Diffeomorphism of open intervals in $\mathbb{R}$ with specified values I know two open intervals on $\mathbb{R}$ are diffeomorphic to each other.  My question is if I have a intervals $(a-\varepsilon,b+\varepsilon)$ and $(c-\delta,d+\delta)$, is there a diffeomorphism $\phi:(a-\varepsilon,b+\varepsilon)\to(c-\delta,d+\delta)$ so that $\phi\in C^{(r)}$ (given an $r>0$) with the property that $\phi(a)=c$ and $\phi(b)=d$?
 A: Claim: for any finite sequences $x_1<x_2<\dots<x_n$ and $y_1<y_2<\dots,y_n$ there is a $C^\infty$-diffeomorphism $f:\mathbb R\to\mathbb R$ such that $f(x_j)=y_j$ for all $j$. 
(Your statement follows by considering $a-\varepsilon, a, b, b+\varepsilon$ and $c-\delta,c,d,d+\delta$.)
Proof of the claim: Let $$\delta=\frac{1}{3}\min_{1\le j\le n-1}(x_{j+1}-x_{j})$$ Let $g:\mathbb R\to \mathbb R$ be a continuous piecewise affine function with the following properties


*

*strictly increasing 

*affine on each interval $[x_j-\delta,x_j+\delta]$ 

*$g(x_j)=y_j$ for all $j$


Such a function is easy to find. Next, let $\psi:\mathbb R\to [0,\infty)$ is a function which is 


*

*$C^\infty$-smooth 

*even, that is, $\psi(-x)=\psi(x)$

*supported on $[-\delta,\delta]$ 

*satisfies $\int_{\mathbb R} \psi=1$


Such a mollifier is also easy to find. Finally, let $f=g*\psi$ and observe that 


*

*$f$ is $C^\infty$-smooth

*$f'=g'*\varphi$ is strictly positive

*$f(x_j)=y_j$, because $g$ is affine in $\delta$-neighborhood of $x_j$, and the linear term cancels out due to the symmetry of $\psi$.

