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.

The existence of bump functions is one of the more significant differences between smooth and analytic/real analytic functions.

• All right. I'll trust you and Wikipedia on the existence of this bump function. Can you name a course in which I can learn these things in detail? Thanks! Feb 27, 2013 at 21:53
• @TuringMachine Bump functions pop up in lots of places. I first learned up about them from a course in smooth manifolds. They also come up lots in PDE and most advanced analysis courses. Feb 27, 2013 at 22:03
• @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
• Hm, I see. Looks like one of those things which you practice for some time and it then becomes "second nature". I'll look more into that some day. Thanks for the help! Feb 28, 2013 at 13:56