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Let $f(z)=\sum\limits_{n=0}^{\infty} a_n z^n$ be an analytic function in $|z|\leq R$, such that for all $z$, $|f(z)|\leq M$. Let $z_0$ be one of $f$'s zeros such that for all other zeros $z$, $|z_0|\leq |z|$. Let's write $d=|z_0|$. Show that $d\geq \frac{R|a_0|}{M+|a_0|}$.

I tried several directions here. For example, I used Cauchy's theorem to write out:

$$f(z_0)=\int\limits_{|z|=R}\frac{f(z)}{z-z_0}dz$$

such that: $$0=|f(z_0)|=\left|\space \int\limits_{|z|=R}\frac{f(z)}{z-z_0}dz\right|\leq 2\pi R\cdot \frac{M}{R-d}$$

I also tried looking at the circle $|z|<d$, such that since $z_0$ is $f$'s closest circle then, for all $|z|<d$, $f(z)\neq 0$. I used Cauchy's theorem here to show that:

$$\forall |z|<d, 0<|f(z)|=\left|\space\int\limits_{|\xi |=d}\frac{f(\xi)}{\xi-z}d\xi\right|\leq 2\pi d \cdot \frac{M}{d-|z|}$$

This didn't yield any results.

In order to get $a_0$ into this equation, I tried to say that this is in particular true for $z=0$, such that:

$$|a_0|=|f(0)|\leq 2\pi d\cdot \frac{M}{d}$$

But this didn't yield any results.

What would be a good direction for this type of question?

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  • $\begingroup$ First assume $R = M = 1$. Use the Schwarz lemma and an automorphism of the unit disk. Then generalise. $\endgroup$ – Daniel Fischer Feb 1 '17 at 15:14
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    $\begingroup$ Just to clarify: Your summation goes from $n = 1...\infty$, but you mention $a_0$ in your bound. Do you mean to sum from $n=0 .... \infty$? $\endgroup$ – Badam Baplan Feb 9 '17 at 19:07
  • $\begingroup$ Also, it would be helpful to know where in your studies of complex analysis you are. $\endgroup$ – Badam Baplan Feb 9 '17 at 20:46
  • $\begingroup$ May I suggest to correct this question? $f(z)=\sum_{n=1}^{\infty}a_nz^{n}$ leaves no space for $a_0$ and it is not clear if $f(0)=0$ is the result of a typo or a proper input. $\endgroup$ – rtybase Feb 10 '17 at 21:11
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I will assume $d<R$, otherwise $$d \geq R > \frac{R|a_0|}{M+|a_0|}$$ is a trivial case. Now, let's look at this function: $$g(z)=\frac{f(z)-a_0}{z}=a_1+\sum_{k=2}^{\infty}a_kz^{k-1} \tag 1$$ which has a removable singularity at $z=0$. Additionally: $$|g(z_0)|\leq |a_1|+\sum_{k=2}^{\infty}|a_k||z_0^{k-1}|=|a_1|+\sum_{k=2}^{\infty}|a_k|d^{k-1}$$

From Cauchy's estimate and Taylor series, for $f(z)$: $$|a_k|=\left|\frac{f^{(k)}(0)}{k!}\right|=\frac{1}{2\pi}\left|\int_{C_{R}} \frac{f(z)}{z^{k+1}} dz\right|\leq \frac{M}{R^k}$$ As a result: $$|g(z_0)|\leq \frac{M}{R}+\frac{Md}{R^2}+\frac{Md^2}{R^3}+...=\frac{M}{R}\left(1+\frac{d}{R}+\frac{d^2}{R^2}+...\right)=\frac{M}{R}\frac{R}{R-d}=\frac{M}{R-d}$$

Then from (1): $$|g(z_0)|=\frac{|a_0|}{d} \Rightarrow d=\frac{|a_0|}{|g(z_0)|}\geq \frac{|a_0|}{\frac{M}{R-d}}=\frac{|a_0|(R-d)}{M}$$ Or $$dM\geq R|a_0|-d|a_0| \Leftrightarrow d(M+|a_0|)\geq R|a_0| \Leftrightarrow d \geq \frac{R|a_0|}{M+|a_0|}$$

