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A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the Haar wavelet, replace each adjacent pair $(u,v)$ byThe approximation coefficient $a$ is $\sqrt{2}$ times the pair's arithmetic mean; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.

For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt{2},7\sqrt{2})$ and detail coefficients $(-\sqrt{2},-\sqrt{2})$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform isThis is an orthogonal matrix transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared Euclidean norm:The inverse function for each pair is $u=(a+d)/\sqrt{2}$, $v=(a-d)/\sqrt{2}$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical energy. ^[ref]

Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a step function on $n$ equal cells of $[0,1]$, its normalized L2 inner product coefficients are the discrete coefficients divided by $\sqrt{n}$. For our four values they are $(5,-2,-1/\sqrt{2},-1/\sqrt{2})$, as the Haar projection example shows. The same normalization is used by the wavelet regression estimator on $[0,1]$.

Under independent Gaussian noise with a common variance, orthogonality preserves that coefficient noise variance and independence. It does not make every arbitrary correlated signal independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.

A dyadic partition of $[0,1]$ at level $J$ consists of $2^J$ equal cells of length $2^{-J}$. Each cell splits into two cells at the next level. Use left-closed, right-open cells, adding the endpoint $1$ to the final cell when a pointwise representative is needed. These nested partitions underlie the Haar scaling functions and Haar wavelets.

At each point there is at most one supported Haar wavelet per scale. If a detail coefficient is below $\delta$, its value there is at most $\delta 2^{j/2}/\sqrt{L}$. Summing these bounds up to level $K$ gives the displayed geometric estimate. Selecting $K$ so the final term is of the order of the modulus of continuity controls all omitted fine details, even though arbitrary deletions from a uniformly convergent series need not preserve convergence.

The Haar projection at level $J$ is the orthogonal projection onto the functions constant on level-$J$ dyadic intervals. On each cell it replaces $f$ by its cell average. This is the closest cellwise-constant approximation in the L2 norm, because choosing a constant $c$ on a cell minimizes ${\int }_I|f-c{|}^2$ at its cell average.

On $[0,1]$, take the step function with values $2,4,6,8$ on successive quarter-intervals. At level zero, its Haar projection is the constant $5$. At level one, it has values $3$ and $7$ on the two halves. At level two, it recovers the four original values exactly. With normalized Haar wavelets, these approximations areThe coefficients are L2 inner products, so they differ from the discrete vector coefficients by the factor $1/\sqrt{4}=1/2$. Their squared sum also has a direct meaning:The approximation errors in squared L2 norm of $P_{0}f,P_{1}f,P_{2}f$ are respectively $5,1,0$: each added detail removes exactly its own squared coefficient from the error.

The Haar projection can be expressed using level-$J$ Haar scaling functions, or through a coarse scaling family and the Haar wavelets at intermediate levels. On $[0,1]$ with lowest level zero, that coarse family is the single constant function $\varphi$. On the whole real line, use the integer translates of a coarse Haar scaling function instead; one global constant is not square-integrable. Refinement improves the representation while retaining the earlier coefficients.

The Haar wavelet is a signed contrast between two adjacent half-intervals:It has compact support, zero integral, and unit L2 norm: $\int \psi =1/2-1/2=0$ and $\int |\psi {|}^2=1$. The zero integral is its one vanishing moment. Its Haar scaling function is $\varphi =1_{[0,1)}$, which records a coarse value rather than a contrast. Endpoint conventions do not alter the L2 inner products.

The normalized copies ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ compare the halves of the dyadic interval $[k{2}^{-j},(k+1)2^{-j})$. Narrower copies are taller so that their L2 norms stay one. Thus a coefficient identifies both a location, through $k$, and a scale, through $j$.

For a simple step function $f$ equal to $2$ on $[0,1/2)$ and $6$ on $[1/2,1)$, its coarse coefficient is $⟨f,\varphi ⟩=4$, while its first contrast is $⟨f,\psi ⟩=-2$. Thus $f=4\varphi -2\psi$. The negative coefficient raises the second half and lowers the first. Finer contrasts vanish because $f$ is constant on each half. A jump aligned with this dyadic interval boundary therefore needs only two nonzero coefficients; jumps at other positions can require details at many scales.

