Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question was motivated by this one. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Define $f^n(x)=f\circ f\circ\cdot\cdot\cdot\circ f$, $n$ times, $f^0(x)=x$ and $k\geq 2$ a integer. Suppose that $f$ satisfies the equation: $$\tag{1}\sum_{n=0}^k(-1)^n\binom{k}{k-n}f^{k-n}(x)=0$$

Note that for each fixed $a\in\mathbb{R}$, $f_a(x)=x+a$ satisfies $(1)$. Are the only solutions of $(1)$ of the form $f_a$ for some $a$? Also note that $f_a$ is an solution of the equation $$\tag{2}(f-I)^k(x)=a$$

The same question applies here: Is the only solution of $(2)$ of the form $f_a$? If we let $a$ vary in $(2)$, we get a family of equations. Is that family of equations equivalently to $(1)$?

share|improve this question
    
Note that you changed your notation in $(1)$: we usually define $f^{(m)}$ as the derivative of order $m$ of the function $f$. –  Pedro Tamaroff Mar 13 '13 at 1:23
    
@PeterTamaroff, thank you, let me fix it. –  Tomás Mar 13 '13 at 11:35
    
It is plain to observe that $f$ must be an increasing bijection onto $\Bbb{R}$. But except the case $n=2$, where the solution was already dealt in the link above, I have no idea what the answer would be. –  sos440 Mar 13 '13 at 13:44

3 Answers 3

I first tried to modify the proof at the link in the question, but soon I realized that the proof holds because of the particular form satisfied in the case $n = 2$.

Instead, I was able to deduce a partial answer:

Proposition. Let assume that a continuous function $f : \Bbb{R} \to \Bbb{R}$ satisfies $$ \sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} f^k (x) = a, \quad \forall x \in \Bbb{R} \tag{1}$$ for some real number $a \in \Bbb{R}$ and some integer $n \geq 1$. If we further assume that $f$ has at least one fixed point, then $f$ is the identity function.

We first observe the following properties.

Lemma 1. Let $f$ be a continuous function satisfying $(1)$. That is, let $f$ satisfy the assumption of Proposition except that $f$ has a fixed point. Then $f$ is an increasing bijection from $\Bbb{R}$ onto $\Bbb{R}$.

Proof. Assume $f(x) = f(y)$. Then

$$ x = (-1)^n a - \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} f^k (x) = (-1)^n a - \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} f^k (y) = y $$

and $f$ is injective. In particular, $f(\Bbb{R})$ is an open interval.

To prove the surjectivity of $f$, assume that $f$ is not surjective. Then either $\sup f$ or $\inf f$ is finite, and in any cases we can find a real number $\alpha \in \Bbb{R}$ and a sequence of real numbers $x_j$ such that $|x_j| \to \infty$ and $f(x_j) \to \alpha$ as $j \to \infty$. But this implies

\begin{align*} \infty = \lim_{j\to\infty} \left| x_j \right| = \lim_{j\to\infty} \left| (-1)^n a - \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} f^k (x_j) \right| = \left| (-1)^n a - \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} f^{k-1}(\alpha) \right| < \infty, \end{align*}

a contradiction. Therefore $f$ is surjective. In particular, $f$ is a homeomorphism and it is either increasing or decreasing.

To prove that $f$ is increasing, assume that $f$ is decreasing. Then $(-1)^{k-1} f^{k}$ is decreasing for any $k \geq 1$. Thus the right-hand side of

$$ x = (-1)^n a + \sum_{k=1}^{n} \binom{n}{k} (-1)^{k-1} f^k (x) $$

is also decreasing, a contradiction. This completes the proof. ■

Lemma 2. Suppose that $f : \Bbb{R} \to \Bbb{R}$ be an continuous increasing bijection, so that it extends to a continuous bijection $\bar{f}$ from $\bar{\Bbb{R}} = [-\infty, \infty]$ onto $\bar{\Bbb{R}}$. Then for any $x \in \bar{\Bbb{R}}$, we have the following trichotomy:

  1. If $x < f(x)$, then $f^k (x)$ converges to the smallest fixed point of $\bar{f}$ greater than $x$.
  2. If $x > f(x)$, then $f^k (x)$ converges to the greatest fixed point of $\bar{f}$ smaller than $x$.
  3. If $x = f(x)$, then $f^k (x)$ converges to $x$.

Proof. Since $\bar{f} : \bar{\Bbb{R}} \to \bar{\Bbb{R}}$ is continuous, the set $F = \{ \bar{f}(x) = x : x \in \bar{\Bbb{R}} \}$ is a closed set containing both $\infty$ and $-\infty$. If $F = \bar{\Bbb{R}}$, there is nothing to prove and assume that $F^c \neq \varnothing$. Since $F^c$ is open in $\Bbb{R}$, we can decompose it as a countable union of disjoint open intervals $U_j$. Note that on each $U_j$, either $f(x) > x$ identically for all $x \in U_j$ or $f(x) < x$ identically for all $x \in U_j$ by continuity of $f$.

