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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 $(2)$ & $(4)$.

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  • 1
    $\begingroup$ Have you heard of the Schwarz-Pick lemma yet? $\endgroup$ – Daniel Fischer Dec 11 '14 at 16:21
  • $\begingroup$ Yes...But how we can apply for options (2) & (4)? $\endgroup$ – Empty Dec 11 '14 at 16:45
  • $\begingroup$ But for option (d) f(z)=z-1 exists satisfying the given conditions.I am confused... $\endgroup$ – Nitin Uniyal Aug 21 '15 at 15:10
  • $\begingroup$ Is your $f$ maps $D$ onto $D$ ? $\endgroup$ – Empty Aug 21 '15 at 17:29
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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

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

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  • $\begingroup$ sir what do you mean by ' Hyperbolic Distance '? $\endgroup$ – Empty Dec 12 '14 at 6:10
  • $\begingroup$ 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 $z,w\in D$. $\endgroup$ – Daniel Fischer Dec 12 '14 at 9:32
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Dec 12 '14 at 9:32
  • $\begingroup$ 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 $|f(1/4)|\ge 1/4$ ? $\endgroup$ – Empty Mar 8 '15 at 17:39
  • $\begingroup$ @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 $0$, $\endgroup$ – Daniel Fischer Mar 8 '15 at 18:07

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