Its jump discontinuities make the Haar wavelet particularly transparent for changes between neighbouring regions, while limiting approximation of smooth variation. A Daubechies wavelet uses a more elaborate pattern to obtain additional vanishing moments and, at sufficiently high order, smoother basis functions. The elementary Haar refinement identity has coefficients $h_0=h_1=1/\sqrt{2}$ and explains how the coarse and fine descriptions fit together.

The historical source is Alfred Haar's ^[ref]. Chapter III, beginning on page 361, constructs the constant function and these localized piecewise-constant orthogonal functions. Its original endpoint assignments differ from the half-open convention used here, without changing L2 inner products.

Let $\psi$ be an integrable function with at least one vanishing moment. For a unit jump discontinuity $f(x)=H(x-x_0)$, where $H$ is the Heaviside step function, substitution into the coefficient givesThe signed integral of the whole analyzing function is zero, but the jump discontinuity cuts through its support of a function and usually leaves an unbalanced partial integral. At fixed relative position $\theta$ with nonzero partial integral, the coefficient decays as $\sqrt{a}$, slower than the $a^{q+1/2}$ Taylor remainder bound on a smooth region. “Edges survive” means this relative contrast: the normalized coefficients still tend to zero as $a\to 0$.

For an original exact comparison, use the smooth polynomial $f(x)=x+x^2$ at $b=0$. The Haar wavelet $h$ and the second-difference analyzing function $\eta$ giveThe linear term cancels for $\eta$; it remains for $h$. For the fixed unit jump discontinuity $H(x)$, choose $b=-a/4$ so that $\theta =1/4$ at every scale. Then

Position matters. If an analyzing function's support of a function lies entirely on one constant side of the jump discontinuity, its coefficient is zero. Even a crossing can cancel: $\eta$ has zero partial integral at $\theta =3/2$. A jump discontinuity on a dyadic interval boundary can likewise escape finer Haar wavelets whose supports do not cross it. These examples describe particular functions and normalizations, not a universal ranking of wavelets for signal processing.

Let $H_{n,k}(s)=2^{n/2}(1_{(0,1/2)}-1_{(1/2,1)})(2^{n}s-k)$ for $n\ge 0$ and $0\le k<2^n$. Integrating these Haar wavelets gives tents of height $2^{-n/2-1}$ supported on the corresponding dyadic intervals. For independent standard normal random variables $Z_0,Z_{n,k}$, the displayed series, summed level by level, is exactly the succession of midpoint displacements in the Lévy construction of Brownian motion. Its almost sure convergence is uniform, and its limit is standard Brownian motion. This gives a continuous function from Gaussian coefficients even though the unintegrated unit-variance Haar wavelet series diverges in L2 space.

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 wavelet $\psi$ generates an orthonormal basis of $L^2(\mathbb{R})$ throughThe factor $2^{j/2}$ compensates for the shrinking support of a function, keeping the L2 norm equal to one. Orthogonality makes different coefficients separate squared-norm contributions, and completeness allows the wavelet series to recover any square-integrable function in the L2 norm.

For the Haar wavelet, wavelets at one scale have disjoint supports. At different scales their dyadic intervals are either disjoint or nested. On the support of the finer wavelet, the coarser one is constant, while the finer one has zero integral, so their L2 inner product is zero. Together with the unit L2 norm, this explains orthonormality. On the bounded interval $[0,1]$, add the constant Haar scaling function and use $j\ge 0$, $0\le k<2^j$; the constant function is not an $L^2(\mathbb{R})$ basis element on the whole line.

Enumerate the orthonormal basis consisting of $H_0$ and the Haar wavelets $H_{n,k}$ as $(e_j{)}_{j\ge 1}$, and enumerate the corresponding independent standard normal random variables as $(A_j{)}_{j\ge 1}$. Set $c_j=⟨h,e_j{⟩}_{\mathcal{H}}$. The Parseval identity for a Hilbertian basis givesThe partial sums $T_m={\sum }_{j=1}^m{c}_j{A}_j$ form a martingale for $\sigma (A_1,\dots ,A_m)$, since the next summand is independent of the past and has mean zero. Moreover,The martingale convergence theorem for a martingale bounded in L2 space proves convergence almost surely and convergence in L2. Equivalently one may apply the Kolmogorov convergence theorem to the summable variances $c_j^2$.

