# Complex Analysis - Cauchy's Inequality

• Mar 29th 2010, 12:47 PM
slevvio
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
• Mar 29th 2010, 02:02 PM
Opalg
Quote:

Originally Posted by slevvio
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.

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.
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$|f^{(n)}(0)| = \left| \frac{n!}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz \right| \le \frac{2 \pi R n!}{2 \pi} \text{sup}_{|z| = R}\left| \frac{f(z)}{z^{n+1}} \right|$ 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.
• Mar 29th 2010, 02:12 PM
slevvio
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?