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Cauchy's Estimate

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    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.
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    ReputationRep:
    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.)
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    (Original post by 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?
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    ReputationRep:
    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.)
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    Hmm, looks a bit heavy-handed but I can't see anything else so perhaps that's the best to be done. Thanks again, Zhen. :yy:

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