Actually one can resort to the two-scale equation in multiresolution analysis. Perform Fourier transformation on both side of $\phi(t)=\phi(2t)+\phi(2t-1)$, it turns out that
$\hat\phi(\omega)=\frac{1}{2}\hat\phi(\frac{1}{2}\omega)+\frac{1}{2}\hat\phi(\frac{1}{2}\omega)e^{-\frac{i\omega}{2}}$ , that is, $\frac{\hat{\phi}(2\omega)}{\hat{\phi}(\omega)}=\frac{1+e^{-i\omega}}{2}=m(\omega)$ Therefore $\hat\phi(\omega)=\prod_{k=1}^{\infty}m\left(\frac{\omega}{2^k}\right)=\prod_{k=1}^{\infty}\frac{1+e^{-i2^{-k}\omega}}{2}\hat\phi(0)$
My question is, how to calculate this limit? Thank you~