# Does there exist a compactly supported smooth function on $\mathbb{R}$ with all derivatives bounded by 1? [closed]

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.

• If all derivatives are bounded by 1, the Taylor series converges. This makes the function analytic, which is inconsistent with compact support. – Michael Renardy Feb 10 at 18:52
• @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. – user44191 Feb 10 at 19:32
• Yes, you are right. – Michael Renardy Feb 10 at 20:57
• 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. – Michael Renardy Feb 10 at 21:02

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.