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So I am working on the following problem.

Suppose that $f$ is entire and $n$ is a fixed positive integer. If $$I_k:=\left[\frac{2(k-1)\pi}{n},\frac{2k\pi}{n}\right],$$ for $k=1,2,\dots,n$ and $$\alpha_k:=\sup_{\theta\in I_k}|f(e^{i\theta})|,$$ prove that $|f(0)|\leq (\alpha_1\alpha_2\cdots\alpha_n)^\frac{1}{n}.$

I began by noting that $m(I_k)=\frac{2\pi}{n}\forall k.$ Then I used the Mean Value Property to get $$|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi} |f(e^{i\theta})|d\theta=\frac{1}{2\pi}\sum_{k=1}^n \int_{I_k} |f(e^{i\theta})|d\theta \leq \frac{1}{2\pi}\sum_{k=1}^n \frac{2\pi \alpha_k}{n}=\sum_{k=1}^n\frac{\alpha_k}{n}.$$

Now I can see that what I have on the RHS is the arithmetic mean, but in general $$\left(\prod_{i=1}^n \alpha_i\right)^\frac{1}{n}\leq\sum_{i=1}^n\frac{\alpha_i}{n}.$$ That seems to suggest that my route is not producing the sharpest bound.

Next, I though maybe I could use Jensen's to try and force the geometric mean to appear, but I haven't produced anything useful as of yet.

Any advice is greatly appreciated.

user21820
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Scott
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  • The expected (easier) way to prove it is probably the method used here. – Daniel Fischer Jul 14 '15 at 11:51
  • In my class we didn't cover Schwarz-Christoffel formula, but I am working through a prelim packet and the curriculum changes over time. Thanks for the link. I will have to read up on that as well. – Scott Jul 14 '15 at 12:59
  • I meant the method from the answer. No such thing as Schwarz-Christoffel is needed. Just the plain old maximum modulus principle and the right idea. – Daniel Fischer Jul 14 '15 at 13:01
  • Oh, sorry about that. I just woke up and am still a bit groggy. I'll try and work the problem that way as well. Can never have too many solutions. – Scott Jul 14 '15 at 13:05

1 Answers1

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If $f(0)=0$ there is nothing to prove. From now on assume $f(0)\ne0$.

Let $a_1,\dots,a_m$ be the zeroes of $f$ in $\{|z|<1\}$. Suppose first that $f(z)\ne0$ if $|z|=1$. By Jensen's formula $$ \log|f(0)|=\sum_{i=1}^m\log|a_k|+\frac{1}{2\,\pi}\int_0^{2\pi}\log|f(e^{i\theta})|\,d\theta\le\frac1n\sum_{k=1}^n\log|\alpha_k|, $$ which gives the desired inequality.

If $f(z)=0$ for some $z$ with $|z|=1$, take a sequence $r_n\to1$ such that $f$ does not vanish on $\{|z|=r_n\}$, apply the above argument on the disk of radius $r_n$ and let $n\to\infty$.

user21820
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Julián Aguirre
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  • Wouldn't it be better to use something other than "$a$" which is so alike to "$α$"? – user21820 Jul 14 '15 at 11:51
  • Thanks for the help. I don't believer we covered this formula in my class (by Jensen's I meant the inequality involving convex functions), but I just read through the statement of the theorem (and proof) and your solution looks good. I appreciate it. – Scott Jul 14 '15 at 13:02