some property of bijective polynomial on the unit disc
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From: quasi on 6 Aug 2010 05:10 On Fri, 06 Aug 2010 03:39:51 EDT, master1729 <tommy1729(a)gmail.com> wrote: >> On Thu, 5 Aug 2010 04:59:47 -0700 (PDT), achille >> <achille_hui(a)yahoo.com.hk> wrote: >> >> >On Aug 5, 7:05 pm, master1729 <tommy1...(a)gmail.com> >> wrote: >> >> >> >> but z = 1 gives (z+1)^2 = 4 ?? >> >> >> >> 4 is not within the unit disk ? >> >> >> >You can always scale and add a constant to (z+1)^2 >> to >> >make the range of P(unit disk) falls within the unit >> disk. >> > >> >In fact, when one say a function f(z) is bijective >> on the >> >closed unit disk D, there are several possible >> interpretations: >> > >> >f is bijective between D and the range f(D), and >> > 1) the range can be anything. >> >OR 2) the range f(D) is a subset of D. >> >OR 3) the range f(D) = D. >> >> For the original problem, I guess the interpretation >> of "bijective" >> was intended to be (1) above. >> >> In fact, are there any polynomials f(z) other than >> >> f(z) = c z, where |c| = 1 >> >> which are literally bijective (i.e., using >> interpretation 3) on the >> unit disk? >> >> quasi > >i was assuming 3 all the time. > >does that make proof different ? > >now im confused ... It suffices that f is injective as a map from the interior of D to C. It follows that f' is nonzero in the interior of D, hence all zeros of f' have absolute value at least 1. But then the product of those zeros also has absolute value at least 1. Now just express the product of the roots of f' (including multiplicity) in terms of the coefficients of f'. quasi
From: master1729 on 7 Aug 2010 11:28 achille wrote : > When I first see the question, my first response is > to > interpret it as (3). But then I remember the only > biholomorphic functions that map unit disk onto > itself > are the Mobius transform, interpretation (3) doesn't > make sense. which brings me to a basic question : how to find the functions that map a certain differentiable closed curve Q biholomorphic onto itself. and more intresting would be to add the condition that f(z) has only one fixpoint .. or prove that such an f(z) doesnt exist. e.g. for Q = unit circle f(z) = parabolic moebius transform. ( as achille already noticed )
From: achille on 7 Aug 2010 20:32 On Aug 8, 3:28 am, master1729 <tommy1...(a)gmail.com> wrote: > achille wrote : > > > When I first see the question, my first response is > > to > > interpret it as (3). But then I remember the only > > biholomorphic functions that map unit disk onto > > itself > > are the Mobius transform, interpretation (3) doesn't > > make sense. > > which brings me to a basic question : > > how to find the functions that map a certain differentiable closed curve Q biholomorphic onto itself. > > and more intresting would be to add the condition that f(z) has only one fixpoint .. or prove that such an f(z) doesnt exist. > > e.g. for Q = unit circle f(z) = parabolic moebius transform. ( as achille already noticed ) I assume your Q doesn't intersect itself, ie. a simple closed curve. By Jordan-Schonflies theorem, Q separate the plane into two regions, one bounded and other unbounded. Q is the boundary of both regions. Furthermore, the bounded region Q_int and unbounded region Q_ext are homeomorphic to the inside and outside of unit circle respectively. Apply Riemann mapping theorem to Q_int, there is a biholomorphic map \phi from Q_int onto the open unit disk D. So your problem transform to 1) find the Riemann mapping \phi 2) find g(z) biholomorphic map D onto D, extend it to the closed unit disk \bar{D} and check whether it has only one fixed point. 3) verify one can extend \phi to \bar{Q_int} onto \bar{D}, the closed unit disk. I have no idea about (1) nor (3). However for (2), we know g(z) are just Moebius transforms and it is easy to check if g is not the identity, then g has either no or two fixed point on the unit circle S^1. REF: http://en.wikipedia.org/wiki/Jordan%E2%80%93Sch%C3%B6nflies_theorem http://en.wikipedia.org/wiki/Riemann_mapping_theorem
