Assume $f(x)\in C^{\infty}[0,1]$, $f(1)=1$. For an integer $n$, $f^{(k)}(0)=0,\,\forall 0\le k<n$. Define $\|f^{(k)}\|_\infty=\sup\limits_{x\in[0,1]}|f^{(k)}(x)|$, then there exists a constant M unrelated to both f and n, such that $\sum\limits_{k=1}^n\frac{1}{\sqrt[k]{\|f^{(k)}\|_\infty}}<M$.
I met this inequality in a book without proof (it's called "Borel inequality" in the book), I guess that for any $n$, the sum is controlled by the case $f(x)=x^n$, thus we could take $M=e$, but I'm unable to prove it. Could anyone help me?
EDIT
By some computation, my conjecture over dominance of $x^n$ is wrong.
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iSSin
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1In this question I asked about why the derivatives of a smooth bump function grow so fast, maybe in the answers you could find some ideas. – Joako Nov 02 '24 at 03:09
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@Joako Thanks for your idea, yet this question might require more precise estimation. – iSSin Nov 02 '24 at 04:43