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Topic: One-to-one function (Read 6447 times) |
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Michael Dagg
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One-to-one function
« on: Jan 31st, 2006, 9:19am » |
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Let D be the open unit disk and f one-to-one such that f: D -> C and f(z) + f(-z) = 0.
Show that there is a function g: D -> C that is also one-to-one such that f(z) = sqrt(g(z^2)).
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Michael Dagg
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Icarus
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Re: One-to-one function
« Reply #1 on: Jan 31st, 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: One-to-one function
« Reply #2 on: Jan 31st, 2006, 3:27pm » |
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Ha ha, no. It was meant to accompany the other one-to-one function problem given by cain to get a dialog going. Proceed if you must, however.
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Michael Dagg
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Icarus
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Re: One-to-one function
« Reply #3 on: Jan 31st, 2006, 4:04pm » |
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Alright, I'll bite.
Define g(z2) = f2(z). Since f(-z) = -f(z), we have f2(z) = f2(-z), so g is well-defined. Since every element of D has a square root also in D, g is defined on all of D.
If g(z2) = g(w2), then f2(z) = f2(w), and f(z) = +/- f(w) = f(+/-w). Since f is injective, z = +/- w. Therefore z2 = w2. 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: One-to-one function
« Reply #4 on: Jan 31st, 2006, 7:08pm » |
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on Jan 31st, 2006, 4:04pm, Icarus wrote:| Note that this also works when f is even. |
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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
f2(z) = a0 + a1z2 + a2z4 + . . .
is analytic on D, then limsup an1/n < 1, so
g(z) := a0 + a1z + a2z2 + . . .
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 f2(z) = g(z2).
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
h2(z) = g(z)/z,
and it suffices to check
f(z) := zh(z2) ( = sqrt[g(z2)] )
is injective. Suppose f(z)=f(w). Squaring, we have g(z2)=g(w2), 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: One-to-one function
« Reply #5 on: Feb 1st, 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 square-root extractions.
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| « Last Edit: Feb 2nd, 2006, 1:02am by Michael Dagg » |
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Michael Dagg
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