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I'm looking for a source for the following result.

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 $c$.

Suppose that we have a bound on the fourth derivative of $f$ at the critical points of $f$. I've been told that given a bound on the fourth derivative of $f$ you can convert that into a lipschitz bound on the derivative of the gradient, $\nabla f$. This gives you a lower bound on the diameter of a ball contained in the $A(c)$ for a critical point $c$ of $f$.

I'm looking for a reference and official statement of this result.

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  • $\begingroup$ The gradient takes one derivative of $f$ in every direction. The derivatives of the gradient take another. The derivatives of that take another: 4th derivatives of $f$. A bound of these 4th derivatives of $f$ controls how rapidly the derivatives of the gradient change. If they don't change much, and start at zero at the critical point, then they don't get big quickly. There is nothing tricky here: just the old idea that the derivative is the rate of change, I think. $\endgroup$
    – Ben McKay
    Apr 29, 2016 at 21:27
  • $\begingroup$ I guess what I'm wondering is how does the constant that lower bounds the diameter of a ball contained in $A(c)$ depend on the dimension $d$ and the Lipschitz constant $\alpha$. $\endgroup$
    – Blake
    Apr 29, 2016 at 23:24

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