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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.

The H1 space is the first-order Sobolev space whose elements and first weak derivatives belong to L2 space. It is a Hilbert space with inner product given by the sum of the L2 inner products of the functions and of their first weak derivatives.

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.

The mass matrix is the Gram matrix of a finite element basis in the L2 inner product. Its coefficient quadratic form is the squared L2 norm of the represented trial function, so it is a positive-definite matrix for a linearly independent basis.

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.

Solutions to all three questions of the finite element methods for partial differential equations course of the University of Oxford Part C examination, sat on 6 June 2023. Source: original released PDF, bundle pages 325–328; paper pages 1–4. The original numbering is retained; question statements are not reproduced.

All function spaces and bilinear forms below are real. Write $(p,q{)}_0={\int }_{\Omega }pq$ for the L2 inner product, and $\Vert p{\Vert }_1^2=\Vert p{\Vert }_0^2+\Vert \nabla p{\Vert }_0^2$ for the first-order Sobolev norm; the notation applies componentwise to vector fields. We use the Poincaré inequality explicitly allowed by the paper. On an arbitrary unbounded Lipschitz domain that inequality would need an additional assumption. The convergence assertions use the usual full conforming finite element spaces, or vector subspaces containing the indicated interpolants; merely being a vector subspace of the stated piecewise polynomial functions is insufficient. Part 1(d) explains the intended meaning of positivity for its nonsymmetric matrix.

The Fourier sine basis of the unit square is $e_{j\ell }(x,y)=2\sin (j\pi x)\sin (\ell \pi y)$, $j,\ell \ge 1$, normalized in the L2 inner product. It satisfies the zero Dirichlet boundary conditions and $\Delta e_{j\ell }=-{\pi }^2(j^2+{\ell }^2)e_{j\ell }$. Expand the given initial datum as $u_0=\sum a_{j\ell }{e}_{j\ell }$, where $\sum |a_{j\ell }{|}^2=\Vert u_0{\Vert }_2^2$. The solution isFor $\kappa \le 2{\pi }^2$, every exponent is nonpositive. The series converges in $L^2$ for $t\ge 0$, tends to $u_0$ in $L^2$ as $t\downarrow 0$ by dominated convergence, and differentiates termwise for $t>0$ because the Gaussian high-frequency decay dominates every polynomial derivative factor. This constructs a classical positive-time solution and the appropriate mild solution at the possibly rough initial time.

Its norm and the difference of two such solutions satisfyExistence, uniqueness in the natural $L^2$ solution class, and continuous dependence follow. For uniqueness one can project any weak solution onto each sine mode, obtaining the same scalar linear equation and initial coefficient. Thus this is a Hadamard well-posed problem, with a contractive strongly continuous semigroup in the stated range.

To prove pointwise decay from merely $L^2$ data, use $|e_{j\ell }|\le 2$ and the Cauchy-Schwarz inequality:For $t\ge 1$, factor out $e^{2(\kappa -2{\pi }^2)t}$; the remaining sum is bounded by ${\sum }_{j,\ell \ge 1}{e}^{-2{\pi }^2(j^2+{\ell }^2-2)}<\infty$. ThereforeAt equality the first sine mode is stationary, so strict inequality is essential for decay of every datum. This is the Dirichlet heat-reaction threshold on the unit square. The equation is actually well posed on every fixed finite time interval for any fixed real $\kappa$; beyond the threshold it has a growing first mode and loses long-time contractivity. The restricted well-posedness assertion in the question is therefore sufficient rather than necessary.

Work with real functions and $f\in L^2(0,1)$. To give the variational problem a definite domain, choose the homogeneous Dirichlet boundary conditions used in the next part. The natural form space is the zero-trace Sobolev space $H=H_0^1(0,1)$, with norm $\Vert v{\Vert }_H=\Vert v^{\prime }{\Vert }_2$. The Sobolev trace operator is well defined because one-dimensional $H^1$ functions have absolutely continuous representatives. For example $v(0)={\int }_0^{1}v(x)dx-{\int }_0^1(1-x)v^{\prime }(x)dx$, so its endpoint trace is bounded in the $H^1$ norm; the other endpoint has the analogous estimate. The zero-trace space is closed in the Hilbert space $H^1$, and the Poincaré inequality $\Vert v{\Vert }_2\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_2$ makes the chosen derivative norm equivalent to its full norm. Thus $H$ is a Hilbert space.

