- #1
Unconscious
- 74
- 12
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.
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.