1. ## nondegenerate critical points

$I$ is an open interval. Let $f:I\to\mathbb{R}$ be a function of class $C^1$ and $K\subset I$ compact. If all critical points of $f$ in $K$ are nondegenerate then there exists only finitely many of them in $K$.

I've proven that if $c\in I$ is a nondegenerate critical point of the differentiable function $f:I\to\mathbb{R}$ then there exists $\delta>0$ such that $c$ is the only critical point of $f$ in $(c-\delta,c+\delta)$. This should be helpful.

2. Originally Posted by math9
$I$ is an open interval. Let $f:I\to\mathbb{R}$ be a function of class $C^1$ and $K\subset I$ compact. If all critical points of $f$ in $K$ are nondegenerate then there exists only finitely many of them in $K$.

I've proven that if $c\in I$ is a nondegenerate critical point of the differentiable function $f:I\to\mathbb{R}$ then there exists $\delta>0$ such that $c$ is the only critical point of $f$ in $(c-\delta,c+\delta)$. This should be helpful.
If you have, indeed, proven this fact, then finishing should be quite straight-forward. Use the fact that, for a compact set, every open cover has a finite subcover. That should get you the finite number of nondegenerate critical points.

3. Originally Posted by Ackbeet
If you have, indeed, proven this fact, then finishing should be quite straight-forward. Use the fact that, for a compact set, every open cover has a finite subcover. That should get you the finite number of nondegenerate critical points.
What open cover should I use?

4. Actually, maybe a different approach would be better, come to think of it. How about this:

You've essentially prove that every nondegenerate critical point is isolated. Since K is compact, it is closed and bounded. If there were an infinite number of nondegenerate critical points, they would have to accumulate somewhere. But they can't, because they're isolated.

How does that work?

5. Originally Posted by Ackbeet

You've essentially prove that every nondegenerate critical point is isolated. Since K is compact, it is closed and bounded. If there were an infinite number of nondegenerate critical points, they would have to accumulate somewhere. But they can't, because they're isolated.

How does that work?
Mmmm.. I don't know... I don't think this reasoning is correct. Consider the set $P=\{0\}\cup\{1/n;n\in\mathbb{N}\}$. It is easy to see that all points of $P$, except for 0, are isolated. Yet they accumulate at 0.

6. Yeah, I think you're right in post # 5. Your set P is compact since it's a subset of the reals, closed, and bounded. You could probably come up with a function for which your set P (minus 0) is the set of nondegenerate critical points, even. So, it looks like my post # 4 isn't going to work. I think this is a job for Plato or Opalg. I'm quite certain one of them can help you better than I.

7. Originally Posted by math9
$I$ is an open interval. Let $f:I\to\mathbb{R}$ be a function of class $C^1$ and $K\subset I$ compact. If all critical points of $f$ in $K$ are nondegenerate then there exists only finitely many of them in $K$.

I've proven that if $c\in I$ is a nondegenerate critical point of the differentiable function $f:I\to\mathbb{R}$ then there exists $\delta>0$ such that $c$ is the only critical point of $f$ in $(c-\delta,c+\delta)$. This should be helpful.
I think I need to know a bit more about the terminology here. The function is said to be of class $C^1$ (so it has one continuous derivative), but the usual definition of nondegeneracy uses the second derivative. So are you actually assuming that the function is twice differentiable (at the critical points, at any rate), or is there another definition of nondegeneracy?

Example: The function $f(x) = x^3\sin\frac1x$ (with f(0)=0) is of class $C^1$, but it has infinitely many critical points in the interval [–1,1]. All of them are nondegenerate except possibly for the one at the origin. The function is not twice differentiable at the origin, so how are you going to define the critical point there to be degenerate?

8. Originally Posted by Opalg
So are you actually assuming that the function is twice differentiable (at the critical points, at any rate), or is there another definition of nondegeneracy?
Yes, I'm assuming that the function is twice differentiable at all of its critical points.

9. Originally Posted by math9
Yes, I'm assuming that the function is twice differentiable at all of its critical points.
Okay, then you can get a proof by contradiction. Suppose that there are infinitely many critical points. By compactness, there exists a convergent set of such points, say $x_n\to y$. Then $f'(x_n)=0$ for each n, and by continuity of $f'$ it follows that $f'(y) = 0$. Thus y is a critical point. But that contradicts what you have already proved, namely that each critical point is isolated.

Hmm, I don't seem to have mentioned the second derivative at all in that argument, having previously fretted over it. I think it must implicitly come somewhere in the proof that critical points are isolated.

10. Originally Posted by Opalg
Hmm, I don't seem to have mentioned the second derivative at all in that argument, having previously fretted over it. I think it must implicitly come somewhere in the proof that critical points are isolated.
Many thanks! Actually, I do use the hypothesis that $f''(x)\neq 0$ to prove that nondegenerate critical points are isolated.