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

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: $$f(x)=x+2+\frac1{x+1}$$

My approach is by construction.

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

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

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. $$2f_0(-1)=f_0(0)=f_1(0)$$

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. $$f_0'(0)=f_1'(0)=0$$

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