I want to construct two $C^2$-maps $f,g:[0,3]\rightarrow[0,1]$ with the following properties:
1. $f(x)+g(x)=1$ for all $x\in[0,3]$
2. $f$ has support included in $[0,2)$ and $g$ has support in $(1,3]$
What I thought:
The support of a map $\phi$ is $\overline{\{x\in[0,3]:\phi(x)\neq 0\}}$. Thus $f$ must be zero for $x\geq2-\epsilon$ and $g$ must be zero for $x\leq1+\delta$ for $\epsilon,\delta>0$.
My best guess was to define two piecewise linear functions $f$ and $g$ that are constantly $1$ at $x\leq 1+\delta$ and $x\geq 2-\epsilon$ respectively. However these are not twice differentiable at the borderpoints $x=2-\epsilon$ and $1+\delta$.
Is there a better example? If not, how do I make these piecewise linear maps $C^2$?
You can take: if $x\in [1+\epsilon,2-\epsilon]$, $g(x)= \frac{\int_{1+\epsilon}^x(t-1-\epsilon)^2(2-\epsilon-t)^2dt}{\int_{1+\epsilon}^{2-\epsilon}(t-1-\epsilon)^2(2-\epsilon-t)^2dt}$,
and, if $x \leq 1+\epsilon$, $g(x)=0$
and, if $x \geq 2-\epsilon$, $g(x)=1$
Then, we take: $f(x)=1-g(x)$ for all $x \in [0,3]$.
Then $f(x)+g(x)=1$ for all $x$.
Let $A=\int_{1+\epsilon}^{2-\epsilon}(t-1-\epsilon)^2(2-\epsilon-t)^2dt$
$A>0$
For $x \in ]1+ \epsilon,2- \epsilon[$, $g'(x)=\frac{(x-1- \epsilon)^2(2-\epsilon-x)^2}{A}$.
So, $\lim_{x \mapsto (1+ \epsilon)^+} g'(x)=0$.
$g''(x)=\frac{1}{A}(2(x-1-\epsilon)(2- \epsilon-x)^2-2(x-1-\epsilon)^2(2-\epsilon-x))$
So $\lim_{x \mapsto (1+ \epsilon)^+} g''(x)=0$
And: $\lim_{x \mapsto (1+ \epsilon)^-} g'(x)=0$
$\lim_{x \mapsto (1+ \epsilon)^-} g''(x)=0$
So, $g$ is $C^2$ at $x=1+ \epsilon$.