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 $f^{(0)}(x) = f(x)$.


1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$ Feb 27, 2013 at 21:53
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    $\begingroup$ @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. $\endgroup$
    – JSchlather
    Feb 27, 2013 at 22:03
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    $\begingroup$ @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.) $\endgroup$
    – awwalker
    Feb 28, 2013 at 4:39
  • $\begingroup$ 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! $\endgroup$ Feb 28, 2013 at 13:56

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