From: master1729 on 8 Aug 2010 11:56 achille wrote : > On Aug 8, 3:28 am, master1729 <tommy1...(a)gmail.com> > wrote: > > achille wrote : > > > > > When I first see the question, my first response > is > > > to > > > interpret it as (3). But then I remember the only > > > biholomorphic functions that map unit disk onto > > > itself > > > are the Mobius transform, interpretation (3) > doesn't > > > make sense. > > > > which brings me to a basic question : > > > > how to find the functions that map a certain > differentiable closed curve Q biholomorphic onto > itself. > > > > and more intresting would be to add the condition > that f(z) has only one fixpoint .. or prove that such > an f(z) doesnt exist. > > > > e.g. for Q = unit circle f(z) = parabolic moebius > transform. ( as achille already noticed ) > > I assume your Q doesn't intersect itself, ie. a > simple closed curve. > > By Jordan-Schonflies theorem, Q separate the plane > into two regions, > one bounded and other unbounded. Q is the boundary of > both regions. > Furthermore, the bounded region Q_int and unbounded > region Q_ext are > homeomorphic to the inside and outside of unit circle > respectively. > > Apply Riemann mapping theorem to Q_int, there is a > biholomorphic map > \phi from Q_int onto the open unit disk D. So your > problem transform > to > > 1) find the Riemann mapping \phi > 2) find g(z) biholomorphic map D onto D, extend it to > the closed > unit disk \bar{D} and check whether it has only > nly one fixed point. > 3) verify one can extend \phi to \bar{Q_int} onto > \bar{D}, the closed > unit disk. > > I have no idea about (1) nor (3). However for (2), we > know g(z) are > just Moebius transforms and it is easy to check if g > is not the > identity, then g has either no or two fixed point on > the unit > circle S^1. > > REF: > http://en.wikipedia.org/wiki/Jordan%E2%80%93Sch%C3%B6n > flies_theorem > > > > > http://en.wikipedia.org/wiki/Riemann_mapping_theorem i was aware of that. indeed (1) and (3) are the problem. hence my question. in fact one could replace the unit circle in steps (1)(2)(3) with another simple closed curve ... ( as you very likely know ) but that might not help a lot. why is it again that moebius is the only mapping unit circle to unit circle biholomorphicly ? which functions map biholomorphicly in - say - the interior of an ellips ? what simple closed curves can not have functions mapping biholomorphicly their interior onto themselves and be entire at the same time ? the method of mapping first to the unit circle ( as achille discribed ) seems to imply ( by riemann mapping theorem ) that the number of freedoms for solutions is around 3. perhaps old hat for riemann fanatics ...
From: achille on 8 Aug 2010 20:47
On Aug 9, 3:56 am, master1729 <tommy1...(a)gmail.com> wrote: > > why is it again that moebius is the only mapping unit circle to unit circle biholomorphicly ? > The key is maximum/minimum modulus principle plus the fact that non-constant holomorphic functions has isolated zeros. If g is holomorphic on the open unit disk D with the property |g(z)|->1 whenever |z|->1, then by maximum modulus principle, |g| <= 1 on D. If g doesn't have a zero in D, then apply minimum modulus principle to g ( = maximum modulus principle on 1/g ), we get |g| >= 1 on D. This implies |g| and hence g is constant on D. It is easy to see the number of zeros of g is finite. If you cancel each zero by dividing it with an Moebius transform, one can use minimum modulus principle again to what's remain to conclude g is the product of finitely many Mobius transforms. If you want g to be biholomorphic, then by inspecting the degree of the map g(z) on circle |z| = 1-eps, we can conclude the number of Mobius transform factor is 1. BTW, when you say 'mapping unit circle to unit circle biholomorphicly', do you mean 1) open unit disk OR 2) only the circle (ie. boundary of the unit disk). If it is (2), the map g one seek is one defined and holomorphic over a tubular neighbour of the circle and bijective when one restrict it to the unit circle. One can extend it to a holomorphic function over D using Cauchy integral formula and with minor modification, above argument still applies. > which functions map biholomorphicly in - say - the interior of an ellips ? > No idea as I don't know WANT to construct the Riemann map between unit circle and ellipse. > what simple closed curves can not have functions mapping biholomorphicly their interior onto themselves and be entire at the same time ? > No idea again ;-) > the method of mapping first to the unit circle ( as achille discribed ) seems to imply ( by riemann mapping theorem ) that the number of freedoms for solutions is around 3. > Yes. In fact, for any simply connected open set U, the degree of freedom for Riemann map \phi : U -> D, itself is 3. It is Henri Poincaré who first proved that \phi is essentially unique in the sense: if z0 is an element of U and \theta is an arbitrary angle, then there exists precisely one \phi with \phi(z0) = 0 AND \phi'(z0) / | \phi'(z0) | = exp(i\theta). |