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!

  • 2
    $\begingroup$ This is a very interesting question! $\endgroup$
    – Nick Alger
    Oct 22, 2021 at 0:34
  • $\begingroup$ I agree with Nick Agler. This is an interesting problem. I though about it and you can reduce it to the case $a_i = x_i$ for all $i=1, \ldots , k-2$ using compositions of analytic diffeomorphisms $\endgroup$ Oct 27, 2021 at 22:27

2 Answers 2


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


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 $f(a_k)=b_k, k=1,...,n$.

  • $\begingroup$ Excellent, thank you for the detailed answer! $\endgroup$
    – Mauro
    Oct 30, 2021 at 17:47
  • $\begingroup$ Cool. For lemma 4, you can build a continuous piecewise linear function along the circle, where the function takes value $b_i$ at the points $a_i$, and the nodes of the function are halfway in-between the $a_i$. Then mollify it by convolving with a small bump function, to turn it into a smooth diffeomorphosm. (Perhaps this is standard, but it took me some thought to figure out) $\endgroup$
    – Nick Alger
    Oct 30, 2021 at 19:43
  • $\begingroup$ @NickAlger: Right. $\endgroup$ Oct 30, 2021 at 20:29

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 $x_i$.

  • $\begingroup$ Could you explain why does your basic linear algebra (the 3rd paragraph) argument works? Why would you get positive constants $c_i$? Also, what happens to your solution if you have a sequence where one of the $\delta_i$'s converges to zero? It appears that the limit would be an analytic function $\phi$ vanishing on the interval $I_i$, which is impossible. $\endgroup$ Oct 31, 2021 at 11:18
  • 1
    $\begingroup$ @Moishe Kohan: The matrix of the linear system is close to the identity, so invertible. It is not dependent on $\delta_i$. Now, the solution is $A^{-1}\cdot \delta$. Recall that $\delta$ is a vector with entries $>0$. The solution is continuous in $A$. For $A$ close enough to $I$. it is close enough to $\delta$, and so positive. $\endgroup$
    – orangeskid
    Oct 31, 2021 at 11:55
  • $\begingroup$ @MoisheKohan Suppose you got a negative constant $c_i$. Then for sufficiently small $\sigma$, the function must go down in the ith interval. But this is a contradiction: the function must go up because the constants are chosen to solve the interpolation problem. Also, this system is solvable because the system is diagonally dominant for small $\sigma$ $\endgroup$
    – Nick Alger
    Oct 31, 2021 at 13:38
  • $\begingroup$ @Nick Alger: The way I think about it: the values that I can obtain on those intervals form a positive cone. To show that it is the cone of vectors with positive entries. it is enough to show that I can produce vectors in the cone close enough to the generators of the closed cone, which are the $e_i$'s. Similarly would work with points inside a closed polytope, if taking about convex hulls, rather than conic hulls( the last part of the answer). $\endgroup$
    – orangeskid
    Oct 31, 2021 at 17:13
  • $\begingroup$ A more delicate problem is the existence of an positive analytic probability density function such that a given countable interval partition of $\mathbb{R}$ has prescribed measures. $\endgroup$
    – orangeskid
    Oct 31, 2021 at 18:15

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