Now let $x \neq f(x)$. Then $x \in U_j$ for some $j$. Assume first that $x < f(x)$. Then for any $x < y \leq f(x)$, we have $f(y) > f(x) \geq y$ and we have $[x, f(x)] \subset U_j$. In particular, $f(x) \in U_j$. Now this argument can be repeatedly applied to yield that $f^k (x) \in U_j$ for all $k$. Since $f^k(x)$ is monotone increasing, it must converge to some point $\alpha \in \bar{\Bbb{R}}$. It is plain to check that $\alpha = \sup U_j$ and $\alpha$ is a fixed point of $\bar{f}$, proving the first option of the trichotomy. The second option follows exactly the same manner and the third option is trivial. ■

As an immediate consequence, we obtain the following corollary:

Corollary. If $f : \Bbb{R} \to \Bbb{R}$ is a continuous increasing bijection having at least one fixed point, then for any $x \in \Bbb{R}$, either $f^k (x)$ or $f^{-k} (x)$ converges to a fixed point of $f$.

Proof. Let $U_j$ be as in the proof above. If $x \in U_j$, then $f^k (x)$ and $f^{-k} (x)$ converges to $\sup U_j$ and $\inf U_j$. Since $f$ has a fixed point, either $\sup U_j$ or $\inf U_j$ is finite. ■

Now we are ready to prove the proposition.

Proof of Proposition. Define

$$g(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} (-1)^{n-1-k} f^k (x). $$

It is plain to observe that $g \circ f - g = a$. Let $F$ be the set of fixed points of $f$. Then $F$ is non-empty by assumption. If we pick a point $\alpha \in F$, then

$$0 = g(f(\alpha)) - g(\alpha) = a.$$

In particular, we have

$$ g = g\circ f = g \circ f^{k} \quad \forall k \in \Bbb{Z}. $$

Thus if $n = 1$, then $g(x) = x$ and this identity immediately yields $f(x) = x$. This shows that we may assume $n \geq 2$.

In view of the corollary above, for each $x \in \Bbb{R}$, either $f^{k}(x)$ or $f^{-k}(x)$ converges to a point of $F$. Thus there exists $\alpha \in F$ such that

$$g(x) = g(\alpha) = 0,$$

where the last inequality holds by the definition of $g$ together with the assumption $n \geq 2$. Putting together, we showed that if $(1)$ holds for some $n \geq 2$ and $a \in \Bbb{R}$, then $(1)$ also holds for $n-1$ instead of $n$ and $a = 0$. Therefore by the mathematical induction, we obtain the desired conclusion. ■

share|improve this answer

Let $P\in\Bbb R[X]$ be a polynomial of degree${}<k$, such that its polynomial function $p:\Bbb R\to\Bbb R$ is nowhere decreasing (in other words $p'(x)\geq0$ for all $x\in\Bbb R$; this obviously requires $\deg P$ to be odd). Then $p^{-1}$ is well defined and continuous. Define $f:x\mapsto p(p^{-1}(x)+1)$ which is also continuous, and satisfies $f^n(p(x))=p(x+n)$ for all $n$. Now $$ \sum_{i=0}^k(-1)^i\binom k{k-i}f^{k-i}(p(x)) =\sum_{i=0}^k(-1)^i\binom kip(x+k-i) =\Delta^k(p)(x+k) =0 $$ for all $x\in\Bbb R$, where $\Delta$ is a finite difference operator defined by $\Delta g: x\mapsto g(x)-g(x-1)$, which decreases the degree of polynomial functions, so that $\Delta^k(p)=0$ identically.

Since $p$ is surjective this shows $f$ satisfies your requirement and that the answer to your first question is negative for $k\geq4$. The simplest counterexample is $f:x\mapsto (\sqrt[3]x+1)^3$.

share|improve this answer
    
Interesting. Please, let me ask you two things. 1 - Can you spot the error in the proof of @sos440? 2 - Any suggestion about the case $k=3$? Thank you. –  Tomás Mar 19 '13 at 12:06
    
I'm not sure there is an error in the proofs by @sos440, since all the significant statments there require that $f$ have a fixed point, and the function defined in my answer never have a fixed point. –  Marc van Leeuwen Mar 19 '13 at 12:31
1  
As for the case $k=3$, the method described here gives counterexamples defined on $\Bbb R_{\geq0}$, for instance $x\mapsto(\sqrt x+1)^2=x+2\sqrt x+1$, but not functions defined on all of $\Bbb R$. I think I can prove that there aren't any continuous non-affine functions $f:\Bbb R\to\Bbb R$ that satisfy $x-3f(x)+3f^2(x)-f^3(x)=0$ for all $x$. –  Marc van Leeuwen Mar 19 '13 at 12:40
    
You are right, I forgot he assumend a fixed point. –  Tomás Mar 19 '13 at 12:49
1  
(+1) This is quite simple yet suggesting! –  sos440 Mar 19 '13 at 13:21

Consider a case $k = 2$: $$f(f(x)) - 2f(x) + x = 0$$ Let $f(x) = x + g(x)$, then $$g(x) = g(x + g(x))$$ For example, $$g(\frac{x}{n}) = g(\frac{x}{n} + g(\frac{x}{n})) = ... = g(x + g(\frac{x}{n}))$$

Assuming that $g$ is contiguous at $0$, $g(x + g(0)) = g(0)$ for any $x$, that is, $g$ is a constant indeed. The similar mechanics works for other $k$ as well.

I don't know how to approach it without a contiguity assumption.

share|improve this answer
    
The function $f$ is continuous. –  Tomás Mar 12 '13 at 22:07
    
Can you explain me how do you prove that $g(\frac{x}{n})=g(x+g(\frac{x}{n}))$? –  Tomás Mar 12 '13 at 22:39
    
If you think that "The similar mechanics works for other $k$ as well", you should demonstrate it. It doesn't work in the same way, see my answer. –  Marc van Leeuwen Mar 19 '13 at 13:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.