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Proof of analytic function being bounded inside the region if bounded on the boundary

  1. Nov 29, 2011 #1
    This problem is from Mathematical methods for physicists by Arfken, problem 6.4.7.

    A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.

    The hint is to consider w(z) = 1/f(z).

    I have tried to solve this in this way using Gauss' mean value theorem. Suppose, there is a point within C where |f(z)|≥ M. As |f(z)|≤ M on C, there has to be a local maximum of |f(z)| within C. We call that point of local maximum z°. Then we can take a circle with a however small radius r around z°. From Cauchy's integral theorem,

    f(z°) = (1/2πi)∫c[f(z)/(z-z°)] dz (here c is the small circle of radius r)

    Now taking, z = z° + r*exp(iθ)
    f(z°) = (1/2π)∫ f(z° + r*exp(iθ)) dθ

    Taking the modulus, |f(z°)| = 1/2π |∫ f(z° + r*exp(iθ)) dθ| ≤ 1/2π ∫ |f(z° + r*exp(iθ))| dθ
    (in all cases above, the integral is taken from 0 to 2π).

    Now, m(f) = 1/2π ∫ |f(z° + r*exp(iθ))| dθ is the mean value of |f(z)| on the circle of radius r. From above, we rewrite,
    |f(z°)| ≤ m(f).

    Again, there must be at least one point z1 on the circle where |f(z1)| is larger than m(f) or the |f(z)| is equal to m(f) for all the points on the circle. So, |f(z1)| ≥ m(f) ≥ |f(z°)|. We can take r as small as we want to make z1 as nearer to z°. So, there is point z1 in the neighborhood of z° for which, |f(z1)| ≥ |f(z°)|, so |f(z°)| can't be a local maximum, and thus there is no point inside C such that |f(z)| ≥ M.

    Now, my problem is that, I didn't use the condition f(z) ≠ 0 within C for this proof. So, there should be a problem with my solution, but I couldn't find any. Now, thinking about the hint, taking w(z) = 1/f(z), we don't know whether w(z) is continuous on C or not. For, f(z) = 0 for some point on the contour C, w(z) is not continuous and we can't take Cauchy integral of w(z) on C to be zero. But, for f(z), as it is continuous on C and analytic inside, it's contour integral on C would be zero from Cauchy-Goursat theorem. I think I could use this to solve the problem, but I failed to find any path to do so. I would appreciate it very much if someone can point out where I am doing wrong or give some more hints to solve this. I am sorry for the bad format of equations as I am not yet used to using latex.
     
    Last edited: Nov 29, 2011
  2. jcsd
  3. Nov 29, 2011 #2
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    Its late so maybe I am missing something but you don't need f≠0. This follows from the maximum modulus principle that a holomorphic function on a compact subset of ℂ attains its maximum value on the boundary.
     
  4. Nov 29, 2011 #3
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    But, in the text, the right next problem is,

    If f(z) = 0 within the contour C, show that the foregoing result (the statement of the previous problem) does not hold and that it is possible to have |f(z)| = 0 at one or more points in the interior with |f(z)| > 0 over the entire bounding contour.

    As it says, it does not hold if f(z) = 0 then there have to be some problem with the text then. Any clue?
     
  5. Nov 29, 2011 #4
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    I don't see how that contradicts your first statement:

    A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.

    I think what the question is trying to get at is the "minimum modulus principle." That is if f is never zero then |f| attains its minimum on the boundary as well.
     
  6. Nov 29, 2011 #5
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    The first problem asks to show that |f(z)|≤ M if f(z)≠0 within C and the second one asks to show that, |f(z)|≤ M does not hold if f(z)= 0 within C.

    If the f(z) has its minimum on the boundary, how can it help me to show that |f(z)| is bounded within the region? Thanks for you help, I really appreciate your kind replies.
     
  7. Nov 29, 2011 #6
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    I think the question is written wrong. If |f| > 0 on the boundary but f = 0 at some point inside this doesn't contradict the fact f has its "maximum modulus" on the boundary.
     
  8. Nov 29, 2011 #7
    Re: Proof of analytic function being bounded inside the region if bounded on the boun

    I think the question doesn't want to arrive at maximum modulus principle, it is asking to proof that a function is bounded within the region if its bounded on the boundary. Am I missing something here?
     
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