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Topic: Analytic problem (Read 9808 times) 

Felix.R
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Analytic problem
« on: Jun 12^{th}, 2007, 2:32pm » 
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Suppose g is analytic on abs(z)<=1 and let abs(g(z)) be maximized for abs(z)<=1 at z0 with abs(z0)=1. Prove that g'(z0) is not zero unless g is constant.


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Michael Dagg
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Re: Analytic problem
« Reply #1 on: Jun 13^{th}, 2007, 9:52am » 
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You should be able to solve this using Schwarz. Notice that g(z) < 1 in z < 1 by the maximum principle.


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Felix.R
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Re: Analytic problem
« Reply #2 on: Jun 15^{th}, 2007, 10:51pm » 
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Is it because the function g is not constant that the maximum principle gives the statement you made? Actually this is not that easy to see for one thing. I think. I am not sure what to do with Schwarz's lemma because it does not appear to lead me to a function composed with g that I can work with. So I don't understand how I can solve it using Schwarz. Explain if you will.


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Michael Dagg
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Re: Analytic problem
« Reply #3 on: Jun 17^{th}, 2007, 8:57pm » 
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Yes. You are at liberty to make some assumptions here. For example, you may assume that g(z0) = z0, z0 = 1. Noting that g(z) < 1 in z < 1, and so if we take g(0) = 0 then by Schwarz we have g(z) <= z, z < 1.

« Last Edit: Jun 17^{th}, 2007, 9:00pm by Michael Dagg » 
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Felix.R
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Re: Analytic problem
« Reply #4 on: Jun 18^{th}, 2007, 6:36pm » 
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I do not follow all this. I understand assuming g(1) = 1 and g(0) = 0 but doesn't this require that we also assume that g(0) not zero? Also, I can't use the bar character for absolute value I guess because my language pack seems to display character codes on this web site for that symbol. I have no idea why.

« Last Edit: Jun 18^{th}, 2007, 6:37pm by Felix.R » 
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Michael Dagg
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Re: Analytic problem
« Reply #5 on: Jun 18^{th}, 2007, 7:07pm » 
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I just lost a fair amount of typed text here. I am certainly annoyed by expired pages messages.


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Michael Dagg
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Re: Analytic problem
« Reply #6 on: Jul 10^{th}, 2007, 9:58am » 
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Did you solve this problem?


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Felix.R
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Re: Analytic problem
« Reply #7 on: Jul 11^{th}, 2007, 9:32am » 
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Alas, only partially to a degree with some missing pieces using a Taylor series. We appreciate help!


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Michael Dagg
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Re: Analytic problem
« Reply #8 on: Jul 30^{th}, 2007, 6:56pm » 
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That leads me to guess what you mean but you might be on to something. Is that series that of g(z)  g(z0) ? See what you can do with this: Show that g'(1) >= 1 and take g(0) = a, a <> 0. Now consider h(z) = (g(z)  a)/(1  a bar g(z)).


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Michael Dagg
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Re: Analytic problem
« Reply #9 on: Sep 15^{th}, 2007, 8:34am » 
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Did you try to work with this?


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Michael Dagg
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Re: Analytic problem
« Reply #10 on: Sep 29^{th}, 2007, 7:16pm » 
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What happened to you on this problem? This is a nice problem by the way  not too hard but interesting, but maybe you solved it and forgot about us? I would like to see what you came up with.

« Last Edit: Sep 29^{th}, 2007, 7:30pm by Michael Dagg » 
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Sameer
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Re: Analytic problem
« Reply #11 on: Sep 29^{th}, 2007, 7:21pm » 
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Yea seems like he is MIA..


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Ghost Sniper
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Re: Analytic problem
« Reply #12 on: Dec 13^{th}, 2007, 10:51am » 
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[redacted] im right // excuse me? SMQ

« Last Edit: Dec 13^{th}, 2007, 11:06am by SMQ » 
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Michael Dagg
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Re: Analytic problem
« Reply #13 on: Jan 23^{rd}, 2008, 11:35pm » 
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>> [redacted] im right >> excuse me? SMQ Being curious, I had wondered what this exchanged was referring to but apparently I had miss it at some point  perhaps Ghost had proposed solution but took it back (?). Still, a pretty cool problem, and in spite of my remarks it has gone unsolved  on this forum nevertheless.


