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This construction adapts compactly supported orthonormal wavelets to an interval by finite changes near its endpoints. It preserves local support, polynomial reproduction, nested approximation spaces and orthogonal detail spaces. The resulting basis contains coarse scaling functions, unchanged interior wavelets, and finitely many boundary wavelets per endpoint and per scale, without forcing periodic or zero boundary values.

Assume the scaling function Fourier transform and the MRA low-pass filter are $C^{p-1}$, and $m^{(k)}(\pi )=0$ for $k<p$. For every nonzero integer $j$, write $j=2^r\ell$ with $\ell$ odd. Iterating the scaling refinement equation at $2\pi j+t$ supplies a factor $m(\pi \ell +t/2^{r+1})$. All its derivatives of order less than $p$ vanish at $t=0$, by periodicity. The product rule gives ${\hat{\phi }}^{(k)}(2\pi j)=0$ for $k<p$.

An interval-adapted wavelet basis is an orthonormal basis of $L^2[0,1]$ made of a finite coarse scaling function family and resolution-indexed wavelets. A localized construction modifies only a bounded number of functions near each endpoint at every level. Boundary modification must preserve nested refinement spaces and the required vanishing moments; simple restriction of a whole-line basis does not do so.

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.

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.

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.

A multiresolution analysis organizes a function into successively finer approximation spaces. In the usual orthonormal construction on $L^2(\mathbb{R})$, these are closed linear subspaces $(V_j{)}_{j\in \mathbb{Z}}$ withRescaling relates the levels: $f(x)\in V_j$ if and only if $f(2x)\in V_{j+1}$. Integer translates of a scaling function form an orthonormal basis of $V_0$. The detail space is the orthogonal complement of $V_j$ inside $V_{j+1}$, giving $V_{j+1}=V_j\oplus W_j$. Coarse information and new details therefore fit together without overlap in squared L2 norm.

For the Haar wavelet, $V_j$ consists of square-integrable functions constant on cells of length $2^{-j}$. Splitting every cell into two halves creates one new contrast per old cell, the Haar refinement identity. On the bounded example $[0,1]$, the first three spaces have dimensions $1,2,4$: a whole-interval mean, two half-interval means, and four quarter-interval means. The Haar projection diagram displays those three resolutions. This bounded construction starts at level zero; the displayed intersection condition refers to the full family on the real line.

The usefulness is a controlled way to increase resolution. Earlier coarse information remains available while new wavelet coefficients identify what the finer description adds. The wavelet approximation complexity connects this extra resolution to both approximation error and the number of coefficients being estimated. ^[ref]

An orthonormal multiresolution analysis is a sequence of closed subspaces $V_j\subset L^2(\mathbb{R})$ satisfying $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, ${\bigcap }_j{V}_j=\{0\}$, and $f\in V_j$ iff $f(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$, and its scaled translates ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ span $V_j$. Thus $V_j$ contains information at resolution about $2^{-j}$.

The detail space $W_j=V_{j+1}\cap V_j^{\perp }$ is generated by the corresponding wavelet functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. The orthogonal splitting $V_{j+1}=V_j\oplus W_j$ iterates toAccordingly a square-integrable signal has a coarse-and-detail wavelet series,and its wavelet projection onto $V_J$ keeps the coarse coefficients and details below level $J$. Increasing $J$ adds finer-scale oscillations and localized features. On the real line there are infinitely many translates at each level; for a bounded observation interval, with an appropriate interval or periodic construction, the relevant approximation dimension satisfies $d_J\asymp 2^J$. For the Haar wavelet on a unit interval this is exactly $2^J$, corresponding to a separate mean on each dyadic subinterval.

This hierarchy gives a concrete meaning to wavelet approximation complexity. Coarse resolution uses few coefficients and can miss sharp features; fine resolution uses more coefficients and can fit noise. By orthogonality, projection bias is the energy of the omitted details,For an orthonormal sequence model with independent coefficient noise of variance ${\sigma }^2/n$, estimating the $d_J$ retained coefficients without shrinkage givesThe first term decreases with resolution while the second increases, making the bias-variance tradeoff explicit. Equivalently, a fixed orthogonal regression projection of rank $d_J$ has effective degrees of freedom $d_J$ and prediction optimism $2{\sigma }^2{d}_J/n$ under independent equal-variance errors.

