Proving Alternating Derivatives with Induction in Mathematical Analysis I

[Unconscious]

Hi forum.
I'm trying to prove a claim from Mathematical Analysis I - Zorich since some days, but I succeeded only in part.
The complete claim is:

$$\left\{\begin{matrix} f\in\mathcal{C}^{(n)}(-1,1) \\ \sup_{x\in (-1,1)}|f(x)|\leq 1 \\ |f'(0)|>\alpha _n \end{matrix}\right. \Rightarrow \exists x_1,...,x_n \in (-1,1) : f^{(n)}(x_i)f^{(n)}(x_{i+1})=\pm (-1)^i $$

for $i=1,...,n-1$.

In natural language: there exists an $\alpha _n$ such that for every $f$ satisfying the properties listed, then we have that $f^{(n)}(x)$ alternates its sign in $n$ points inside $(-1,1)$.

I thought that induction can solve the problem.
So, I started proving the claim for $n=2$, and this can be done simply by reductio ad absurdum using Taylor expansion at order 2 with Lagrange remainder.

I can't start the induction chain, supposing that it's true for $n$ and proving it by $n+1$.
The book suggests me to use this property:

$$\inf_{x\in I}|f(x)| \leq \frac{1}{|I_2|}\left(\inf_{x\in I_1}|f(x)|+\inf_{x\in I_3}|f(x)|\right)$$

which is valid for every interval (open or closed) $I\subset (-1,1)$ partitioned as $I=I_1 \cup I_2 \cup I_3$ ($I_1$, $I_2$ and $I_3$ with no points in common except at most the boundary points).
An example of partition can be $I=[a,b]\cup [b,c]\cup [c,d]$.

Any hint?

Thanks in advance.

 

[Erland]

This problem is not stated in a clear way. At least, it is not clear to me what is assumed and what is the conclusion to be proved.

 

[Unconscious]

it is not clear to me what is assumed and what is the conclusion to be proved.

Property listed (assumptions):

$$\left\{\begin{matrix} f\in\mathcal{C}^{(n)}(-1,1) \\ \sup_{x\in (-1,1)}|f(x)|\leq 1 \\ |f'(0)|>\alpha _n \end{matrix}\right.$$

What we have to deduce from the assumptions:

$f^{(n)}$ alternates its sign in $n$ points in $(-1,1)$.

 

[Erland]

Property listed (assumptions):

$$\left\{\begin{matrix} f\in\mathcal{C}^{(n)}(-1,1) \\ \sup_{x\in (-1,1)}|f(x)|\leq 1 \\ |f'(0)|>\alpha _n \end{matrix}\right.$$

What we have to deduce from the assumptions:

$f^{(n)}$ alternates its sign in $n$ points in $(-1,1)$.

But this is not true even for $n=1$. No matter how big $\alpha_1>0$ is, there are always functions satifying the assumptions, with the given $\alpha_1$, such that the conclusion is false.

For example, take:
$$f(x)=\frac2\pi\arctan( \frac\pi2(\alpha_1+1)x).$$
Then, $f'$ is positive everywhere, and thus does not change sign anywhere. Yet, the assumptions hold, with $f'(0)=\alpha_1+1$.

 

[Unconscious]

The statement has sense only for $n=2$ on, because:

f' changes its sign in one point

has no sense.
It is clearer if we look at the statement in its math language and not in its natural language, in particular the expression after the "$\Rightarrow$" in #1.

 

[Erland]

OK, sorry for the misunderstanding about changing/alternating sign. I think I understand what you mean now. You mean:

For each integer $n\ge 2$ there exists an $\alpha_n\in \Bbb R$, such that for all functions $f$ who satisfy
$$\left\{\begin{matrix} f\in\mathcal{C}^{(n)}(-1,1) \\ \sup_{x\in (-1,1)}|f(x)|\leq 1 \\ |f'(0)|>\alpha _n\,, \end{matrix}\right.$$
there are $x_1,\, x_2,\,\dots x_n\in \Bbb R$ such that $-1<x_1<x_2<\dots <x_n<1$ and such that $f^{(n)}(x_i)f^{(n)}( x_{i+1})<0$ for $i=1,\,2,\,\dots n-1$.

This is what you mean, right?

Not so easy, it seems to me.

 

[Unconscious]

Yes, this is what we have to prove.
For $n=2$ all right, but for $n>2$...

 

[Erland]

Well, if we have $-1<x_1<x_2<\dots<x_n<1$ such that $f^{(n)}(x_i)f^{(n)}(x_{i+1})<0$ for $i=1,\,2,\dots,n-1$, then it follows easily by the Mean Value Theorem that there are $y_1, \,y_2,\dots,y_{n-1}$ such that $-1<x_1<y_1<x_2<y_2<\dots<y_{n-1}<x_n<1$, with $f^{(n+1)}(y_i)f^{(n+1)}(y_{i+1})<0$ for $i=1,\,2,\dots,n-2$. But we need two more such $y_i$:s. Can we prove that there are $y_0<x_1$ and $y_n>x_n$ extending this sequence of points with alternating sign of $f^{(n+1)}$, or that the conclusion still holds if this is not true?

 

[Unconscious]

This is exactly my proof attempt!
We have done the same mind-path, and I can’t go further from this point.
So, I can’t answer your last question because I don’t know the answer.

 

[Erland]

In order to get a proof by induction to work, I think we need bounds for the derivatives, independent of $f$. I think it suffices to give bounds for the derivatives not everywhere in $(-1,1)$, but at separate points with arbitrary density, where the bound may depend upon the density.

I think the following might be useful, which is proved by induction on $n$:

If $f$ satisfies the first two conditions in post #6, and $m \ge n$, then, to every open subinterval $I\subset (-1,1)$ of length $3^{n-m}$, there exists a $c\in I$ such that
$$|f^{(n)}(c)|\le 2^n3^{\frac{n(2m-n+1)}2}.$$

 

[Unconscious]

I'm thinking on this.
If I have news I'll post them here.

Thanks for your time.