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Let $P$ be a large number. Let $[a,b]$ be a fixed interval. Is it possible to construct a function $w$ with the following properties?: $w$ is smooth, the support of $w$ is contained in $I = [a- 1/P, b + 1/P]$, $w(x) \geq 1$ on $[a, b]$ and $$ \int_{\mathbb{R}}|w^{(n)}(x)| dx \leq C_n P^T $$ where $w^{(n)}$ is the $n$ the derivative, $C_n$ is a positive constant depending only on $n$ and $T$ is some positive number that does not depend on $P$ or $n$.

For example, if I take a function that is $0$ if $x \notin I$, $1$ if $x \in [a,b]$ and connect by line the two points $(a-1/P, 0)$ and $(1, a)$ and similarly for the other side, then this function is not smooth (piecewise differentiable) and would satisfy the above inequality. Though this only gives a piecewise continuous function with the desired property, is there an elaborate way to make a smooth function with the desired properties somehow?

Maybe the condition is too strong and it's not possible... Any comments and clarifications are appreciated. Thank you.

This question is similar to my previous question Existence of a smooth function that approximates a characteristic function of an interval with certain property. This is because I only realized afterwards that what I actually wanted to know was this question.

Edit: The original question was only $C_n$ in the inequality and this was answered by Lee Moos before I could make the intended edit to the question

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    $\begingroup$ Since $w(b) \geq 1$ and $w(b+1/P)=0$ there is a point $c$ between $b$ and $b+1/P$ where $|w'(c)| \geq P$. Apply this argument to $w'$ between $c$ and $b+1/P$ to find $d$ where $|w''(d)| \geq P^2$ and then, by induction, $\|w^n\|_\infty \geq P^n$. But then the integral of $|w^{n+1}|$ between $b$ abd $b+1/P$ is at leat $P^n$. $\endgroup$
    – Giorgio Metafune
    Commented Oct 24, 2022 at 11:35
  • $\begingroup$ I get that $\|w^{(n)}\|_{\infty} \geq P^n$. But if this only happens in a very small neighbourhood of length $P^{-n}$ then the integral may still be bounded by $O(1)$? $\endgroup$
    – Johnny T.
    Commented Oct 24, 2022 at 13:01
  • $\begingroup$ It implies the same lower bound for the L1 norm of the next derivative. $\endgroup$
    – Giorgio Metafune
    Commented Oct 24, 2022 at 13:33
  • $\begingroup$ I'm struggling to see it. could you please elaborate? $\endgroup$
    – Johnny T.
    Commented Oct 24, 2022 at 14:08
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    $\begingroup$ You have $$P^n \leq |w^n (c_n)|=|w^n(c_n)-w^n (b+1/P)| \leq \int_{c_n}^{b+1/P} |w^{n+1}| \leq \int_{\mathbb R} |w^{n+1}|.$$ $\endgroup$
    – Giorgio Metafune
    Commented Oct 24, 2022 at 15:50

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