# Convergence of local stable manifolds

This question is a kind of local version of a previous post (MO224171).

Let $\ f_n \$ be a sequence of Morse functions on $\mathbb{R}^d$, converging, in the $C^\infty$-topology, to a limit Morse function $\ f$: $$f_n \to f \ .$$

At any critical point $\ p\$ of the limit function $\ f$, it can be proved that there exists an open neighborhood $\ U\$ such that, for $n \geq n_0$, $$f_{n|U} \mbox{ has a unique critical point, p_n, of the same Morse index than } p.$$

Moreover, the sequence of these critical points $p_n$ converges to $p$.

On the other hand, let $\ W^s(p,f)\$ stand for the stable manifold associated to $p$: if $\tau_t(x)$ denotes the integral curve of the gradient of $f$ passing through $x$ at $t=0$, then

$$W^s(p , f) := \left[ \ x \in \mathbb{R}^d \ \colon \ \lim_{t \to \infty} \tau_t (x) = p \ \right] .$$

If $V$ is a neighborhood of $p$, we can also consider the local stable manifold:

$$W^s_V(p , f) := W^s(p , f_{|V}) .$$

My question is:

Does there exist a neighborhood $V$ of $p$ such that the local stable manifolds $\ W^s_V(p_n , f_n)\$ converge (in some adequate sense) to $W^s_V(p , f)$?

For example, such that the Hausdorff distance $d_H (W^s_V(p_n , f_n) , W^s_V(p , f))$ converge to zero?

I've read in Wikipedia that local stable manifolds vary continuously in a neighborhood of $f$, but couldn't find a proof. Probably this is not difficult for the well versed in dynamical systems or differential topology, but I cannot arrange the proof of the Stable Manifold Theorem to work for a family.

• I think the answer is yes, and $C^2$ convergence is enough.There is a box nbd of $p$ of the form $D^s\times D^u$ such that $W^s( p; f )\cap D^s\times D^u$ and $W^s( p_n; f_n )\cap D^s\times D^u$ for $n$ large enough are graphs of maps $F: D^s\to D^u$ resp. $F_n: D^s\to D^u$; moreover $F_n\to F$ uniformly with second order derivatives on $D^s$. The same local convergence of $W^s$ as graphs happens then at any $x\in W^s(p;f)$. (Of course, no hope of global convergence, e.g. Hausdorff distance, as shown by easy 2 dimensional examples). – Pietro Majer Nov 24 '15 at 20:44
• Thanks! But, how can you assure that all the $F_n$ are defined in the same $D^s$? (I looked at a proof of the Stable Manifold Theorem, and tried to arrange precisely this idea, but I couldn't avoid the $D^s_n$ of the $F_n$ from becoming arbitrary small!). – José Navarro Nov 25 '15 at 10:34
• This is a key point of course. If you keep track of the radius of $D^s$ in the proof of the stable manifold theorem, e.g. via the graph transform, you can check there is a uniform radius. – Pietro Majer Nov 25 '15 at 13:02
• Or here, via transversality. link.springer.com/chapter/10.1007/1-4020-4266-3_01#page-1 – Pietro Majer Nov 25 '15 at 13:06
• ops, sorry the latter is not free; here's a free version dm.unipi.it/~abbondandolo/preprints/montreal.pdf – Pietro Majer Nov 25 '15 at 13:12

Yes, $p$ is a hyperbolic fixed point of the gradient flow, and local (un)stable manifolds of a hyperbolic fixed persist smoothly under small perturbations (which your functions $f_n$ are, relative to $f$ for $n$ large). I don't have a basic reference at hand, but this is a special case of a normally hyperbolic invariant manifold $M$ consisting of a single point, $M = \{p\}$, see for the NHIM result theorem 4.1(f) in Hirsch, Pugh, Shub: "Invariant manifolds", Springer LNM 538 (1977). The $C^r$ (here $r = \infty$) closeness should be viewed in the way already commented by Pietro Majer: you view $W^{u,s}(p;f|_V)$ as graphs, and then the graphs of $W^{u,s}(p;f_n|_V)$ converge to these in $C^r$ norm.
Moreover, this local result can be extended to any compact neighborhood of $p$ that does not contain other fixed points: simply note that the extended (un)stable manifold is obtained under the gradient flow for a fixed, finite time $t$. Smooth dependence of flows on parameters implies that the flows $\Phi_n^t$ converge to $\Phi^t$, hence the (un)stable manifolds converge too. (Near another fixed point $q$ the problem is that the flow time to pass it blows up.)