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        general >> complex analysis >> One-to-one function
        
(Message started by: Michael_Dagg on Jan 31^st, 2006, 9:19am)

Title: One-to-one function
Post by Michael_Dagg on Jan 31^st, 2006, 9:19am

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)).

Title: Re: One-to-one function
Post by Icarus on Jan 31^st, 2006, 3:19pm

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?

Title: Re: One-to-one function
Post by Michael_Dagg on Jan 31^st, 2006, 3:27pm

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.

Title: Re: One-to-one function
Post by Icarus on Jan 31^st, 2006, 4:04pm

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.

Title: Re: One-to-one function
Post by Eigenray on Jan 31^st, 2006, 7:08pm

on 01/31/06 at 16:04:34, 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
f²(z) = a₀ + a₁z² + a₂z⁴ + . . .
is analytic on D, then limsup a_n^1/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.

Title: Re: One-to-one function
Post by Michael_Dagg on Feb 1^st, 2006, 10:31am

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.