Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is anyone familiar with Converse to Rouche's Theorem?

  1. Jul 17, 2006 #1
    Hello,

    I was wondering if anyone is familiar with the converse to Rouche's Theorem published in a paper titled "A Converse to Rouche's Theorem," in American Mathematcal Monthly, Vol. 89, No. 5 (May, 1982), pp. 302-305 by David Challener and Lee Rubel?

    Inquisitively,

    Edwin
     
  2. jcsd
  3. Jul 18, 2006 #2
    I'd be kind of curious as to how it's worded, since a straight converse wouldn't be true. For example the converse of this version wouldn't be true:

    Original Theorem:

    If the complex-valued functions f and g are holomorphic inside and on some closed contour C, with |g(z)| < |f(z)| on C, then f and f + g have the same number of zeros inside C, where each zero is counted as many times as its multiplicity.

    (from wikipedia)

    Converse Theorem:

    If the complex-valued functions f and g are holomorphic inside and on some closed contour C, and f and f + g have the same number of zeros inside C, where each zero is counted as many times as its multiplicity, then |g(z)| < |f(z)| on C.

    Take for example f(x) = g(x) = x. Then f(x) = x and (f + g)(x) = 2x both have one zero, but it is not true that |g(z)| < |f(z)| on C.
     
  4. Jul 18, 2006 #3
  5. Jul 19, 2006 #4

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    as observed above, the strict converse of rouche is false.

    zeroes of holomorphic functioins are computed by the argument principle, i.e. by integrating a certain closed differential form, df/f for analytic f, around a loop missing all zeroes of f.

    the fundamental theorem is that two loops yield the same integrals for all closed differential forms iff they are "homologous".

    rouches theorem has a strong convexity hypothesis that guarantees two paths are homologous, namely that for all instants t of time, the two path points r(t), s(t), lie in some common hyperplane missing the origin.

    this condition is certainly not necessary for homology of the paths. In fact I cannot think of a natural converse of rouche that has remotely the same condition as rouche uses. i will look at the reference if possible.
     
  6. Jul 19, 2006 #5

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    well its not a converse to (my version of) rouche at all. the conclusion is not a condition on the paths since the path is fixed as the unit circle, a convex curve.

    but i probably do not know what rouches theorem says. ill look it up.
     
  7. Jul 19, 2006 #6

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    no im right, ropuche is a statement about an arbitrary loop, plus a convexity condition on the values, which translates into a convexity conditioin on the image of the path under the two functions.

    i.e. given a loop and a function defined on and inside the loop, with no zeroes on the loop, look at the image of the loop under the function. this is a loop not passing through the origin,m and the number of zeroes of the original function inside the loop equals the winding number of this loop around the origin, i.e. essentially the integral of dw/w, which translates back under f to the integral of df/f.


    so the converse is saying something like, if the original path is already convex, then the only way the image paths can have the same winding numbers is if the functions also satisfy some kind of convexity property.

    but it is much weaker than that since it is proved only for the unit circle. still it seems sort of interesting.

    so rouche says here is condition on the values of two functions on a path that guarantees the same winding number. the "converse" says there is at least one path such that the condition is actually necessary. not much of a converse, but still interesting.
     
  8. Jul 19, 2006 #7

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    a more inetersting converse would expand the list of admissible loops or prove that it cannot be expanded.
     
  9. Jul 19, 2006 #8

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    take two analytic functions defiend on a nbhd of the unit discd and asume they wrap the unit circle the same number of times around 0.

    what can you say about the relation between the values of the functions? that is the gist of it i guess.
     
  10. Jul 19, 2006 #9

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    well i dont know what a balshke product is, although of course it was defiend right there, but the esterman "strengthening" of rouches theiorem sems to me to be exactly the original theorem. i.e. the hypothesis still just means geometrically that the two values f(z), g(z) are in the same open hyperplane missing zero, for all points of the circle, so the straight line homotoyp shows they HVE THE same number of zeroes.

    the generalization is some kind of generalized hyperplane?
     
  11. Jul 19, 2006 #10
    I was able to print out the full text...Challener and Rubel proved the following theorem.

    "Theorem 1. Suppose f and g are analytic in a |z| <= 1+epsilon,
    Epsilon > 0, with no zero's on |z| = 1. If the Z_f and Z_g are the number of zero's of f and g in |z| < 1, then: Z_f = Z_g if and only if there exist finite Blaschke products A, B of the same order such that
    |Af + Bg| < |f| + |g| on |z| = 1" (Challener; Rubel 304).

    Challener and Rubel included the following Lemma in their proof:

    "Lemma 1 given f, g analytic in |z| <= 1 + epsilon, epsilon > 0, f and g having no zeros on |z| = 1, then there exist finite Blaschke products A, B such that |Af + Bg| < max{|f|,|g|} on |z| = 1." (Challener;Rubel 303)

    Challener and Rubel stated

    "Our proof rests on the following fact proved by Helson and Sarason in [4].

    Fact: Any continuous unimodular function y on |z| = 1 is the uniform limit of finite Blaschke products." (Challener;Rubel 303)

    I think that they also prove that Rouche's Theorem doesn't admit a converse in general, but I'm not sure....

    I imagine that if one can prove that "Any continues unimodular function y on |z - k| = C, is the uniform limit of finite Blaschke products, where C is any real number >= 1, k is any complex number," then one can maybe extend the proof created by Challener and Rubel to any circular disc in the complex plane.

    Mathwonk Wrote:

    Code (Text):
    a more interesting converse would expand the list of admissible loops or prove that it cannot be expanded.
    I would be very interested in learning the answer to that question that Mathwonk posed. Can anyone tell, based on the information above, if such a converse, mentioned by Mathwonk in the quote above, can be constructed?

    Inquisitively,

    Edwin

    Source of Quotes

    David Challener and Lee Rubel, The American Mathematical Monthly, Vol. 89, No. 5. (May, 1982) pp. 302-305.

    4. H. Helsom and D. Sarason, Past and future, Math. Scand.,21 (1967) Lemma on page 9.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is anyone familiar with Converse to Rouche's Theorem?
  1. Rouche's theorem (Replies: 2)

Loading...