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1. Let $f\in C^{(n)}\left(]-1,1[\right)$ and $\sup \limits_{-1<x<1} |f(x)|\leq 1$. Let $m_k(I)=\inf \limits_{x\in I} |f^{(k)}(x)|$, where $I$ is an interval (by interval here I mean closed, open or half-open intervals) 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 $$m_k(I)\leq \dfrac{1}{\mu}\left(m_{k-1}(I_1)+m_{k-1}(I_3)\right);$$

b) If $I$ has length $\lambda$, then $$m_k(I)\leq \dfrac{2^{k(k+1)/2}k^k}{\lambda^k};$$

2. Let $f$ be twice differentiable on $\mathbb{R}$. Let $M_k=\sup \limits_{x\in \mathbb{R}}|f^{(k)}(x)|$ for $k=0,1,2$. Then $M_1\leq \sqrt{2M_0M_2}$.

3. Let $f$ is differentiable $p$ times on $\mathbb{R}$ and the quantities $M_0$ and $M_p=\sup \limits_{x\in \mathbb{R}}|f^{(p)}(x)|$ are finite, then the quantities $M_k=\sup \limits_{x\in \mathbb{R}}|f^{(k)}(x)|$, $1\leq k\leq p$, are also finite and $$M_k\leq 2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$

Question: I was able to solve problems 1 and 2 but I have issues with problem 3. The hint to this problem says that we need to use problem 1b), 2) and induction.

So I suppose that we need to do induction on $p$ and the base case $p=2$ is just problem 2. I cannot handle the induction step.

I would be very grateful if someone can show the solution. I have spent some time on this problem and also created the post on MSE but it did not help me.

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    $\begingroup$ With $C=\sqrt 2$, show first that $M_k \le C^k M_0^{\frac{1}{k+1}} M_{k+1}^{\frac{k}{k+1}}$ by induction on $k$. $\endgroup$
    – Giorgio Metafune
    Commented Mar 29, 2021 at 7:14
  • $\begingroup$ @GiorgioMetafune, I think that I got how to prove it but I guess there is one subtle moment. More precisely, in this proof we are implicitly using the fact that $M_k$ are are finite. But we do not know this in advance. $\endgroup$
    – RFZ
    Commented Mar 30, 2021 at 1:17
  • $\begingroup$ Yes, true, but this follows from this argument. Call $C=\|f\|_\infty+\|f^p\|_\infty$; by Taylor's formula the polynomial $$P(h)=\sum_{k=1}^{p-1} \frac{f^k(x)}{k!}h^k$$ satisfies $|P(h)| \le 2C(1+|h|^p) \le 4C$ if $|h| \leq 1$. Since all norms on a finite dimensional space are equivalent, then $|f^k(x)| \le K$, with $K$ independent of $x$. $\endgroup$
    – Giorgio Metafune
    Commented Mar 30, 2021 at 8:45
  • $\begingroup$ @GiorgioMetafune, I am confused, so please let me ask you some questions. I will try to expand your last comment (please correct me if I'm wrong): for any $x\in \mathbb{R}$ and $0<|h|\leq 1$ by Taylor's formula we have: $f(x+h)=f(x)+f'(x)h+\dots+\dfrac{f^{p-1}(x)}{(p-1)!}h^{p-1}+\dfrac{f^{(p)}(\xi)}{p!}h^p$ for some $\xi$ in between $x$ and $x+h$. Denote $P(h):=f(x+h)-f(x)-\dfrac{f^{(p)}(\xi)}{p!}h^p=\sum \limits_{k=1}^{p-1}\dfrac{f^{(k)}(x)}{k!}h^k$ and the LHS can be bounded in the following way: $|P(h)|\leq |f(x+h)|+|f(x)|+|f^{(p)}(\xi)|\leq 2C.$ How it follows that $|f^{(k)}(x)|\leq K$? $\endgroup$
    – RFZ
    Commented Mar 30, 2021 at 14:27
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    $\begingroup$ By assumption you have $M_0(I), M_p(I) \le C$ for every unit interval $I$ and, by the argument above applied to each $I$, $M_k(I) \le C_k$, for every $I$. $\endgroup$
    – Giorgio Metafune
    Commented Apr 4, 2021 at 7:22

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