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Felix.R
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 Analytic problem   « on: Jun 12th, 2007, 2:32pm » Quote Modify

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 13th, 2007, 9:52am » Quote Modify

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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Michael Dagg
Felix.R
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 Re: Analytic problem   « Reply #2 on: Jun 15th, 2007, 10:51pm » Quote Modify

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 17th, 2007, 8:57pm » Quote Modify

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 17th, 2007, 9:00pm by Michael Dagg » IP Logged

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Michael Dagg
Felix.R
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 Re: Analytic problem   « Reply #4 on: Jun 18th, 2007, 6:36pm » Quote Modify

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 18th, 2007, 6:37pm by Felix.R » IP Logged
Michael Dagg
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 Re: Analytic problem   « Reply #5 on: Jun 18th, 2007, 7:07pm » Quote Modify

I just lost a fair amount of typed text here. I am
certainly annoyed by expired pages messages.
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Michael Dagg
Michael Dagg
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 Re: Analytic problem   « Reply #6 on: Jul 10th, 2007, 9:58am » Quote Modify

Did you solve this problem?
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Michael Dagg
Felix.R
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 Re: Analytic problem   « Reply #7 on: Jul 11th, 2007, 9:32am » Quote Modify

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 30th, 2007, 6:56pm » Quote Modify

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
Michael Dagg
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 Re: Analytic problem   « Reply #9 on: Sep 15th, 2007, 8:34am » Quote Modify

Did you try to work with this?
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Michael Dagg
Michael Dagg
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 Re: Analytic problem   « Reply #10 on: Sep 29th, 2007, 7:16pm » Quote Modify

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

I would like to see what you came up with.
 « Last Edit: Sep 29th, 2007, 7:30pm by Michael Dagg » IP Logged

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Michael Dagg
Sameer
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 Re: Analytic problem   « Reply #11 on: Sep 29th, 2007, 7:21pm » Quote Modify

Yea seems like he is MIA..
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple

Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
Ghost Sniper
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 Re: Analytic problem   « Reply #12 on: Dec 13th, 2007, 10:51am » Quote Modify

[redacted] im right

// excuse me?  --SMQ
 « Last Edit: Dec 13th, 2007, 11:06am by SMQ » IP Logged

*sob* I miss my mommy... *blows nose* huh, I'm on? oh right.

(thinks to self) Time for my speech to these college kids.

"Reason is more important than all emotions..."
Michael Dagg
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 Re: Analytic problem   « Reply #13 on: Jan 23rd, 2008, 11:35pm » Quote Modify

>> [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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Michael Dagg
Eigenray
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 Re: Analytic problem   « Reply #14 on: Mar 14th, 2008, 6:43pm » Quote Modify

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 1,

g(1+w) - 1 = wn (C + O(w)),

where C = g(n)(1)/n! 0.  Writing w=reit, 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 = (2k-Arg[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 21st, 2008, 1:59pm » Quote Modify

Geometrically nice!

derivative at 1 is actually positive (not just non-zero)!
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Michael Dagg
Eigenray
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 Re: Analytic problem   « Reply #16 on: Mar 21st, 2008, 2:50pm » Quote Modify

Actually, we can think of it like a Lagrange multiplier problem:

|f|2(z0) = z0 for some 0,

since otherwise we could increase |f| by moving along the unit circle (or in the -z0 direction, if < 0).  Since

|f|2 = 2f bar[f'],

we get that

f'(z0) = t z0/bar[f(z0)]

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 22nd, 2008, 11:49pm » Quote Modify

Pardon, but how is |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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Eigenray
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 Re: Analytic problem   « Reply #18 on: Mar 23rd, 2008, 12:40am » Quote Modify

I'm thinking of h(x,y) = |f(x+iy)|2 = u2 + v2 just as a real-valued 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
= (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 h = 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 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 23rd, 2008, 12:40am by Eigenray » IP Logged
TJMann
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 Re: Analytic problem   « Reply #19 on: Mar 24th, 2008, 9:59am » Quote Modify

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 24th, 2008, 1:56pm » Quote Modify

This is just because h 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 = <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 z0.  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 z0.

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 25th, 2008, 3:55pm » Quote Modify

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 21st, 2014, 5:50pm » Quote Modify

You have to use Schwarz Equation !
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