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Let $A=[0,1]$ and $C=\{0\}\cup\{\frac{1}{n},\ n\in\mathbb{N}\}$.

i) Is there a function $f:A\rightarrow\mathbb{R}$ such that $f\in C^{r}(A)$, $r\geq 2$ and the set of critical "Values" of $f$ is $C$?

ii) Is there a function $f:A\rightarrow\mathbb{R}$ such that $f\in C^{1}(A)$ and the set of critical "Values" of $f$ is $-C\cup C$?

This is a problem from a course of differential topology that i did last year. Is related with Morse theory. I couldn't figure out any good solution for it. I appreciate some help.

EDIT: A critical value is image of a critical point, i.e. if $f'(x)=0$ then $f(x)$ is a critical value.

Thanks

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    To be sure: $C$ is the set you describe at the top of your post, while $C^r$ is the set of continuously differentiable functions of order $r$?2012-09-29
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    you are right..2012-09-29
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    You can obtain example (ii) by gluing two copies of example (i). To be more precise, suppose $f:[0,1]\to [0,1]$ satisfies (i) and additionally $f(0)=0=f'(0)$. Then the odd extension $f(-x)=-f(x)$ is in $C^1[-1,1]$ and satisfies (ii). Of course, $[-1,1]$ can be changed to $[0,1]$ by a linear transformation.2012-09-30
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    (ii) is clearly wrong because $-C\not\subset A$.2012-10-04

2 Answers 2

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The answer to question (i) is no, the answer to question (ii) is yes. (Thanks to Tomas for pointing out a mistake in my previous argument.)

(ii) Existence of the $\mathcal{C}^1$ function. Let $F:[0,1]\to[0,1]$ be a $\mathcal{C}^1$ function such that $F(0)=0$, $F(1)=1$, $F'(0)=F'(1)=0$, and such that $F'(x)>0$ for all $x \in (0,1)$. (E.g., one could use $F(x) = \sin^2 \frac{\pi x}{2}$.)

Now let $a_n$ be a strictly decreasing positive sequence with $a_1 = 1$ and $\lim\limits_{n\to\infty} a_n = 0$. Define the function $f$ on the interval $[a_{n+1},a_n]$ by $$f(x) = \frac{1}{n+1} + \left(\frac1n - \frac{1}{n+1}\right) F\left(\frac{x-a_{n+1}}{a_{n}-a_{n+1}}\right)$$ for all $n$. I.e., $f$ is an affine version of $F$ with $f(a_n) = \frac1n$ and $f(a_{n+1}) = \frac1{n+1}$. This already gives us a (strictly increasing) $\mathcal{C}^1$ function $f:(0,1] \to \mathbb{R}$ with critical values $\{ \frac1n:n \in \mathbb{N} \}$. Extending it to $[0,1]$ by $f(0)=0$ makes the function continuous. Now we just have to find a sequence $(a_n)$ for which this function satisfies $\lim\limits_{x\to 0} f'(x) = 0$. Then the odd extension of $f$ to $[-1,1]$ is $\mathcal{C}^1$ and has critical values $C \cup (-C)$.

Letting $M = \sup F'$, we get for all $x\in[a_{n+1},a_n]$ that $$0

(i) Non-existence of the $\mathcal{C}^2$ function. Assume that $f$ is $\mathcal{C}^2$ on $[0,1]$ with critical values $C$. Then there exists a sequence of disjoint intervals $I_n = [a_n,b_n]$ with critical points $a_n$, $b_n$, no critical points in $(a_n,b_n)$, and $f(a_n) \ge \frac1n$ or $f(b_n) \ge \frac1n$. (This is shown by an elementary, but not quite trivial argument.) This implies that $$|f(a_n)-f(b_n)| \ge \frac1n - \frac1{n+1}= \frac1{n(n+1)}\ge \frac{1}{2n^2}.$$ By the Mean Value Theorem there exists $c_n \in (a_n,b_n)$ with $$|f'(c_n)|\ge \frac{1}{2n^2(b_n-a_n)}.$$ Since $f'(a_n) = 0$, another application of the Mean Value Theorem to $f'$ shows that there exists $d_n \in (a_n,c_n)$ with $$|f''(d_n)| \ge \frac{1}{2n^2(c_n-a_n)(b_n-a_n)} \ge \frac{1}{2n^2(b_n-a_n)^2}.$$ By assumption $f$ is $\mathcal{C}^2$, so $|f''| \le M$ for some constant $M$, implying that $$b_n-a_n \ge \frac{1}{n\sqrt{2M}}.$$ This gives $\sum\limits_{n=1}^\infty (b_n-a_n) = +\infty$. However, since these intervals are mutually disjoint subsets of $[0,1]$, we have $\sum\limits_{n=1}^\infty (b_n-a_n) \le 1$, which is the desired contradiction.

