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Topic: Interesting Bounded Analytic Map (Read 2797 times) 

Michael Dagg
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Interesting Bounded Analytic Map
« on: Aug 7^{th}, 2006, 2:35pm » 
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Pick n = 2006 distinct points in the unit disc, z_{i} in D, and prescribe arbitrarily n = 2006 complex numbers w_{i} in C. When can one find a bounded analytic function f on the disc, with sup norm <= 1, such that f(z_{i}) = w_{i}?

« Last Edit: Aug 7^{th}, 2006, 2:36pm by Michael Dagg » 
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Eigenray
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Re: Interesting Bounded Analytic Map
« Reply #1 on: Aug 8^{th}, 2006, 7:53pm » 
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For n=1, this is easy: iff w_{1} < 1. For n=2, it's not too hard. If w_{1}=w_{2}, then the requirement is again just w_{1} < 1. Otherwise, by the open mapping theorem we need w_{1}, w_{2} to both lie in the open unit disc D. Now the answer is the SchwarzPick lemma: Consider the Blaschke factor g_{w}(z) = (wz)/(1  w'z), where w' is the conjugate of w. If w is in D, it's easily verified that g_{w} is an analytic involution of D, preserving the boundary, and interchanging the points w and 0. So if f takes z_{i} to w_{i}, then the composition F = g_{w0}fg_{z0} has F(0)=0, and F(g_{z0}(z_{1})) = g_{w0}(w_{1}). 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 w_{0}  w_{1}/1w_{0}'w_{1} = g_{w0}(w_{1}) < g_{z0}(z_{1}) = z_{0}z_{1}/1z_{0}'z_{1}, 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 < r^{2} < r.


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Eigenray
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Re: Interesting Bounded Analytic Map
« Reply #2 on: Aug 9^{th}, 2006, 5:00am » 
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This is apparently known as the NevanlinnaPick problem, and the condition is that the nxn "Pick matrix" (1  w_{i}' w_{j})/(1  z_{i}' z_{j} ) be positive semidefinite. (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 1w^{2} 1 1 1 1/(1r^{2}) 1/(1+r^{2}) 1 1/(1+r^{2}) 1/(1r^{2}), and indeed this matrix is positive semidefinite exactly when w < r^{2}. 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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Michael Dagg
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Re: Interesting Bounded Analytic Map
« Reply #3 on: Aug 9^{th}, 2006, 10:14am » 
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Indeed. It is a classical problem. Even wellknown 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 FBlaschke 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 9^{th}, 2006, 12:39pm by Michael Dagg » 
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Michael Dagg
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Re: Interesting Bounded Analytic Map
« Reply #4 on: Aug 9^{th}, 2006, 1:02pm » 
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While one is tuned in on the SchwarzPick lemma, a related "Topics" problem whose study provides insight into working with holomorphic maps on complex Banach manifolds is Pseudodistance 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 9^{th}, 2006, 1:25pm by Michael Dagg » 
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