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?

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