# A qualifying exam problem involving Schwarz lemma

This is a problem in the book "Berkeley Problems in Mathematics", which I think the solution given is wrong, can someone help?

The following problem appeared in Spring 1991.

Let the function $f$ be analytic in the unit disc, with $|f(z)|\leqslant 1$ and $f(0)=0$. Assume that there is a number $r$ in $(0,1)$ such that $f(r)=f(-r)=0$. Prove that

$$|f(z)|\leqslant |z|\left| \frac{z^2-r^2}{1-r^2z^2}\right|.$$

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I would try to consider the function $g(z)=f(z)\,\frac{1-r^2z^2}{z^2-r^2}$, which is holomorphic on the disc because both $r$ and $-r$ are roots of $f$, so we can write $f(z)=(z^2-r^2)h(z)$, with $h$ holomorphic. –  Matemáticos Chibchas Jan 19 '13 at 15:42
Using your observation, the question will become very trivial. The solution given in the book involving multiple application of Schwarz lemma, and I think the steps given are also wrong. –  KWO Jan 19 '13 at 16:01

## 1 Answer

Note that the right hand side of the inequality is equal to $1$ on the unit circle. Then use the maximum modulus principle on the holomorphic function $$\frac{f(z)(1-r^2z^2)}{z(z^2-r^2)}.$$

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thanks for your help. –  KWO Jan 20 '13 at 15:26
I think there is a mistake in your argument, the analyticity breaks down at points r and -r hence the maximum modulus principle can not be applied. –  KWO Jan 20 '13 at 16:42
@KWO Since $f$ vanishes at $\pm r$ the quotient is well defined and holomorphic. This is the same principle that you also use for $f(z)/z$ in the Schwarz lemma. –  WimC Jan 20 '13 at 16:51