Prove that $ \inf_{-1 < x< 1}|f^{k}| \leq \frac{2^{k(k+1)/2}k^k}{\lambda^k} $ for $|f_{-1 < x < 1}(x)| \leq 1$ and $\lambda$ is length of $x_1 - x_0$

Question

Let

$$ |f_{-1 < x < 1}(x)| \leq 1 $$

$\lambda $ is length of conventional interval $I$ in $(-1, 1)$.

Prove that on this interval:

$$ \inf_{x \in I}|f^{k}| \leq \frac{2^{k(k+1)/2}k^k}{\lambda^k} $$

Where $k$ is k-th derivative, and $f^{k-1}(x)$ for $1 ...k-1$ is continuous function.

Since if function is not monotonic on this interval, it's derivative $\inf_{x \in i}|f'(x)| = 0$, so let's consider the critical case when function and all it's derivatives until $k$ are monotonically increasing (decreasing) functions.

Using Lagrange's formula for Taylor polynomial's reminder

$$ f(x) = f(x_0) + f'(x_0)(x-x_0) + ... \frac{f^k(x_0)(x-x_0)^k}{k!} + \frac{f^{k+1}(\xi)(x - x_o)^{k+1}}{(k+1)!}$$

Take $x_0$ at $i$ left point, and choosing x at $i$ right point, for $k=0$

$$ f(x_o) + f'(\xi)\lambda \leq 1 $$

Since we consider increasing function, $f(x_0)$ minimum value is $-1$

$$ inf|f(x)| \leq 2/\lambda $$

This leads to formula:

$$ \inf_{-1 < x< 1}|f^{k}| \leq \frac{2^{k+1}k!}{\lambda^k} $$

EDIT

However, we can't just interchange derivative value of some order at point with obtained bound from above for $|\inf f^k(x)|$

Other thoughts

This may be somehow related to $e$. At $n\to \infty$ $\frac{(n+1)^{(n+1)}}{(n+1)!} * \frac{n!}{n^{n}} \to e $

Take a look at different variation of Taylor polynomial reminder's formula:

$$ r_k = \frac{1}{(k+1)!}f^{k+1}(\xi)(x - \xi)^k(x - x_0) $$

So if we could bound $(x-\xi)^k \leq(1 - \frac{1}{k})^k \to e^{-1}$, then $k^k$ starts to make sense, as we increase some previous term by $e * (k+1)$

So if we devide distance $ x - x_0$ by 2 at each term, starting with $\lambda$, it looks very close.

It's quite difficult to understand in parts. You say "Since if function is not monotonic on this interval", but you never mention $f$ being monotonic before then. Is the first line saying that $|f|$ is bounded by $1$ on $-1\le x \le 1$? — πr8, Jan 02 '17 at 21:24
Also, how do $\lambda, x_0, x_1$ become a part of the bound, when they're not involved in the definition of the $\inf$? — πr8, Jan 02 '17 at 21:26
@πr8, it's also known that function is continuous and all it's derivatives until k exist, but otherwise it would make any sense. — Joe Half Face, Jan 02 '17 at 21:29
So is the question: If $f$ has derivatives up to and including $f^{(k)}$, and $|f|\le1$ for $x\in[-1,1]$, then show that $\inf_{-1 < x< 1}|f^{(k)}| \le \frac{2^{k(k+1)/2}k^k}{\lambda^k}$? — πr8, Jan 02 '17 at 21:34
@πr8 on $x \in I$, where $-1<I<1$ and $\lambda$ is length of $I$ — Joe Half Face, Jan 02 '17 at 21:38
I'm a little confused by your notation: do you mean $I=[-1,1]$, or $I\subset[-1,1]$? When you say $x\in I$, that means $I$ is a set, but writing $-1<I<1$ isn't something you can say about a set. — πr8, Jan 02 '17 at 21:41

Prove that $ \inf_{-1 < x< 1}|f^{k}| \leq \frac{2^{k(k+1)/2}k^k}{\lambda^k} $ for $|f_{-1 < x < 1}(x)| \leq 1$ and $\lambda$ is length of $x_1 - x_0$

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