Does there exist a bounded function of class $C^{\infty}$ such that all of its derivatives at $0$ coincide with the derivatives of $e^{-x} +x+1$?

Consider the function $g:\mathbb{R}\to\mathbb{R}$ where $g(x) = e^{-x} + x + 1$.

Does there exist a bounded function $f:\mathbb{R}\to\mathbb{R}$ of class $C^{\infty}$ such that $f^{(n)}(0) = g^{(n)}(0)$ for all $n = 0, 1, 2, \ldots$?

Here, $f^{(n)}(x)$ denotes the nth derivative of $f$ at $x$, and $f^{(0)}(x) = f(x)$.

• Are you familiar with partitions of unity? (en.wikipedia.org/wiki/Partition_of_unity) Feb 27, 2013 at 20:00
• @Blumenthal Type this [this]${}$(en.wikipedia.org/wiki/Partition_of_unity) in order to get this Feb 27, 2013 at 20:02
• @ABlumenthal Nope. I've never seen that before. Does that offer an immediate answer to this? Feb 27, 2013 at 20:04
• It should be as simple as adding smooth bump functions to $e^{-x}+x+1$. Feb 27, 2013 at 20:36
• @JSchlather Is that a simple thing to do? Is the Whitney extension theorem an overkill? Feb 27, 2013 at 20:38

Let $f : \mathbb{R} \to \mathbb{R}$ be a smooth function that is identically $1$ on a neighborhood of $0$, and has compact support. Look into bump functions for more on this. Then $fg = g$ on a neighborhood of $0$, so that $(fg)^{(n)}=g^{(n)}$ as desired. But $fg$ has compact support, so it is bounded.
• @TuringMachine It's easy (at least in $\mathbb{R}$) to get bump functions. Let $f(x)=e^{-1/x^2}$ for $x >0$ and $f=0$ for $x\leq 0$. Claim: $f$ is smooth. Now look at $h(x)=f(x)f(1-x)$, which is smooth and supported in $[0,1]$. Now integrate to get a smooth CDF, which equals $0$ on $x<0$ and constant for $x>1$. Normalize to get $h(x)=1$ for $x>1$. Let $j=h(x)h(3-x)$. This is smooth, supported in $[0,3]$, and satisfies $j=1$ on $(1,2)$. Hope this helps! (For $\mathbb{R}^n$, consider rotating $j$ about an axis.) Feb 28, 2013 at 4:39