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Topic: Diameter of f(D) (Read 6411 times) 

Eigenray
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Diameter of f(D)
« on: Dec 3^{rd}, 2005, 5:12pm » 
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Suppose f : D > C is holomorphic, where D is the unit disc. If d = sup_{z,w in D} f(z)f(w) is the diameter of f(D), show that 2f'(0) < d, and that equality holds precisely when f is linear.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #1 on: Dec 9^{th}, 2005, 7:32pm » 
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Wow. This is a nice find. Am I allowed to up some hint for a counterexample?

« Last Edit: Dec 9^{th}, 2005, 7:35pm by Michael Dagg » 
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Eigenray
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Re: Diameter of f(D)
« Reply #2 on: Dec 10^{th}, 2005, 6:00pm » 
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I don't understand what you mean.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #3 on: Dec 11^{th}, 2005, 7:38am » 
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I had a counterexample hint in mind, which is what I meant, but here is a better hint: use a compactness argument to find a (closed) disk of diameter d which contains f(D). If its center is c, then (2/d)(f  c) maps D into D. Then apply Schwarz.


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Eigenray
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Re: Diameter of f(D)
« Reply #4 on: Dec 11^{th}, 2005, 1:31pm » 
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But if, say, f(D) is an equilateral triangle with diameter (side length) d, then the smallest disk containing it has diameter 2d/sqrt(3) > d.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #5 on: Dec 19^{th}, 2005, 1:22pm » 
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Indeed  your tactful advisory correct. Note that F(z) = f(z)  f(z) maps D into D, the disk centered at 0 of radius d, and F(0) = 0, so Schwarz is applicable to F/d.


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cain
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How did you think up that F?


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Michael Dagg
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Re: Diameter of f(D)
« Reply #7 on: Dec 21^{st}, 2005, 12:32pm » 
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The short answer is that Eigenray made me think it up. What can you say about that F? (I see the mapping problem you posted.)


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Eigenray
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Re: Diameter of f(D)
« Reply #8 on: Dec 21^{st}, 2005, 6:21pm » 
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on Dec 19^{th}, 2005, 1:22pm, Michael_Dagg wrote:Note that F(z) = f(z)  f(z) maps D into D, the disk centered at 0 of radius d, and F(0) = 0, so Schwarz is applicable to F/d. 
 That takes care of the odd coefficients, but what about the even ones? That is, Schwarz tells you that if equality holds, then F(z) = dz, i.e., f(z) = d/2 z + g(z^{2}) for some g.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #9 on: Dec 29^{th}, 2005, 5:46am » 
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I've tried out a few things and haven't come up with the result for equality. It is an interesting problem as it first made me think Polya, Landau and Toeplitz. Counterexample still sticks out at me but I don't see it at the moment.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #10 on: Jan 3^{rd}, 2006, 3:38pm » 
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Do you have a result for equality?


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Eigenray
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Re: Diameter of f(D)
« Reply #11 on: Jan 3^{rd}, 2006, 8:21pm » 
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Alas, no. The inequality is given as an exercise in Stein & Shakarchi's Complex Analysis. It then says "moreover, it can be shown that equality holds precisely when f is linear," but no reference is given.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #12 on: Jan 4^{th}, 2006, 7:26am » 
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My suspecion gave light: after a closer search I did find the problem in Polya and Szego's PROBLEMS AND THEOREMS IN ANALYSIS,III.239. They reference a 1907 paper of Landau and Toeplitz in a now defunct journal that our library does not have (Arch.der Math.und Physik,ser.3,vol.11, pp.302307), but I suspect it will also be found in Landau's multivolume COLLECTED WORKS, which (mirabili dictu) our library does have. There's no guarantee that they discuss the case of equality, but Landau was so meticulous that I suspect they did. Anyway, Polya and Szego's solution for the inequality is very similar to the one I had found, namely, via consideration of the function f(z)f(z) ;but they do not consider equality.

« Last Edit: Jan 4^{th}, 2006, 7:32am by Michael Dagg » 
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Michael Dagg
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I have a solution to this problem that I post as a scanned image. The quality is not very good but I would welcome its tagging by one of the gladiators here. I introduce the oscillation F(z) = f(z)  f(z) (third one) subsequent to the first two by LandauToeplitz. It turns out (as you might expect, although the proof is a bit subtle) that all these oscillations are equal, but the proof I wrote skirts this matter. Actually, LandauToeplitz worked only with functions that are holomorphic in a neighborhood of the closed unit disk. It takes more work to relax that to mere holomorphy in the open unit disk, where, e.g., one can no longer speak of the oscillation on the unit circle as before. Please alert me to any typos.


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Eigenray
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Re: Diameter of f(D)
« Reply #14 on: Jan 27^{th}, 2006, 4:58pm » 
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Looks good! Thanks. For something this long, I think LaTeX is just easier. Unfortunately, "This attachment causes the attachments directory to exceed it's [sic] maximum capacity by approximately 58 kilobytes and cannot be uploaded." So it may be found here.