Wavelets also permit local rather than only global complexity control. Thresholding small estimated detail coefficients, for example by soft thresholding, retains strong local structure while suppressing fine-scale noise elsewhere. A sparse retained coefficient set can represent localized features much more economically than keeping every coefficient through a high level. Its locations and scales matter as well as its count. If that set is selected using the same noisy responses, however, selection adds adaptivity: its raw coefficient count is not automatically its covariance-based degrees of freedom. Resolution or threshold tuning should therefore account for prediction error, using an appropriate risk estimate or cross-validation, rather than only decreasing training error.

Let $\varphi$ be the scaling function and $\psi$ an orthonormal wavelet, with ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. For $f\in L^2(\mathbb{R})$, the wavelet series at a coarse level $j_0$ isThe series converges in the L2 norm. Its coarse part captures broad variation and successive wavelet levels add finer detail. Truncating at levels $j<J$ produces the orthogonal projection onto $V_J$; Parseval identity makes its squared approximation error the sum of the omitted squared coefficients. Merely square-integrable functions need not have pointwise convergence.

For the Haar wavelet, use the cell convention $I_{j,k}=(k{2}^{-j},(k+1)2^{-j}]$ and let $I_j(x)$ be the unique such interval containing $x$. The Haar scaling functions are $2^{j/2}{1}_{I_{j,k}}$, so the Haar projection isFor a merely locally integrable function this cell-average formula remains defined, even if a global $L^2$ orthogonal projection is unavailable. If $\Vert f^{\prime }{\Vert }_{\infty }=L$, the mean value theorem gives Lipschitz continuity and, since $|t-x|\le 2^{-j}$ on the cell,Thus the requested bound holds with $\boxed{c=\Vert f^{\prime }{\Vert }_{\infty }}$, by the Haar projection error for a Lipschitz function.

To estimate a density, replace each scaling or wavelet coefficient $\int f{\varphi }_{j,k}$ or $\int f{\psi }_{j,k}$ by the sample average of that basis function. The finite-resolution Haar density estimator is thereforeIt is the histogram on the dyadic cells, is nonnegative, and integrates to one. Its expectation is $K_{j}f(x)$. A density with bounded derivative is bounded: the Lipschitz density height bound gives $f(x{)}^2\le L$ by integrating the triangular lower envelope $(f(x)-L|t-x|{)}_{+}$. The case $L=0$ cannot be a probability density on the whole real line. Hence $B=\Vert f{\Vert }_{\infty }<\infty$.

The cell probability is $p_j(x)={\int }_{I_j(x)}f\le B{2}^{-j}$. The sample count has a binomial distribution, soThe Cauchy-Schwarz inequality bounds the mean centered absolute deviation by the square root of this variance. Adding the projection bias givesChoose $j_n=\lfloor ({\log }_{2}n)/3\rfloor$, so $2^{j_n}\asymp n^{1/3}$. The two terms have the same order, andThis is a bias-variance tradeoff: refining the cells decreases the averaging bias but increases the sampling variance.

The Haar projection averages this normal distribution density on each cell; the finer cells follow its variation more closely. A Haar density estimator replaces these exact cell probabilities by empirical frequencies and therefore also introduces sampling noise.

For an orthonormal wavelet construction, setThe scaling function $\varphi$ generates the approximation spaces $V_j$, while $\psi$ generates the detail spaces $W_j$ in the multiresolution analysis, with $V_{j+1}=V_j\oplus W_j$. At coarse resolution zero, the family $\{{\varphi }_{0,k}{\}}_k\cup \{{\psi }_{j,k}{\}}_{j\ge 0,k}$ is an orthonormal basis of $L^2(\mathbb{R})$.

For the given square-integrable function the coefficients and wavelet series areFor real wavelets these are the real L2 inner products; use complex conjugates for a complex basis. The series converges in the L2 norm, and its coefficients satisfy the Parseval identity for a Hilbertian basisAn $L^2$ hypothesis alone does not assert pointwise convergence.