Differentiability of $p,q$ implies continuity on the compact interval. Hence there are finite constants $p_{\max },q_{\max }$ and a positive $p_{\min }$ with $p_{\min }\le p\le p_{\max }$ and $0\le q\le q_{\max }$. Define the symmetric bilinear formThe Cauchy-Schwarz inequality and Poincaré inequality give all the estimates required by the Lax-Milgram theorem:That theorem states that a continuous coercive bilinear form on a Hilbert space and a bounded linear functional determine a unique weak solution. Completeness, continuity, coercivity and boundedness have now all been verified. Therefore there is a unique $u\in H$ with $a(u,v)=⟨f,v⟩$ for every $v\in H$.

The energy version of the given variational problem is $I(v)=a(v,v)-2⟨f,v⟩$. Its exact variation isBoundedness of $a$ makes the last term quadratic in $\Vert w{\Vert }_H$, so the Fréchet derivative is $DI(v)[w]=2a(v,w)-2⟨f,w⟩$. Its vanishing for every test function is preciselyTaking compactly supported smooth tests and integrating in the distributional sense shows that this is the Euler-Lagrange equation $-(p{u}^{\prime }{)}^{\prime }+qu=f$. Conversely that weak equation gives the displayed stationary condition. The stationary point is actually the unique global minimizer, becauseEquality occurs only when $v=u$. This proves the variational equivalence, including existence and uniqueness rather than only a formal derivative calculation.

For precision about the literal L2 inner product $⟨Lv,v⟩$, use the flux-domain realization of a one-dimensional elliptic operatorFor such $v$, weak integration by parts and its zero endpoint values give $⟨Lv,v⟩=a(v,v)$. On the larger form space $H$, the same energy is interpreted by the weak operator pairing. The minimizer belongs to $D(L)$ automatically: its weak equation gives $(p{u}^{\prime }{)}^{\prime }=qu-f\in L^2$, while $p{u}^{\prime }\in L^2$. Thus its equation also holds as an equality of $L^2$ functions. This domain does not assume bounded $p^{\prime }$, which mere differentiability would not supply. If $p\in C^1$ one may equivalently use $D(L)=H^2\cap H_0^1$, but that stronger coefficient assumption is not needed for the proof here.

Fix a bounded tile in Euclidean space $\Omega$ with labelled flat boundary faces. For each face label $s$, encode an assembly of $n$ copies by an involutive real symmetric signed permutation matrix $M_s$: a paired interface contributes a two-by-two swap, an unglued face with a Neumann boundary condition contributes $+1$ on the diagonal, and an unglued face with a Dirichlet boundary condition contributes $-1$. Use the same face-identification convention on each copy. If a second assembly has matrices $N_s$ and there is an invertible constant matrix $C$ satisfyingthen the map $u=(u_1,\dots ,u_n)\mapsto Cu$ is an isomorphism between corresponding Laplace-Beltrami operator eigenspaces. This is the transplantation theorem.

To prove it, restrict an eigenfunction to the tile copies and identify each copy with $\Omega$. Along a face, let $a$ be its vector of boundary values and $b$ its vector of outward normal derivatives. At a glued interface the values agree and the outward derivatives are opposite. At an unglued face with a Neumann boundary condition the derivative vanishes, and at a face with a Dirichlet boundary condition the value vanishes. All three cases are encoded byThe intertwining identities give $N_s(Ca)=Ca$ and $N_s(Cb)=-Cb$, so the transplanted function satisfies the second assembly's interface and boundary conditions. On each tile its Laplace-Beltrami operator is the same constant linear combination of the original Laplacians, hence it has the same eigenvalue. Across an interface, matching values and opposite normal derivatives make it a weak solution; elliptic regularity, or the allowed reflection principle, makes it a genuine smooth solution there. At corners the operator is understood by its finite-energy, Friedrichs extension; the same matching argument applies to its quadratic form domain. Applying $C^{-1}$ reverses the construction. Thus dimensions of all eigenspaces agree, including multiplicities, and the two specified boundary-value spectra coincide.

The requested pair exists with explicitly specified mixed boundary conditions and polygonal boundary. Take four copies $Q_0,Q_1,Q_2,Q_3$ of an octagonal regular polygon in Euclidean space. Label four alternate sides $a,b,c,d$ in cyclic order; the four intervening sides always have a Neumann boundary condition. The labelled sides are disjoint even at their endpoints. Parameterize every side in the positive boundary direction of the reference octagon. When pairing same-labelled sides, identify equal parameters. This produces a locally flat surface with boundary: a neighbourhood of an interior point of an interface consists of two half-disks glued along their diameters. All tile vertices remain boundary points, with boundary angles $3\pi /4$ or $3\pi /2$; there are no interior cone points.

