Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 340 (Topics in Mathematics of Information)

An orthonormal wavelet is a function whose integer translates and dyadic dilatesform an orthonormal basis of the Hilbert space $L^2(\mathbb{R})$. Thus their pairwise inner products are ${\delta }_{j,j^{\prime }}{\delta }_{k,k^{\prime }}$, and their closed linear span is the whole space. Equivalently, every $f\in L^2(\mathbb{R})$ has the norm-convergent expansionThe factor $2^{j/2}$ preserves the $L^2$ norm under dilation.

Let $P_j$ be the orthogonal projection onto $V_j$. The dilation property and the orthonormal basis property give the orthonormal basis ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ of $V_j$. The periodized nonnegative functionhas period one and ${\int }_0^{1}F=\Vert \phi {\Vert }_2^2=1$, by Tonelli theorem. For a function $g\in L^2(\mathbb{R})$ with compact support contained in a bounded interval $I$, Cauchy-Schwarz inequality and the Parseval identity giveas $j\to -\infty$. The last step uses absolute continuity of the integral of a locally $L^1$ function, since the interval's length tends to zero. If $f\in {\bigcap }_j{V}_j$, thenFunctions with compact support are dense in $L^2$, so $f=0$. ConsequentlyThis proof needs neither the density axiom nor any compact support assumption on the scaling function.

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.

With $N$ vanishing moments, a wavelet annihilates every polynomial of degree below $N$. A Taylor polynomial then explains small wavelet coefficients on smooth parts of a signal, and efficient best N-term approximation. Compact support localizes coefficients near a feature, limits the number of boundary interactions, and permits a finite filter implementation. Increasing the number of vanishing moments while keeping an orthonormal basis generally requires a larger support of a function: a finite orthonormal filter with $N$ vanishing moments needs at least $2N$ taps, and the minimal-support Daubechies wavelet has support length $2N-1$ in the standard normalization.

The Haar wavelet, $\psi ={\chi }_{[0,1/2)}-{\chi }_{[1/2,1)}$, has one vanishing moment, unit support length and discontinuities. It is inexpensive and particularly suitable for piecewise constant data with sharp jumps. A Daubechies wavelet of a larger order has more vanishing moments and a longer finite filter; sufficiently large orders also provide greater regularity. It is useful when smooth trends should yield small coefficients, although the wider support of a function can spread a jump across more coefficients. Choose Haar for compact jump localization; choose a higher-order Daubechies wavelet for smooth polynomial structure. More vanishing moments alone does not make every low-order wavelet highly differentiable.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int \phi (x)e^{-ix\xi }dx$. Nesting and the orthonormal basis of $V_1$ give the scaling refinement equationHere $h_k=⟨\phi ,\sqrt{2}\phi (2\cdot -k)⟩$. Compact support makes these inner products zero for all but finitely many $k$, so $m$ is a trigonometric polynomial. Expanding the orthogonality of the integer translates of $\phi$ in the orthonormal basis of $V_1$ yieldsConsequentlyThis holds everywhere because $m$ is continuous. The invoked properties are nesting, dyadic dilation, orthonormal integer translates, and compact support; mere finite-energy refinement would not imply the quadrature mirror filter identity.

The density axiom and continuity of the Fourier transform at zero imply $|\hat{\phi }(0)|=1$. Here is a proof that avoids assuming a normalization of the integral. Choose a nonzero $f$ whose Fourier transform has bounded support. The MRA projection Fourier identity, with no aliasing once $j$ is sufficiently large, isSince $\phi \in L^1$, its Fourier transform is continuous and bounded, so this tends to $|\hat{\phi }(0){|}^2\Vert f{\Vert }_2^2$. Density and nesting give $P_{j}f\to f$ in $L^2$. Thus $|\hat{\phi }(0)|=1$, in particular it is nonzero. Evaluating the scaling refinement equation in frequency at zero now giveswhich is stronger than the requested absolute-value equality. Multiplying the scaling function by a constant of modulus one normalizes its integral to one without changing $m$.

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.

