# structural stability

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.

• 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 – Niyazi Mar 13 '11 at 4:18
• 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 – ivo Mar 13 '11 at 10:12
• 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. – Niyazi Mar 13 '11 at 12:35
• 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 – ivo Mar 13 '11 at 12:43
• Ivo, the domain and range of your map is not clear, and you should give motivation and context to your question. – Benoît Kloeckner Mar 14 '11 at 9:09

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 $f$.
• Equivalently, define $S$ to be the set of points where $f$ is not locally a homeomorphism. – passerby Mar 14 '11 at 1:37