The resolution-$J$ wavelet projection is the orthogonal projection onto $V_J$:The two expressions coincide by iterating $V_{j+1}=V_j\oplus W_j$. Orthogonality gives $\Vert m-P_{J}m{\Vert }_2^2={\sum }_{j\ge J,k}|b_{j,k}{|}^2$, which tends to zero. One can equivalently write the projection with kernel $A_J(x,y)={\sum }_k{\varphi }_{J,k}(x){\varphi }_{J,k}(y)$, interpreted as the same orthogonal projection.

In fixed-design nonparametric regression, the observations satisfywith independent errors. In the usual Gaussian version ${\epsilon }_i\sim N(0,{\sigma }^2)$. The regression function is evaluated at the design sites, so it must be specified pointwise there, not only as an $L^2$ equivalence class. For $x_i=i/n$, the observations concern $m$ on $[0,1]$. They cannot identify its values outside this interval. Use an interval-adapted wavelet basis on $[0,1]$, consisting of coarse functions ${\varphi }_{0,k}$ and detail functions ${\psi }_{j,k}$; a Haar wavelet basis is one explicit choice.

Here is a wavelet regression estimator valid for every $n$. Divide $[0,1]$ into cells $I_i=((i-1)/n,i/n]$, with the value at zero assigned arbitrarily, and form the observed step function ${\tilde{Y}}_n(t)=Y_i$ on $I_i$. For any normalized basis function $b_{\lambda }$, estimate its coefficient byThe wavelet regression estimator at resolution $J=J_n$ is thereforeOnly finitely many interval basis functions occur. Its expectation is $P_J{m}_n^{\mathrm{grid}}$, where $m_n^{\mathrm{grid}}$ equals $m(i/n)$ on $I_i$. Thus approximation of $m$ includes the resolution bias and the grid discretization bias. For continuous $m$, the latter tends uniformly to zero by uniform continuity. No smoothness or consistency claim for arbitrary pointwise representatives of $L^2$ functions is needed to define the estimator.

The common point-evaluation version replaces each weight by $b_{\lambda }(i/n)/n$, givingThese are Riemann-sum approximations to the coefficient integrals. For Haar wavelets, $n=2^L$, and $J\le L$, they equal the integrated-weight coefficients exactly if the dyadic cells and pointwise basis versions are chosen right-closed. Each retained basis function is then constant on every sampling cell. This gives a concrete implementation by the discrete wavelet transform. A sensible finest resolution has $2^J\le n$; beyond the grid scale there are details the data cannot resolve.

For wavelet coefficient thresholding, retain the coarse coefficients but discard detail coefficients dominated by noise. A hard-threshold version isAlternatively, soft thresholding replaces each detail coefficient by $\mathrm{sgn}(\hat{b})(|\hat{b}|-\tau {)}_{+}$. The unthresholded wavelet regression estimator retains all details at the chosen resolution; thresholding retains evidence for local features while suppressing small fluctuations.

The thresholds can be derived from a simultaneous wavelet coefficient noise bound. Write $q_{\lambda }^2={\sum }_i{w}_{\lambda i}^2$. Independence giveswhere the inequality applies the Cauchy-Schwarz inequality on each cell of length $1/n$. In the dyadic Haar implementation, $q_{\lambda }^2=1/n$ exactly. For Gaussian errors, the normalized coefficient noise is standard normal. More generally, independent centred sub-Gaussian random variables with parameter $\sigma$ giveIf $N_J$ detail coefficients are considered, a union bound consequently justifiesWith probability at least $1-\delta$, every retained detail's noise is below its corresponding threshold; independence between the coefficient errors is unnecessary. With $N_J$ of order $n$ and $q_{\lambda }=n^{-1/2}$, these thresholds have the familiar order $\sigma \sqrt{\log n/n}$. Coefficients larger than their noise level represent potential signal, and a coefficient zero in the discretized mean is not selected on this simultaneous event. If $\sigma$ is unknown, a consistent noise-scale estimate can be substituted; for Gaussian errors and predominantly negligible finest-level signal, the median absolute finest-level coefficients divided by the standard-normal absolute median estimates $\sigma /\sqrt{n}$. Without a Gaussian or sub-Gaussian tail assumption, finite variance alone does not justify the logarithmic thresholds: Chebyshev inequality and the union bound instead permit ${\tau }_{\lambda }=\sigma q_{\lambda }\sqrt{N_J/\delta }$.

