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

Things I have tried do not seem to work and I am stuck. I would be grateful for some help.

Edit: Second Part:

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

It is suggested that induction and part a) be used. For $k=1$ the assertion holds because of the mean value theorem and the supremem on the values of the $f$ on $(-1,1)$.

Omran Kouba
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user110503
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1 Answers1

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(a) It is sufficient to prove the case $k=1$, because we can then apply the result to $f_{k-1}$. So, in what follows we suppose that $k=1$.

(b) We May suppose that $f'$ does not vanish on $I$, because otherwise there in nothing to prove. By the intermediate value property, we conclude that $f'$ keeps a constant sign on $I$. Replacing $f$ by $-f$ if necessary, we may suppose also that $f'(x)>0$ for every $x\in I$.

(c) So we reduced the problem to the situation where $f$ is strictly increasing on $I=[a,b]$ and $I_1=[a,\alpha]$, $I_2=[\alpha,\beta]$, $I_3=[\beta,b]$ with $-1<a\leq\alpha\leq\beta\leq b<1$.

Now, $$ (\beta-\alpha)m_1(I)\leq (\beta-\alpha)m_1(I_2)\leq\int_\alpha^\beta f'(t)dt=f(\beta)-f(\alpha)=m_0(I_3)-f(\alpha)\tag{1} $$ Now $f$ is increasing on $I_1$ so $f(x)\leq f(\alpha)$ for $x\in I_1$ or $-f(\alpha)\leq-f(x)\leq|f(x)|$. This shows that $-f(\alpha)\leq m_0(I_1)$, so $(1)$ implies that $$\mu\, m_1(I)\leq m_0(I_1)+m_0(I_3)$$ where $\mu= \beta-\alpha=|I_2|$. This concludes the proof.$\qquad\square$

Second Part.

Let $I=[a,b]$ be a sub-interval of $(-1,1)$ of length $\lambda$. Taking $I_1=[a,a]$, $I_2=I$ and $I_3=[b,b]$ we get $$m_1(I)\leq \frac{1}{\lambda}(|f(a)|+|f(b)|)\leq \frac{2}{\lambda}$$ which is the desired inequality for $k=1$ and any sub-interval $I$ of length $\lambda$.

Suppose the result is proved for $k-1$. Let $I=[a,b]$ be a sub-interval of $(-1,1)$ of length $\lambda$. Define $$ I_1=\left[a,a+\frac{k-1}{2k}\lambda\right],\quad I_2=\left[a+\frac{k-1}{2k}\lambda,b-\frac{k-1}{2k}\lambda\right],\quad I_3=\left[b-\frac{k-1}{2k}\lambda,b\right] $$ so that, $|I_1|=|I_3|=\dfrac{k-1}{2k}\lambda$ and $|I_2|=\dfrac{\lambda}{k}$. Thus using the induction hypothesis and the first part we get $$\eqalign{ m_k(I)&\leq \frac{k}{\lambda}\left(m_{k-1}(I_1)+m_{k-1}(I_3)\right)\cr &\leq \frac{k}{\lambda}\left(\frac{2^{k(k-1)/2}(k-1)^{k-1}}{\left(\frac{\lambda(k-1)}{2k}\right)^{k-1}} +\frac{2^{k(k-1)/2}(k-1)^{k-1}}{\left(\frac{\lambda(k-1)}{2k}\right)^{k-1}} \right)\cr &=\frac{ 2^{k(k+1)/2}k^k}{\lambda^k} } $$ which is the desired inequality for $k$.$\qquad\square$

Omran Kouba
  • 28,772