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.

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 $a = 0$.
