Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$

Solution 1:

I believe Julien's proof is right and is also elegant, although there are a lot of typos in his write up. I'll rewrite it here (as Community Wiki), since there seems to be some doubt about that. Of course, this doesn't answer Julien's request for a different proof, but I see nothing wrong with his proof

I introduce the notation $Z(h)$ for the number of zeroes of $h$ on $(-1,1)$.

Let $f$ be as stated. If $f$ vanishes on $(-1,1)$ we are done so, without loss of generality, let $f(x)$ be positive on $(-1,1)$.

For any positive integer $k$, define $$ g:=x\to \frac{f(x)}{(1-x^2)^k} \mbox{ for } |x|<1, \mbox{ and } 0 \mbox{ otherwise} $$ L'Hospital's rule shows that $g$ is in $C^\infty(\mathbb{R},\mathbb{R})$, and $g$ is identically zero outside $(-1,1)$.

Choose $k$ such that $$ 0 = f(-1) \ < \ f\left(-\frac{1}{2}\right) (1-\frac14)^{-k} \ >\ f(0)\ < \ f\left(\frac{1}{2}\right) (1-\frac14)^{-k} \ > \ f(1) = 0. $$ So $$ 0 = g(-1) \ < \ g(-1/2) \ >\ g(0)\ <\ g(1/2)\ >\ g(1) = 0.$$

Then $g'$ must be positive somewhere on $(-1,-1/2)$, negative somewhere on $(-1/2,0)$, positive somewhere on $(0,1/2)$ and negative somewhere on $(1/2,1)$. Thus $Z(g^{(1)}) \geq 3$, and the result holds for $g$ (instead of $f$), with $n=1$.

We will show that $Z(f^{(2k+1)}) \geq 2k+3$.

Lemma Let $p(x)$ be identically $0$ outside $(-1,1)$ and let $a$ be a real number not in $(-1,1)$. Set $q(x) = p(x) (x-a)$. Then $Z(q^{(n+1)}(x)) \geq Z(p^{(n)}(x))+1$.

Proof Let the zeroes of $p^{(n)}$ be $-1 < r_1 < r_2 < \cdots < r_m < 1$ and set $r_0=-1$ and $r_{m+1}=1$. Then $(x-a)^{n+1} p^{(n)}(x)$ vanishes at all the $r_i$, so by Rolle's theorem, $Z(\frac{d}{dx} ( (x-a)^{n+1} p^{(n)}(x) )) \geq m+1$. But $\frac{d}{dx} ( (x-a)^{n+1} p^{(n)}(x) ) = (x-a)^n q^{(n+1)}$, and $(x-a)^{n}$ doesn't vanish on $[-1,1]$, so $Z(q^{(n+1)}) \geq m+1$ as desired. $\square$

Basically, the lemma says that throwing in a non vanishing linear factor doesn't make the Rolle's theorem bound worse.

Now set $h_0 = g$, $h_1 = (1-x) h_0$, $h_2 = (1+x) h_1$, $h_3 = (1-x) h_2$, $h_4 = (1+x) h_3$ etcetera, until $h_{2k} = (1-x^2)^k h_0 = f$. Applying the lemma repeatedly gives $$Z(f^{(2k+1)}) \geq Z(h_{2k-1}^{(2k)} )+1 \geq Z(h_{2k-2}^{2k-1})+2 \geq \cdots \geq Z(h_0^{(1)}) + 2k \geq 3+2k$$ as desired.

