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Michael Dagg
Senior Riddler     Gender: Posts: 500 Interesting Bounded Analytic Map   « on: Aug 7th, 2006, 2:35pm » Quote Modify

Pick n = 2006 distinct points in the unit disc, zi in D, and
prescribe arbitrarily n = 2006 complex numbers wi in C.

When can one find a bounded analytic function f on
the disc, with sup norm <= 1, such that

f(zi) = wi?
 « Last Edit: Aug 7th, 2006, 2:36pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler      Gender: Posts: 1948 Re: Interesting Bounded Analytic Map   « Reply #1 on: Aug 8th, 2006, 7:53pm » Quote Modify

For n=1, this is easy: iff |w1| < 1.

For n=2, it's not too hard.  If w1=w2, then the requirement is again just |w1| < 1.  Otherwise, by the open mapping theorem we need w1, w2 to both lie in the open unit disc D.  Now the answer is the Schwarz-Pick lemma:

Consider the Blaschke factor
gw(z) = (w-z)/(1 - w'z),
where w' is the conjugate of w.  If w is in D, it's easily verified that gw is an analytic involution of D, preserving the boundary, and interchanging the points w and 0.  So if f takes zi to wi, then the composition
F = gw0fgz0
has F(0)=0, and F(gz0(z1)) = gw0(w1).

Moreover, f : D -> D iff F : D -> D.  In this case, the Schwarz lemma (i.e., the maximum modulus principle applied to F(z)/z on the disc |z|<r, as r->1) gives
|w0 - w1|/|1-w0'w1| = |gw0(w1)| < |gz0(z1)| = |z0-z1|/|1-z0'z1|,
and this is also sufficient since we may take F to be multiplication by the appropriate constant (and indeed, in the case of equality, this is the unique solution).

The case n=3 is harder (and I'm guessing a solution to this case would generalize for any n).  It's necessary for the above condition to hold for any pair of specified values, but not sufficient.  For example, if we require
f(0) = w,  f(r) = f(-r) = 0,
then the above only shows that we need |w| < r, but in fact by Jensen's lemma we need at least |w| < r2 < r. IP Logged
Eigenray
wu::riddles Moderator
Uberpuzzler      Gender: Posts: 1948 Re: Interesting Bounded Analytic Map   « Reply #2 on: Aug 9th, 2006, 5:00am » Quote Modify

This is apparently known as the Nevanlinna-Pick problem, and the condition is that the nxn "Pick matrix"
(1 - wi' wj)/(1 - zi' zj )
be positive semi-definite.  (Pick's paper is available here, if you speak German.)

Applying this to my previous example, f(0)=w, f(r)=f(-r)=0, the matrix is
1-|w|2    1     1
1    1/(1-r2)  1/(1+r2)
1   1/(1+r2)   1/(1-r2),
and indeed this matrix is positive semidefinite exactly when |w| < r2.  So Jensen's formula happens to give a sharp result in this case.

Interesting.  At least the answer gives some hint for how to prove it... IP Logged
Michael Dagg
Senior Riddler     Gender: Posts: 500 Re: Interesting Bounded Analytic Map   « Reply #3 on: Aug 9th, 2006, 10:14am » Quote Modify

Indeed.

It is a classical problem. Even well-known as it is, it may not
pop up in ones studies until years later -- one generally
gets introduced to it in "Topics" of analysis seminars.

The work you did is quite good -- noting F-Blaschke
and MMP applied to   F(z)/z   over  |z| < r  gave you what you
needed.

References:

John Garnett, Bounded Analytic Functions, Academic Press
Peter Duren, Theory of H^p Spaces, Dover
 « Last Edit: Aug 9th, 2006, 12:39pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
Michael Dagg
Senior Riddler     Gender: Posts: 500 Re: Interesting Bounded Analytic Map   « Reply #4 on: Aug 9th, 2006, 1:02pm » Quote Modify

While one is tuned in on the Schwarz-Pick lemma, a
related "Topics" problem whose study provides insight
into working with holomorphic maps on complex Banach
manifolds is

Pseudo-distance on a domain U \subset C^n can be
defined by Caratheodory distance and Kobayashi distance.

Show that the two distances agree if U is a bounded
convex domain.

Reference:

M. Jarnicki, P. Pflug (1993), Invariant Distances and
Metrics in Complex Analysis, de Gruyter Expositions in
Mathematics, Walter de Gruyter.
 « Last Edit: Aug 9th, 2006, 1:25pm by Michael Dagg » IP Logged

Regards,
Michael Dagg

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