ f(\omega^nz)|$$ $$ \leq |f(\omega z)|+|f(\omega^2z)|+\cdots+ |f(\omega^nz)| \leq n $$

Therefore $$|g(z)| \leq n |z|^n$$

Which is the desired result.

I can not show the second part of the theorem. I assume there's some small point that I'm missing or something. I've tried a lot of things, but I've failed.

Please give me a hint or something.


For the second part:

Using your $$g(z)= n \cdot \frac{f^{(n)}(0)}{n!} z^n + \cdots$$ and function $$h(z) = \begin{cases} \displaystyle \frac{g(z)}{z^n}, & \mbox{if } z \neq 0 \\ \displaystyle \frac{f^{(n)}(0)}{(n-1)!}, & \mbox{if } z = 0 \end{cases}$$

Suppose that $|g(z)| = n |z|^n$. Then |h(z)|=1 for some $z_0$ in your domain. What does the maximum modulus principle say about $h$?

  • $\begingroup$ Hm, I've already considered the maximum modulus principle. Does it say anything useful? Remember that $h(z)$ is not necessarily from $D$ to $D$, unless we prove it is. So, we can't say that $|h(z)|<1$. $\endgroup$ – math.n00b May 7 '14 at 18:15
  • $\begingroup$ That's true, but by your inequality $|h(z)/n| \le 1$ on the boundary. Max Mod would give that $h(z)/n$ is constant. $\endgroup$ – Chris C May 7 '14 at 20:03

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