## An estimate for holomorphic function

Suppose ${f: \mathbb{D}\rightarrow {\mathbb R}}$ is a non-negative function on the unit disk. Suppose ${u}$ is a holomorphic function such that ${|u|\leq f}$ on ${\mathbb{D}}$. Can we draw an upper bound for ${|u(0)|}$?

After discussing with Yin-Tat about this problem, we’ve come up with the following

Theorem 1 (Yin Tat’s estimate, 2011) With the above assumptions, suppose furthermore ${f\leq C}$, then we have

$\displaystyle |u(0)|\leq \frac{C(C|a|+f(a))}{C-|a|f(a)}$

for all ${a\in \mathbb{D}}$.

Corollary 2 In particular, if ${f}$ attains its minimum at ${z_0\in \mathbb{D}}$ with ${f(z_0)=m}$, then

$\displaystyle |u(0)|\leq \frac{C(C|z_0|+m)}{C-m|z_0|}.$

To begin with, let’s recall

Theorem 3 (Schwarz lemma) If ${u}$ is a holomorphic function on the unit disk such that ${|u|\leq 1}$ and ${u(0)=0}$, then

$\displaystyle |u(z)|\leq |z| \text{ on }\mathbb{D}.$

Proof: ${u/z}$ can be extended to a holomorphic function on ${\mathbb{D}}$. By maximum principle, (first on ${B_r(0)}$ and then take ${r\rightarrow 1}$), ${|u/z|\leq 1}$. $\Box$

Proof: (Proof of Yin-Tat’s estimate)
Suppose first that ${C=1}$. We want to prove

$\displaystyle |u(0)|\leq \frac{|a|+f(a)}{1-|a|f(a)}. \ \ \ \ \ (1)$

For technical reason, let us assume also that ${|f(z)|<1}$ for all ${z}$. (This can be done by changing ${f}$ to ${\tilde f=\alpha f}$ for ${\alpha<1}$ first. After proving (1) for ${\tilde f}$, we then let ${\alpha \rightarrow 1}$. ) Take any ${a\in \mathbb{D}}$ and let ${m=f(a)}$ (not necessarily minimum) and ${u(a)=l}$, so ${|l|\leq m<1}$.

Take ${\psi (z)=\frac{z+a}{1+\overline a z}}$ and ${\phi (w)=\frac{w-l}{1-\overline l w}}$. Then ${\psi(0)=a}$ and ${\phi(l)=0}$. The composition ${v=\phi\circ u \circ \psi }$ is a function from the unit disk to itself (i.e. ${|v|\leq 1}$) such that ${v(0)=0}$. We then apply the Schwarz lemma to deduce that

$\displaystyle |v(z)|\leq |z|$

for ${z\in \mathbb{D}}$. Note that ${\psi (-a)=0}$, so

$\displaystyle |v(-a)|=|\phi \circ u(0)|\leq |-a|=|a|$

$\displaystyle \Rightarrow \left|\frac{u(0)-l}{1-\overline l u(0)}\right|\leq |a|.$

Rearranging and applying triangle inequality,

$\displaystyle |u(0)|\leq \frac{|a|+|l|}{1-|a||l|}\leq \frac{|a|+m}{1-|a|m}.$

We have shown (1). (Note: the denominator is nonzero. )

In the general case, by replacing ${f}$ with ${\tilde f=f/C}$, ${u}$ with ${\tilde u=u/C}$ and ${m}$ with ${\tilde m=m/C}$ in the above estimate, we have

$\displaystyle \frac{|u(0)|}{C}=|\tilde u(0)|\leq \frac{|a|+m/C}{1-|a|m/C}.$

So

$\displaystyle |u(0)|\leq \frac{C(C|a|+m)}{C-m|a|}.$

$\Box$

Remark 1 Earlier, I have derived a bound ${|u(0)|\leq f(a)+C\tanh^{-1}(a)}$ (by using Cauchy integral formula, etc, interested readers may see Yin Tat’s facebook). Let me show that the bound in Yin Tat’s estimate is better than this one. First suppose ${C=1}$, and for simplicity ${a=|a|\geq0 }$ is real, we have to show for ${m=f(a)}$ fixed,

$\displaystyle m+\tanh^{-1}a\geq \frac{m+a}{1-ma}. \ \ \ \ \ (2)$

This is equivalent to

$\displaystyle \tanh^{-1}a\geq \frac{a(1-m^2)}{1-ma}.$

For ${a=0}$ this is true, and differentiating ${\tanh^{-1}a- \frac{a(1-m^2)}{1-ma}}$ gives

$\displaystyle \frac{1}{1-a^2} -\frac{1-m^2}{(1-ma)^2}$

which is easily seen to be non-negative. Substitute ${\tilde m=m/C}$ in (2) will prove the case for general ${C}$.

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### 9 Responses to An estimate for holomorphic function

1. Yin Tat Lee says:

It looks like 12.5 in real and complex rudin

2. KKK says:

Basically, all estimate involving Schwarz lemma or alike uses normalization like this.

3. edisychan says:

The equation after “Note that {\psi(-a)=0}, so” should be “{|v(-a)|=|\phi\circ u(0)|}…” right?

• edisychan says:

gosh, i should type $\psi(-a)=0$ and $“{|v(-a)|=|\phi\circ u(0)|$…

• edisychan says:

gosh^2, I am such a noob. I should type $\psi(-a)=0$ and $“|v(-a)|=|\phi\circ u(0)|$
Hope this time I am correct.

• edisychan says:

=.=…
$|v(-a)|=|\phi\circ u(0)|$… [Ok, get it, thanks! -KKK (By the way your icon is very cool, I hope I can have this one too ^.^)]

• edisychan says:

haha thx, btw can you add me to the list of authors?