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Let $f$ be a holomorphic function on D = $ ( z\in C : |z| <1 ) $ such that $ | f(z)|\leq1$. Let $ g : D: \rightarrow C $ be such that

$ g(z) = \frac{ f(z)} {z} $ if $z\in D $, $ z\neq 0$ and $ g(0) = \ f' (0) $ .

I have to select which are the correct options.

1) g is holomorphic (Seems correct by definition)

2) $ |g(z)|\leq 1$ for all $ z\in D$.

3) $ |f'(z)|\leq 1$ for all $ z\in D$.

4) $ |f'(0)|\leq 1$.

The solution set says all four are correct. Please suggest.

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    @LeonidKovalev I suppose these are past year exam questions.2012-06-11

1 Answers 1

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The solution is wrong: Consider $f(z) = z^n$ for (3)...

The rest is an application of the maximum modulus principle to $g$ and is correct. But we do need to have $f(0) = 0$ for $g$ to be holomorphic, as was already hinted at in the comments by Leonid Kovalev. If this additional information on $f$ is not given in the exercise statement, then all points stated there are wrong the first three points stated there are wrong: For (1) and (2) consider the constant function $f(z) = 1$.

(4) turns out to be right just from the assumption $|f(z)|\le 1$ alone. This follows from Cauchy's formula for the derivative. For any $0 we have

$f'(z) = \frac{1}{2\pi i}\oint_{|z| = 1-r} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta$

In particular

$|f'(0)| \le \frac{1}{2\pi}\oint_{|z| = 1-r} \frac{|f(\zeta)|}{(1-r)^2} \, d\zeta \le \frac{1}{1-r}$

for all $0. Letting $r\to 0$ gives $|f'(0)| \le 1$.
Remark: In the case were $f(0) = 0$, you might want to check out the Schwarz Lemma.

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    @preeti: I have expanded my answer to give you a counterexample for (1) and (2) and have written down a proof for (4). Exam or no exam, the solution is wrong and the problem statement most probably is missing the additional condition $f(0) = 0$ - unless this is supposed to be a trap for the students.2012-06-11