Let $\ f_n \ $ be a sequence of Morse functions on $\mathbb{R}^d$, adequately converging (in the $C^2$-topology, say) 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)\ $ stand for the stable manifold associated to $p$: $$ W^s(p) := \left[ \ x \in \mathbb{R}^d \ \colon \ \lim_{t \to \infty} \tau_t (x) = p \ \right] , $$ where $\tau_t(x)$ denotes the integral curve of the gradient of $f$ passing through $x$ at $t=0$. Analogously, let $\ W^s(p_n)\ $ stand for the stable manifold associated to $p_n$ (and the gradient of $f_n$)

My question is:

*Do the stable manifolds $\ W^s(p_n)\ $ converge (in an adequate sense) to $W^s(p)$?*

For example, *does the Hausdorff distance $d_H (W^s(p_n) , W^s(p))$ converge to zero?*

In dimension 1, the biggest stable manifolds are intervals, which are delimited by critical points, and it is not difficult to check that this distance $\ d_H ( W^s(p_n) , W^s(p))\ $ converges to zero.

In dimension 2, analyzing the integral curves that separate stable manifolds, I also think we can prove $d_H(W^s(p_n), W^s(p))$ goes to zero.

Nevertheless, in the general case ($d \geq 3$), this method seems to fail.

I am working with a colleague on a quite far topic, but we have come to this problem. We are both totally new in this area, although our intuition is that the question above should have an affirmative answer, at least for a wide class of nice functions.