# Prove that a $1$-periodic function $\phi$ is constant if $\phi(\frac{x}{2}) \phi(\frac{x+1}{2})$ is a constant multiple of $\phi$

For $0 < x< \infty$, let $\phi (x)$ be positive and continuously twice differentiable satisfying:

(a) $\phi (x+1) = \phi (x)$

(b) $\phi(\frac{x}{2}) \phi(\frac{x+1}{2}) = d\phi(x),$ where $d$ is a constant.

Prove that $\phi$ is a constant.

I am trying to answer this as the first step in the proof of Euler's reflection formula. I was am given the hint "Let $g(x) = \frac{\mathrm{d}^2}{\mathrm{d}x^2} \log \phi (x)$ and observe that $g(x+1)=g(x)$ and $\frac{1}{4}(g(\frac{x}{2}) + g(\frac{x+1}{2})) = g(x)$"

Firstly, I don't understand how they get the second bit of the hint, and then even assuming that I'm still not sure what to do. Any help would be much appreciated. Thanks.

• Take logarithms in (b) and differentiate twice. Then choose an $x$ such that $g$ has a local extremum at $x$. – Daniel Fischer Dec 24 '14 at 21:20
• @DanielFischer Ah yes. I see how they get the second part of the hint now. By property (b), we know that g(0)=g(1). So by Rolle's theorem, we know that $\exists x_0 \in (0, 1) : g'(c) = 0$. What am I supposed to do from here? And am I actually allowed to use Rolle's theorem here as we're only told that $\phi$ is twice continuously differentiable, and we'd require it to be three times differentiable for this. – user112495 Dec 24 '14 at 22:03
• You don't need Rolle's theorem. The continuity and periodicity of $g$ ensure that $g$ has local extrema in $[0,1]$. You want to show that $g\equiv 0$. – Daniel Fischer Dec 24 '14 at 22:06
• @DanielFischer Sorry, I'm still not sure how to go about doing that. – user112495 Dec 24 '14 at 22:28

Taking logarithms in (b), we obtain the identity

$$\log \phi\biggl(\frac{x}{2}\biggr) + \log \phi\biggl(\frac{x+1}{2}\biggr) = \log \phi(x) + \log d.\tag{1}$$

Differentiating $(1)$ twice, that becomes

$$\frac{1}{4}\Biggl(g\biggl(\frac{x}{2}\biggr) + g\biggl(\frac{x+1}{2}\biggr)\Biggr) = g(x).\tag{2}$$

So $g$ is a continuous function with period $1$ that satisfies the relation $(2)$. By periodicity, we can assume $g$ is defined on all of $\mathbb{R}.$

We want to show that $\phi$ is constant, so in particular that $g\equiv 0$. Choose $x_1,x_2 \in [0,1]$ so that

$$g(x_1) = \min \{ g(x) : x\in [0,1]\};\qquad g(x_2) = \max \{ g(x) : x\in [0,1]\}.$$

Since $g$ is continuous and $[0,1]$ is compact, such points exist. Since $g$ has period $1$, $g(x_1)$ is also the global minimum that $g$ attains on $\mathbb{R}$, and $g(x_2)$ the global maximum.

By $(2)$, we have

$$g(x_1) = \frac{1}{4}\Biggl(g\biggl(\frac{x_1}{2}\biggr) + g\biggl(\frac{x_1+1}{2}\biggr)\Biggr) \geqslant \frac{1}{4}\bigl(g(x_1) + g(x_1)\bigr) = \frac{1}{2}g(x_1),$$

so $g(x_1) \geqslant 0$. The same argument shows $g(x_2) \leqslant 0$, hence $g\equiv 0$, as desired.

Therefore, it follows that

$$\biggl(\frac{d}{dx}\log \phi\biggr)(x) \equiv a = \text{const},$$

and hence

$$\log \phi(x) = ax+b$$

and $\phi(x) = e^{ax+b}$ for some constants $a,b\in\mathbb{R}$. It remains to show that $a = 0$.