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$f:\mathbb R \to \mathbb R $ continuous, with a point of odd period, implies existence of a point of even period

This is the question. I can't prove it. It's an exercise to prove Sarkovskii theorem, but it on I have to do this part and I'm ready.

Arturo Magidin
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    In http://math.stackexchange.com/questions/2901/period-of-3-implies-chaos there are useful links – Ross Millikan Oct 28 '11 at 18:17
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    What does "it on I have to do this part and I'm ready" mean? and what are you ready for? – Gerry Myerson Oct 28 '11 at 21:51
  • http://math.arizona.edu/~dwang/BurnsHasselblattRevised-1.pdf – Angela Pretorius Oct 29 '11 at 07:21
  • Strictly speaking, a fixed point is a point with odd period, so we should change "a point of odd period" to "a point of odd period that is not a fixed point". Otherwise set $f$ to be the identity, then every point is of odd period. – Alp Uzman Mar 16 '15 at 18:26

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The answer to this question appear in the book dynamical systems by Robert Devaney. In this book, Devaney give a proof for Sarkovskii theorem, may be you will see this proof and take the idea for you exercice. I hope this be useful.

Bastian Galasso-Diaz
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