# Smooth function with equibounded family of derivatives

By $\mathcal{C}^{\infty}(\mathbb{R})$ we denote the space of smooth functions $\mathbb{R}\rightarrow \mathbb{R}$. Also, by $\mathrm{supp}(f)$ we denote the closure $\mathrm{Cl}(f^{-1}(\mathbb{R}_{\ne 0}))$. I'm trying to build a function $f\in \mathcal{C}^{\infty}(\mathbb{R})$ such that the family of higher derivatives $\{f^{(k)}: k\in \mathbb{N}\}$ is equibounded, i.e. there exists $c>0$ such that for each $k$ the image $f^{(k)}(\mathbb{R})\subset (-c,c)$. Furthermore, I want $\mathrm{supp}(f)$ to be a compact set. Is it possible?

Thus I want to give an important example in functional analysis. I'll be grateful for any hints!

I know the example of a smooth function $g$ such that $\mathrm{supp}(f)=[-1,1]$ but the family of derivatives is not equibounded! (Really, let $f(x)=e^{-\frac{1}{x^2}}$ for $x>0$ and $f(x)=0$ for $x\le 0$. Let $g(x)=f(1-x^2)$. Well, $g\in \mathcal{C}^{\infty}(\mathbb{R})$ and $\mathrm{supp}(g)=[-1,1]$).

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Well, there's $f \equiv 0$. I think that's the only one. – Daniel Fischer Oct 25 '13 at 17:24
Of course, I meant $f\ne 0$ – user74574 Oct 25 '13 at 17:24

A nonzero function in $C_c^\infty(\mathbb{R})$ cannot have equibounded derivatives. Without loss of generality, suppose $f$ has support in $[-1,1]$. Suppose you have $\lvert f^{(k)}(t)\rvert \leqslant C$ for some $k > 0$.

Then you have

$$\lvert f^{(k-1)}(t)\rvert = \left\lvert \int_{-1}^t f^{(k)}(s)\,ds\right\rvert \leqslant C\cdot (t+1)$$

for $-1 \leqslant t \leqslant 0$, and similarly $\lvert f^{(k-1)}(t)\rvert \leqslant C(1-t)$ for $0\leqslant t \leqslant 1$. And further

$$\lvert f^{(k-2)}(t)\rvert \leqslant \int_{-1}^t C(s+1)\,ds \leqslant \frac{C}{2}(t+1)^2$$

for $-1 \leqslant t \leqslant 0$, continuing, you get

$$\lvert f^{(k-m)}(t)\rvert \leqslant \frac{C}{m!}(t+1)^m,$$

and similarly for $0 \leqslant t \leqslant 1$. Altogether,

$$\lvert f(t)\rvert \leqslant \frac{C}{k!}$$

for all $t$. If the $f^{(k)}$ are equibounded, that estimate forces $f \equiv 0$ by letting $k\to\infty$.

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