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The normalized Haar scaling functions and wavelets obeyThis orthogonal two-by-two transformation gives the multiresolution analysis decomposition $V_{j+1}=V_j\oplus W_j$. Iterating it expresses a Haar approximation either through level-$j$ cell averages or through coarse averages and all wavelet details at lower levels. The identities remain valid locally for coefficient integrals of locally integrable functions, without a global $L^2$ assumption.

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.

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.

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 $L^2(\mathbb{R})$.

The half-open translates $[-\pi ,\pi )+2\pi k$ partition the real line. Exactly one summand is therefore nonzero in the periodization of $|f{|}^2$, giving ${\sum }_k|f(t+2\pi k){|}^2=1$.

Choose the Shannon scaling mask to be the $2\pi$-periodic extension of $1_{[-\pi /2,\pi /2)}$. On the support of $f$, its product with $f$ is $1_{[-\pi /2,\pi /2)}(t)=f(2t)$; outside that support both sides vanish. Thus the scaling refinement equation holds in frequency. The Fourier coefficients give its precise filter normalization:These coefficients are square summable, so the interpretation in part (b) applies.

The Fourier inversion theorem gives the Shannon scaling functionThis is the normalized sinc function. Its Fourier transform is taken in the Plancherel theorem sense: the Shannon scaling function belongs to $L^2$ but not $L^1$, so its Fourier integral is not absolutely convergent. The inverse integral above is an ordinary integral because the rectangular transform belongs to $L^1$. The resulting refinement spaces are exactly the $L^2$ functions with Fourier transform supported in $[-2^j\pi ,2^j\pi ]$; their union is dense and their intersection is zero. Thus this example also gives a full multiresolution analysis.

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 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.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ 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 givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ 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 $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe 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 $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, 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: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 $1+O((2^{n}h{)}^2)$ 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 $t\ne 0$. 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.

Put $W_j=V_{j+1}\cap V_j^{\perp }$, the detail space of the multiresolution analysis. The Meyer-Mallat theorem gives a function $\psi \in W_0$ whose integer translates form an orthonormal basis of $W_0$. Dilations then give an orthonormal basis of each $W_j$, and the density and trivial-intersection axioms implyFor an explicit filter construction, write the scaling refinement equation $\phi (x)=\sqrt{2}{\sum }_k{h}_k\phi (2x-k)$. With a quadrature mirror filter one may take $g_k=(-1{)}^k\overline{h_{1-k}}$ and $\psi (x)=\sqrt{2}{\sum }_k{g}_k\phi (2x-k)$. A harmless constant sign or integer translation gives an equivalent orthonormal wavelet.

The answer is no with the integer-translation and dilation conventions of this multiresolution analysis. The orthonormal basis axiom forces $V_0$ to consist of $L^2$ functions constant on the cells $[k-1/2,k+1/2)$. Dilation forces $V_1$ to consist of functions constant on $[k/2-1/4,k/2+1/4)$. However, the proposed scaling function changes value at $1/2$, in the interior of the $V_1$ cell $[1/4,3/4)$. It is therefore not in $V_1$, contradicting nesting:Changing values at endpoints makes no difference in $L^2$. The usual Haar wavelet instead uses the scaling function ${\chi }_{[0,1)}$; an arbitrary half-unit translation does not preserve the required refinement grid.

At zero, the partial products are $m(0{)}^n$. Their assumed convergence and $|m(0)|=1$ force $m(0)=1$ and $\Theta (0)=1$. Uniform convergence on compact sets makes $\Theta$ continuous. The identity with $k=0$ gives $\Theta \in L^2$ with squared norm $2\pi$. Define $\phi$ by the inverse Fourier transform in $L^2$. The Plancherel theorem and the other given integral identities giveDefine $V_j$ as the closed span of ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$. These functions form an orthonormal basis of $V_j$, and dyadic dilation gives the scale axiom. Shifting the infinite product givesso the finite Fourier series of $m$ gives a scaling refinement equation and $V_j\subset V_{j+1}$. The coarse-scale projection argument of part 1(b) proves the trivial-intersection axiom.

For density, take a function $f$ with bounded Fourier support. The MRA projection Fourier identity gives, for large $j$,Uniform convergence of $\Theta (2^{-j}\xi )$ to one on that support proves the limit. Since $P_j$ is an orthogonal projection, $\Vert f-P_{j}f{\Vert }_2^2=\Vert f{\Vert }_2^2-\Vert P_{j}f{\Vert }_2^2\to 0$. Such functions are dense in $L^2$, givingThus all the multiresolution analysis axioms hold. Under the strong convergence and orthogonality assumptions supplied here, the additional nonvanishing condition is not needed in this last verification; it is useful when establishing those assumptions from a filter.

For the multiresolution analysis $V_j$ and its detail spaces $W_j$, take the completed tensor product ${\mathcal{V}}_j=V_j\otimes V_j$. Since $V_{j+1}=V_j\oplus W_j$,The three tensor-product wavelets areTaking products of members of the two one-dimensional families gives an orthonormal basis of each respective completed tensor product. Density and the coarse-scale trivial intersection yieldas an orthonormal basis of $L^2({\mathbb{R}}^2)$. The normalization is $2^j$ in two dimensions. On the square, tensorize the interval-adapted wavelet basis and include the finite coarse scaling function part.

An orthonormal wavelet is $\psi \in L^2(\mathbb{R})$ such that $\{2^{j/2}\psi (2^{j}x-k){\}}_{j,k\in \mathbb{Z}}$ is an orthonormal basis. A multiresolution analysis is a nested family of closed spaces $V_j$ with trivial intersection, dense union, dyadic scaling, integer-translation invariance of $V_0$, and a generator $\varphi$ whose integer translates form an orthonormal basis of $V_0$. The Meyer-Mallat theorem says every such analysis has an orthonormal wavelet whose translates span $V_{j+1}\ominus V_j$.

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.

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.

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.

From a one-dimensional multiresolution analysis, the simultaneous-scale two-dimensional detail space splits as $(W_j\otimes V_j)\oplus (V_j\otimes W_j)\oplus (W_j\otimes W_j)$. Its generating wavelets are $\psi \otimes \phi$, $\phi \otimes \psi$, and $\psi \otimes \psi$. Their normalized dilates are $2^j\Psi (2^j\cdot -k)$ for $k\in {\mathbb{Z}}^2$. Include a coarse scaling family when working on a bounded square.

An orthonormal multiresolution analysis separates coarse scaling functions from progressively finer wavelet details. On a bounded interval, the approximation space at level $J$ has dimension of order $2^J$. Higher resolution reduces projection bias but estimates more noisy coefficients. In an orthonormal Gaussian sequence model with coefficient noise variance ${\sigma }^2/n$, risk is the omitted squared coefficient sum plus ${\sigma }^2{d}_J/n$. Selecting or shrinking individual detail coefficients yields spatially adaptive complexity; a data-selected subset need not have effective degrees of freedom equal to its raw selected count.