Suppose we have a sequence, { f\_n }, of symplectic diffeomorphisms of R^{2n} converging to a function f. By f\_n converging to f I mean: f\_n converges to f in C^k(B\_r) for every r > 0, where B\_r is the ball centered at the origin of radius r. Suppose k > 0. 


Question: Is f is diffeomorphism? Certainly, f is symplectic and hence a local diffeomorphism, and so my question is really: is f invertible?


Interestingly enough, this is not true if f\_n is not symplectic (e.g. take f\_n = x/n, which converges to 0 in C^k(B\_r) for every r > 0).


Thanks!



Is the limit of symplectic diffeomophisms a diffeomorphism?
No, $f$ doesn't have to be invertible. It has to be injective, even in the volume-preserving category rather than the symplectic category. None of the singular values of $Df$ can go to 0, because some other singular value would go to $\infty$. This shows that it is locally injective, and then globally it is not possible to make the locally injective pieces overlap.


But even when $n=1$, it does not have to be surjective. In this case the symplectic condition says just that $f\_n$ is area-preserving. You can reshape larger and larger circles centered at $0$ to non-concentric circles with the same area that lie in the upper half plane. So the image of $f$ could be the upper half plane. Actually, I think that the image of $f$ can be any open topological disk with infinite area.


