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An hamiltonian system with $n$ degree of freedom is said to be completely integrable when there exists an system $f_1,\ldots,f_n$ of first integrals mutually Poisson-commuting, such that $df_1(x),\ldots,df_n(x)$ are linearly independent for any $x$ in a dense subset.

Consequently any hamiltonian system with one degree of freedom, whose non fixed points constitute a dense subset, is completely integrable, (infact in this case just its hamilton function gives a maximal set of Poisson-commuting first integrals which are independent on a dense subset).

Now in a paper I read that any hamiltonian system with one degree of freedom is completely integrable, because there should exist always a smooth function which is Poisson-commuting with the Hamilton function and whose regular points are every-where dense.

I have some difficulty in understanding this statetement, so is there someone who could give me an hint?
My guess is that, given an Hamilton function $H$ on $(M,\omega)$ whose regular points constitute an open subset $U$, if it is possible to prove that there exists a smooth function $\delta$ vanishing on $U$ and whose regular points are dense in $M\setminus\overline{U}$, then we could take $H+\delta$.

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Isn't this function the Hamiltonian itself? –  Fabian May 2 '11 at 17:41
    
sure, but only when the non-fixed points of the hamiltonian system constitute a dense subset. –  Giuseppe Tortorella May 2 '11 at 19:28
    
My question is about the remaining case. –  Giuseppe Tortorella May 2 '11 at 19:28
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up vote 1 down vote accepted

I know this is very late, but what you are concerned about is if the set of points where $dH = 0$ contains an open set. If you consider the interior of this set $dH=0$ and take any function $f$ whose support is contained in here, then $H + f$ commutes with $H$. By choosing $f$ generically, $H+f$ shouldn't have any critical points in this set, and I believe you are done.

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