Existence of real analytic diffeomorphisms with prescribed values on a finite set My question is the following: given a finite set $ F = \{ x_1, \dots, x_k \} \subset \mathbb{R} $ such that $x_i < x_{i+1}$ for all $i = 1, \dots, k-1$ and numbers $a_1, \dots, a_k \in \mathbb{R}$ such that $ a_i < a_{i+1}$ for $i=1,\dots,k-1 $, is it possible to find a real analytic diffeomorphism $f: \mathbb{R} \to \mathbb{R}$ such that $f(x_i) = a_i$?
I know that it's easy to find real analytic functions satisfying these conditions (for instance, using polynomial interpolation), but is there a  way to get at least one that is a diffeomorphism (i.e. with $f'>0$ and surjective)? Thanks!
 A: Yes, this is possible.
Here is a proof. I omitted some of the arguments which are standard. Let me know if you need help reconstructing these.
In what follows, $C^i(M,N)$ is the space of $C^i$-smooth maps between $C^i$-smooth manifolds with topology of $C^i$-convergence on compacts.
Lemma 1. The subgroup of diffeomorphisms $Diff(S^1)$ is open in $C^1(S^1,S^1)$. (The same holds with $S^1$ replaced by a general compact manifold.)
Lemma 2. Let $z_0,...,z_n\in {\mathbb C}$ be distinct points and $w_0,...,w_n\in {\mathbb C}$ be arbitrary points. Then there exists a complex polynomial $p$ of degree $\le n$ such that $p(z_k)=w_k, k=0,...,n$. Moreover, such polynomial is unique and its coefficients depend continuously on $(w_0,...,w_n)$.
Lemma 3. (A form of Weierstrass approximation theorem, see e.g answers to this question.) $C^\omega(S^1, {\mathbb R}^2)$ is dense (in $C^1$-topology, of course)  in $C^1(S^1, {\mathbb R}^2)$.
Lemma 4. Suppose we are given two finite ordered $n$-element subsets
$A=\{a_0,...,a_n\}$ and $B=\{b_0,...,b_n\}$ in $S^1$ with the same cyclic order. Then there exists a $C^1$-smooth (even infinitely smooth) diffeomorphism $h: S^1\to S^1$ sending $a_k$ to $b_k$, $k=0,...,n$.
Lemma 5. In the setting of Lemma 4, there exists a real-analytic diffeomorphism $g: S^1\to S^1$ such that $g|_A= h|_A$.
Proof. Using Lemma 3, construct a sequence $(h_m)$ in $C^\omega(S^1, {\mathbb R}^2)$ which $C^1$-converges to $h$, where $h$ is as in Lemma 4. In particular, for each $k$, $(b_k- h_m(a_k))\to 0$ as $m\to \infty$. Using Lemma 2, for each $m$, find a polynomial $p_m$ of degree $\le n$ such that $p_m(a_k)= b_k- h_m(a_k)$ for each $k$. In particular, the sequence $(p_m)$, regarded as maps in $C^1(S^1, {\mathbb C})$, converges to $0$ as $m\to\infty$.  Set $f_m:= h_m + p_m$. Then for all $m$,
$$
f_m|_A= h|_A,
$$
each $f_m$ is real-analytic, and $f_m\to h$ in $C^1(S^1,  {\mathbb C})$. Lastly, set
$$
g_m:= \frac{f_m}{|f_m|}, m\in {\mathbb N}. 
$$
Again, $g_m$ is real-analytic, $g_m|_A= h|_A$  and $g_m\to h$ in $C^1(S^1,S^1)$. Lemma 1 implies that each $g_m$ is a diffeomorphism $S^1\to S^1$ for all large $m$. qed
Lastly:
Theorem 1. Suppose that $x_1< x_2< ...< x_{n}$ and $y_1<y_2< ... < y_{n}$ are real numbers. Then there exists a real-analytic diffeomorphism $f: {\mathbb R}\to  {\mathbb R}$ such that $f(x_k)=y_k, k=1,...,n$.
Proof.  Consider the stereographic projection $q: S^1\to {\mathbb R}\cup \{\infty\}$; it is real-analytic, actually, linear-fractional. Set $a_k:=q^{-1}(x_k), b_k:= q^{-1}(y_k), k=1,...,n$. Set $a_0=b_0=q^{-1}(\infty)$, the north pole of the unit circle. Then the
tuples $(a_0,...,a_n), (b_0,...,b_n)$ appear in the same cyclic order on $S^1$. By Lemma 5, there is a real-analytic diffeomorphism
$g: S^1\to S^1$ such that $g(a_k)=b_k, k=0,...n$. In particular, $g(a_0)=a_0=b_0$.
Lastly, set
$$
f:= q\circ g \circ q^{-1}.
$$
This map, when restricted to the real line, is real-analytic and satisfies $f(x_k)=y_k, k=1,...,n$.  qed
With more work, one can prove (using similar ideas):
Theorem 2. Suppose that $M$ is a compact connected real-analytic manifold of dimension $\ge 2$. Then the group of real-analytic self-diffeomorphisms of $M$ is $n$-transitive for every $n<\infty$, i,e. given any finite subsets of distinct points $A=\{a_1,...,a_n\}$ and $B=\{b_1,...,b_n\}$ in $M$, there is a real-analytic diffeomorphism $f: M\to M$ such that  $f(a_k)=b_k, k=1,...,n$.
A: Consider the cone $\mathcal{C}$ of functions of the form
$$\phi(t)=\sum_{i=1}^N \alpha_i \exp \left(-\frac{(t-\mu_i)^2}{2 \sigma_i^2}\right) $$
where $\alpha_i, \sigma_i > 0$, and $\mu_i \in \mathbb{R}$.
Let $x_1 \cdots < x_k$ be fixed real numbers, $\delta_1$, $\ldots$, $\delta_{k-1}$ be positive numbers. Let us show that there exists a functions $\phi$ in $\mathcal{C}$ such that
$$\int_{x_{k}}^{x_{k+1}} \phi(t) dt = \delta_k$$
Indeed, for every $i$, $1\le l \le k-1$, consider a function $\phi_i$ of the form $\phi_i = \frac{1}{\sqrt{2\pi} \sigma_i} \exp \left(-\frac{(x-\beta_i)^2}{2 \sigma_i^2}\right)$ ( a Gaussian function), such that $\int_{I_i} \phi_i \simeq 1$, and $\int_{I_j}\phi_i \simeq 0$ for $j\ne l$. (indeed, just take $\mu_i\in (x_i, x_{i+1})$, and $\sigma_i>0$ small enough).
If the above approximations are good enough, we will find a positive combination  of the $\phi_l$ that is $\phi(t) = \sum_{i=1}^{k-1} c_i \phi_i$ ( the $c_i$ will be approximately $\delta_i$) such that
$$\int_{I_i} \phi= \delta_i$$
( basic linear algebra)
Now, we are almost done. Say we want $f(x_i)= y_i$, $i=1, \ldots, k$.  Take first $\epsilon>0$ small enough such that
$$\delta_i \colon = (y_{i+1}-y_i) - \epsilon (x_{i+1}-x_i) > 0$$
for $i=1, \ldots, k-1$. Now consider $\phi$ a positive linear combination of $k-1$ Gaussians such that
$$\int_{x_i}^{x_{i+1}}\phi(t) d t = \delta_i$$
for all $1\le i \le k-1$. We have therefore
$$\int_{x_i}^{x_{i+1}} (\epsilon + \phi(t)) dt = y_{i+1} - y_i$$
Let
$$f(x) = c +\epsilon x +  \int_{0}^t \phi(t) dt$$
where $c$ is a constant.
Note that we have $f(x_{i+1})-f(x_i) = y_{i+1}-y_i$, for all $1\le i \le k-1$. Now choose $c$ such that $f(x_1) = y_1$. We conclude that $f(x_i) = y_i$, for all $1\le i \le k$. We have  $f'(x)>0$ for all $x$, $f$ is analytic, $\lim_{x\to -\infty} f(x) = -\infty$, $\lim_{x\to +\infty} f(x) = +\infty$.  Therefore, $f$ is an analytic diffeomorphism of $\mathbb{R}$.
$\bf{Added:}$ We can actually make the function $f$ a polynomial.  We only need to get some positive polynomials $\phi_i$ with integral $\simeq 1$ in $(x_i, x_{i+1})$ and $0$ on all $(x_j, x_{j+1})$ for $j\ne i$.  Indeed, consider $P_i$ a polynomial that approximate an almost stair continuous function that is $0$ on the $(x_j, x_{j+1})$, and a constant on $(x_i, x_{i+1})$. Then take $\phi_i = \alpha_i P_i^2$. Then a positive combination
$$\phi= \sum_{i=1}^{k-1} c_i \phi_i$$
satisfies
$$\int_{x_i}^{x_{i+1}} \phi(t) dt = y_{i+1} - y_i- \epsilon(x_{i+1}-x_i)$$
Take now
$$f(x) = c+ \epsilon x + \int_0^x \phi(t) dt$$
a polynomial, and a diffeomorphism of $\mathbb{R}$.
$\bf{Added:}$ The first approach can be understood better if we pose the problem: given points $-\infty = x_0 < x_1 < \ldots < x_k< x_{k+1} = + \infty$, and $m_0$, $\ldots$, $m_k > 0$, $\sum_{i=0}^k m_i=1$ does there exist a probability measure with density a convex combination of Gaussians such that the measure of the interval $I_k=(x_k, x_{k+1})$ is $m_k$.  The intuition is: the density $\phi$ is a convex combination of Gaussians $\phi_i$, each with a high spike on the interval $(x_i, x_{i+1})$, the coefficient is approximately $m_i$.  Now the cumulative distribution function will be an analytic diffeo from $\mathbb{R}$ to $(0,1)$ with prescribed values at $x_i$.