A finite linear combination of independent normal random variables is normal. Its characteristic function is $\mathbb{E}{e}^{iu{T}_m}=\exp (-u^2{\sum }_{j\le m}{c}_j^2/2)$. Passing to the limit identifiesFor $h=0$ this is the constant zero random variable. Each fixed $h$ has its own permissible null set; simultaneous pathwise convergence for all $h\in \mathcal{H}$ is unnecessary.

Let $S_N={\sum }_{n=0}^N{\sum }_{k=0}^{2^n-1}{a}_{n,k}{H}_{n,k}$. The assumed orthonormality of the Haar wavelets givesEnumerate the coefficients level by level as $A_1,A_2,\dots$. They are independent and identically distributed standard normal random variables, so the squares have mean one. The strong law of large numbers impliesConsequently,A convergent sequence in a Hilbert space must be bounded in norm. Hence this random Haar wavelet series diverges in $\mathcal{H}$ almost surely.

Put $L=2\pi$ and represent the circle by $[0,L)$. The full orthonormal basis of Haar wavelets includes the constant $H_{\ast }=L^{-1/2}$ and, for every dyadic interval $I$ of length $\ell =L{2}^{-j}$, the normalized detailTake half-open intervals and the corresponding values at their boundaries, so every point belongs to exactly one interval at each level. The constant basis function is necessary: if it were omitted, even $f\equiv 1$ could not satisfy the asserted convergence. We use the full conventional Haar wavelet system in the threshold sum.

Let $A_{J}f$ be the function equal to the mean of $f$ on each level-$J$ dyadic interval. Comparing the means on a parent and its two children proves the Haar refinement identity directly: the difference between the child mean and the parent mean is the appropriate value of $c_I{H}_I$. Inductively this gives the Haar projectionWriting ${\omega }_f(s)={\sup }_{\mathrm{dist}(x,y)\le s}|f(x)-f(y)|$ for the modulus of continuity, the cell-average formula yieldsFor a detail on a cell $I$, its integral is zero. Subtracting $f(x_I)$ for any $x_I\in I$ therefore gives the coefficient estimateIn particular $|c_I|\le 2\Vert f{\Vert }_{\infty }\sqrt{\ell }$, so for every $\delta >0$ only finitely many details can have $|c_I|\ge \delta$. The hard-threshold sumis thus a well-defined finite sum. Its chosen details need not form a complete set of levels; this is why convergence of the Haar projections alone does not prove the result.

Fix $\eta >0$. Choose $J\ge 1$ so that ${\omega }_f(L{2}^{-J})\le \eta$. All finer details satisfy $|c_I|\le \eta \sqrt{L}{2}^{-j/2}$ at level $j\ge J$. For sufficiently small $\delta$, every nonzero coefficient at levels below $J$, and the constant coefficient if nonzero, is retained. Also choose the unique integer $K\ge J$ such thatFor $j>K$ we have $|c_I|\le \eta \sqrt{L}{2}^{-j/2}<\delta$, so no such detail is retained. Consequently $T_{\delta }f$ differs from $A_{K+1}f$ only by omitted details at levels $J$ through $K$. At any point there is at most one supported detail per level, and an omitted detail has magnitude at most $\delta /\sqrt{L{2}^{-j}}$. The geometric bound for omitted Haar details givesSince $\Vert A_{K+1}f-f{\Vert }_{\infty }\le \eta$, we concludeThe right side can be made arbitrarily small, proving uniform hard-threshold convergence of normalized Haar expansions:The square-root scaling imposed by the specified $L^2$ normalization is what makes the omitted contributions a controlled geometric series.

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 ${\tau }_{\lambda }=\sigma q_{\lambda }\sqrt{N_J/\delta }$.