The factor $1+e^{-i\xi }$ has a simple zero at $\pi$, and $L(\pi )\ne 0$ makes the zero of $m$ there exactly order $N$. The quadrature mirror filter construction gives a high-pass symbol $m_1(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase, andThe normalized scaling function has $\hat{\phi }(0)=1$, so $\hat{\psi }$ has a zero of exactly order $N$ at zero. Compact support of the Daubechies wavelet permits differentiation under its Fourier transform:HenceThere are exactly $N$ vanishing moments, rather than just at least $N$.

Normalize $\hat{\phi }(0)=1$, as in the displayed product convention. Iterating the scaling refinement equation gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$. The finite-product identityhas the value one at zero by continuity. ThereforeWrite $M=\sup |L|$. Since $L(0)=1$, we have $M\ge 1$. For $|\xi |>1$, take $J=\lceil {\log }_2|\xi |\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi {|}^{{\log }_{2}M}$. For the remaining factors, the trigonometric polynomial satisfies $|L(t)|\le 1+C|t|$ on $|t|\le 1$, henceUsing $|(1-e^{-i\xi })/(i\xi )|\le \min (1,2/|\xi |)$ givesThe strict hypothesis permits $\epsilon >0$ with $N-{\log }_{2}M\ge \alpha +1+\epsilon$. The given Fourier-decay criterion proves uniform Hölder regularity of exponent $\alpha$. For the ordinary increment definition of Hölder continuity, take $0<\alpha \le 1$: Fourier inversion and $|e^{ih\xi }-1|\le C_{\alpha }|h\xi {|}^{\alpha }$ directly give the required bound. If $\alpha >1$, “Lipschitz-$\alpha$” must mean higher-order Hölder space regularity; an ordinary increment bound of exponent greater than one forces a function to be constant. An unnormalized scaling function contributes its constant phase to the product formula.

Write $c_m=⟨f,g_m⟩$. A linear N-term approximation fixes the indices independently of $f$, normally the first $N$ in a prescribed ordering:A best N-term approximation chooses the indices using $f$: retain $N$ coefficients of largest absolute value, resolving ties arbitrarily, and set $f_N^n={\sum }_{m\in {\Lambda }_N(f)}{c}_m{g}_m$. The Parseval identity shows why this choice is optimal:For any fixed index set, the orthogonal projection coefficients minimize the error; optimizing the set then means discarding the smallest squared coefficients. Thus “linear” requires a specified ordering, while “nonlinear” refers to the data-dependent selection.

Use the usual localized, compact support construction of an interval-adapted wavelet basis, including boundary wavelets with the stated vanishing moments, and order the linear N-term approximation by increasing resolution. Also interpret a piecewise polynomial function as having finitely many pieces. These conventions matter: regularity and vanishing moments alone, or an arbitrary enumeration, do not establish the asserted rates.

At scale $j$, a wavelet whose support lies in one polynomial piece has zero coefficient because $p<q$. Only a bounded number $B$ of wavelets per scale can meet a partition point. Their $L^1$ norms are bounded by $C{2}^{-j/2}$, and $f$ is bounded. Thus $|c_{j,k}|\le C_f{2}^{-j/2}$ andRetain the fixed number of coarse scaling function coefficients and every nonzero coefficient through level $J$. This uses at most $B(J+1)+B_0$ terms, leaving squared error at most $C{2}^{-J}$. The optimal best N-term approximation is no worse; choose $J$ proportional to $N$ to obtain $O({\omega }^N)$ for some $\omega \in (0,1)$. In contrast, retaining all wavelets through level $J$ costs $\asymp 2^J$ terms. Choosing the last complete level before $N$ givesThese are squared errors; the corresponding $L^2$ errors are $O({\omega }^{N/2})$ and $O(N^{-1/2})$.

For a function smooth on finitely many closed pieces with finite one-sided derivatives, integration by parts gives its Fourier series coefficients, for $n\ne 0$,where $J_r=f(x_r+)-f(x_r-)$ includes the jump of the periodic extension at the endpoint when necessary. In particular $|c_n|\le C/|n|$. Keeping frequencies $|n|\le K$, with $N\asymp K$, gives squared error $O({\sum }_{n>K}{n}^{-2})=O(N^{-1})$.

