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Let, $D=\{z\in \mathbb C:|z|<1\}$. Which are correct?

  1. there exists a holomorphic function $f:D \to D$ with $f(0)=0$ & $f'(0)=2$.

  2. there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=\dfrac{3}{4}$ & $f'\left(\dfrac{2}{3}\right)=\dfrac{3}{4}$.

  3. there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$ & $f'\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$

  4. there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}$ & $f'\left(\dfrac{1}{4}\right)=1$.

With the help of Schwarz lemma & its applications, we find that $(1)$ is false & $(3)$ is true.

But, I can not think about options $(2)$ & $(4)$.

Empty
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2 Answers2

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We want to determine whether for given $a,b,c,d$, there exists a holomorphic $f\colon D \to D$ with

  1. $f(a) = b$, and
  2. $f'(c) = d$.

A typical way to attack such a problem is the Schwarz-Pick lemma, resp. its differential version

$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\rvert^2}\tag{1}$$

for $z\in D$ when $f\colon D\to D$ is holomorphic, and if we have equality at one point, then $f$ is an automorphism of $D$.

In our case, we must check whether

$$\lvert d\rvert \leqslant \frac{1 - \lvert f(c)\rvert^2}{1-\lvert c\rvert^2}\tag{2}$$

for some holomorphic $f\colon D\to D$ with $f(a) = b$. If $(2)$ doesn't hold for any such $f$, then $(1)$ tells us that no $f$ with the prescribed properties exists, and if there is such an $f$ that $(2)$ holds, we often get enough restrictions from $(2)$ that constructing a function with the desired properties or a proof that no such function exists are easier.

In case 4., $f\colon D\to D$ with $f\left(\frac{1}{2}\right) = - \frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$, the Schwarz-Pick lemma tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \geqslant \frac{1}{4}$ since the hyperbolic distance between $f\left(\frac{1}{4}\right)$ and $-\frac{1}{2}$ can be at most equal to the hyperbolic distance between $\frac{1}{4}$ and $\frac{1}{2}$. On the other hand, $(2)$ tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \leqslant \frac{1}{4}$ in order to have the right hand side $\geqslant 1$. The only point in $D$ satisfying both requirements is $-\frac{1}{4}$, so if an $f$ with the desired properties exists, we must have $f\left(-\frac{1}{4}\right) = -\frac{1}{4}$, and since equality holds in $(1)$ then, it follows that $f(z) = -z$. But then we have $f'\left(\frac{1}{4}\right) = -1$, so there is no holomorphic $f\colon D \to D$ with $f\left(\frac{1}{2}\right) = -\frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$.

For case 2., $f\colon D\to D$ with $f\left(\frac{3}{4}\right) = \frac{3}{4}$ and $f'\left(\frac{2}{3}\right) = \frac{3}{4}$, the Schwarz-Pick lemma is not as effective. From it, we obtain the bounds $$\frac{2}{3} \leqslant \left\lvert f\left(\frac{2}{3}\right)\right\rvert \leqslant \sqrt{\frac{7}{12}},$$

which don't narrow down the possibilities for $f$ much. However, with so much space to play, we suspect that such an $f$ exists. To find one, we move the fixed point of $f$ to $0$ and consider $g = T_{3/4}\circ f \circ T_{-3/4}$, where

$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}\cdot z}.$$

We want $f$ to "shrink the unit disk towards $\frac{3}{4}$", so we make the ansatz $g(z) = c\cdot z$ for some $c\in (0,1)$ which we want to determine so that $f'\left(\frac{2}{3}\right) = \frac{3}{4}$. So we try

$$f(z) = T_{-3/4}\left(c\cdot T_{3/4}(z)\right).$$

We have $T_{3/4}(2/3) = -\frac{1}{6}$, hence

$$f'\left(\frac{2}{3}\right) = T_{-3/4}'\left(-\frac{c}{6}\right)\cdot c \cdot T_{3/4}'\left(\frac{2}{3}\right).$$

