# Does there exist $f:D\to D$ such that $f(0)=1/2$ and $f(1/2)=7/8$?

This is the problem I'm working on:

Let $D=\{z:|z|<1\}$. Does there exist a holomorphic function $f:D\to D$ such that $f(0)=1/2$ and $f(1/2)=7/8$?

We're near the end of our course so I have pretty much any tool at my disposal, such as Riemann's mapping theorem or the Schwarz lemma. These were also both just recently covered, so I'm thinking one or both should be used for the problem.

My attempts are mostly of this form: define $\phi:D\to D$ by $\phi(z)=\frac{7/8-z}{1-7z/8}$. Then we have $\phi(7/8)=0$, so $(\phi\circ f\circ f)(0)=0$ (I'm going to write the function $\phi f^2$), and by the Schwarz lemma we have $|\phi f^2(z)|\leqslant|z|$ for all $z$, and $|\phi f^2(0)|\leqslant1$. On the other hand we can calculate

$$1\geqslant|\phi f^2(0)|=|\phi'(7/8)|\cdot|f'(1/2)|\cdot|f'(0)|$$

which implies $|f'(1/2)|\cdot|f'(0)|\leqslant\frac{1}{|\phi'(7/8)|}=\frac{15}{64}$. However, I don't know what else I could say from here.

Can anyone tell me if I'm on the right track, or give a hint?

• Suppose such an $f$ exists, and look at $$z \mapsto \frac{f(z) - 1/2}{1 - \frac{1}{2}f(z)}.$$ – Daniel Fischer Dec 11 '16 at 12:49
• @DanielFischer that was actually one of the other things I tried. Following a similar thought process as my example, I arrived at $|f'(0)|<4/3$, with the equality being strict by Schwarz lemma. But I didn't know where to go from here either – Alex Mathers Dec 11 '16 at 12:52
• @DanielFischer actually I just figured it out. Thank you! – Alex Mathers Dec 11 '16 at 12:55

You're using the right idea, but you're overcomplicating things. We don't need a derivative in there. Supposing such an $f$ exists, we look at

$$g \colon z \mapsto \frac{f(z) - \frac{1}{2}}{1 - \frac{1}{2} f(z)}.$$

This is then a holomorphic function $D \to D$, with $g(0) = 0$. By the Schwarz lemma, we must have $\lvert g(z)\rvert \leqslant \lvert z\rvert$ for all $z \in D$. But

$$g\biggl(\frac{1}{2}\biggr) = \frac{\frac{7}{8} - \frac{1}{2}}{1 - \frac{7}{16}} = \frac{2}{3},$$

which contradicts the Schwarz lemma. Hence such an $f$ doesn't exist.

• Yep, that was what I realized when I made my comment. I was trying to use the estimates on the derivatives too much. Thanks again. I'll wait a while before accepting the answer – Alex Mathers Dec 11 '16 at 12:59

Use the Schwarz-Pick Lemma (which follows from the Schwarz Lemma): $$\left|\frac{f(z_1) - f(z_2)}{1-\overline{f(z_2)}f(z_1)}\right| \leq \left|\frac{z_1 - z_2}{1-\overline{z_2}z_1}\right|$$ with $z_1=1/2$ and $z_2=0$. Then $$\frac{2}{3}=\left|\frac{\frac{7}{8} - \frac{1}{2}}{1-\frac{1}{2}\cdot \frac{7}{8}}\right| \leq \frac{1}{2}$$ and we get a contradiction.