6
$\begingroup$

I am studying for a complex analysis exam tomorrow, and I am trying my hand at this problem:

Suppose $f$ is analytic in $\mathbb{D}$ and $|f(z)|\leq 1$ in $\mathbb{D}$ and $f(0) = 1/2$. Prove that $|f(1/3)|\geq 1/5$.

Hint: use the invariant form of Schwarz's Lemma.

As I understand it, the hint refers to the result that for any holomorphic $f:\mathbb{D}\to\mathbb{D}$, we have $$\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right|\leq\left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$$ for any $z_1,z_2\in\mathbb{D}$. Taking $z_1=0$ and $z_2=\frac{1}{3}$, this gets us $$\frac{|f(\frac{1}{3})-\frac{1}{2}|}{\;\left|1-\overline{f(\frac{1}{3})}\frac{1}{2}\right|\;}\leq\frac{1}{3}$$ but try as I might, I cannot manipulate this to produce the desired result. I'm worried I'm not understanding something correctly, and the exam is tomorrow, so any quick help would be much appreciated.

P.S. I found this exercise here, where it is problem 5.


Edit: Aha, I have an idea! We know that for any $a\in \mathbb{D}$, the map $\phi_a:\mathbb{D}\to\mathbb{D}$ defined by $$\phi_a(z)=\dfrac{a-z}{1-\overline{a}z}$$ sends $\phi_a(0)=a$ and $\phi_a(a)=0$. Observe that $$\phi_{\frac{1}{2}}(\tfrac{1}{3})=\frac{\frac{1}{2}-\frac{1}{3}}{1-\frac{1}{3}\cdot\frac{1}{2}}=\frac{\;\frac{1}{6}\;}{\frac{5}{6}}=\frac{1}{5}.$$ So if we can prove that any other map $f:\mathbb{D}\to\mathbb{D}$ with $f(0)=\frac{1}{2}$ is obtained from $\phi_a$ in a way that can only increase $|f(\frac{1}{3})|$, then we are done.

However, this would seem to be in the opposite direction from what I would expect: one of the consequences of Schwarz's lemma is that any holomorphic map from $\mathbb{D}$ to $\mathbb{D}$ is a contraction in the hyperbolic metric. So distances should be getting smaller.

$\endgroup$

1 Answer 1

2
$\begingroup$

The inequality should be (you put the bar on the wrong factor) $$\frac{\left| f (1/3)-1/2 \right|}{\left|1-1/2f(1/3) \right|} \leq \frac{1}{3} $$ which simplifies to $$9\left| f(1/3)-1/2 \right|^2 \leq \left| 1-1/2f(1/3) \right|^2 .$$ Now, using the identity $|\zeta|^2=\zeta \bar{\zeta}$ and expanding we find $$9(|f(1/3)|^2-\Re [f(1/3)]+1/4) \leq 1-\Re f(1/3)+1/4 |f(1/3)|^2 $$ Writing $f(1/3)=x+yi$, and multiplying by 4 gives $$36x^2+36y^2-36x+9 \leq 4-4x+x^2+y^2, $$ which simplifies further to

$$35x^2+35y^2-32x+5 \leq 0 $$

Completing the square gives

$$\left( x -\frac{16}{35} \right)^2+y^2 \leq \left( \frac{9}{35} \right)^2 $$

The points $(x,y)$ within this circle which is closest to the origin is

$$x=\frac{16-9}{35}=\frac{7}{35}=\frac{1}{5},y=0$$ Which gives $|f(1/3)| \geq 1/5$.

$\endgroup$
1
  • $\begingroup$ Ah, it took me a while to understand what you meant by "wrong" - I thought you meant that I misstated the result, but now I realize you're pointing out that I should have switched the roles of $0$ and $\frac{1}{3}$. The rest of your answer looks good, thanks so much for your help! I am still registering so it'll be just a bit more before I can upvote / accept your answer. $\endgroup$
    – complexist
    May 6, 2014 at 6:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .