Construct a diffeomorphism that extends the identity map

Question

Is there a diffeomorphism $f:[1,+ \infty) \rightarrow [1,3)$, such that $f$ restricted on $[1,2]$ is the identity map? I think its true, but I don't know how to construct one.

Answer 1

If you want a smooth diffeomorphism you're going to have to get non-analytic functions involved (since the series expansion of $f$ about any point in $[1,2]$ will be the identity function, which is not $f$). Fear not - we can still find an explicit solution.

The first port of call is the famous $$\eta(x) = \cases{ e^{-1/x} & if $x>0$ \\ 0 & if $x \le 0$,}$$ which is a smooth function (check this!) that is increasing on $[0,\infty)$ and converges to $1$. Using this to smoothly interpolate between the identity and the constant 3, we get a smooth function

$$ f(x) = x(1 - \eta(x-2)) + 3\eta(x - 2) = x + (3-x)\eta(x-2).$$

You can check that

$$f'(x) = 1 - \eta(x-2)\frac{x^2 - 3x + 1}{x^2 - 4x + 4}>0.$$

Answer 2

It suffices to find an increasing diffeomorphism $g : (2, \infty) \to (2,3)$ such that $\lim \limits _{ \begin {array} {c} x \to 2 \\ x>2 \end{array} } g'(x) = 1$ (this last condition will allow you to glue $g$ to the identity while maintaining a continuous derivative of the resulting function). There aren't many functions with a simple explicit description that squeeze an interval of infinite length into one of finite length; the usual suspects are $\arctan, \Bbb e^{-x}, \frac {2x} {1+x^2}$... You'll also have to compose with a rescaling and a translation in order to land in $(2,3)$. The choice $g(x) = 3 - \Bbb e ^{2-x}$ does the job.

Therefore, $f(x) = \left\{ \begin{array} {ll} x, && x \in [1,2] \\ 3 - \Bbb e ^{2-x}, && x \in (2, \infty) \end{array} \right.$ is the desired diffeomorphism. (Note that it is not twice-derivable, though, i.e. it is a diffeomorphism as in real analysis, but not as in differential geometry.)

Answer 3

Define $$f(x) = \cases{ x,x\in [1,2] \\ x+xe^{-1/(x-2)},\,x>2.}$$ Then $f\in C^\infty([1,\infty))$ and strictly increases to $\infty$ as $x\to \infty.$ So $f$ is a diffeomorphism with the desired property.