« Last Edit: Jan 27^{th}, 2006, 5:03pm by Eigenray » 
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Icarus
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Re: Diameter of f(D)
« Reply #15 on: Jan 28^{th}, 2006, 11:03am » 
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I tried to convert this too, but ran into the same problem.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #16 on: Jan 31^{st}, 2006, 8:53am » 
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That's nice Eigenray. I noticed the attachment limitation also when I tried to put up a png file that has better quality instead of the jpeg.

« Last Edit: Jan 31^{st}, 2006, 8:55am by Michael Dagg » 
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Michael Dagg
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Re: Diameter of f(D)
« Reply #17 on: Feb 1^{st}, 2006, 11:48am » 
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This problem is really good food (that is, the thought of it might be addictive)  one idea I want to add, which is hard to resist, and something you likely already know, e.g. : Let d = diam f(D). Fix (a) in D and put W(z) = (za)/(1a bar z). Then by Schwarz we have (f(z)  f(a))/W(z) <= max((f(w)  f(a))/ W(w) over {w = 1}, <= d. Letting z approach (a), this gives f'(a)(1  a^2) <= d. Take a=0, then f'(0) <= d. This result is evident (albeit contrived differently) from the solution I posted. Similar arguments give the (best possible, as equality holds for conformal automorphisms) inequality f'(a) <= (1  f(a)^2)/(1  a^2) for every (a) in D, and every holomorphic selfmap f of D. You'll find this in the famous 1912 Math. Annalen paper of Caratheodory and maybe even in Schwarz' original papers (and most graduatelevel CA textbooks), however, providing a construction (e.g., the function W or otherwise) may take some doing. (Previously I mentioned that I had tried out a few things and this is one of such  made complete sense even without knowing W).

« Last Edit: Feb 2^{nd}, 2006, 1:45am by Michael Dagg » 
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Paul_W
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Re: Diameter of f(D)
« Reply #18 on: Feb 3^{rd}, 2006, 2:55pm » 
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It would certainly be tempting to instead write W(z) = 2 (za) / (1  a bar z) then get 2 f'(0) <= d. But, there is a problem with that.


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Michael Dagg
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Re: Diameter of f(D)
« Reply #19 on: Feb 4^{th}, 2006, 12:16pm » 
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I think you meant to write W2(z) = W(z)/2 . But neither are plausible and if one actually understands the problem the use of them would not be considered. Getting inequality by way of some acceptable W is possible but getting equality this same way would require you come up with f in some linear form (and then when combined with W ) applicable to Schwarz. I can't say it is not possible but say it will likely not be easy (e.g., I never could get any useful bounds going this route but I still think about it).

« Last Edit: Feb 4^{th}, 2006, 6:40pm by Michael Dagg » 
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Paul_W
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Re: Diameter of f(D)
« Reply #20 on: Feb 6^{th}, 2006, 11:37am » 
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The numbering of the theorems (1,2) in handwritten one differs from the numbering in the pdf file (3,4). Does it matter (because the last sentence of the handwritten one references Theorem 1, which would be 3 in the pdf) or does the result (5) actually follow from Lemma 1?


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Eigenray
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Re: Diameter of f(D)
« Reply #21 on: Feb 6^{th}, 2006, 2:23pm » 
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Whoops... forgot about that. < \newtheorem{lemma}[thm]{Lemma}  > \newtheorem{lemma}{Lemma}


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Paul_W
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Re: Diameter of f(D)
« Reply #22 on: Feb 6^{th}, 2006, 6:13pm » 
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Honestly when I asked that question, I did not know the answer but now after reading it umpteen times it is themorm 1. Interesting is the obvious circle of iff's. Hardly got this far in complex analysis but it is real analysis is what I mostly know.


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shiba_san
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Re: Diameter of f(D)
« Reply #23 on: Aug 28^{th}, 2006, 3:14am » 
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a way to solve this one when the function is injective is to compute exactly the area of f(D) in terms of the a_i's, where f(z) = a_0 + a_1 z + a_2 z^2 + ... This is not hard, just a change of variable and notice that the determinant of jacobian matrix of f at the point z is precisely f'(z)^2. The conclusion immediatly follows if one uses the fact that a disk is the shape with minimal diameter with a given area. Nevertheless it does not works that well if f is not supposed injective, but I found the idea was nice ..


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shiba_san
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Re: Diameter of f(D)
« Reply #24 on: Aug 28^{th}, 2006, 3:30am » 
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Suppose that d is the diameter of f(D). Consider g(z)=f(z)a, where a is a constant chosen such that g(D) is included in the disk of diameter d, centered at the origin. If one calls C the unit circle and because g(z) <= d/2 on the unit circle we get that: f'(0) = g'(0) = 1/(2 pi) int_C g(z)/z^2 dz <= 1/(2 pi) 2 pi (d/2) = d/2 Hence d >= 2 f'(0) ??


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