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

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.

-
Was there anything about $f(0)=0$ in the problem? (Aside: who teaches complex analysis via multiple choice questions?) – user31373 Jun 10 '12 at 20:44
@LeonidKovalev re your aside: probably someone who doesn't have a TA to grade the homework. – Robert Israel Jun 11 '12 at 7:29
@RobertIsrael Neither do I (but this is even more besides the point). – user31373 Jun 11 '12 at 13:47
@LeonidKovalev I suppose these are past year exam questions. – srijan Jun 11 '12 at 15:40

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 $0<r<|z|<1$ 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 $0<r<1$. 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.

-
Thanks for your reply. But the question is from an exam test held few months ago. No additional condition of f(0)= 0 is given. – preeti Jun 10 '12 at 21:06
@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. – Sam Jun 11 '12 at 6:59