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This is a follow-up of the question

Is there a bound on the length of the longest Morse trajectory?.

Consider the following setup. Let $(M,g)$ be a (complete) Riemannian Hilbert manifold and $f$ a Palais-Smale Morse function which is bounded below. In the cited question Pietro Majer essentially states that: Let $a$ be a real number. Then, for every critical point $x$ of $f$ there is a neighborhood $U_x$ of $x$ and a constant $C,$ s.t. whenever $\gamma:[0,T]\rightarrow M$ is a Morse trajectory (i.e. $\dot{\gamma}(t)=-\nabla f(\gamma(t))$) with $f(\gamma(0))\leq a$ and $\gamma([0,T])\subseteq U_x$, then the length of $\gamma$ is bounded by $C$.

When Pietro's answer came online, I thought I had a proof for the statement above. Unfortunately I found a gap in my argument which I could not fix. Any ideas?

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It's quite a standard fact, not difficult though a bit technical. Of course it comes from the hyperbolic structure of the flow near its equilibrium points, and from the existence of a Lyapounov function for the flow (the function itself), which in turn this gives the existence of an isolating neighborhood (in the language of Conley theory). You may find it useful to check this, a paper of some years ago where I myself needed these length estimates (see section 5.1)

http://onlinelibrary.wiley.com/doi/10.1002/cpa.1012/pdf

You may also find interesting these lectures.

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  • $\begingroup$ These estimates were exactly what I was looking for! (The paper was already in the drawer of my desk...) Thank you! $\endgroup$
    – Orbicular
    Commented Dec 1, 2010 at 15:38

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