This question already has an answer here:

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.


marked as duplicate by davidlowryduda May 2 '16 at 20:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ 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. $\endgroup$ – user296602 May 2 '16 at 19:57
  • $\begingroup$ @T.Bongers Yes, that does indeed work. Thank you. $\endgroup$ – davidlowryduda May 2 '16 at 20:00
  • $\begingroup$ You're very welcome. I've been looking for a duplicate, and can't find one. $\endgroup$ – user296602 May 2 '16 at 20:01
  • $\begingroup$ Here $\endgroup$ – Daniel Fischer May 2 '16 at 20:34
  • $\begingroup$ This version of the question and zhw's answer are better than the linked duplicate. $\endgroup$ – Nick Alger Jul 11 '17 at 7:42

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 $\phi \equiv 0.$

  • $\begingroup$ Ah yes, in fact the Taylor series error would necessarily vanish. Of course! $\endgroup$ – davidlowryduda May 2 '16 at 20:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.