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How can I prove (preferrably without the use of any heavy theorems) the existence of a smooth function $\mu\!:\mathbb{R}\rightarrow\!\mathbb{R}$ with properties $\mu(0)\!>\!\varepsilon$, $\:\mu_{[2\varepsilon,\infty)}\!=\!0$, $\:-1\!<\!\mu'\!\leq\!0$?

I'm guessing bump functions should come in handy, but I don't know how to bound the derivative.

This is needed in the proof of Morse handle attachment theorem: Banyaga & Hurtubise, p.65.

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You should modify a smooth cutoff function: – Siminore Sep 6 '12 at 15:04
up vote 1 down vote accepted

The function $$f(x)=\begin{cases}e^{-\frac 1 {\sqrt x}}&x>0\\0&x\le 0\end{cases}$$ is smooth. Note that for $x>0$ we have $f'(x) = \frac12 x^{-\frac32}e^{-\frac1{\sqrt x}}$ and $f''(x) = \frac14x^{-3}(1-3\sqrt x) e^{-\frac1{\sqrt x}}$. Therefore $0\le f'(x)\le f'(\frac19)= \frac{27}{2e^3}$.

Now set $\mu(x) = a\cdot f(b\cdot(2\epsilon - x))$ with $a=\frac76e^3\epsilon>0$ and $b=\frac1{18\epsilon}>0$. We obtain

  • $\mu(x)=0$ for $x\ge 2\epsilon$
  • $\mu'(x)= -a b f'(b\cdot(2\epsilon - x))$ is $\le 0 $ and $\ge -ab\cdot \frac{27}{2e^3} = -\frac78>-1$.
  • $\mu(0)=a f(2b\epsilon) = \frac76e^3\epsilon f(\frac19)=\frac76\epsilon>\epsilon$
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@Eitzen: In general, for any $a\!>\!0$ and $r\!\in\!\mathbb{N}_0$, we have $$\lim_{x\downarrow0}\, (e^{-1/x^a})^{(r)}=0,$$ where $(r)$ is the $r$-th derivative, correct? Thank you for your help! – Leon Sep 10 '12 at 11:02
Yes, the reason is that (e.g. by induction) the derivatives are of the form $P_r(x^{-1},x^{-1-\alpha})e^{-1/x^\alpha}$ with a polynomial $P_r(X,Y)$ and the exponential is stronger than any polynomial – Hagen von Eitzen Sep 10 '12 at 11:41

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