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Math Help - Complex Analysis - Cauchy's Inequality

  1. #1
    Senior Member slevvio's Avatar
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    Complex Analysis - Cauchy's Inequality

    Hello everyone, I was wondering if anyone could give me a hand with this question.

    Let f be an entire function satisfying |f(z)| \le K|z^2| for |z| \ge 1, where K is a positive constant. Show using Cauchy's inequality, that f^{(n)}(0) = 0 for  n > 2 and deduce that f is a quadratic polynomial.

    Ok f is entire so is defined and analytic on all of the complex plane. So it is analytic on D(0,R), some big disc.

    Then by Taylor's theorem, so has a power series expansion f(z) = a_0 + a_1 z + a_2 z^2 + \ldots for all z in C, and we can use Cauchy's integral formula for derivatives to get

    |f^{(n)}(0)| = \left| \frac{n!}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz \right| \le  \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R}\left| \frac{f(z)}{z^{n+1}} \right| by the estimation lemma. I understand it all up until this point.

    However , nowthe solution I have says that

     \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le Rn!\frac{KR^2}{R^{n+1}}

    which I think tells me that

    (*) \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le \frac{KR^2}{R^{n+1}} but I have no idea where this comes from! Does anyone have an idea how to help?

    For reference Cauchy's inequality in my notes says if f analytic on an open disc D(z_0,R) and f is bounded on the circle |z-z_0 | = r < R with |f(z)| \le M for some M on |z-z_0| = r, then for  n \in \mathbb{N},

    |f^{(n)}(z_0)| \le \frac{Mn!}{r^n}.

    If anyone could help me how to use this to get (*) it would be much appreciated
    Last edited by slevvio; March 29th 2010 at 12:41 PM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by slevvio View Post
    Let f be an entire function satisfying |f(z)| \le K|z^2| for |z| \ge 1, where K is a positive constant. Show using Cauchy's inequality, that f^{(n)}(0) = 0 for  n > 2 and deduce that f is a quadratic polynomial.

    .
    .
    .

    |f^{(n)}(0)| = \left| \frac{n!}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz \right| \le  \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R}\left| \frac{f(z)}{z^{n+1}} \right| by the estimation lemma. I understand it all up until this point.

    However , now the solution I have says that

     \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le Rn!\frac{KR^2}{R^{n+1}}

    which I think tells me that

    (*) \text{sup}_{|z| = R} \left|\frac{f(z)}{z^{n+1}} \right| \le \frac{KR^2}{R^{n+1}} but I have no idea where this comes from! Does anyone have an idea how to help?
    So you know that |f^{(n)}(0)| \leqslant  Rn!{\textstyle\sup_{|z| = R}}\Bigl| \frac{f(z)}{z^{n+1}} \Bigr|. You also know that, on the circle |z|=R, |f(z)|\leqslant K|z|^2 = KR^2 and (obviously) |z^{n+1}| = R^{n+1}. Therefore |f^{(n)}(0)| \leqslant  \frac{Rn!KR^2}{R^{n+1}} = \frac{n!K}{R^{n-2}}. If n>2 then the right side tends to 0 as R\to\infty. Therefore f^{(n)}(0) = 0 for all n>2.
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  3. #3
    Senior Member slevvio's Avatar
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    ahh this is of course if R is very big, so its bigger than 1, so the inequality comes into play from the start!

    Thanks very much

    but i was worried about one little thing, how do i know that i can talk about

    \text{sup}_{|z|=R} \left| \frac{f(z)}{z^{n+1}} \right|? What if this function is not bounded on the circle?
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