Constructing a $C^2$ function

Question

I would like to find a $C^2$ function $g:\mathbb{R}\rightarrow[0,1]$ such that

$g|_{[-1,1]}=1$ and $g(x)=0$ for $|x|>2.$

I know that being $C^2$ means that the second derivative of $g$ must exist and be continuous, but I find it hard to find one that satisfies the condition. I think a polynomial wouldn't work here but something with a cosine might, although I'm not sure how.

http://math.stackexchange.com/questions/197431/construction-of-cut-off-function — Tsemo Aristide, Sep 11 '16 at 15:36

Robert Z · Accepted Answer · 2016-09-11T16:17:41.743

0

Consider the polynomial $$P(x)=(6x^2+3x+1)(1-x)^3$$ then $P([0,1])=[0,1]$, $P(0)=1$, $P(1)=0$, $P'(0)=P''(0)=P'(1)=P''(1)=0$. Then a function $g$ which satisfies your conditions is $$g(x) = \left\{\begin{array}{ll} 1 & \text{for } x\in[-1,1]\\ P(|x|-1) & \text{for } x\in[-2,-1]\cup[1,2]\\ 0 & \text{otherwise} \end{array}\right. $$

edited Sep 11 '16 at 16:17

answered Sep 11 '16 at 15:53

Robert Z

145,942

Can I make $g$ a piecewise function, with $g=P$ for $-2<x<-1$ and $1<x<2$? – Harmonica Sep 11 '16 at 15:57
Yes but you have to fix the argument. See my edited answer. – Robert Z Sep 11 '16 at 16:06
Why do you have to fix the argument? For the continuity? – Harmonica Sep 11 '16 at 16:08
Because $x-1$ send $[1,2]$ in $[0,1]$. Moreover the graph of $P(|x|)$ symmetric with respect to $y$ axis. – Robert Z Sep 11 '16 at 16:15
How do I know that $g$ is $C^2$? – Harmonica Sep 14 '16 at 13:37
@Harmonica Because the limit of $g''(x)$ at $x=1,-1,2,-2$ from the left and frome the right is equal (it is 0). – Robert Z Sep 14 '16 at 13:41

score 0 · Answer 2 · answered Sep 11 '16 at 16:49

Define

$$f(t) = 2\sin^2(\pi t)\cdot\chi_{[-2,-1]}(t) - 2\sin^2(\pi t)\cdot\chi_{[1,2]}(t).$$

Then $f\in C^1(\mathbb R),$ $f$ is odd, and $\int_{-2}^{-1}f(t)\,dt = 1.$ Now set

$$g(x) = \int_{-\infty}^x f(t)\, dt.$$

Then $g$ satisfies the requirements.

Constructing a $C^2$ function

2 Answers2