Use half-open intervals and setThe Haar scaling functions and Haar wavelets areFor fixed $j$, each scaling function has squared integral $2^j{2}^{-j}=1$, and different $k$ have disjoint supports, proving that this family is orthonormal. Each wavelet also has norm one and integral zero. At a fixed level, distinct wavelets have disjoint supports. At different levels, their dyadic supports are either disjoint or the finer support lies in a half on which the coarser wavelet is constant. The finer wavelet has integral zero, so their inner product vanishes. Finally, a wavelet with $l\ge 0$ is supported inside a unit interval; it is orthogonal to its containing ${\phi }_{0,k}$ because its integral is zero, and to all other unit scaling functions because their supports are disjoint. Thus both the scaling family at each fixed level and the complete stated Haar family are orthonormal sets.

The Haar refinement identity isIt shows that the closed scaling spaces $V_j$ satisfy $V_{j+1}=V_j\oplus W_j$, where $W_j$ is the closed span of the level-$j$ wavelets. The union of the $V_j$ is dense in $L^2(\mathbb{R})$: continuous compactly supported functions can be approximated in $L^2$ by their dyadic cell averages, and such continuous functions are dense in $L^2$. Hence the orthonormal Haar family is also a complete orthonormal basis.

For a locally integrable function, define the compact-support coefficient integralsThese exist without requiring $f\in L^2$. The two formulas for the Haar approximation areFor $j=0$ the wavelet sum is empty. For every fixed $j$, these sums are locally finite, so they make sense pointwise and locally in $L^1$. The refinement identities and their orthogonal two-by-two coefficient transformation establish equality by induction, even for merely locally integrable $f$. If $f\in L^2$, this is also the orthogonal projection onto $V_j$.

In particular, for $\delta =2^{-j}$ and $x\in I_k=[k\delta ,(k+1)\delta )$,Now impose the symmetry and monotonicity conditions. The limit zero and decrease on $[0,\infty )$ imply $0\le f(x)\le f(0)$ for $x\ge 0$. For $k\ge 0$, the cell average and every value inside $I_k$ lie between $f((k+1)\delta )$ and $f(k\delta )$. ThereforeSumming over the positive half-line telescopes and gives a bound $\delta f(0)$. Reflection maps each dyadic cell to another dyadic cell up to endpoints. Since $f$ is even, its Haar approximation is even almost everywhere as well, so the negative half-line has the same error. ConsequentlyThe argument proves integrability of the difference, even if $f$ itself is not integrable on the whole line. It is the symmetric monotone case of the Haar approximation error for a function of bounded variation.

In the Gaussian white noise model, observe the entire pathwhere $W$ is standard Brownian motion and the deterministic drift belongs to L2 space on $[0,1]$. Equivalently, $dY(t)=f(t)dt+n^{-1/2}dW(t)$. For every deterministic $h\in L^2[0,1]$, the observed stochastic integral satisfiesThe noise $W(h)=\int hdW$ is an isonormal Gaussian process. In particular its variance is $\Vert h{\Vert }_2^2$. Gaussian white noise is interpreted through these integrals, rather than as an ordinary random function with a pointwise value at every time. No smoothness of $f$ is needed.

Here is the Gaussian maximum bound without independence. Put $M={\max }_{1\le i\le N}|g_i|$. It is integrable since it is bounded by ${\sum }_i|g_i|$. For every $\lambda >0$,The second inequality uses only the moment-generating function of the standard normal distribution, so independence is unnecessary. Jensen inequality gives $e^{\lambda EM}\le E{e}^{\lambda M}$, henceMinimizing at $\lambda =\sqrt{2\log (2N)}$ proves

For the dyadic partition, let $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$, $0\le k<2^J$, putting the endpoint $1$ into the final interval. The Haar scaling functionsform an orthonormal basis of the space $V_J$ of functions constant on each interval. The Haar wavelets can be written asFor $J\ge 1$, the constant function ${\varphi }_{0,0}=1$ together with ${\psi }_{j,k}$, $0\le j<J$, is another orthonormal basis of $V_J$. Indeed, each successive level splits a cell's two constants into their sum and difference; the number of basis elements is $1+{\sum }_{j=0}^{J-1}{2}^j=2^J$. Thus the Haar projection is the orthogonal projectionOn cell $I_{J,k}$ its value is $2^J{\int }_{I_{J,k}}f(t)dt$. The first formula also applies to $J=0$.