Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Let $f : ]0, +\infty[ \rightarrow ]0, +\infty[$ be a function of class $\mathscr{C}^2$ such that the function $\ln(f)$ is convex and satisfies $f(1) = 1$ and $f(x+1) = xf(x)$ for all $x > 0$. The function $S(x) = \ln\left(\frac{f(x)}{\Gamma(x)}\right)$ is 1-periodic and convex. Deduce that $f = \Gamma$.
Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where
$$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$
Let $f : ]0, +\infty[ \rightarrow ]0, +\infty[$ be a function of class $\mathscr{C}^2$ such that the function $\ln(f)$ is convex and satisfies $f(1) = 1$ and $f(x+1) = xf(x)$ for all $x > 0$. The function $S(x) = \ln\left(\frac{f(x)}{\Gamma(x)}\right)$ is 1-periodic and convex.
Deduce that $f = \Gamma$.