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Let us consider a bump function $\phi: \mathbb{R} \longrightarrow \mathbb{R}$, smooth, with compact support. The most common examples are built from the function $$ \psi(x) = \begin{cases} \exp ( \frac {1}{x^2 - 1}) & \lvert x \rvert < 1 \\ 0 & \text{otherwise} \end{cases}.$$ Although this function behaves very well, its derivatives become arbitrarily large inside the unit ball. I wonder --- do there exist bump functions with "small" derivatives?

Stated more precisely, does there exist a bump function $\phi \in C^\infty_c(\mathbb{R})$ and some $M < \infty$ such that $\phi^{(k)} \leq M$ for all $k$th derivatives?

Intuitively, I feel the answer is no, as a sort of cost of vanishing entirely.

davidlowryduda
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    Here's a sketch that I think should work: Suppose $f(0) = 0$. The bound on the first derivative gives $|f(x)| \le Mx$. The bound on the second derivative gives that the first derivative is at most $Mx$, so that $|f(x)| \le \frac 1 2 M x^2$. The third derivative gives $\frac 1 {3!} M x^3$, and so on. Conclude that $f$ is identically zero. –  May 02 '16 at 19:57
  • @T.Bongers Yes, that does indeed work. Thank you. – davidlowryduda May 02 '16 at 20:00
  • You're very welcome. I've been looking for a duplicate, and can't find one. –  May 02 '16 at 20:01
  • Here – Daniel Fischer May 02 '16 at 20:34
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    This version of the question and zhw's answer are better than the linked duplicate. – Nick Alger Jul 11 '17 at 07:42
  • The result holds if by smooth you mean $C^k$ for positive integer k (not infinity). –  May 13 '20 at 00:30

1 Answers1

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If that bound held, then $\phi$ would be equal to its Taylor series everywhere (by considering the LaGrange form of remainder in Taylor's theorem), no matter where we center the series. In particular, we could center the series at a point where all derivatives are $0.$ That would give $\phi \equiv 0.$

zhw.
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