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Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$ be a smooth function and consider the $d$-manifold $M = \{(x, f(x)): x \in \mathbb{R}^d\} \subset \mathbb{R}^{d+1}$. Let $P$ be a property of some subset of the critical points of $M$. Consider the negative of the gradient vector field $V$, that is at each point we consider the $(x, f(x)) \in M$ we assign the vector $V(x,f(x))=(−\nabla f(x),−|\nabla f|^2)$. This generates a flow $F: M \times \mathbb{R} \rightarrow M$ on $M$ and a set of integral curves that travel along the manifold toward the critical points of $M$. Consider a critical point $c \in M$ and define the attractor set $A(c) = \{p \in M: \exists t \in \mathbb{R}, F(p, t) = c\}$ of $c$. Let $\mathcal{C}$ be the set of critical points with property $P$. Consider the set $\mathcal{A} = \bigcup_{c \in \mathcal{C}} A(c)$, the set of all points that travel along integral curves to critical points in $\mathcal{C}$. Let $\mu$ be the volume measure on $M$. How could I compute the ratio $\frac{\mu(\mathcal{A})}{\mu(M)}$ and possibly lower bound it by some constant?

Edit: Fixed vector field definition according to Ben's correction.

Edit: Assume that $f > 0$ everywhere and has at least one critical point. I'm interested in any way we can approach this problem, so make any further assumptions you'd like on $f$.

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  • $\begingroup$ Since $f$ might have no critical points, the only lower bound (in absence of other information) that will hold for all such problems is zero. What data do you want your lower bound to depend on? Keep in mind that if $f(x)$ gets large negative as $|x|$ gets large, gradient flow will tend to move you away from critical points. $\endgroup$
    – Ben McKay
    Commented Apr 18, 2016 at 20:28
  • $\begingroup$ So $f > 0$ everywhere and assume that it has at least 1 local minima. In general it's nonconvex. I'd like the lower bound to depend on properties like $d$ and the number of critical points with the property $P$ $\endgroup$
    – Blake
    Commented Apr 18, 2016 at 20:31
  • $\begingroup$ You can still make small valleys and large valleys in a smooth mountain range, and there is no bound on how much or how little of the rain water falls into some chosen valleys, if all you know is how many valleys you get to choose: xkcd.com/681 $\endgroup$
    – Ben McKay
    Commented Apr 18, 2016 at 20:46
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    $\begingroup$ @Ryan Budney: wouldn't you need that bound to be relative to the eigenvalues of the Hessian of $f$ at the critical points? Otherwise, you could simply take any bounded $f$ and then rescale it to make its fourth derivative as small as you like without changing the basins of attraction. $\endgroup$
    – Jaap Eldering
    Commented Apr 23, 2016 at 16:23
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    $\begingroup$ I am afraid that this web site has the fundamental problem that if people don't suddenly see a clear insight into your problem that is at the same time a fun insight for them to pursue, then they lose interest immediately. I suspect that Ryan Budney (who is a great source of help to many people on this site) just didn't have anything exciting come to his mind. That doesn't mean it isn't a good problem, or that he isn't clever enough to make progress with it, but only that it didn't immediately set his brain on fire. This site depends on the rare instances when we set brains on fire. $\endgroup$
    – Ben McKay
    Commented Apr 29, 2016 at 21:32

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