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Let $f$ be an analytic map defined on $D=\{z:|z|<1\}$ such that $|f(z)|\le 1\forall z\in D$. Then which of the following statements are true?

  1. There exists $z_0\in D$ such that $f(z_0)=1$.

  2. The image of $f$ is an open set.

  3. $f(0)=0$

  4. $f$ is constant.

Well first I wonder how can $|f|$ assume the value $1$? 4 is correct I guess, and 2 may be correct by the Open Mapping Theorem, but I am not able to figure out the other options.

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What is the question? Which of the alternatives that may hold or which of the alternatives that certainly hold? –  mrf Jun 4 '12 at 21:44
    
edited mrffffffffffff –  Bunuelian Trick Jun 4 '12 at 21:45
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2 Answers

up vote 2 down vote accepted

Hint: Consider the functions $f_1(z) = 1/2$ and $f_2(z) = z$. Both satisfy the assumptions.

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1) Is not true. $f$ could be a constant function taking value $\frac{1}{2}$ for example. However, if $f$ is not constant, by the maximum modulus principle, $f(z) \neq 1$ for all $z \in D$.

2) Again if $f$ is not constant, by the open mapping theorem, $f(D)$ is an open set.

3). Again constant functions are counterexamples. However if $f(0) = 0$, then there is a lot more that can be said of $f$ by the Schwarz Lemma.

4) The identity function $f(z) = z$ shows that $f$ need not be constant.

None of the above are necessarily true.

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