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I need to prove the following:

If $a\gt 0$, $n\in \mathbb{N}-\{0\}$, and if $f$ analytic on $(0, 2a)$ with $$|f^{(n)}(0)| = \frac{n!}{a^n}\cdot \sup\left\{|f(z)|\mid : |z|=a\right\},$$ then $f(z) = rz^n$ for some $r\in \mathbb{C}$.

I have tried to borrow ideas from the proof of Liouville's Theorem and go backwards. However, the problem doesn't quite seem to fit the hypothesis of Liouville's Theorem. Any suggestion would be appreciated. Thanks!

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Is there a question? Or are you just letting us know something? –  Arturo Magidin Jan 30 '11 at 5:23
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@lively: Please present your question in a more readable form. –  anonymous Jan 30 '11 at 5:24
    
Thanks Arturo! everything until 'then' is the hypothesis. I have to prove f(z)=rz^n. Thanks... –  lively Jan 30 '11 at 5:32
    
@lively: Then please edit your question, and make it an actual question. If it is homework (which your "I have to prove" suggests) then tag it with the [homework] tag. In any case, please phrase it like a question (not an assignment or order), and state what you have tried, or where you are stuck, or why you are stuck. –  Arturo Magidin Jan 30 '11 at 5:35
    
Sorry Arturo if my question looked like an order; that was, in no way, my intent. :-S –  lively Jan 30 '11 at 6:32

1 Answer 1

Put for $z\neq 0$: $g(z):=\frac{f(z)}{z^n}$. We can show thanks to Cauchy's integral formula that for $|z|<a$ $$\left|g(z)-\frac{f^{(n)}(0)}{n!}\right|\leq \frac{|z|}{(a-|z|)a^{n+1}}\sup_{|\xi|=a}|f(\xi)|,$$ so we can put $g(0):=\frac{f^{(n)}(0)}{n!}$. $g$ is still analytic and by hypothesis we have $$|g(0)|=\left|\frac 1{2\pi i}\int_{C(0,a)}\frac{f(\xi)}{\xi^{n+1}}d\xi\right|=\sup_{|\xi|=a}|g(\xi)|,$$ and by maximum principle we get that $g$ is constant on $B(0,a)$, hence on $B(0,2a)$.

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