# Functional equation involving gamma function

Recently, I found the following functional equation: $$n^{nx-1}\cdot\prod_{k=0}^{n-1}{\Gamma{\left(x+\frac{k}{n}\right)}}=\Gamma{(nx)}\cdot\prod_{k=1}^{n-1}{\Gamma{\left(\frac{k}{n}\right)}}$$ Now my question: I've read, that the gamma function is defined as solution of the following three functional equations/conditions: $$f(x+1)=x\cdot f(x) \space \forall x\in\mathbb R^+$$ $$f(1)=1$$ $$\ln(f(x)) \space \text {is convex \forall x\in\mathbb R^+}$$ And, due to the Bohr-Mollerup theorem, the gamma function is the only function satisfying those three conditions.

Is it equivalent to the following definition?: $$f(x+1)=x\cdot f(x)\space \forall x\in\mathbb R^+$$ $$f(1)=1$$ $$n^{nx-1}\cdot\prod_{k=0}^{n-1}{f{\left(x+\frac{k}{n}\right)}}=f{(nx)}\cdot\prod_{k=1}^{n-1}{f{\left(\frac{k}{n}\right)}}\space \forall x\in\mathbb R^+,n\in\mathbb N$$ Meaning, is the gamma function also the only solution to those three functional equations?

These equations do not determine the Gamma function.

Here are some solutions, other than $\Gamma$:

\begin{align*} f_1(x) &= \begin{cases} \Gamma(x) & x\in\mathbb{Q} \\ 0 & x\not\in\mathbb{Q} \end{cases} \\ f_2(x) &= \begin{cases} 2^{\alpha}\cdot\Gamma(x) & \text{if } x=\alpha\sqrt2+\beta; \,\,\alpha,\beta\in\mathbb{Q} \\ \Gamma(x) & x\not\in\mathbb{Q} \end{cases} \\ f_3(x) &= e^{\cos(2\pi x)} \cdot \Gamma(x) \end{align*}

The third example shows that assuming continuity is not sufficient. The assumption about the logarithmic convexity is so important.

• I believe that your last example incorrect - it doesn't satisfy the $n=2$ case of the equation. – Alexander Shamov Oct 19 '14 at 6:55
• Yes, the third one is wrong. – user141614 Oct 19 '14 at 12:53

As user141614 pointed out correctly, there are other solutions to this equation, so from now on I assume that $f$ is continuous.

The only continuous real-valued $f$ that satisfies the equations is $\Gamma$.

However, it will follow from the proof that by dropping the requirement $f(1) = 1$ we get more interesting solutions, like, e.g., $f(x) = \Gamma(x) (1 - \cos 2 \pi x)^{\alpha}, \alpha > 0$.

Proof. The first thing we show is that $f$ cannot have zeroes.

$f(x) = 0$ for some $x \in \mathbb{N}$, together with the functional equation, would contradict the requirement that $f(1) = 1$.

Assume that $f(x) = 0$ for some $x \notin \mathbb{N}$. Then $f(nx \operatorname{mod} 1) = 0$ for all $n \in \mathbb{N}$. This cannot happen because the orbit $\{nx \operatorname{mod} 1, n \in \mathbb{N}\}$ is either dense or contains an integer.

Now denote $L(x) := \log f(x) - \log \Gamma(x)$. $L$ is a $1$-periodic continuous function that satisfies the equation $$L(nx) = \sum_{k=0}^{n-1} L \left( x + \frac{k}{n} \right) + C_n,$$ where $C_n$ are constants.

Now consider the Fourier series of $L$: $L(x) \sim \sum_{m} L_m e^{2 \pi i m x}$. When rewritten in terms of $L_m$, the equation above becomes $$\sum_m L_m e^{2 \pi inmx} = \sum_m L_m e^{2 \pi imx} \sum_{k=0}^{n-1} e^{2 \pi imk/n} + C_n$$ Note that $\sum_{k=0}^{n-1} e^{2 \pi imk/n}$ is $0$ if $m$ is not divisible by $n$, and $n$ otherwise. So finally we get $L_{nk} = \frac{1}{n} L_k, k \in \mathbb{Z} \setminus 0$. In other words, $$L(x) = L_0 - L_1 \log(1 - e^{2 \pi i x}) - L_{-1} \log (1 - e^{-2 \pi i x})$$

This can only be continuous if $L_1 = L_{-1} = 0$.