An orthonormal multiresolution analysis consists of closed subspaces $V_j\subset L_2(\mathbb{R})$, $j\in \mathbb{Z}$, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L_2(\mathbb{R})$, ${\bigcap }_j{V}_j=\{0\}$, and the dilation equivalence $h\in V_j⟺h(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$. Henceis an orthonormal basis of $V_j$. Set $W_j=V_{j+1}\ominus V_j$, the orthogonal complement of $V_j$ inside $V_{j+1}$. A function $\psi$ whose integer translates form an orthonormal basis of $W_0$ produces a complete orthonormal wavelet system $\{2^{j/2}\psi (2^j\cdot -k):j,k\in \mathbb{Z}\}$: different $W_j$ are orthogonal, their sum contains every difference $V_b\ominus V_a$, and density together with the zero intersection exhausts $L_2(\mathbb{R})$.

Here is the filter construction that guarantees such a wavelet. Nesting puts $\varphi$ in $V_1$, so expansion in its orthonormal basis gives the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }_2$. Define the MRA low-pass filter and its complementary quadrature mirror filter bySplitting the orthonormal translates and Fourier periodization identity at $2t$, as in part (b), gives $|m(t){|}^2+|m(t+\pi ){|}^2=1$ almost everywhere. Consequently the rows ofhave length one and are orthogonal; the matrix is unitary. Put $b_n=(-1{)}^n\overline{a_{1-n}}$ and $\psi ={\sum }_n{b}_n\varphi (2\cdot -n)$, so $q(t)=\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$.

To see completeness as well as orthogonality, identify $V_1$ with its fine-scale coefficient sequences and take their Fourier series. Group the frequencies $t$ and $t+\pi$ over a half-period. The two rows of the displayed matrix are the normalized coarse and complementary channels; their unitarity is exactly an onto isometry from these two channels to the fine-scale coefficient space. Translates of $\varphi$ and $\psi$ correspond to the Fourier basis in the respective channels. They therefore jointly form an orthonormal basis of $V_1$. The orthogonal complement channel gives the wavelet completion of a multiresolution filter, namely the integer translates of $\psi$ span $W_0$. This proves the asserted relation with an orthonormal wavelet.

Use the Fourier transform convention $f(t)=\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$ and the Plancherel theorem normalization $\Vert \varphi {\Vert }_2^2=(2\pi {)}^{-1}\Vert f{\Vert }_2^2$. For a general $L_2$ scaling function, the transform means its $L_2$ extension; an ordinary absolutely convergent integral is not required.

A change of variables gives $\widehat{\varphi (2\cdot -n)}(t)=\frac{1}{2}{e}^{-int/2}f(t/2)$. Thus the scaling refinement equation transforms to $f(t)=m(t/2)f(t/2)$, or $f(2t)=m(t)f(t)$, with the $2\pi$-periodic MRA low-pass filter $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The periodicity is necessary; allowing an arbitrary quotient $f(2t)/f(t)$ would not express refinement by integer translates.

For orthonormality, let $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. This is a $2\pi$-periodic $L_1$ function, since $f\in L_2$. The Plancherel theorem givesThese inner products equal ${\delta }_{j\ell }$ precisely when the Fourier coefficients of $P$ agree with those of one. The uniqueness of Fourier coefficients in L1 proves the equivalenceFor completeness, the converse refinement implication is also valid with the natural $L_2$ series interpretation. Suppose this periodization identity and $f(2t)=m(t)f(t)$ hold for a measurable $2\pi$-periodic $m$. Splitting the periodization at $2t$ into even and odd translates givesHence $m$ is bounded and belongs to $L_2$ of a period. Write $m=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$ in $L_2$ and let $m_N$ denote the symmetric partial sums. Since $P=1$,After substituting $t/2$, this proves convergence of the transformed refinement sums to $f$ in $L_2$. Inverting the Fourier transform proves $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$ in $L_2$. Thus the two spatial conditions and the two Fourier conditions are equivalent, with the periodic mask and convergence conventions made explicit.

