Solving $f(x+1)=xf(x)f\left(\frac1x\right)$ Recently, while scribbling around, I came up with the functional equation $f(x+1)=xf(x)f\left(\frac1x\right)$. It is somewhat similar to the functional equation for which the Gamma function is a solution, since $\Gamma(x+1)=x\Gamma(x)$, but it has the extra factor of $f\left(\frac1x\right)$.
If we introduce the restriction that $x\geq 1$, then I noticed that there is at least one solution (with this restriction put in place). That solution turns out to be, amazingly, $f(x)=\{x\}$, or the fractional part of $x$.
Edit: In fact, if $x\geq 1$, then $f(x)=\{x+a\}$ is a solution for all $a\in\mathbb{R}$ (I'm using the convention that $\{x\}=x-\lfloor x\rfloor$, where $\lfloor x\rfloor$ is defined as the greatest integer that is lower than $x$, therefore $a$ can be negative as well).
I have no idea how to approach this problem at all. The only thing that I was able to deduce is that for $n\in\mathbb{N}$, the following holds:
$$f(n)=f(1)\cdot (n-1)!\prod_{k=1}^{n-1}f\left(\frac1k\right)$$
I don't know if this helps in any way, but it might be a good start. 
Knowing that when $x$ is restricted to be greater or equal to $1$, $\{x\}$ is a solution, I wonder if there can even exist a solution that is continuous. However, I don't know how to prove or disprove this, nor do I know how to find any solutions for the equation.
Is there a way to at least gain insight on how a possible solution for the equation must behave?
 A: I'll assume the restriction that $x \geq 1$ in the functional equation and consider functions with domain $\mathbb{R}^+$.
By applying the functional equation repeatedly we see that for any $x \geq 1$ and positive integer $n$, $f(x+n)=f(x)\prod_{k=0}^{n-1}(x+k)f(\frac{1}{x+k})$. Therefore if $f$ is defined on $(0, 2)$ this will extend uniquely to all of $\mathbb{R}^+$.
For arbitrary $f$ on $(0, 2)$, its extension will satisfy the functional equation. Note that $f(x)=\{x\}$ is a "nice" solution because the $(x+k)$ and $f(\frac{1}{x+k})$ terms cancel out above and it is periodic with degree $1$. In fact, if we define any $f$ with
$$f(x)=\begin{cases} x & 0 < x < 1 \\ g(x) & 1 \leq x < 2 \end{cases}$$
for arbitrary $g$, then $f$ extends to $g$ periodized with period $1$ on $[1, \infty)$. This will be continuous iff $g(1)=g(2)=1$. For example, the function $f(x) = \min(x, 1)$ is a continuous solution to the functional equation.
By the functional equation, $f(2) = f(1+1) = f(1)^2$. Any function $f$ that is continuous on $(0, 2)$ and such that $\lim_{x \to 2} f(x) = f(1)^2$ will extend continuously to $\mathbb{R}^+$. Of course $f(x)=0$ is a trivial continuous solution. Another solution from extending the constant function $f(x)=1$ on $(0, 2)$ to $\mathbb{R}^+$ is
$$f(x) = \begin{cases} 1 & x < 1 \\ \frac{\Gamma(x)}{\Gamma(1+\{x\})} & x \geq 1 \end{cases}$$
You can construct any continuous solution with this method.
The Bohr-Mollerup theorem suggests that you might find a more interesting family of solutions by considering log-convexity and the boundary condition $f(1)=1$, but I didn't see a nice way to use this.
Also, note that if $f$ and $g$ satisfy the functional equation, then $h = f/g$ satisfies $h(x+1) = h(x)h(1/x)$. This functional equation will again extend from $(0, 2)$ to $\mathbb{R}^+$ by $h(x+n) = h(x)\prod_{k=0}^{n-1}h(\frac{1}{x+k})$ for $x \geq 1$ and positive integer $n$, and will have the same continuity condition as above. Maybe with some stricter conditions this functional equation or $H = \log h$ satisfying $H(x+1) = H(x) + H(1/x)$ might have some interesting solutions.