Solution 2:

• Lemma: Let f a such function and $P\in \mathbb{R}[X]$ a polynomial and $deg(P)=d$ without zeros on $(-1,1)$. Then: $$\forall n\in \mathbb{N}, Z((fP)^{(n+d)}) \geq Z(f^{(n)})+d $$

Proof. By induction it's sufficient to prove $\forall n\in \mathbb{N}$, $\forall \alpha \in \mathbb{R}$\ $(-1,1)$ : $$Z((g)^{(n+1)}) \geq Z(f^{(n)})+1 $$ where $g(x)=(x-\alpha)f(x)$ ,

Moreover, $g^{(n+1)}(x)=(x-\alpha)^{-(1-n)}\frac{d}{dx}\Bigl((x-\alpha)^{n}f^{(n)}(x)\Bigr)$. Then $g^{(n+1)}$ vanishes at least once again on (-1,1). QED

• Let p an integer such that
  $$ \Bigl(\frac{3}{4}\Bigr)^p < \min\left(\frac{f(\frac12)}{f(0)},\frac{f(-\frac12)}{f(0)}\right) $$ and $$ g(x)=(1-x^2)^{-p}f(x),\qquad g(-1)=g(1)=0 $$ Then $$ g(-1)<g(-1/2),\qquad g(-1/2)>g(0),\qquad g(0)<g(1/2),\qquad g(1/2)>g(1). $$ Thus, $g'$ vanishes at least three times on (-1,1), by the lemma we get $$ Z(f^{(n+2p)}) \geq Z(g^{(n)})+2p $$ and we have (with $n=1$) $$ Z(f^{(1+2p)}) \geq 3+2p $$ Therefore, $f^{(m)}$ for $m\geq 2p+1$ has at least $m+2$ zeros on $(-1,1)$

Solution 3:

$\def\cH{\mathcal{H}}\def\RR{\mathbb{R}}$A bit more than the question asks for is true. Let $\cH$ be the class of function $f: \RR \to \RR$ for which $f^{(n)}(-1) = f^{(n)}(1) = 0$ for all $n$. Note that, for $f \in \cH$, the sequence $Z(f^{(n)})$ is weakly increasing. I will show that $Z(f^{(n)})-n \to \infty$. Let's define $L(f) = \lim_{n \to \infty} Z(f^{(n)})-n$.

We recall the lemma from my other answer:

Lemma Let $a \in (-\infty, -1] \cup [1, \infty)$ and $p \in \cH$. Put $q(x) = (x-a) p(x)$. Then $Z(q^{(n+1)}) \geq Z(p^{(n)})+1$.

We can rewrite this as $Z(q^{n+1}) - (n+1) \geq Z(p^{(n)}) - n$ and take the limit to deduce.

Corollary 1 With notation as in the lemma, $L(q) \geq L(p)$.

Using this repeatedly

Corollary 2 For any positive integers $A$ and $B$, we have $L(f) \geq L(f (1+z)^{-A} (1-z)^{-B})$.

Now, suppose for the sake of contradiction that $L(f)<\infty$. Replacing $f$ by one of its derivatives, we may assume that $Z(f^{(n)}) = n +c$ for all $n$. Let the zeroes of $f$ be $-1< r_1 < r_2 < \cdots < r_c< 1$ and without loss of generality suppose $f$ is positive on $(-1,r_1)$. But then if we choose $A$ and $B$ very large with $(A-B)/(A+B) \in (-1,r_1)$, then $g(z) : = f (1+z)^{-A} (1-z)^{-B})$ will have positive second derivative at $(A-B)/(A+B)$, so $g'$ has at least $3$ zeroes in $(-1, r_1)$, as well as at least one zero in each of $(r_1,r_2)$, $(r_2, r_3)$, ..., $(r_{c-1},r_c)$, $(r_c, 1)$, so $Z(g') \geq c+3$ and $L(g) \geq c+2$.

This contradicts corollary $2$.

In a different question, the OP asks about functions with $f^{(n)}(-1) = f^{(n)}(1) = 0$ for $n \geq 1$, $f(-1)=0$ and $f(1)=1$. Clearly, if $f$ is such a function then $f' \in \cH$, so the bound also applies to such functions. In the comments to that question, there were a bunch of off by $1$ errors, so I thought I'd write up an answer to show that $L(f)$ is infinite anyway, and the off by $1$'s don't matter.