Scaling function ($\phi$)

A scaling function generates the approximation space $V_0$ of a multiresolution analysis by its integer translates. In an orthonormal construction those translates form an orthonormal basis, and ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ generate $V_j$. Integrable orthonormal scaling functions have $|\hat{\phi }(0)|=1$, as follows from density and the MRA projection Fourier identity.

Let a real measurable Fourier transform equal one on $|t|<2\pi /3$, vanish on $|t|\ge 4\pi /3$, and satisfy the displayed complementary identity on the positive transition band. On a fundamental interval, either only the central translate is nonzero, or exactly the two complementary translates contribute. Thus orthonormal translates and Fourier periodization holds. The periodic mask defined on $[-\pi ,\pi ]$ by $m(t)=f(2t)$ for $|t|<2\pi /3$ and $m(t)=0$ otherwise satisfies the scaling refinement equation in Fourier form. A concrete even transition is $f(t)=\cos (\pi (3|t|/(2\pi )-1)/2)$ on the transition bands. For complex-valued $f$, ordinary squares cannot replace absolute squares: transition values $i$ and $\sqrt{2}$ have squares summing to one but squared moduli summing to three.

With $\hat{\varphi }(\xi )=\int \varphi (x)e^{-ix\xi }dx$ in the Plancherel theorem sense, the integer translates of $\varphi \in L^2(\mathbb{R})$ are orthonormal exactly whenThe Fourier coefficients of the periodized energy are the translate inner products. Uniqueness of Fourier coefficients in L1 therefore proves both directions. No absolute integrability assumption on $\varphi$ is required.

If a periodic unimodular function $a$ and its inverse have absolutely summable Fourier coefficients, replacing ${\hat{\phi }}_0$ by $\hat{\phi }=a{\hat{\phi }}_0$ preserves the orthonormal integer-translate span and integrability of an integrable scaling function. In physical space this is an absolutely summable combination of integer function translations. The new MRA low-pass filter is $m(t)=a(2t)m_0(t)/a(t)$. A rough phase can therefore preserve the multiresolution analysis while changing MRA low-pass filter regularity.

LetThe Fourier coefficients of $\theta$ are absolutely summable, so $a$ belongs to the Wiener algebra, as does $a^{-1}$. The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ implies $a(2t)=a(t)a(t+\pi )$. Starting with a compactly supported Daubechies wavelet having $p\ge 3$ vanishing moments, the periodic phase change of a scaling function gives $m(t)=a(t+\pi )m_0(t)$, while the canonical high-pass construction leaves $\hat{\psi }(2t)=e^{-it}\overline{m_0(t+\pi )}{\hat{\phi }}_0(t)$ unchanged.

At a dyadic point $t_0=2\pi k/2^j$, the terms of the difference quotient with $n\ge j$ and $2^n|h|\le 1$ each contribute $1+O((2^{n}h{)}^2)$. Their number tends to infinity, their total error is bounded, and the remaining tail contributes a bounded amount. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, so neither $\theta$ nor $a$ has a finite derivative there. These points are dense. Away from the isolated zero of $m_0$ near $\pi$, multiplication by its nonzero smooth value cannot remove this nondifferentiability. Thus the new MRA low-pass filter is not differentiable throughout any neighborhood of $\pi$, despite integrability of the scaling function and unchanged vanishing moments. Ordinary higher derivatives at $\pi$ cannot be inferred, although the corresponding Peano zero survives.

Nesting in an orthonormal multiresolution analysis expresses a scaling function in the finer-scale orthonormal basis. The associated low-pass symbol is $m(\xi )=2^{-1/2}{\sum }_k{h}_k{e}^{-ik\xi }$, and $\hat{\phi }(2\xi )=m(\xi )\hat{\phi }(\xi )$. If $\phi$ has compact support, only finitely many coefficients $h_k$ are nonzero. With $\hat{\phi }(0)=1$, iteration gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$ when the limit exists.

For an orthonormal scaling function satisfying $\phi (x)=\sqrt{2}{\sum }_k{h}_k\phi (2x-k)$, its low-pass Fourier series symbol is $m(\xi )=2^{-1/2}{\sum }_k{h}_k{e}^{-ik\xi }$. The scaling refinement equation becomes $\hat{\phi }(2\xi )=m(\xi )\hat{\phi }(\xi )$. The symbol is initially an almost-everywhere defined periodic function; the assumption that the scaling function is Lebesgue integrable alone does not make it smooth. The orthonormal translates give the quadrature mirror filter identity.

For an orthonormal wavelet refinement filter, ${\sum }_k{h}_k\overline{h_{k-2n}}={\delta }_{n0}$ is equivalent to the displayed identity. A high-pass filter can be chosen with coefficients $g_k=(-1{)}^k\overline{h_{1-k}}$. In frequency, its symbol is $e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase. The two channels split an approximation space into a coarser approximation and its orthogonal complement.

For a scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, set $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The quadrature mirror filter identity makes the two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ a unitary matrix, where $q(t)=-e^{-it}\overline{m(t+\pi )}$. The complementary coefficients $b_n=(-1{)}^n\overline{a_{1-n}}$ yield $\psi ={\sum }_n{b}_n\varphi (2\cdot -n)$. In a multiresolution analysis, coefficient-space completeness shows its translates form an orthonormal basis of $V_1\ominus V_0$, hence its dilations and translates form an orthonormal wavelet basis of the whole space.

Use $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$ and let $P_j$ be the orthogonal projection onto the closed span of the orthonormal translates and dilates of a scaling function. If $\hat{f}$ is supported in $[-R,R]$ and $R<\pi 2^j$, thenIndeed, the Plancherel theorem writes the coefficient against ${\phi }_{j,k}$ as $2^{-j/2}(2\pi {)}^{-1}\int \hat{f}(\xi )\overline{\hat{\phi }(2^{-j}\xi )}{e}^{ik{2}^{-j}\xi }d\xi$. With $\xi =2^j\theta$, these are $2^{j/2}$ times the Fourier series coefficients of $\hat{f}(2^j\theta )\overline{\hat{\phi }(\theta )}$ on $[-\pi ,\pi ]$. The Parseval identity proves the formula. It shows that continuity and unit modulus at zero imply density of the refinement spaces, and conversely that density forces this unit modulus when the Fourier transform is continuous at zero.