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Cauchy's Estimate

I'm having trouble pinning down what should be a straightforward question.

Q: Let R>0 and f be an entire function such that

|f(z)| \leq |z|^n

for

|z| > R

. Show that f is a polynomial of degree at most n.

I have a hint that says it would be helpful to apply Cauchy's estimate. So I fix R>0 and choose S>R. Then f is analytic on B(0,S) and

|f(z)| \leq |S|^n

for |z|>R. Then, applying Cauchy's estimate for any z0 in the annulus gives

|f^{(n)}(z_0)| \leq \frac{|S|^n n!}{S^n} = n!

. Which implies f is poly of max degree n (as n'th derivative is constant).

But what about all those points inside B(0,R)? How do I deal with them? :holmes:

Thanks in advance.

Reply 1

Zhen Lin

12

Use Cauchy's integral formula to establish a bound on the function inside the disc, by integrating around a contour sufficiently large. (I think that's how it goes. If not I'll check my work and see how I went about it.)

Reply 2

Kolya

OP

17

Zhen Lin

Use Cauchy's integral formula to establish a bound on the function inside the disc, by integrating around a contour sufficiently large. (I think that's how it goes. If not I'll check my work and see how I went about it.)

I'm not entirely clear what you're getting at. Could we let, say,

w=z+2R

, and the conclude from Cauchy's integral formula (with

\gamma (t) = Se^{it}

for suitably large S>R) that

2\pi i f(w) = \int _{\gamma } \frac{f(z)}{2R} \ dz

? Which gives us

\int _{\gamma } f(z) \leq 4\pi i R|z|^n

as an upper bound for the integral inside the disc. Is that enough to conclude that f itself is bounded inside the disc? Is that what you were suggesting?

Reply 3

Zhen Lin

12

Well, this is what I had in mind. Let

\gamma(t) = w + S e^{i t}

, where

|w| < R

and S is chosen so that

|\gamma(t)| > R

for all t. Then

\displaystyle f^{(m)} (w) = \frac{m!}{2 \pi i} \oint_{\gamma} \frac{f(z)}{(z - w)^{n+1}} \, dz

, so

\displaystyle |f(w)| = \frac{m!}{2 \pi} \left| \int_{-\pi}^{\pi} \frac{f(w + S e^{i t})}{S^{m+1} e^{i (m + 1) t}} i S e^{i t} \, dt \right| \le \frac{m!}{2\pi S^m} \int_{-\pi}^{\pi} \left| f(w + S e^{i t}) \right| \, dt

, which you can now bound above by

2 \pi S^{n - m} \left( 1 + \frac{R}{S} \right)^{n}

. So for

m > n

, taking

S \to \infty

allows us to see that

f^{(m)}(z) = 0

for

|z| < R

as well. (Actually, this gives the result for all z, since a holomorphic function constant on some open subset of a connected domain is globally constant.)