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Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}\setminus (-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least $n+1$ zeros on $(-1,1)$

I have already post this exercice in Mathematics Stack Exchange and I have succeed to prove it with the following lemma :

Lemma: Let $f$ be a smooth function vanishing outside $(-1,1)$ and $P\in \mathbb{R}[X]$ a polynomial of degree $\deg(P)=d$ without zeros in $(-1,1)$. Then: $$\forall n\in \mathbb{N}, Z((fP)^{(n+d)}) \geq Z(f^{(n)})+d,$$ where $Z(g)$ the number of zeros of $g$ on $(-1,1)$


In fact, we don't understand why, after reading, this mysterious zero will appear. Plus, it is not really highlighted the role of the compactness of the support which is essential here.

It would be very interesting to have an another proof explaining the appearance of this additional zero.

References

  1. The exercise in Mathematics Stack Exchange
  2. Oral exercice of ENS Paris (2011)

NB: I'm still a student but I am convinced that this is an interesting question that I hope deserves its place here.

Thank you in advance for your time,

Julien.

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    $\begingroup$ I would appreciate it if you show us the details of the proof of the lemma, I still do not understand the sketch you gave on MSE. $\endgroup$ Feb 13, 2014 at 14:40
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    $\begingroup$ You argue by induction, and see that it is enough that if $|\alpha|\ge1$ then $g^{(n+1)}$ has at least one more zero in $(-1,1)$ than $f^{(n)})$, where $g(x)=(x-\alpha)f(x)$, for which you mention the identity $$ g^{(n+1)}(x)=(x-\alpha)^{-(1-n)}\frac{d}{dx}\Bigl((x-\alpha)^{n}f^{(n)}(x)\Bigr).$$ I do not see how this identity allows you to conclude that $g^{(n)}$ has that extra zero. $\endgroup$ Feb 13, 2014 at 19:00
  • $\begingroup$ (Hmm... I seem to have lots of indices off, I'll repost if needed.) $\endgroup$ Feb 13, 2014 at 20:23
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    $\begingroup$ Yes, that is the claim. How do you derive it from the identity? You also need the extra zero to lie in $(-1,1)$. $\endgroup$ Feb 13, 2014 at 20:58
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    $\begingroup$ @AndresCaicedo I also think Julien's proof works; I took the liberty of writing up a CW version of it at math.SE which spells out the proof of the lemma, corrects some interchanges of $n$ and $n+1$, and uses consistent notation throughout. (The original proof had $(f,g)$ inside the proof of the lemma corresponding to $(g,f)$ outside the lemma's namespace.) $\endgroup$ Feb 14, 2014 at 20:06

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Unfortunately I haven't been able to find an intuitive explanation for this behavior or even an alternative proof. But I've played a bit with your proof and I'm posting my version here in the hope that it might help someone else to find an explanation.

Definition

  1. Let $\mathcal{S} = \{f \in C^\infty([-1, 1])\colon \forall n\in \mathbb{N}\colon f^{(n)}(-1) = f^{(n)}(1) = 0 \}$.
  2. For $n\in\mathbb{N}$ and $\alpha\in\mathbb{R}\setminus[-1, 1]$ let $D_\alpha^n = \frac{1}{(x + \alpha)^{n-1}}\frac{d}{dx}(x+\alpha)^n$.
  3. For $f\in\mathcal{S}$ let $Z(f)\in \mathbb{N}\cup \{\infty\}$ be the number of zeros of $f$ in $(-1, 1)$.

Lemma

  1. $Z(D^n_\alpha f) \ge 1 + Z(f)$.
  2. $\left((x+\alpha)f\right)^{(n)} = D_\alpha^n f^{(n-1)}$.

Proof

  1. $f$ (and thus $(x+\alpha)^n f$) has $Z(f)+2$ zeros on $[-1, 1]$. By Rolle's theorem, $D_\alpha^n f$ has at least $Z(f)+1$ zeros in $(-1, 1)$.
  2. $$((x+\alpha)f)^{(n)} = nf^{(n-1)} + (x + \alpha)f^{(n)} = \frac{1}{(x+\alpha)^{n-1}}\frac{d}{dx}\left((x+\alpha)^n f^{(n-1)}\right) = D_\alpha^n f^{(n-1)}$$ $\square$

Theorem

For all $f\in\mathcal{S}$ there is an $n\in\mathbb{N}$ such that $Z(f^{(n)})> n + 1$.

Proof

If $f$ has a zero in $(-1, 1)$, we are done. Otherwise assume that $f>0$ in $(-1, 1)$ and let $\alpha = \frac{6}{5}$ and $\beta = \frac{4}{5}$. Note that $|\alpha^2 - \beta^2|<1$. Let $g_k = \frac{f}{(x^2 - \alpha^2)^{2k}}$. Then

  • $g_k(-1) = 0$,
  • $g_k(-\beta) \to \infty$,
  • $g_k(0) \to 0$,
  • $g_k(\beta) \to \infty$, and
  • $g_k(1) = 0$

for $k\to\infty$. Hence, for $k$ large enough, $g_k$ has at least two local maxima and one local minimum in $(-1, 1)$. Thus, $Z(g_k') > 2$.

By factoring $x^2 - \alpha^2 = (x-\alpha)(x+\alpha)$ and applying the lemma $4k$ times, we see $$f^{(4k+1)} = ((x^2 - \alpha^2)^{2k}g_k)^{(4k+1)} = D_{-\alpha}^{4k+1}D_\alpha^{4k}\cdots D_{-\alpha}^{3}D_\alpha^{2}g_k'.$$ Hence, $Z(f^{(4k+1)}) \ge 4k + Z(g_k') > 4k + 2$. $\square$

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