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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?

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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

1 Answer 1

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.)

This function is continuous on the unit circle as the roots are in the open unit disk.

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.

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