If $f(f(x)) = x+1, f(x+1) = f(x) + 1$, is it true that $f(x) = x + 1/2$?

If $f(f(x)) = x+1, f(x+1) = f(x) + 1$, where $f: \Bbb R \rightarrow \Bbb R$ is real-analytic, bijective, monotonically increasing, is it true that $f(x) = x + 1/2$?
I have tried to represent $f(x)$ as power series in a neighborhood of arbitrary $x_{0}$ and $y_{0} = f(x_{0})$. It's obvious that we can choose such a neighborhood $U_{0}(x_{0})$ that it's image by $f$ is $U_{1}(y_{0})$. Then, we can try to find the power series of $f(f(x))$, which is equal to $x+1$. But I cannot see, how such an equation shows that $f(x) = x + 1/2$.

Solution 1:

The answer is no. All that can be said is that $f$ is conjugate to the shift $x\mapsto x+1/2$. More precisely:
Theorem A real analytic function $\newcommand{\R}{{\mathbb R}}f:\R\to\R$ satisfies $f(f(x))=x+1$ for all real $x$ if and only if it can be written $$\tag{1}f(x)=h\left(h^{-1}(x)+\tfrac12\right)$$ where $h:\R\to\R$ is real analytic and satisfies $h(x+1)=h(x)+1$ and $h'(x)>0$ for all real $x$.
Remarks: 1. (1) is equivalent to $h^{-1}\circ f\circ h(x)=x+\frac12$, that is $h$ conjugates $f$ to the shift $x\mapsto x+\frac12$.
2. The Theorem is also valid for continuous functions instead of real analytic functions. The statement "$h'(x)>0$" has to be replaced by "$h$ strictly increasing".
3. An analogous theorem holds for real analytic $f$ satisfying $f^{\circ n}(x)=x+1$ where $f^{\circ n}$ denotes the repeated composition of $n$ copies of $f$.
Proof: The sufficieny of (1) had already been checked in previous answers or comments: $$f(f(x))=h(h^{-1}(x)+1)=h(h^{-1}(x))+1=x+1.$$ So let us prove the necessity as well. Recall from the comments to the question that $f(x+1)=f(x)+1$ for all $x$ because both expressions are equal to $f(f(f(x)))$ and that $f$ is necessarily bijective with inverse $f^{-1}(x)=f(x-1)=f(x)-1$. Since $f$ is continuous and bijective, it is also strictly monotonous. As $f(x+1)=f(x)+1$, it must be strictly increasing.

In order to motivate the construction, let us discuss the uniqueness of $h$ in (1), if it exists. We claim that $\tilde h=h\circ \phi$ with a strictly increasing real analytic $\phi:\R\to\R$ satisfying $\phi(x+\frac12)=\phi(x)+\frac12$ works as well as $h$. Indeed $$\tilde h(\tilde h^{-1}(x)+\tfrac12)=h(\phi(\tilde h^{-1}(x)+\tfrac12))=h(\phi(\tilde h^{-1}(x))+\tfrac12)=h(h^{-1}(x)+\tfrac12).$$ As $g(x)=h(x)-1$ is 1-periodic, it can be written $g(x)=u(x)+v(x)$ where $u(x)=(g(x+\frac12)+g(x))/2$ is $1/2$-periodic and $v(x)=(g(x)-g(x+\frac12))/2$ satisfies $v(x+\frac12)=-v(x)$. Hence $h=\psi+v$, where $\psi(x)=x+u(x)$ satisfies $\psi(x+\frac12)=\psi(x)+\frac12$. Observe that $\psi(x)=(h(x)+h(x+\frac12))/2$ also has a positive derivative and hence an inverse function $\psi^{-1}$. Going over to $\tilde h=h\circ \psi^{-1}$, we arrive at some $\tilde h$ also working in (1) but such that $\tilde v(x)=\tilde h(x)-x=v\circ \psi^{-1}(x)$ satisfies $\tilde v(x+\frac12)=-\tilde v(x)$.
In summary: If some $h$ with (1) exists then there also exists an $h$ satisfying additionally $h(x)=x+v(x)$ where $v(x+\frac12)=-v(x)$.