Here are the complete gluing and boundary conditions. An entry $ij$ pairs the corresponding sides of $Q_i,Q_j$. The indicated unpaired sides have a Dirichlet boundary condition; every other unpaired side has a Neumann boundary condition.Both assemblies are compact and connected. The Laplace-Beltrami operator is the nonnegative operator $P=-\Delta$ associated with the finite-energy quadratic form $\int |\nabla u{|}^2$, with zero Sobolev trace on the specified Dirichlet arcs. Thus the mixed boundary condition is defined without requiring classical differentiability at a corner or at a Dirichlet–Neumann junction.

The four signed permutation matrices for $D$ areFor $B$ they areUse the Hadamard matrixThe intertwining relations can be checked on all columns at once. Writing the columns of $C$ as $h_0,h_1,h_2,h_3$, direct application of the displayed $N_s$ givesFor the four intervening faces both assemblies have the identity face matrix, so these intertwining relations hold as well. Also $C^{T}C=C^2=4I$. This four-tile mixed-boundary transplantation between a disk and a nonorientable surface gives a bijection of every pair of eigenspaces, preserving eigenvalues and multiplicities.

There is also a direct argument on the quadratic form domains that handles all boundary corners. The orthogonal matrix $U=C/2$ acts on tile restrictions, preserves their L2 inner product and the sum of their Dirichlet energies, and commutes with taking a Sobolev trace. The condition $M_{s}u=u$ on each face encodes equality of the two traces at a paired interface and vanishing trace at an unpaired Dirichlet face. Intertwining takes these conditions exactly to $N_s(Uu)=Uu$. Consequently $U$ maps the whole form domain onto the other form domain and preserves its quadratic form. The associated self-adjoint operators are unitarily equivalent, proving isospectrality for the stated mixed boundary conditions without any assumption of smoothness at a Dirichlet–Neumann junction.

We now verify the actual topology using the topology of disk tiles glued along disjoint boundary arcs. Because the gluing arcs in each tile are disjoint, the resulting surface with boundary is a thickening of its gluing graph, with one vertex for each closed disc tile and one edge for each paired arc. One can construct this homotopy equivalence by taking a small central closed disc in each tile and narrow strips to its paired arcs. What remains is a collar neighbourhood of the boundary, which retracts onto this connected system of closed discs and strips. Shrinking each central closed disc and each strip then gives the gluing graph. Equivalently, begin with a spanning tree of the gluing graph: its closed disc tiles joined along boundary arcs form one closed disc, and every remaining pair of boundary arcs adds one band. Thus for $n$ tiles and $e$ paired arcs the fundamental group is a free group of rank $e-n+1$, and the Euler characteristic is $n-e$.

For $D$, the gluing graph is the pathGluing a new closed disc to a closed disc along one boundary arc again gives a closed disc; applying this three times proves that $D$ is homeomorphic to a closed disc. In particularFor $B$, the five paired arcs give the graph edges $13,12,02,01,23$. It is connected, and henceHere $F_2$ is the free group on two generators, so $B$ is not simply connected. To decide orientation of a surface, give every reference tile its counterclockwise orientation. Our same-parameter gluing preserves boundary directions, so a global orientation requires opposite orientation signs on paired tiles: the two induced boundary orientations at an interior interface must be opposite. Such signs exist exactly when the gluing graph is bipartite. In $B$ the cyclehas length three. Following it forces the sign to reverse three times and return to its starting tile with the opposite sign. This proves that $B$ is a nonorientable surface. It has nonempty boundary because every intervening octagon side remains unpaired. Thus $D$ is a simply-connected orientable surface with boundary, whereas its isospectral partner $B$ is neither simply connected nor orientable, exactly as required, for the specified mixed boundary problem.

The boundary qualification is essential. With smooth boundary and the same pure Dirichlet boundary condition or pure Neumann boundary condition on both surfaces, the following existing obstruction applies. Polygonal corners and the mixed boundary conditions in the construction above place it outside that obstruction.

With the interpretation of compact smooth Riemannian surfaces with the same pure Dirichlet boundary condition or pure Neumann boundary condition, the printed simply-connected/nonorientable pair cannot exist. The two-dimensional heat trace has the following expansion (the smooth-boundary coefficients are given in the heat-coefficient formulas, Theorems 5.1–5.2):where the boundary sign is minus for Dirichlet and plus for Neumann. The Gauss-Bonnet theorem identifies the constant term with $\chi (M)/6$. Isospectrality therefore forces equal Euler characteristics. A compact connected simply-connected surface with boundary is a disk and has $\chi =1$. A connected nonorientable surface with $c\ge 1$ crosscaps and $b\ge 1$ boundary components has $\chi =2-c-b\le 0$. Thus no such smooth pure-Dirichlet or pure-Neumann pair exists. This is the Euler characteristic from the heat trace of a bordered surface.

