Construction of cut-off function I want to know the example of cut-off function. 
$\phi \in C^2 ([0,\infty))$ satisfies the followings : 
(1) $\phi(x) =1$ on $[0,r]$
(2) $\phi(x) =0$ on $ x > 2r$
(3) $- C r^{-1} \phi^{1/2}(x) \leq \phi ' (x) \leq 0$ on $r \leq x \leq 2r $
(4) $| \phi '' (x) | \leq C r^{-2}$ on $r \leq x \leq 2r$
Is there an explicit example about cut-off function ? 
Thank you in advance. 
 A: Let
$$f(x)=\begin{cases}\exp(-x^{-1})&x>0\\0&x\le 0\end{cases}$$
Then $f'(x)=x^{-2} f(x)$, $f''(x)=(-2x^{-3} + x^{-4}) f(x)$.
Note that $f$ and all its derivatives are continuous also at $x=0$ because the exponential dominates the powers of $x$. Also note that $f$ is strictly increasing from 0 to 1 on $[0,1]$. Define
$$\psi(x)=\frac{f(x)}{f(x)+f(1-x)}.$$
Note that


*

*$x>0$ or $1-x>0$, hene the denominator is positive

*$\psi(x)+\psi(1-x)=1$.

*$0\le \psi(x)\le 1$

*If $x\le0$ then $\psi(x)=0$ and if $x\ge1$ then $\psi(x)=1$

*$\phi$, $\phi'$, $\phi''$ are continuous. Since $\phi$ is constant on $(-\infty,0]\cup[1,\infty)$, they are also bounded.


For $0<x<1$, we calculate $\psi'(x)=\frac{f'(x)(f(x)+f(1-x))- f(x)(f'(x)-f'(1-x))}{(f(x)+f(1-x))^2}=\frac{f'(x)f(1-x)- f(x)f'(1-x)}{(f(x)+f(1-x))^2}$.
The numerator is $f'(x)f(1-x)- f(x)f'(1-x)=(x^{-2}+(1-x)^{-2})f(x)f(1-x)$. This is strictly positive, hence $\psi$ is strictly increasing on $[0,1]$. 
Finally for $x\ge0$ let $$\phi(x)=\psi^2\left(2-\frac xr\right).$$
Clearly, $\phi$ is continuous and conditions (1) and (2) are satisfied.
For $r<x<2r$, we have $$\tag{i}\phi'(x)=-\frac1r\psi\left(1-\frac xr\right)\psi'\left(1-\frac xr\right).$$ Since $\psi'$ is bounded, we see that $\phi'(x)\ge -\frac1rC\sqrt{\phi\left(1-\frac xr\right)}$ for some constant $C$ (independent of $r$). And of course, the factors $\psi$ and $\psi'$ in $(i)$ are nonnegative, hence $\phi'(x)\le 0$.
Moreover, $$\phi''(x) = \frac1{r^2} \psi'\left(1-\frac xr\right)^2+\frac1{r^2}\psi\left(1-\frac xr\right)\psi''\left(1-\frac xr\right)$$
and since $\phi$, $\phi'$, $\phi''$ are bounded, we have $|\psi''(x)|\le Cr^{-2}$ for some constant independent of $r$.
Remark: Our $\phi$ is not only $\in C^2$, but in fact in $C^\infty$ (but not in $C^\omega$)
