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Topic: Onetoone function (Read 6170 times) 

Michael Dagg
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Onetoone function
« on: Jan 31^{st}, 2006, 9:19am » 
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Let D be the open unit disk and f onetoone such that f: D > C and f(z) + f(z) = 0. Show that there is a function g: D > C that is also onetoone such that f(z) = sqrt(g(z^2)).


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Icarus
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Re: Onetoone function
« Reply #1 on: Jan 31^{st}, 2006, 3:19pm » 
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That is obvious. Did you want us to prove that g is analytic on D as well, or that there exists a single analytic branch of sqrt for which the formula holds?


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Michael Dagg
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Re: Onetoone function
« Reply #2 on: Jan 31^{st}, 2006, 3:27pm » 
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Ha ha, no. It was meant to accompany the other onetoone function problem given by cain to get a dialog going. Proceed if you must, however.


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Icarus
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Re: Onetoone function
« Reply #3 on: Jan 31^{st}, 2006, 4:04pm » 
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Alright, I'll bite. Define g(z^{2}) = f^{2}(z). Since f(z) = f(z), we have f^{2}(z) = f^{2}(z), so g is welldefined. Since every element of D has a square root also in D, g is defined on all of D. If g(z^{2}) = g(w^{2}), then f^{2}(z) = f^{2}(w), and f(z) = +/ f(w) = f(+/w). Since f is injective, z = +/ w. Therefore z^{2} = w^{2}. So g is injective as well. Note that this also works when f is even.


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Eigenray
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Re: Onetoone function
« Reply #4 on: Jan 31^{st}, 2006, 7:08pm » 
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on Jan 31^{st}, 2006, 4:04pm, Icarus wrote:Note that this also works when f is even. 
 Yes, but there aren't many even functions injective on D. Note also that if f is analytic, then g will be also. For, if f^{2}(z) = a_{0} + a_{1}z^{2} + a_{2}z^{4} + . . . is analytic on D, then limsup a_{n}^{1/n} < 1, so g(z) := a_{0} + a_{1}z + a_{2}z^{2} + . . . is also analytic on D. But more interesting is the converse: If g : D > C is an injective analytic function with g(0)=0, then there exists an injective analytic f : D > C with f^{2}(z) = g(z^{2}). For, the function g(z)/z is analytic and (since g is injective) it has no zeros in D. Thus it has an analytic square root, say h^{2}(z) = g(z)/z, and it suffices to check f(z) := zh(z^{2}) ( = sqrt[g(z^{2})] ) is injective. Suppose f(z)=f(w). Squaring, we have g(z^{2})=g(w^{2}), so by injectivity of g, z= +/ w. But as f is clearly odd, f(z)=f(z) implies f(z)=0, which implies z=0, again by injectivity of g. Thus z=w.


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Michael Dagg
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Re: Onetoone function
« Reply #5 on: Feb 1^{st}, 2006, 10:31am » 
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Showing that g is holomorphic is not difficult  try finish up. The form with square roots is (when f is the identity function) at variance with the fact that there is no holomorphic square root of the identity function in D, and zz covers D. What I mean is that there is no holomorphic (or even continuous) function s in D such that s(zz)=z for all z in D. I think this is not the converse but the same problem with the notation reversed (roles of f and g interchanged), unless, of course, I am thinking about a different problem. For the reason I've indicated above, I prefer to refrain from writing sgrt(g(z^2)) (although I didn't when I stated the problem but of course it is part of the problem) and verbalizing "the square root of g(z^2)", and be content to write f^2(z) = g(z^2). This function f does what's wanted; but one would not want to misleadingly suggest existence of continuous squareroot extractions.

« Last Edit: Feb 2^{nd}, 2006, 1:02am by Michael Dagg » 
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