# I Boundedness of derivatives

#### 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?

• Delta2

#### 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:

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.

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#### 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.

• Greg Bernhardt

#### 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.

#### 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.