Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

Cauchy's Estimate

Announcements Posted on
Competition: win a karting session for you and seven mates! 24-07-2015
  1. Offline

    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.
  2. Offline

    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.)
  3. Offline

    (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?
  4. Offline

    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.)
  5. Offline

    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:

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: May 27, 2010
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Have you ever posted a picture of your food on social media?
Results and Clearing

SQA results chat

Come talk about your results here

new on tsr

Join the American Society

Whether you're from the USA or just love it!

Study resources
x

Think you'll be in clearing or adjustment?

Hear direct from unis that want to talk to you

Get email alerts for university course places that match your subjects and grades. Just let us know what you're studying.

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.