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Let $f:\mathbb R^n\to\mathbb R$ be a Morse function with uniform nondegenerate Hessian at critical points, i.e. for some $\delta>0$ $$ \forall x \in \{\nabla f=0\}\;\forall \xi\in\mathbb R^n: \quad |\langle \xi, \nabla^2 H(x)\xi \rangle | \geq \delta |\xi|^2 . $$ Edit: Liviu Nicolaescu pointed out a condition to ensure the sublevel sets of $\{f\leq c\}$ to be diffeomorphic to the sphere for $c$ large enough. Therefore, we impose $f$ to have at least linear radial growth at infinity $$ \langle x , \nabla f(x) \rangle \geq A |x| - B > 0 ,\quad A,B>0 . $$ Especially, the above two conditions ensure that $f$ has only finitely many critical points.

Let w.l.o.g. $0$ be a local minima of $f$ and let $\Phi_s(x)$ be the negative gradient flow wrt. $f$, i.e. $$ \dot \Phi_s(x) = -\nabla f(\Phi_s(x)) \quad\text{and}\quad \Phi_0(x)= x . $$ In addtion $\Omega$ denotes the basin of attraction for $0$ or in other words just the stable manifold of $0$, i.e. $$ \Omega = \{ x : \Phi_s(x) \to 0 \text{ for } s\to \infty \} $$ Does $f$ satisfy Neumann boundary condition on $\partial\Omega$ in the sense that the following integration by parts hold $$ \int_\Omega (-\Delta f)\; g \; dx = \int_\Omega \nabla f \cdot \nabla g \; dx \quad\text{for all $g$ such that} \quad \int_\Omega |\nabla f| \; |\nabla g| \; dx < \infty\quad ? $$

Strategy so far:

  • If $f$ is Morse-Smale, then $\partial \Omega$ is the union of stable manifolds heteroclinic connected to $0$
  • for the integration by parts only the (n-1)-dimensional stable manifolds of saddles of index 1 are relevant.
  • hence $\mathit{H}^{n-1}$ almost all $x\in \partial \Omega$ lie on a stable manifold of a 1-saddle and there the proof follows by contradiction and the definition of $\Omega$. Hereby $H^{n-1}$ denotes the (n-1)-dimensional Hausdorff measure.

Is it necessary for $f$ to be Morse-Smale?
Is there some soft argument?
What are relevant references?

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In your situation, is $f$ already given, and you look for conditions on $g\in C^1(\Omega)$ (that is, on its decay at $\partial \Omega$) in order that the formula holds? –  Pietro Majer Feb 5 '13 at 17:03
    
I tried to clarify the question. If you want I could give some more background. The focus is not on the noncompatness of $\mathbb R^n$, but more on the detailed structure of $\partial \Omega$ especially in the neighborhood of critical points of index larger 1. Here I expect the boundary of $\Omega$ to be nonsmooth with corners. –  Frederic Feb 5 '13 at 19:52

1 Answer 1

up vote 0 down vote accepted

What troubles me is the noncompactness of $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. The boundary $\partial \Omega$ could be noncompact in a rather unpleasant way, difficult to control. For example, there might exists a sequence of critical points $p_\nu$ of index $1$ going to $\infty$ as $\nu\to \infty$ such that there exists a gradient flow trajectory from $p_\nu$ to $0$ for any $\nu$. In this case $\partial\Omega$ will have an $(n-1)$-dimensional stratum for each $p_\nu$ making it hard to predict.

On the other hand, in the compact case, i.e., gradient flows on compact manifolds there is a beautiful paper of Harvey and Lawson that describes one instance when an integration by parts holds.

Edit Another issue you might neet to confront on a noncompact manifold is that of the multiplicities of the top strata of $\partial \Omega$. If $p$ is a critical point of index $1$ and there are infinitely many gradient trajectories from $p$ to $0$ they you would be hard-pressed to give a mening of the multiplicity of the stratum of $\partial \Omega$ corresponding to $p$.

In the compact case this multiplicity is defined by assigning in a certain canonical fashion a sign $\pm 1$ to each such connecting trajectory and then adding up all these signs. Clearly, if there are infinitely many connecting trajectory this procedure does not make sense.

This is a serious problem that appears in symplectic Floer homology. As far as I know, this is dealt with by taking advantage of the peculiarities of each concrete situation.

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I thought this case is ruled out, by the assumption of growth at infinity, i.e. $f(x)\to \infty$ whenever $|x|\to \infty$. Does this not ensure that there is not such a sequence of critical points going to infinity? –  Frederic Feb 5 '13 at 16:21
1  
well,no, think e.g. to $f(x):=\sqrt{x^2+1}/2+\sin x$. –  Pietro Majer Feb 5 '13 at 16:58
    
Also the function $x^2(2-\cos x)$. –  Liviu Nicolaescu Feb 5 '13 at 18:48
    
Sorry, you are right. But since I don't want to have trouble with noncompactness of $\mathbb R^n$ let's assume that $f$ has at least linear growth at infinity. I will adjust the question. –  Frederic Feb 5 '13 at 19:33
    
The big problem is that you can still have infinitely many critical points. You need to impose a condition that prohibits this from happening. –  Liviu Nicolaescu Feb 5 '13 at 20:17

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