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general >> complex analysis >> Analytic problem
(Message started by: Felix.R on Jun 12th, 2007, 2:32pm)

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.

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.

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.

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.

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.

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.

Title: Re: Analytic problem
Post by Michael_Dagg on Jul 10th, 2007, 9:58am
Did you solve this problem?

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!

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)).

Title: Re: Analytic problem
Post by Michael_Dagg on Sep 15th, 2007, 8:34am
Did you try to work with this?

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.

Title: Re: Analytic problem
Post by Sameer on Sep 29th, 2007, 7:21pm
Yea seems like he is MIA..

Title: Re: Analytic problem
Post by Ghost Sniper on Dec 13th, 2007, 10:51am
[redacted] im right

// excuse me?  --SMQ

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.

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.

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)!

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.

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)?

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.

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}

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.

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.

Title: Re: Analytic problem
Post by Nourhan on May 21st, 2014, 5:50pm
You have to use Schwarz Equation !



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