A best N-term approximation cannot improve this worst-case rate. For example, $f={\chi }_{[0,1/2)}$ has $|c_n|=1/(\pi |n|)$ on odd nonzero $n$ and zero on even nonzero $n$. Even the optimal selection therefore leaves a squared tail comparable to $N^{-1}$. Thus for the class with jumps,or $N^{-1/2}$ in the $L^2$ norm. Individual functions with no jumps, or additional cancellation, can converge faster. “Smooth except at finitely many points” must include controlled one-sided smoothness; smoothness merely on open pieces permits pathological behavior near their endpoints.

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.

Use localized tensor-product wavelets and finitely many bounded polynomial pieces, with the boundary wavelets adapted as in part (b). A rectifiable smooth curve of length $\ell$ meets $O(1+\ell 2^j)$ dyadic squares of side $2^{-j}$: subdividing an arclength parametrization into pieces of length at most $2^{-j}$ covers it by that many balls, each meeting only a bounded number of squares. Enlarging squares by the fixed support diameter preserves the count.

Each normalized two-dimensional wavelet has $L^1$ norm $O(2^{-j})$. A coefficient meeting the curve is therefore $O(2^{-j})$, and the total squared energy of these coefficients at level $j$ is $O(2^{-j})$. If $p<q$, all other coefficients vanish by the vanishing moments. Keeping the curve coefficients through level $J$ costs $O(2^J)$ and leaves squared error $O(2^{-J})$.

The printed part (e) does not repeat $p<q$. The bound still holds for any fixed polynomial degree when $q\ge 1$: on a smooth piece, a Taylor polynomial in the variable carrying a wavelet gives coefficient size $O(2^{-(q+1)j})$. There are $O(4^j)$ such coefficients, so their squared energy is $O(2^{-2qj})$. Retain all coefficients through $K=\lfloor J/2\rfloor$, and curve coefficients through $J$. The cost is $O(4^K+2^J)=O(2^J)$ and the omitted squared energy isThe best N-term approximation is at least as good as this selection, provingThis argument covers the unqualified finite-degree clause without adding an unnecessary restriction $p<q$.

Under the usual definition, the null space property relative to $S$ is $\Vert h_S{\Vert }_1<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. It characterizes recovery by basis pursuit of every vector supported in $S$. Indeed, if it holds, then for any such $x_0$ and any nonzero $h\in \ker A$,Conversely, failure for some $h$ makes $x_0=-h_S$ and $z=h_{S^c}$ feasible with the same measurements and $\Vert z{\Vert }_1\le \Vert x_0{\Vert }_1$. ThusFor the single fixed vector in the printing, necessity is false with that definition. For example,has $\ker A=\mathrm{span}\{(1,1,1)\}$. Along its feasible line, the objective is $|1+t|+|-1+t|+|t|$, uniquely minimized at $t=0$, although the null space property would require $2<1$.

The correct fixed-sign null space condition isSufficiency follows from the supporting-line inequality for the absolute value. Necessity follows by considering $x_0\pm th$ for small positive $t$, when all signs on $S$ remain unchanged: a nonpositive directional increase gives either a decrease or another minimizer. If “relative to $S$” was intended to include these signs, this is the precise condition needed.

Let $T$ be the support of a vector $\hat{x}$. If the columns indexed by $T$ were linearly dependent, there would be a nonzero $h$ supported in $T$ with $Ah=0$. For sufficiently small real $t$, the nonzero signs of ${\hat{x}}_T$ remain fixed, soBoth positive and negative $t$ preserve the residual and hence feasibility, even with noise. A nonzero slope contradicts minimality in one direction. A zero slope gives distinct minimizers, contradicting uniqueness. Thus those columns are linearly independent, andThis proves the sparse vector assertion without assuming a noiseless residual.