Piecewise-smooth boundaries introduce corner terms, and mixed boundary conditions change the heat coefficients, as the explicit construction above demonstrates. Moreover, in the usual pure-Neumann polygonal reflection-transplantation setup, the pure Neumann reflection transplantation preserves Euler characteristic argument applies even with corners. The numbers of tiles agree. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$, so these agree too. For each tile-vertex label, vertex classes are orbits of the subgroup generated by incident-face involutions. Their number is the dimension of invariant vectors in that permutation representation, which is unchanged by the intertwiner. Hence the numbers of vertices agree, proving equal $V-E+F$. Thus this standard tiled construction also cannot supply a disk on one side and a nonorientable surface on the other.

For comparison, the following existing construction gives an orientable and a nonorientable pair with the same spectrum for the pure Neumann boundary condition. Both have nontrivial fundamental groups; this is a separate example, not the construction used to meet the simply-connected requirement above. This eight-tile Neumann transplantation across orientability makes the comparison explicit. The finite seed is the affine modulo-eight example in the original orientability construction; the intertwiner and the topology checks are derived below. Take eight copies of a regular Euclidean hexagon and label three pairwise nonadjacent sides $\sigma ,t,u$. Every remaining side is a side with a Neumann boundary condition. Index the copies by $j\in \mathbb{Z}/8\mathbb{Z}$. For the first assembly pair same-labelled faces according tofor the second, usewith all arithmetic modulo eight. A fixed point means an unglued side with a Neumann boundary condition. Pair faces by the reflection convention, matching the same endpoint parameter in the reference tiles. The distinguished sides are disjoint, so no interior vertex singularities are created; the resulting surfaces have polygonal boundary corners.

Let $R{e}_j=e_{j+1}$ on the eight-dimensional tile-index space and setIt commutes with the permutation matrices for $t$ and $\sigma$, because its coefficients are invariant under reversing cyclic order. For $u_1,u_2$, write $C_{ij}=c_{i-j}$, where $c_0=c_4=1$, $c_1=c_7=2$, and all other $c_r$ vanish. The required relation $C{M}_{u_1}=M_{u_2}C$ is exactly $c_r=c_{3r+4}$, verified from these four nonzero entries. Thus all three intertwining relations hold.

The cyclic Fourier vectors diagonalize $C$. Its eigenvalues arenamely $6,2\sqrt{2},2,-2\sqrt{2},-2,-2\sqrt{2},2,2\sqrt{2}$. None is zero, so $C$ is invertible and the two assemblies are isospectral manifolds for the Neumann boundary condition.

We prove their topology as well. Their gluing graphs are connected: the first contains the spanning path $0,1,3,6,2,7,5,4$, and the second contains the spanning cycle $0,1,7,2,6,3,5,4,0$. Each has eight tile vertices and ten paired-face edges. Since tiles are disks and gluing arcs are disjoint, each surface retracts up to homotopy to this multigraph; thus its Euler characteristic is $8-10=-2$ and its fundamental group is a free group of rank three.

An orientation assigns a sign to each reference tile, and a reflection gluing requires opposite signs at the ends of every graph edge. The second graph admits this assignment: take sign $+$ on $\{0,5,6,7\}$ and sign $-$ on $\{1,2,3,4\}$. Every paired face joins opposite signs, proving orientability. In the first graph the cyclehas odd length. Transporting orientation around it reverses the initial sign, proving nonorientability. Hence the first surface is also not simply connected. The second surface is orientable but likewise not simply connected; the disk required in the question is supplied instead by the four-tile mixed-boundary construction above.

The four-tile construction resolves the printed simply-connected requirement under explicit mixed boundary conditions. The eight-tile comparison only concerns orientation of a surface under a pure Neumann boundary condition.

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

Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

A statistical functional is pathwise differentiable relative to chosen statistical paths if its derivative along every path depends only on that path's score function and defines a bounded linear functional on their statistical tangent space. The Riesz representation theorem expresses this derivative as an L2 inner product with a unique element of the statistical tangent space, the canonical gradient. A path family and the associated derivative remainder conditions must both be specified; a formal derivative along one convenient family does not establish differentiability along all paths.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.