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.



Stable manifolds of a sequence of Morse functions
The answer is no. Of course, in a neighbourhood of a critical point,
one has convergence.


Assume that the stable manifolds $W^s(p\_n)$ converge to some set $W$ in the Hausdorff distance.
It can happen that there is a sequence $q\_n\in W^s(p\_n)$ that converges
to a critical point $q\in W$ that lies in the interior of $W$ if $W$ can be viewed as a submanifold of $\mathbb R^d$ with singularities. This can "break" the stable manifold, so $W\setminus W^s(p)$ can be a large set.


Here is an example with $d=2$ and two critical points, assuming that we take the positive gradient flow. Instead of $n$, let $t\in\mathbb R$ be the parameter. In a neighbourhood of $p\_t=(t,1)$, $f\_t$ is given by
$$f\_t(x,y)=\frac{(x-t)^2-(y-1)^2}2\;,$$
so the stable manifold will contain a vertical
interval $\{t\}\times(1-\epsilon,1+\epsilon)$.
In a neighbourhood of the $x$-axis, the function is given by
$$f\_t(x,y)=\frac{y^2-x^2}2-c$$
for some suitable $c>0$. Then there is a critical point $q\_t=(0,0)$ with $W^s(q)=\mathbb R\times\{0\}$.
Finally, assume that $f\_{-t}(x,y)=f\_t(-x,y)$, and that
$$\frac{\partial f\_t(x,y)}{\partial x}\le 0\text{ for $t>0$, $x=0$ and $y\in(0,1)$.}$$


For $t> 0$, the stable manifold of $p\_t$ will descend and then "turn right", so that it becomes asymptotic to the positive $x$-axis.


For $t=0$, the stable manifold will run into the origin and stop there. But $\lim\_{t\searrow 0}W^s(p\_t)$ will be the union of $W^s(p\_0)$ and the positive $x$-axis.


For $t<0$, the behaviour of $W\_s(p,t)$ will be asymptotic to the negative $x$-axis. This tells you that one cannot even define an "extended stable manifold" of $p\_0$ to which $W\_s(p,t)$ will converge both for $t\searrow 0$ and for $t\nearrow 0$.


