# application of the maximum modulus theorem

Let $f$ be holomorphic on the unit disc and continuous on the unit circle. Suppose there is an $M \in \mathbb{R}$ such that $|f(z)| \leq M$ on the unit circle and let $\alpha_1, \alpha_2, ..., \alpha_n$ be zeros of $f$ in the unit disc listed according to multiplicity. Show that $|f(z)| \leq M \frac{|z-\alpha_1| \cdots |z- \alpha_n|}{|1-z \overline{\alpha_1}| \cdots |1-z \overline{\alpha_n}|}$.

Why can't I apply the Maximum Modulus theorem to $f$ directly? Is there something I am missing?

-
The factor $M$ is multiplied with may be less than $1$? – user20266 Aug 5 '12 at 9:29
I think so, there's no mention of it (in our notes) being larger/smaller than 1. That's one big sticking point for me. – buck Aug 5 '12 at 9:48
For example, when $\alpha_j=0$, we should have $|f(z)|\leq Mz^n$, which is a tighter bound than what we get with maximum modulus principle. – Davide Giraudo Aug 5 '12 at 9:52

Such a $M$ exists as the unit circle is compact. Let $M:=\max_{|z|=1}|f(z)|$. The map $$g(z):=f(z)\prod_{j=1}^n\frac{1-z\bar{\alpha_j}}{z-\alpha_j}$$ is holomorphic (since we can write $f(z)=\frac{z-\alpha_n}{1-z\bar{\alpha_n}}g_n(z)$, and continue this process, the cleanest way would be writing the multiplicities, and doing the last step for the multiplicity of the last root.)
If $|z|=1$ and $|a|<1$, then $$\left|\frac{1-z\bar a}{z-a}\right|=\frac{|1-\frac 1za|}{|z-a|}=\frac 1{|z|}=1,$$ hence $g$ is bounded by $M$ in the unit circle.
If $g$ is constant we are done, otherwise we conclude by maximum modulus principle.