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    You just showed that $f'(0)=0$. I cant see why $f\in C^{2}$ on zero or even why $\in C^{1}$ on zero.2012-10-04
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    You are right, there is an additional argument needed. I'll see if this construction is fixable.2012-10-04
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    OK, if all of the piecewise functions (from the lemma) are affine versions of each other, it is not too hard to show that this function will be $\mathcal{C}^1$ (implying a positive answer to (ii)), but it won't be $\mathcal{C}^2$. I'll update my answer later today.2012-10-04
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    OK, fixed the $\mathcal{C}^1$ construction, and showed that $\mathcal{C}^2$ is in fact not possible. Not the most elegant proof in the world, but I hope it is correct this time... :)2012-10-04
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    Just to make it clear. When you change the interval $[-1,1]$ to $[0,1]$ the set of critival values still the same?2012-10-05
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    Yes, you are replacing $f$ by $g(x) = f(2x-1)$. The critical points change, but the critical values stay the same. (This is always true if you precompose with a diffeomorphism. Critical values would change if you postcompose the function with something.)2012-10-05
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    @LukasGeyer : your nonexistence proof hasn’t convinced me yet. You seem to deduce a bound on $|f(a_n)-f(b_n)|$ assuming only that one of $f(a_n)$ or $f(b_n)$ is $\geq \frac{1}{n}$, the other could be anywhere.2012-10-05
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    Since there is no critical point in $(a_n,b_n)$, the critical values at the endpoints have to be different. And the smallest possible difference between two points in $C$, given that one of them is $\ge 1/n$, is $1/n - 1/(n+1)$.2012-10-05
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    What is the "elementary but nontrivial argument"? In a neighbourhood of a critical point $x_n$ with critical value $\frac1n$, we will have $\frac1{n+1}, so critical points are isolated. (Criticals of same value are excluded because they have others between). And the set of critical points is closed. Hence they are discrete (except possibly at the boundary). Apart from possibly a first and a last critical point there is always a "previous" and a "next" one. Thus for almost all $n$, we can take $a_n$ as a critical point with $f(a_n)=\frac1n$ and $b_n$ the "next" critical point.2012-10-05
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    (continued) my "elementary but nontrivial argument" left out the case of $f$ being constant on an interval. So in reality, not critical points are isolated but rather "critcial intervals". This does not destroy the rest of the argument, especially not the intervals $I_n$ in LukasGeyer's proof.2012-10-05
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In each case, the answer is yes. Check out this post by Mate on the Topology Q+A board. To adapt it to your question:

You need this lemma, which isn't so hard to prove if you already know about bump functions:

For any $a,b,c,d \in \mathbb{R}$ with $a, there exists a smooth function $f:[a,b] \rightarrow \mathbb{R}$ such that $f(a) = c, f(b) = d$, and all the derivatives of $f$ vanish at $a$ and $b$.

Now let $r: \mathbb{N} \rightarrow \mathbb{Q}$ be your favorite enumeration the set $C$ (or $-C \cup C$). Extend $r$ to a smooth function $f$ on the (positive) real numbers using the lemma. This function satisfies all your requirements.

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    Hi Adam Saltz. I think your argument dont prove the statement. It is a good argument when the function $f$ is defined on $\mathbb{R}$, but in our case the function is defined on $[0,1]$. I cant see a clean way to adapt the demonstration for this case. If you do, please show me.2012-09-29
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    Indeed, some care needs to be taken with the smoothness of $f$ at the endpoints $0$ or/and $1$.2012-09-30
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    You are both quite right. I misread your question, Kaye and I agree that it is difficult to adapt for the reason LVK states.2012-10-01