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Eigenray
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Re: Analytic problem
« Reply #14 on: Mar 14^{th}, 2008, 6:43pm » 
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I don't think you need anything but the power series expansion. We can assume z_{0}=1, and g(1)=1 (if g is not identically 0). Then for some n 1, g(1+w)  1 = w^{n} (C + O(w)), where C = g^{(n)}(1)/n! 0. Writing w=re^{it}, then for t fixed, Arg [ g(1+w)  1 ] nt + Arg[C] as r 0. If n 2, then for some integer k, we can take t = (2kArg[C])/n (/2, 3/2). Then Arg[g(1+w)1] 0, and therefore for r sufficiently small, g(1+w) > 1, with 1+w in the unit disk.


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Michael Dagg
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Re: Analytic problem
« Reply #15 on: Mar 21^{st}, 2008, 1:59pm » 
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Geometrically nice! I think your proof could be made to show that the derivative at 1 is actually positive (not just nonzero)!


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Eigenray
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Re: Analytic problem
« Reply #16 on: Mar 21^{st}, 2008, 2:50pm » 
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Actually, we can think of it like a Lagrange multiplier problem: f^{2}(z_{0}) = z_{0} for some 0, since otherwise we could increase f by moving along the unit circle (or in the z_{0} direction, if < 0). Since f^{2} = 2f bar[f'], we get that f'(z_{0}) = t z_{0}/bar[f(z_{0})] for some t 0, but we need analycity (analyticality?) of f to get t 0.


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TJMann
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Re: Analytic problem
« Reply #17 on: Mar 22^{nd}, 2008, 11:49pm » 
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Pardon, but how is f^{2}(z_{0}) read? I have never seen del used with complex functions before. Also, I thought analyticity is presupposed if you write f'(z)?


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Eigenray
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Re: Analytic problem
« Reply #18 on: Mar 23^{rd}, 2008, 12:40am » 
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I'm thinking of h(x,y) = f(x+iy)^{2} = u^{2} + v^{2} just as a realvalued function of two real variables. Then h is a vector in ^{2}, which I'm thinking of as the complex number h = h/x + ih/y = (2uu_{x}+2vv_{x}) + i(2uu_{y}+2vv_{y}) = 2(u+iv)(u_{x}+iu_{y}) = 2f bar[f'], using u_{x}=v_{y}, u_{y}=v_{x}. We could also compute h_{z} = (f bar f)_{z} = f_{z} (bar f) + f (bar f)_{z} = f' * bar f, since bar f is antiholomorphic, and use the fact that h = 2 bar[h_{z}]. I don't know if this is standard notation, though. My point was just that from the point of view of multivariable calculus, it's clear that h must be perpendicular to the unit circle, but it's not clear that it can't be the zero vector. That is where we need to think of f as a complex analytic function, not just as a function from ^{2} ^{2}.

« Last Edit: Mar 23^{rd}, 2008, 12:40am by Eigenray » 
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TJMann
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Re: Analytic problem
« Reply #19 on: Mar 24^{th}, 2008, 9:59am » 
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cool. I never thought of it like that and saves lots of space. In your last paragraph, I don't see why the gradient of h is perpendicular to the unit circle though, I think to the level lines of h it is, {(x,y), h(x,y) = const}


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Eigenray
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Re: Analytic problem
« Reply #20 on: Mar 24^{th}, 2008, 1:56pm » 
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This is just because h points in the direction of increasing h. Let v be a unit vector tangent to the unit circle at z_{0}. Then the directional derivative of h in the direction v is just D_{v} h = <h, v>. If this dot product were positive, then h increases as I move along the unit circle in the v direction, contradicting the fact that h is maximized at z_{0}. Similarly, if it were negative, h would increase in the opposite direction. So we must have <h, v> = 0, i.e., h is perpendicular to the unit circle at z_{0}. This is a special case of the Lagrange multiplier method. If we want to maximize f given the constraint g=0, then f must be perpendicular to the level curve g=0, so it must be a scalar times g, i.e., f = g.


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TJMann
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Re: Analytic problem
« Reply #21 on: Mar 25^{th}, 2008, 3:55pm » 
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Thx. I see. I thought you were just speaking in general but see that you were referring to the particulars of the given problem.


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Nourhan
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Re: Analytic problem
« Reply #22 on: May 21^{st}, 2014, 5:50pm » 
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You have to use Schwarz Equation !


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