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The question is the same as in the title. More precisely, does there exist a nonzero smooth mapping $f : \mathbb{R} \to \mathbb{R}$ such that for all $k \in \mathbb{N}$, the $k$-th derivative $f^{(k)}$ satisfies $\sup_{t \in \mathbb{R}} \lvert f^{(k)}(t)\rvert \le 1$ (with $f^{(0)}$ being understood as $f$ itself), and $f$ itself is compactly supported?

I suspects the answer is affirmative, but I can not find a proof at the moment. Here is some elementary observations:

  • If we drop the requirement of compact support, then functions like $\sin(x)$ and $\cos(x)$, as well as suitable linear combinations of them, all satisfy this property. Perhaps one can construct examples out of these functions.

  • Perhaps one can mollify piecewise linear functions satisfying the bound condition for all derivatives except on a finite number of points.

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    $\begingroup$ If all derivatives are bounded by 1, the Taylor series converges. This makes the function analytic, which is inconsistent with compact support. $\endgroup$ – Michael Renardy Feb 10 at 18:52
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    $\begingroup$ @MichaelRenardy Convergence of the Taylor series by itself is insufficient; the series must converge to the function for it to be analytic. For example, $f(x) = e^{-\frac{1}{x^2}}$ has a Taylor series that converges in a neighborhood around each base point, but it is not analytic. $\endgroup$ – user44191 Feb 10 at 19:32
  • $\begingroup$ Yes, you are right. $\endgroup$ – Michael Renardy Feb 10 at 20:57
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    $\begingroup$ However, if all derivatives are bounded by 1, you can use the remainder estimate of Taylor's theorem to show that the Taylor series actually represents the function. $\endgroup$ – Michael Renardy Feb 10 at 21:02
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Just expanding the comment of @MichaelRenardy. Such a function does not exists. Indeed, by Taylor's theorem with the Lagrange form of the remainder, we get for every $a\in\mathbb{R}$ and $x\in\mathbb{R}$ that $$f(x)=f(a)+f'(a)(x-a)+\dots+\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(k+1)}(y)}{(k+1)!}(x-a)^{k+1}$$ for some $y$ between $a$ and $x$.

Now choose $a$ outside the support of $f$ so that all derivatives vanish. Then the bound on the $(k+1)$-st derivative implies that $|f(x)|\leq (x-a)^{k+1}/(k+1)!$ and it remains to let $k$ tend to infinity.

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