Existence of a certain smooth function $\Bbb R\to \Bbb R$ whose derivative takes values in $(-1,0]$

Question

I am reading Morse Theory of Milnor, and in the proof of Theorem 3.2, Milnor says let $\mu:\Bbb R\to \Bbb R$ be a smooth function satisfying $\mu(0)>\epsilon$, $\mu(r)=0$ for $r\geq 2\epsilon$, and $\mu'(r) \in (-1,0]$ for all $r\in \Bbb R$. Here $\epsilon>0$ is a fixed constant. Certainly the first two conditions can be satisfied taking a suitable bump function near $0$, but I can't see how to take $\mu$ such that the third condition holds.

Answer 1

The third is also satisfied by 'taking a suitable bump function', so naturally what you are asking for is details. First the idea: you have $2\epsilon$ space. You need to drop down by some amount $h>\epsilon$ at a speed less than 1. Since this is achievable by straight lines and with room to spare, it should be possible if you don't choose $h$ too big ($h>2\epsilon$.)

Fix $\epsilon$. Consider

$$T(x) = \min(1,\max(1-x,0)).$$ its graph is like ‾\_ where the right angled triangle sits on $[0,1]$. $T(0)=1$. $T'$ is $- 1$ or $0$ except at 2 points. Then setting $$ T_\epsilon(x) = \frac{3\epsilon}2 T\left(\frac{4x}{7\epsilon}\right)$$ we have $T_\epsilon(0)=\frac{3\epsilon}2>\epsilon$ and $\|T_\epsilon'\|_{L^\infty}\le \frac32 \frac47 = \frac67< 1,$ with the correct sign. Further, $T_\epsilon|_{x>7\epsilon/4}\equiv 0$ so $T_\epsilon|_{x>2\epsilon} \equiv 0$. The only issue is that $T_\epsilon$ isn't smooth. But mollifying with a sufficiently small mollifier (a mollifier supported on $[-\epsilon/4,\epsilon/4]$ makes it begin to be zero exactly at $2\epsilon$), or using a partition of unity, the desired smooth function exists.

PS a similar idea works if you wanted a radial function.