An orthonormal multiresolution analysis consists of closed subspaces of a Hilbert space $(V_j{)}_{j\in \mathbb{Z}}$ of the Hilbert space $L^2(\mathbb{R})$ such thatThere is a scaling function $\varphi$ whose integer function translations form an orthonormal basis of $V_0$. Consequently ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ form an orthonormal basis of $V_j$; in particular $V_0$ is invariant under integer function translations.

Let $W_j=V_{j+1}\ominus V_j$ be the orthogonal complement of $V_j$ in $V_{j+1}$. Nesting, density and the trivial intersection imply the orthogonal direct sumFor example, telescoping $V_{J+1}=V_L\oplus {⨁}_{j=L}^J{W}_j$ and letting $L\to -\infty$, $J\to \infty$ gives this decomposition; the corresponding orthogonal projections converge strongly to zero and the identity by the intersection and density properties.

To construct the orthonormal wavelet, use $\varphi \in V_1$ to write the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }^2$ and ${\sum }_n|a_n{|}^2=2$. Put $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The orthonormal translates and Fourier periodization criterion proved in part (b), applied to the scaling refinement equation, yieldsThis is the quadrature mirror filter identity. DefineThe Fourier series of $q$ is $\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$, and $\hat{\psi }(2t)=q(t)\hat{\varphi }(t)$. The two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ form a unitary $2\times 2$ matrix almost everywhere: both have squared norm one and their inner product is zero. This wavelet completion of a multiresolution filter proves not just orthogonality but completeness. Splitting a finer-scale coefficient Fourier series into its values at $t$ and $t+\pi$, this invertible two-channel matrix resolves it into a coarse-scale channel and its complementary channel. Thus the integer function translations of $\psi$ form an orthonormal basis of $W_0$.

The dilated functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ therefore form an orthonormal basis of each $W_j$. The orthogonal direct sum proves the conclusion: every orthonormal multiresolution analysis produces an orthonormal wavelet basis $\{{\psi }_{j,k}:j,k\in \mathbb{Z}\}$ of $L^2(\mathbb{R})$.

An orthonormal multiresolution analysis consists of closed subspaces $(V_j{)}_{j\in \mathbb{Z}}$ of the L2 space $L^2(\mathbb{R})$ with $V_j\subset V_{j+1}$, dense union, intersection $\{0\}$, and the dilation rule $f\in V_j⟺f(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$. Equivalently,form an orthonormal basis of $V_j$. Let $W_j=V_{j+1}\ominus V_j$. Density and the trivial intersection imply $L^2(\mathbb{R})={⨁}_{j\in \mathbb{Z}}{W}_j$.

Nesting gives the scaling refinement equation $\varphi (x)={\sum }_n{a}_n\varphi (2x-n)$, with $h_n=a_n/\sqrt{2}$ the coefficients in the normalized finer-scale basis. Orthogonality of the translates gives ${\sum }_n{a}_n\overline{a_{n-2r}}=2{\delta }_{r0}$. Its Fourier form is the quadrature mirror filter identity $|m(t){|}^2+|m(t+\pi ){|}^2=1$, where $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$.

Set $b_n=(-1{)}^n\overline{a_{1-n}}$ and define $\psi (x)={\sum }_n{b}_n\varphi (2x-n)$ in $L^2$. Its high-pass symbol is $q(t)=-e^{-it}\overline{m(t+\pi )}$. The matrixis unitary almost everywhere. Under the identification of $V_1$ with its finer-scale coefficient space, this two-channel unitary transform splits that space into the translates of $\varphi$ and the translates of $\psi$. The latter are therefore an orthonormal basis of $W_0$. Dilating gives an orthonormal wavelet basisThis wavelet completion of a multiresolution filter explains how the multiresolution analysis supplies a single orthonormal wavelet. The argument uses coefficient-space completeness, not just pairwise orthogonality.

Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

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 $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.