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Title: Analytic problem Post by Felix.R on Jun 12th, 2007, 2:32pm 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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Title: Re: Analytic problem Post by Michael_Dagg on Jun 13th, 2007, 9:52am 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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Title: Re: Analytic problem Post by Felix.R on Jun 15th, 2007, 10:51pm 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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Title: Re: Analytic problem Post by Michael_Dagg on Jun 17th, 2007, 8:57pm 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. |
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Title: Re: Analytic problem Post by Felix.R on Jun 18th, 2007, 6:36pm 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. |
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Title: Re: Analytic problem Post by Michael_Dagg on Jun 18th, 2007, 7:07pm I just lost a fair amount of typed text here. I am certainly annoyed by expired pages messages. |
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Title: Re: Analytic problem Post by Michael_Dagg on Jul 10th, 2007, 9:58am Did you solve this problem? |
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Title: Re: Analytic problem Post by Felix.R on Jul 11th, 2007, 9:32am Alas, only partially to a degree with some missing pieces using a Taylor series. We appreciate help! |
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Title: Re: Analytic problem Post by Michael_Dagg on Jul 30th, 2007, 6:56pm 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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Title: Re: Analytic problem Post by Michael_Dagg on Sep 15th, 2007, 8:34am Did you try to work with this? |
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Title: Re: Analytic problem Post by Michael_Dagg on Sep 29th, 2007, 7:16pm 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. |
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Title: Re: Analytic problem Post by Sameer on Sep 29th, 2007, 7:21pm Yea seems like he is MIA.. |
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Title: Re: Analytic problem Post by Ghost Sniper on Dec 13th, 2007, 10:51am [redacted] im right // excuse me? --SMQ |
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Title: Re: Analytic problem Post by Michael_Dagg on Jan 23rd, 2008, 11:35pm >> [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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Title: Re: Analytic problem Post by Eigenray on Mar 14th, 2008, 6:43pm I don't think you need anything but the power series expansion. We can assume z0=1, and g(1)=1 (if g is not identically 0). Then for some n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1, g(1+w) - 1 = wn (C + O(w)), where C = g(n)(1)/n! http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0. Writing w=reit, then for t fixed, Arg [ g(1+w) - 1 ] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif nt + Arg[C] as r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0. If n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 2, then for some integer k, we can take t = (2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifk-Arg[C])/n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2). Then Arg[g(1+w)-1] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0, and therefore for r sufficiently small, |g(1+w)| > 1, with 1+w in the unit disk. |
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Title: Re: Analytic problem Post by Michael_Dagg on Mar 21st, 2008, 1:59pm Geometrically nice! I think your proof could be made to show that the derivative at 1 is actually positive (not just non-zero)! |
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Title: Re: Analytic problem Post by Eigenray on Mar 21st, 2008, 2:50pm Actually, we can think of it like a Lagrange multiplier problem: http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gif|f|2(z0) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifz0 for some http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 0, since otherwise we could increase |f| by moving along the unit circle (or in the -z0 direction, if http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gif < 0). Since http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gif|f|2 = 2f bar[f'], we get that f'(z0) = t z0/bar[f(z0)] for some t http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 0, but we need analycity (analyticality?) of f to get t http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0. |
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Title: Re: Analytic problem Post by TJMann on Mar 22nd, 2008, 11:49pm Pardon, but how is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gif|f|2(z0) read? I have never seen del used with complex functions before. Also, I thought analyticity is pre-supposed if you write f'(z)? |
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Title: Re: Analytic problem Post by Eigenray on Mar 23rd, 2008, 12:40am I'm thinking of h(x,y) = |f(x+iy)|2 = u2 + v2 just as a real-valued function of two real variables. Then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh is a vector in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif2, which I'm thinking of as the complex number http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifh/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifx + ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifh/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gify = (2uux+2vvx) + i(2uuy+2vvy) = 2(u+iv)(ux+iuy) = 2f bar[f'], using ux=vy, uy=-vx. We could also compute hz = (f bar f)z = fz (bar f) + f (bar f)z = f' * bar f, since bar f is antiholomorphic, and use the fact that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh = 2 bar[hz]. 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif2. |
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Title: Re: Analytic problem Post by TJMann on Mar 24th, 2008, 9:59am 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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Title: Re: Analytic problem Post by Eigenray on Mar 24th, 2008, 1:56pm This is just because http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh points in the direction of increasing h. Let v be a unit vector tangent to the unit circle at z0. Then the directional derivative of h in the direction v is just Dv h = <http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh, 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 z0. Similarly, if it were negative, h would increase in the opposite direction. So we must have <http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh, v> = 0, i.e., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifh is perpendicular to the unit circle at z0. This is a special case of the Lagrange multiplier method. If we want to maximize f given the constraint g=0, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.giff must be perpendicular to the level curve g=0, so it must be a scalar times http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifg, i.e., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.giff = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lambda.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nabla.gifg. |
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Title: Re: Analytic problem Post by TJMann on Mar 25th, 2008, 3:55pm 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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Title: Re: Analytic problem Post by Nourhan on May 21st, 2014, 5:50pm You have to use Schwarz Equation ! |
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