Are there any smooth/analytic solutions to the functional equation $f(x+1)-f(x)=f\left(\frac 1x\right)$?

Question

Inspired by yesterday's closed question with many upvotes, I studied the functional equation $$f(x+1)-f(x)=f\left(\frac 1x\right)\qquad\forall x\in\mathbb R\setminus\{0\}\qquad f\in C^1$$

I have solved the equation by construction, and I will present my solution as an answer. However, my solution only includes the conditions for once-differentiability, and I cannot see any ways to generalize it to infinite-differentiability.

My question is

Are there any smooth/analytic solutions to the functional equation?

Notably, user @John Omielan gave a nearly-analytic solution: $$f(x)=x+2+\frac1{x+1}$$

Answer 1

My approach is by construction.

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Firstly, I will solve the functional equation on $\mathbb R^+$, without considering continuity/differentiability, which I will care about later.

Denote $\phi$ the golden ratio.

Partition $\mathbb R^+$ into

1. $[0,\phi]$
2. $[\phi,1]$
3. $[1,1+\phi]$
4. $[1+\phi,\infty)$

(Call the $n$th partition P$n$)

(Why $\phi$? : My initial idea is to define arbitrary functions on $[0,a]$ and $[1,1+a]$, and use the functional equation to extend the function. It turns out that this allows extension to $[\frac1a,\infty)$. To prevent overlapping of 'arbitrary region' and 'extension region', the critical $a$ satisfies $1+a=\frac 1a\implies a=\phi$.)

In my studies, I found that we can define two arbitrary functions $f_1$ and $f_3$ on P1 and P3 respectively. Then, on P4, we have $$f(x)=f_4(x):=f_3(1+1/x)-f_1(1/x)\qquad x\in [1+\phi,\infty)$$

Furthermore, for $x$ in P2, $$f(x)=f_2(x):=f_4(x+1)-f_3(1/x)=f_3\left(1+\frac1{x+1}\right)-f_1\left(\frac1{x+1}\right)-f_3\left(\frac1x\right)$$

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Secondly, I will solve the functional equation on $\mathbb R^-$.

This case is not analogous to the one above. Partition $\mathbb R^-$ into $[-n,-n-1]$ for $n=0,1,2,\cdots$, and $f(x)=f_{-n}(x)$ on $[-n,-n-1]$.

Since in the functional equation, the arguments are $x$, $x+1$, and $\frac1x$, it is impossible that only one argument is negative. Therefore, $f$ on $\mathbb R^-$ is not completely determined by $f$ on $\mathbb R^+$, and we do have some degrees of freedom on the $\mathbb R^-$.

It turns out that we can define an arbitrary $f_0$.

Then, $$f_{-1}(x)=f_0(x+1)-f_0(1/x)$$ $$f_{-2}(x)=f_{-1}(x+1)-f_0(1/x)=f_0(x+2)-f_0(1/(x+1))-f_0(1/x)$$ $$\cdots$$ $$f_{-n}(x)=f_0(x+n)-\sum^{n-1}_{k=0}f_0\left(\frac1{x+k}\right)$$

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Now let us find the conditions for continuity.

In general, we require

1. $f_{-(n-1)}(-n)=f_{-n}(-n)$ for $n=1,2,3,\cdots$.
2. On $\mathbb R^+$ neighbouring functions have to agree on the boundary.

After a lot of tedious algebra, it turns out that it is required that

1. $f_1(\phi)=0$
2. $f_3(3/2)=2f_3(1)+f_1(1/2)$
3. $2f_0(-1)=f_0(0)=f_1(0)$

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Similarly, for differnetiability,

1. $\phi \cdot f_1'(\phi)=\sqrt5 \cdot f_3'(\phi+1)$
2. $f_3'(3/2)=f_1'(1/2)$
3. $f_0'(0)=f_1'(0)=0$

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To sum up:

If two differentiable functions $\mu:[-1,\phi]$ and $\nu:[1,1+\phi]$ satisfy the following conditions:

1. $\mu(\phi)=0$
2. $\nu(3/2)=2\nu(1)+\mu(1/2)$
3. $2\mu(-1)=\mu(0)$
4. $\phi \cdot \mu'(\phi)=\sqrt5 \cdot \nu'(\phi+1)$
5. $\nu'(3/2)=\mu'(1/2)$
6. $\mu'(0)=0$

then, $$f(x) = \begin{cases} \nu(1+1/x)-\mu(1/x) & x>1+\phi \\ \nu(x) & 1+\phi > x > 1 \\ \nu\left(1+\frac1{x+1}\right)-\mu\left(\frac1{x+1}\right)-\nu\left(\frac1x\right) & 1 > x > \phi \\ \mu(x) & \phi > x > -1 \\ \mu(x+n)-\sum^{n-1}_{k=0}\mu\left(\frac1{x+k}\right) & -n>x>-n-1 \quad (n=1,2,3,\cdots) \end{cases}$$