= Solution
A <wavelet> with $p$ <vanishing moments> satisfies $x^k\psi\in L^1$ and $\int x^k\psi(x)\,dx=0$ for $0\le k<p$. The <moment differentiation of the Fourier transform> theorem states that these weighted integrability hypotheses make $\widehat\psi\in C^{p-1}$, with $\widehat\psi^{(k)}(\xi)=\int(-ix)^k\psi(x)e^{-ix\xi}\,dx$. Hence $\widehat\psi^{(k)}(0)=0$. The <MRA projection Fourier identity>, together with density of the approximation spaces and <continuity> of $\widehat\varphi$ at zero, gives $|\widehat\varphi(0)|=1$. These are the two analytic theorems needed in addition to the <quadrature mirror filter> construction.
For the ordinary <derivative> conclusion, assume also that the <MRA low-pass filter> is $C^{p-1}$ near $\pi$ and $\widehat\varphi$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported <scaling function> with a finite refinement filter. The <wavelet> identity gives
$$
\overline{m(\pi+t)}=\frac{e^{it}\widehat\psi(2t)}{\widehat\varphi(t)}.
$$
The denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the <product rule> and the zero <Taylor polynomial> of $\widehat\psi$ give the <smooth-mask vanishing-moment criterion>
$$
\boxed{m^{(k)}(\pi)=0\quad(0\le k<p),\quad\text{under the stated smoothness hypotheses}.}
$$
\b[The printed integrability hypothesis alone does not guarantee ordinary higher <derivatives>.] It gives only a <continuous> $\widehat\varphi$. Without added smoothness, <Taylor theorem> for $\widehat\psi$ still gives $m(\pi+t)=o(|t|^{p-1})$ for the representative defined by the quotient: a <Peano zero>. This establishes the value at $\pi$, and for $p\ge2$ the first <derivative> there, but it does not automatically establish iterated <derivatives>.
Here is a <lacunary scaling-phase regularity counterexample> for $p\ge3$. Start with a compactly supported <Daubechies wavelet> having $p$ <vanishing moments>, with <scaling function> $\varphi_0$ and finite <MRA low-pass filter> $m_0$. Define the periodic phase
$$
\theta(t)=\sum_{n\ge0}2^{-n}\sin(2^nt),\qquad a(t)=e^{i\theta(t)},\qquad\widehat\varphi(t)=a(t)\widehat\varphi_0(t).
$$
The <Fourier coefficients> of $\theta$ are absolutely summable. The <Wiener algebra> is closed under multiplication and the exponential <series>, so $a$ and $a^{-1}$ have absolutely summable <Fourier coefficients>. Thus $\varphi$ is an absolutely summable combination of <integer> <function translations> of $\varphi_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves <orthogonality> and unit <norm> of its <integer> <function translations>; the inverse phase preserves their complete span. The same approximation spaces therefore give a <multiresolution analysis>.
The identity $\theta(2t)=\theta(t)+\theta(t+\pi)$ gives $a(2t)=a(t)a(t+\pi)$, so the <periodic phase change of a scaling function> produces $m(t)=a(t+\pi)m_0(t)$. In the high-pass formula the phases cancel:
$$
e^{-it}\overline{m(t+\pi)}\widehat\varphi(t)
=e^{-it}\overline{m_0(t+\pi)}\widehat\varphi_0(t).
$$
The associated <wavelet> is therefore exactly the original one, with unchanged <vanishing moments>.
At any dyadic point $t_0=2\pi k/2^j$, the <difference quotient> of the terms with $j\le n\le\log_2(1/|h|)$ is $1+O((2^nh)^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta(t_0+h)-\theta(t_0))/h=\log_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of <differentiability>. Dyadic points are <dense>, and $m_0(\pi+t)$ is nonzero for sufficiently small $t\ne0$. Thus $m$ is not <differentiable> on any neighborhood of $\pi$, so an ordinary second <derivative> at $\pi$, understood as the <derivative> of a locally defined first <derivative>, need not exist. The <Peano zero> remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.
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