# 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!

• This is a very interesting question! Oct 22, 2021 at 0:34
• 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 Oct 27, 2021 at 22:27

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 $$y_1 are real numbers. Then there exists a real-analytic diffeomorphism $$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 $$f(a_k)=b_k, k=1,...,n$$.

• Excellent, thank you for the detailed answer! Oct 30, 2021 at 17:47
• 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) Oct 30, 2021 at 19:43
• @NickAlger: Right. 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 $$x_i$$.

• 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. Oct 31, 2021 at 11:18
• @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. Oct 31, 2021 at 11:55
• @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 $\sigma$ Oct 31, 2021 at 13:38
• @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). Oct 31, 2021 at 17:13
• 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. Oct 31, 2021 at 18:15