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Title: Interesting Bounded Analytic Map Post by Michael_Dagg on Aug 7th, 2006, 2:35pm 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? |
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Title: Re: Interesting Bounded Analytic Map Post by Eigenray on Aug 8th, 2006, 7:53pm 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. |
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Title: Re: Interesting Bounded Analytic Map Post by Eigenray on Aug 9th, 2006, 5:00am This is apparently known as the [link=http://eom.springer.de/N/n066480.htm]Nevanlinna-Pick problem[/link], and the condition is that the nxn "Pick matrix" (1 - wi' wj)/(1 - zi' zj ) be positive semi-definite. (Pick's paper is available [link=http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D37467]here[/link], 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... |
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Title: Re: Interesting Bounded Analytic Map Post by Michael_Dagg on Aug 9th, 2006, 10:14am 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 |
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Title: Re: Interesting Bounded Analytic Map Post by Michael_Dagg on Aug 9th, 2006, 1:02pm 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. |
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