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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.

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]

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.

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.