Now let a real analytic function $f:\R\to\R$ satisfying $f(f(x))=x+1$ be given. We are looking for some $h(x)=x+v(x)$ with $v(x+\frac12)=-v(x)$ satisfying $f(h(x))=h(x+\frac12)$ or equivalently $f(x+v(x))=x+\frac12-v(x)$ or, with $F(x)=x+f(x)$, $$F(x+v(x))=2x+\tfrac12.$$ This motivates our definition $$h(x)=x+v(x)=F^{-1}\left(2x+\tfrac12\right).$$ This is well defined because $F'(x)\geq1$ and it is well known that it defines a real analytic function. We find $h'(x)=2/F'(h(x))>0$ and, via $F(x+v(x))=2x+\frac12$, we obtain that $$\tag{2}f(x+v(x))=x+\tfrac12-v(x)\mbox{ for all }x.$$ We claim that $v(x+\frac12)=-v(x)$ which shows that $f(h(x))=h(x+\frac12)$ and $v$ is 1-periodic and thus completes the proof.
For a proof of the claim, (2) implies that $x+v(x)+1=f(f(x+v(x))=f(x+\frac12-v(x))$ and therefore, replacing $x$ by $x+\frac12$, we find $x+\tfrac32+v(x+\tfrac12)=f(x+1-v(x+\tfrac12))$and thus, using $f(x+1)=f(x)+1$, $$x+\tfrac12+v(x+\tfrac12)=f(x-v(x+\tfrac12)).$$ Hence $F(x-v(x+\frac12))=2x+\frac12=F(x+v(x))$ which implies $-v(x+\frac12)=v(x)$ as wanted.

Solution 2:

This is the idea of @Thomas Andrews:, just adding some details:

Let $h\colon \mathbb{R}\to \mathbb{R}$ a bijection such that $h(x+1) = h(x) +1$. Then the function

$$f\colon x \mapsto h\,(h^{-1}(x) + 1/n) $$

satisfies $f^n(x) = x+1$. Indeed, we have

$$f = h \circ t \circ h^{-1}$$ where $t$ is the translation $x\mapsto x+\frac{1}{n}$. We get

$$f^n = h \circ t^n \circ h^{-1}$$

But note that $t^n = t_1$, the translation $x\mapsto x+1$, and we have $$h\circ t_1 = t_1 \circ h$$ Substituting in the above we get $$f^n = t_1 \circ h \circ h^{-1} = t_1$$

Now, we only need a sufficient supply of $h$ analytic, that commute with $t_1$, but do not commute with the translation $t$. Note that the function $h(x)-x$ has period $1$. So we can start with a period $1$ (but not $\frac{1}{n}$) analytic function $p(x)$ on $\mathbb{R}$, with $p'(x) > -1$, and take $h(x) = x + p(x)$. There were examples produced in the comments, I will leave it here.

$\bf{Added:}$ The function $\bar f\colon \mathbb{R}/\mathbb{Z} \mathbb{R}/\mathbb{Z}$, $\bar f(\bar x) = \overline{f(x)}$ is a diffeomorphism of $\mathbb{R}/\mathbb{Z}\simeq S^1$ preserving orientation, and moreover, an involution ( assume $n=2$ above). Conversely, any preserving orientation involution $f$ of $S^1$ which is not the identity has a lift $f$ satisfying the desired property. So now we have to look for involutions of $S^1$, and we can take consider conjugates of $x\to x+\frac{1}{2}$.

$\bf{Added:}$

This is following the beautiful solution of @Helmut: If $f^n(x)=x+1$, then there exists $h$ diffeomorphism commuting with the translation $x\mapsto x+1$ such that $$h(f(x)) = h(x) + \frac{1}{n}$$ that is $$h \circ f = T_{\frac{1}{n}} \circ h$$

First, note that since $f^n= T_1$, all powers of $f$ commute with $T_1$

Now, define

$$h(x) = \frac{x + f(x) + \cdots + f^{(n-1)}(x)}{n}$$

A simple calculation shows that $$h(f(x)) = h(x) + \frac{1}{n}$$ and $h(x+1) = h(x) + 1$. This is the algebra.

Let's show that $h$ is a diffeomorphism. Note that since $f^n = T_1$, $f$ is bijective, so either $f$ increases or decreases from $-\infty$ to $\infty$. Moreover, $f$ has no fixed points. It follows that $f$ increases, that is $f'(x)>0$, and $\lim_{x\to \pm \infty} f(x)= \pm \infty$. We conclude that $h$ has the same properties, and so is a diffeomorphism.