Since

$$T_w'(z) = \frac{(1-\overline{w}z) +\overline{w}(z-w)}{(1-\overline{w}z)^2} = \frac{1-\lvert w\rvert^2}{(1-\overline{w}z)^2},$$

we compute

$$T_{3/4}'(2/3) = \frac{7/16}{(1/2)^2} = \frac{7}{4};\qquad T_{-3/4}'(-c/6) = \frac{7/16}{(1-c/8)^2} = \frac{28}{(8-c)^2}$$

and find that $c$ should satisfy

$$\frac{49 c}{(8-c)^2} = \frac{3}{4}.$$

Solving the quadratic equation gives the solution

$$c = \frac{122 - 14\sqrt{73}}{3} \approx 0.7946491885181928.$$

Daniel Fischer
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  • sir what do you mean by ' Hyperbolic Distance '? – Empty Dec 12 '14 at 06:10
  • When one takes the disk as a model for the hyperbolic plane, one introduces a different distance - which induces the same topology - such that the automorphisms of the unit disk are isometries with respect to that distance, $$d(z,w) = \operatorname{Ar tanh}\left\lvert \frac{z-w}{1-\overline{w}z}\right\rvert,$$ that is the "hyperbolic distance". The Schwarz-Pick lemma can be formulated as "a holomorphic $f\colon D\to D$ never increases the hyperbolic distance", that is, $d(f(z),f(w)) \leqslant d(z,w)$ for all $z,w\in D$. – Daniel Fischer Dec 12 '14 at 09:32
  • Sorry, I'm used to that being done at the same time as the Schwarz-Pick lemma is treated, so I assumed the hyperbolic distance as something already heard of. – Daniel Fischer Dec 12 '14 at 09:32
  • Sir I could not understand that how $|f(1/4)|\ge 1/4$ ? From your reason on it we have, $\left|\frac{1+2 ,f(1/4)}{2 + f(1/4)}\right| \leq \frac{2}{7}$. But from here how I can show that $|f(1/4)|\ge 1/4$ ? – Empty Mar 08 '15 at 17:39
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    @Panja.S. From the Schwarz-Pick lemma, we obtain $$\left\lvert \frac{f(1/2) - f(1/4)}{1-\overline{f(1/4)}f(1/2)}\right\rvert \leqslant \left\lvert \frac{1/2-1/4}{1-\frac{1}{4}\cdot\frac{1}{2}}\right\rvert.$$ You can insert $f(1/2) = -1/2$ into that, and after some manipulation find that this implies $\lvert f(1/4)\rvert \geqslant 1/4$. On a more conceptual level, the "disks" with respect to the hyperbolic metric (the sets of the form ${ z : d(z,w) < r}$) are also Euclidean disks. However the hyperbolic centre and the Euclidean centre only coincide if the centre is $0$, – Daniel Fischer Mar 08 '15 at 18:07
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    but that doesn't matter here. By the symmetry of the hyperbolic distance, the hyperbolic disk with centre $-1/2$ and hyperbolic radius $d(1/2,1/4)$ is obtained by rotating the hyperbolic disk with centre $1/2$ and the same radius around $0$ by $\pi$ ($180°$), so since the latter disk doesn't intersect ${ z : \lvert z\rvert < 1/4}$, neither does the former. And since $f(1/4)$ lies in the closure of the former disk, it follows that $\lvert f(1/4)\rvert \geqslant 1/4$. – Daniel Fischer Mar 08 '15 at 18:08
  • I like this answer because it shows the underlying thought process. This is much more valuable than computation. – copper.hat Jun 02 '15 at 19:24
  • @DanielFischer Isn't there any shorter way to do this problem?? I'm asking this as this question appeared in an entrance examination in which we cannot afford to spend this much time on any problem. – Shivani Goel Jun 07 '16 at 21:48
  • How do we conclude f(z)=-z after proving f(1/4)=-1/4? Please explain. – Lawrence Mano Jan 06 '21 at 17:51
  • @LawrenceMano We then have $f(1/2) = -1/2$ and $f(1/4) = -1/4$. Thus $g \colon w \mapsto f(-w)$ is a holomoprhic function $D \to D$ with two fixed points. The only such function is the identity, hence $f(z) = g(-z) = -z$ for all $z \in D$. – Daniel Fischer Jan 06 '21 at 18:30
  • Thank you very much. – Lawrence Mano Jan 07 '21 at 00:00
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For Options 2 and 3, Schwarz Lemma does not lead to any contradictions. Therefore, we can explore an alternative approach: seeking functions that meet these criteria. In this regard, the most straightforward choice is a linear function. Assuming a constant derivative, we can perform the necessary calculations to derive linear functions that meet the specified criteria.

For option 2, use $f(z)=\frac{3}{4}\left(z+\frac{1}{4}\right)$.

For option 3, use $f(z)=-\frac{3}{4}\left(z+\frac{1}{4}\right)$.

RENJITH T
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