 Solution

The factor $1+e^{-i\xi }$ has a simple zero at $\pi$, and $L(\pi )\ne 0$ makes the zero of $m$ there exactly order $N$. The quadrature mirror filter construction gives a high-pass symbol $m_1(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase, andThe normalized scaling function has $\hat{\phi }(0)=1$, so $\hat{\psi }$ has a zero of exactly order $N$ at zero. Compact support of the Daubechies wavelet permits differentiation under its Fourier transform:HenceThere are exactly $N$ vanishing moments, rather than just at least $N$.