5

enter image description here

My attempt:- (1)Taking $f(z)=.5$, So, $g(z)= \begin{cases} \frac{.5}{z} & z\neq 0 \\ 0 & z=0 \end{cases} $

So, I can eliminate (1) and (2)

I am trying to apply Schwarz pick lemma for (c), But I am not able to make $|f'(z)|\leq \frac{1-|f(z)|^2}{1-|z|^2}\leq 1$

Please help me.

3 Answers3

2

Take $f(z)=z^2$ for which $|f'(3/4)|=3/2\ge 1$. So (c) is false.

Option (d) is true by using Schwarz Pick theorem,

$|f'(0)|\leq \frac{1-|f(0)|^2}{1-|0|^2}=1-|f(0)|^2\leq 1$

Nitin Uniyal
  • 7,946
1

(3) is false by taking $f(z)=z^2.$

(4) follows from Cauchy's integral theorem: $$f'(0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z^2} dz,$$ where $C$ is a circle of radius $r < 1$. Bound $f'(0)$ using the bound for $f$. Then take $r \to 1^-.$

Dzoooks
  • 2,125
  • 1
  • 11
  • 20
  • $$|f'(0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z^2} dz|\leq \frac{1}{2\pi } \int_C |\frac{f(z)}{z^2}| |dz|\leq \frac{1}{2\pi} \int_C \frac{1}{|z|^2} |dz|$$ –  Jun 13 '19 at 01:30
  • How it is less than $1$? –  Jun 13 '19 at 01:31
  • @Unknownx $|z|=1$ on $C$, and the circumference of $C$ is $2 \pi$. Also, you should replace $|dz|$ with $dz$; you don't have to take absolute values of the differential...I suppose that makes sense, but it's already "positive" in a way... – Dzoooks Jun 13 '19 at 01:31
  • Since function is only defined on $D$. How would take counter integral over a curve which is outside the definition? –  Jun 13 '19 at 01:33
  • Take it on a circle of radius $R<1$ and let $R\rightarrow1$. – eranreches Jun 13 '19 at 01:34
  • @Unknownx Yes, good point! But you can take any circle of radius $r < 1$ and then let $r \to 1^-.$ – Dzoooks Jun 13 '19 at 01:35
  • @ThomasShelby Thanks. Fixed. – Dzoooks Jun 13 '19 at 01:39
0

Let $$f(z)=f(0)+zf'(0)+\frac{z^2}{2!}f''(0)+ . . .$$

$(1)\qquad$ Since $f(z)$ is analytic / holomorphic, so from the above equation we have $$f(0)=0, \qquad g(z)= \frac{f(z)}{z}=f'(0)+\frac{z}{2!}f''(0)+ . . .$$ Therefore $g(z)$ is holomorphic on $D$. $\qquad$(Option $1$ is true)

$(2)\qquad$ $$|g(z)|=|\frac{f(z)}{z}|\le 1 \qquad \forall z \in D$$(given that $|f(z)|\le 1$ and $|z|\lt 1$).$\qquad$(Option $2$ is true)

$(3)\qquad$ By Schwarz Pick Lemma, $$|f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}$$

Given $$|f(z)|\le 1 \qquad \text{and} \qquad |z|\lt 1$$

So $|f'(z)|$ may or may not be less than or equal to $1\quad \forall z\in D.$$\qquad$(Option $3$ is not true)

$(4)\qquad$ $$|g(0)|=|f'(0)|\le 1$$ $\qquad$(Option $4$ is true)

nmasanta
  • 9,222