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This exercise come from Zorich,Mathematical Analysis,I P232 Exercise 6

6 Let $f \in C^{(n)} ( ]-1,1[ )$ and $\sup_{-1<x<1}|f(x)|\leq 1$. let $m_k(I)=\inf_{x\in I}|f^{(k)}(x)|$, where $I$ is an interval contained in $]-1,1[$, show that

a) if $I$ is partitioned into three successive intervals $I_1,I_2$ and $I_3$ and $\mu$ is the length of $I_2$,then $$m_k(I)\leq \frac{1}{\mu}\left(m_{k-1}(I_1)+m_{k-1}(I_3)\right)$$

b) if $I$ has length $\lambda$, then $$m_k(I) \leq \frac{2^{k(k+1)/2}k^k}{\lambda_k}$$

c) there exists a number $\alpha_n$ depending only on $n$ such that if $|f'(0)|\geq \alpha_n$,then the equation $f^{(n)}(x)=0$ has at least $n-1$ distinct roots in $]-1,1[$

Consider the question "c)" Assume $f(x)=\frac{e^x}{e}$, then $f \in C^{\infty}$,and $\sup_{-1<x<1}|f|=\sup_{-1<x<1}\frac{e^x}{e} = 1$, which satisfies the conditions. and if we take $\alpha_n\leq \frac{1}{e}$. then $|f'(0)|=\frac{1}{e}\geq \alpha_n$, but $f^{(n)}=\frac{e^x}{e}$,and $\frac{e^x}{e}=0$ has no roots in $\Bbb{R}$. Dose this a counter-example?

What's wrong with me? Thanks very much.

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Something is missing here. What does $m_k$ have to do with the rest of the problem? – user53153 Feb 26 '13 at 4:34
This is part (c) of a three-part exercise. $m_k$ was used in parts (a) and (b), which are also meant to be helpful in doing part (c). It doesn't directly show up in (c) though. – David Moews Feb 26 '13 at 4:49
@DavidMoews But now I think I have a counterexample. Take a function that looks like $f(x)=x^{1/3}$ but is a little bit nicer at the origin, say the inverse of $\epsilon x+x^3$. It seems that $f^{(3)}<0$ everywhere and $f'(0)$ can be as large as we want. – user53153 Feb 26 '13 at 4:54
@5pm,I have supplemented the a) and b) – Laura Feb 26 '13 at 7:27
@5pm: Rescaling the axes turns this function into the inverse, $h$ say, of $g: x\mapsto x+x^3$. Then, $h'''$ has two zeroes near the origin. – David Moews Feb 26 '13 at 7:54
up vote 3 down vote accepted

Part (c) of this problem wants you to find an $\alpha_n$ that depends only on $n$. Once you have found it, it should work for all $C^n$ functions $f$. So, for example, you might determine that $\alpha_2=100$ worked. Then, to prove that this choice of $\alpha_2$ worked, you would need to prove that for all $C^2$ functions $f$ on $(-1,1)$ with $\sup_{-1<x<1} |f(x)|\le 1$ and $|f'(0)|\ge 100$, $f''$ has at least one root in $(-1,1)$. To express the statement as predicate logic, what you are being asked to prove is that $$ (\forall n\ge 2)(\exists \alpha_n)(\forall f)(f\in C^n(-1,1) \wedge \sup_{-1<x<1} |f(x)|\le 1 \wedge |f'(0)|\ge \alpha_n \rightarrow f^{(n)} \text{ has at least } n-1 \text{ distinct roots in } (-1,1)). $$

You found that, for $f(x)=e^{x-1}$, all of its derivatives have no roots in $(-1,1)$. This is not a counterexample to the exercise because it only shows that you cannot take $\alpha_n\le 1/e$. To find a counterexample to the exercise, you would need to find a method for taking any $K>0$, no matter how large it was, and finding a function $f_K$ with $\sup_{-1<x<1} |f_K(x)|\le 1$ and $|f_K'(0)|\ge K$ so that $f_K^{(n)}(x)$ has $n-2$ or fewer distinct roots in $(-1,1)$.

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I have understood . thanks very much ! – Laura Feb 26 '13 at 13:13

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