# A smooth nonzero function $\mathbb R\to\mathbb R$ with uniformly bounded derivatives tending to zero at infinity?

$\def\scrBp{\mathscr B\boldsymbol.}\def\rD{{\rm D}\kern.4mm}\def\ssp{\kern.4mm}\def\bbR{\mathbb R}$For a certain purpose I invented the Banach space $\scrBp^\infty(\ssp\bbR\ssp)$ defined as follows. The vectors are all smooth functions $x:\bbR\to\bbR$ such that there is $M\in\bbR^+$ with $|\,\rD^kx(t)\,|\le M$ and $\lim_{\,s\to\pm\infty\,}|\,\rD^kx(s)\,|=0$ for all $k\in\mathbb N_0$ and $t\in\bbR$ . The norm of $x$ is the infimum of the set of all these $M$ . Now I am wondering whether this space has any nonzero vectors. Hence the question.

• I doubt so, since neither $e^{-x^2}$-like functions nor functions from $C_c^\infty$ will be in there (the uniform constant breaks these) as far as I can tell. – AlexR Nov 6 '14 at 15:58
• Can a continuous and differentiable function approach some finite limit as $x\to\infty$, without its derivative approaching $0$? – Nishant Nov 6 '14 at 16:13
• @Nishant An example is $t\mapsto t^{-1}\,\sin(t^2)$ . – The-unKnowN Nov 6 '14 at 16:18
• A smooth function with uniformly bounded derivatives is real-analytic, and its Taylor series converges on all of $\mathbb{C}$, so it would be the restriction of an entire function to the real axis. I'm almost sure that means your space is $\{0\}$, but I don't see a proof yet. – Daniel Fischer Nov 6 '14 at 16:18
• In view of Daniel Fischer's comment above, I added the "complex-analysis" tag. – The-unKnowN Nov 8 '14 at 5:04

Choose some $\varphi\in C_{c}^{\infty}\left(\left(-\frac{1}{2\pi},\frac{1}{2\pi}\right)\right)$ and let $f:=\mathcal{F}\varphi=\widehat{\varphi}$ be the Fourier transform of $\varphi$, i.e. $$f\left(\xi\right)=\int_{\mathbb{R}}\varphi\left(x\right)\cdot e^{-2\pi ix\xi}\,{\rm d}x.$$ By the usual properties of the Fourier transform, we conclude that $f\in\mathcal{S}\left(\mathbb{R}\right)$ is a Schwartz-function, which means $$C_{m,n}:=\sup_{x\in\mathbb{R}}\left|x^{m}\cdot\partial^{n}f\left(x\right)\right|<\infty$$ for all $m,n\in\mathbb{N}_{0}$. In particular, this implies (for $\left|x\right|\geq1$): $$\left|\partial^{n}f\left(x\right)\right|=\left|\frac{x\cdot\partial^{n}f\left(x\right)}{x}\right|\leq\frac{C_{1,n}}{\left|x\right|}\xrightarrow[x\to\pm\infty]{}0$$ so that $f$ fulfills the second one of your requirements.
By choosing (e.g.) $\varphi\geq0$ and $\varphi\not\equiv0$, we can also ensure that $f\left(0\right)=\int_{\mathbb{R}}\varphi\left(x\right)\,{\rm d}x>0$, i.e. $f\not\equiv0$ (alternatively, we have $\varphi=\mathcal{F}^{-1}f$, so that $f\not\equiv0$ holds as soon as $\varphi\not\equiv0$ is true).
Finally, we can differentiate "under the integral sign" to get \begin{eqnarray*} \left|\partial^{n}f\left(\xi\right)\right| & = & \left|\int_{\mathbb{R}}\varphi\left(x\right)\cdot\frac{{\rm d}^{n}}{{\rm d}\xi^{n}}e^{-2\pi ix\xi}\,{\rm d}x\right|\\ & = & \left|\int_{\mathbb{R}}\varphi\left(x\right)\cdot\left(-2\pi ix\right)^{n}\cdot e^{-2\pi ix\xi}\,{\rm d}x\right|\\ & \leq & \int_{\mathbb{R}}\left|\varphi\left(x\right)\right|\cdot\left|2\pi x\right|^{n}\,{\rm d}x\\ & \overset{{\rm supp}\left(\varphi\right)\subset\left(-\frac{1}{2\pi},\frac{1}{2\pi}\right)}{\leq} & \int_{-\frac{1}{2\pi}}^{\frac{1}{2\pi}}\left|\varphi\left(x\right)\right|\cdot\left|2\pi x\right|^{n}\,{\rm d}x\\ & \leq & \int_{-\frac{1}{2\pi}}^{\frac{1}{2\pi}}\left|\varphi\left(x\right)\right|\,{\rm d}x\leq\left\Vert \varphi\right\Vert _{L^{1}}, \end{eqnarray*} so that the derivatives of $f$ are also uniformly bounded, which is the first of your desired properties.