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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.

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Why are you interested in this question? Do you want an argument that such a function exists, or do you want an explicit formula? Also, is there a typo in condition (3) for the interval $-2r < x < -r$? On that interval, there must be points where $\phi'(x) > 0$. –  Sam Lisi Sep 16 '12 at 9:29
Thank you for your indication. I want to know the existence. Surely, I have no interest in the existence. But when I read a book, for instance, Lecture on geometric analysis written by P. Li, I can find cut-off argument. But I donot ensure the correctness of cut-off argument. –  Hee Kwon Lee Sep 16 '12 at 9:36
Can you please verify what you mean instead of (3)? As I commented, it is impossible to satisfy it on the region where $2r < x < r$. The tricky part of proving that a cut-off function with these properties exists comes from the differential inequalities (3) and (4). One way to start is to see what functions are extreme cases of (3) and (4). This gives an idea of the type of behaviour the function must have and where the constraints have a chance to be problematic. Have you tried this? Let us know what you have tried to do. –  Sam Lisi Sep 16 '12 at 9:51

1 Answer 1

up vote 4 down vote accepted

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 $C^\omega$)

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