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  • $\begingroup$ Nice. But I'm still not getting how the fact that $z_0$ is the smallest zero helps with anything. Did you have any ideas there? $\endgroup$ – Badam Baplan Feb 12 '17 at 3:06
  • $\begingroup$ @BadamBaplan This property applies to all the zeros, including the smallest one. It's definitely not of any use in my proof, but it could be for other attempts, e.g. exploring the zeros of the analytic functions proofwiki.org/wiki/Zeroes_of_Analytic_Function_are_Isolated ... I obviously didn't try it, so I am only guessing. $\endgroup$ – rtybase Feb 12 '17 at 9:34
  • $\begingroup$ Right right, I was just hoping there would be some very elegant way of proving the inequality that uses the fact that the zero is the smallest $\endgroup$ – Badam Baplan Feb 12 '17 at 17:30
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The Schwarz Lemma is often a good tool to consider when we want to bound (below) the magnitude of a point possessing some property or to bound (above) the derivative at a zero. Our goal here will be to apply the Schwarz lemma to $f$. To do this, we'll apply a couple of transformations to $f$ so that we get a new function $h$ which satisfies the conditions of the lemma, namely

  1. $h$ is analytic on the open unit disk
  2. $h(z) < 1$ for all $|z| < 1$
  3. $h(0) = 0$

By the lemma this will then imply $|h(z)| \leq |z|$ and by our construction of $h$ we will achieve the desired inequality.

First we normalize $f$. Let $$g(z) = \frac{f(Rz)}{M}$$ Since $f$ is analytic on a disk of radius $R$, $g$ is analytic on the unit disk. Similarly the bound of $M$ on $f$ becomes a bound of $1$ on $g$.

Now, we will turn the zero of $f$ at $a_0$ into a zero at the origin. Let $$\phi(z) = \frac{z + z_0/R}{\bar{z_o}z/R + 1}$$

$\phi$ is the Mobius transformation mapping the unit disk into itself, and sending the point $0$ to $z_0/R$. Note that $\phi$ is analytic. It's important to be familiar with Mobius functions. Lars Ahlfors writes in his introductory text that "the rather specialized assumptions of [Schwarz's Lemma] are not essential, but should be looked upon as the result of a normalization". The key to applying Schwarz's Lemma is often normalization by way of Mobius functions.

Finally define $$h(z) = g(\phi(z))$$

Note that $h$ is still analytic on the unit disk and bounded by $1$, since $h$ is the composition of analytic functions on the unit disk, and since $g$ and $h$ have the same image.

We also have now that $h(0) = g(\phi(0)) = g(z_0/R) = f(z_0)/M = 0$

Thus $h$ satisfies the conditions of the Schwarz Lemma, implying $$|h(z)| \leq |z|$$ holds for all $|z| < 1$. Applying this to $z = -z_0/R$, $$|h(-z_0/R)| \leq |z_0|/R$$

On the other hand, by construction $$h(-z_0/R) = g(\phi(-z_0/R)) = g(0) = \frac{f(0)}{M} = \frac{a_0}{M}$$

Putting it all together shows $$\frac{|a_0|}{M} \leq \frac{|z_0|}{R}$$ therefore $$\frac{R|a_0|}{M} \leq |z_0|$$

Noting $M + |a_0| \geq M$ always, we can write $$\frac{R|a_0|}{M + |a_0|} \leq |z_0|$$ which is the inequality slightly weaker inequality you were seeking.

A couple things to note:

  1. we didn't ever need to use the fact that $z_0$ was a smallest zero of $f$
  2. in the last equation we see that the inequality you stated is weaker than the inequality we were able to proof (the difference between $M + |a_0|$ and just $M$ in the denominator).

This makes me wonder where the question came from.

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  • $\begingroup$ Good exposition. +1 $\endgroup$ – Alfred Yerger Feb 9 '17 at 22:33
  • $\begingroup$ Although not very tidy ... e.g. $\frac{|a_0|}{M} \geq \frac{|z_0|}{R}$ and immediate conclusion ... $\endgroup$ – rtybase Feb 9 '17 at 22:42
  • $\begingroup$ I'd appreciate if you have any suggestions to make anything clearer or tidier! I value good exposition and would like to improve at it. $\endgroup$ – Badam Baplan Feb 9 '17 at 22:58
  • $\begingroup$ @BadamBaplan look carefully from "Putting it all together shows" to "therefore" $\endgroup$ – rtybase Feb 9 '17 at 23:02
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    $\begingroup$ oh yup, noticed that right after your comment. thank you! $\endgroup$ – Badam Baplan Feb 9 '17 at 23:18

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