Hi!
Can somebody help me with the following problem. How to prove that map f(x)=asin(x) where a is constant is not structurally C^r stable for any r in N.
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$\begingroup$ ivo, when a\in (0,1), consider h=asin(x)+0.3183. Then compare the fixed points. I can't see why this question is hard to solve $\endgroup$– NiyaziCommented Mar 13, 2011 at 4:18
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$\begingroup$ Niyazi, to prove that f is not structurally stable I need to construct for every c>0 function g such that C^r distance between f and g is less than c and such that there is no top. conjugacy between f and g. In your answer you constructed such function for c=0.3183. This is not enough $\endgroup$– ivoCommented Mar 13, 2011 at 10:12
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$\begingroup$ ivo, it was just an example. use $c/2.$ you just need to shift graphs by $c/2$ then the C^r distance between new function and the old one will be small. The number of fixed points is not preserved. It will show that whatever c>0 is given, one can construct a new close function which is not top. conjugate with the original function. $\endgroup$– NiyaziCommented Mar 13, 2011 at 12:35
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$\begingroup$ Niyazi, if a is between 0 and 1 then our f has only one fixed point right? If I make a small perturbation of f this map has also one fixed point, because the derivation of map x-f(x)-c is always greater than zero, this map will be increasing and therefore we will have only one zero $\endgroup$– ivoCommented Mar 13, 2011 at 12:43
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1$\begingroup$ Ivo, the domain and range of your map is not clear, and you should give motivation and context to your question. $\endgroup$– Benoît KloecknerCommented Mar 14, 2011 at 9:09
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1 Answer
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Let $S$ be the set of points where $f' = 0$. Then $f(S)$ consists of just two points.
This property is preserved by any topological conjugacy, but it won't be true of a generic tiny perturbation of $f$.
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$\begingroup$ Can you please explain why is this property preserved by topological conjugacy since conjugacy is only homeo and it doesn't have to be derivable $\endgroup$– ivoCommented Mar 13, 2011 at 21:07
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$\begingroup$ Equivalently, define $S$ to be the set of points where $f$ is not locally a homeomorphism. $\endgroup$– passerbyCommented Mar 14, 2011 at 1:37