The printed definition of the restricted isometry constant omits the quantifier: the two inequalities must hold for every vector with at most $s$ nonzero entries. The constant is the smallest nonnegative value, or the infimum over positive values; it can be zero.

Let $T=\mathrm{supp}u\cup \mathrm{supp}v$ and $B=A_T^{\ast }{A}_T-I$. The restricted isometry property of order $s+t$ says $|z^{\ast }Bz|\le {\delta }_{s+t}\Vert z{\Vert }_2^2$ for every $z$ supported on $T$. Since $B$ is a Hermitian matrix, the finite-dimensional spectral theorem implies $\Vert B{\Vert }_{2\to 2}\le {\delta }_{s+t}$. Disjoint supports give $⟨u,v⟩=0$, henceThis argument works over $\mathbb{C}$ and avoids the loss of a factor caused by treating real and imaginary parts separately.

The printed bound is false, even for $s=2$ and zero noise. Let $P=I_7-11^T/7$ and choose $A$ with six orthonormal rows spanning $1^{\perp }$, so $A^{T}A=P$. For a vector supported on at most $s$ entries,Take $x_0=(0,0,1,1,1,1,1)$, $s=2$, and $\eta =0$. Every feasible vector is $x_0+t1$, whose ${\ell }^1$ objective is $2|t|+5|1+t|$. Its unique minimizer is $\hat{x}=(-1,-1,0,0,0,0,0)$. YetNo choice of the noise constant repairs this example. Also $s=1$ must be excluded: $A=(1,1)$ has ${\delta }_1=0$, but its two unit-coordinate vectors have identical measurements and both minimize basis pursuit.

The corrected conclusion, for $s\ge 2$, iswith both constants depending on the restricted isometry constant. The order-$s$ threshold itself is valid for $s\ge 2$; it need not be replaced by an order-$2s$ threshold. A precise sharp robust recovery theorem gives, for $\delta ={\delta }_s<1/3$, the admissible constantsThis is the standard stable recovery result, invoked here explicitly; its order restriction and constants are given in arxiv.org/pdf/1302.1236, Theorem 3.3.

For completeness, the mechanism behind a robust null space property proof is as follows. Put $h=\hat{x}-x_0$, let $S$ index the largest $s$ entries of $x_0$, and write $\sigma =\Vert x_{0,S^c}{\Vert }_1$. Feasibility and minimality give the tube and cone inequalitiesIf $A$ has the robust null space property with $0\le \rho <1$ and $\tau >0$, namely $\Vert h_T{\Vert }_2\le \rho \Vert h_{T^c}{\Vert }_1/\sqrt{s}+\tau \Vert Ah{\Vert }_2$ for every $|T|\le s$, thenApply that property again to the largest $s$ entries $T$ of $h$. Sorted coefficients satisfy $\Vert h_{T^c}{\Vert }_2\le \Vert h{\Vert }_1/\sqrt{s}$, because each tail entry is at most $\Vert h{\Vert }_1/s$. ThereforeThis proves the usual two-constant conclusion under the explicitly stated robust null space property and explains why a fixed coefficient one is not supplied by that argument. The sharp order-$s$ theorem above supplies the conclusion for the actual printed threshold, after the necessary corrections.

The total variation seminorm on a domain isThe class is symmetric under negation, so no additional absolute value is needed around the integral. The bounded-variation space isEquivalently, the distributional derivative is a finite vector-valued Radon measure, and the displayed quantity is its total mass in variation. For $f\in W^{1,1}(\Omega )$, it equals ${\int }_{\Omega }|\nabla f{|}_{2}dx$. Thus jumps are allowed, with their interface contributions measured by variation.

Assume $g\in L^2(\Omega )$, as required for the finite-valued data fidelity; extend the total variation seminorm on a domain by $+\infty$ outside $BV(\Omega )$. The zero function is a finite-energy competitor. A minimizing sequence $f_n$ has bounded $\Vert f_n-g{\Vert }_2$ and bounded variation, hence bounded $L^2$ norm. Take a weakly convergent subsequence in $L^2$, with limit $f$. For each fixed test vector field $\zeta$, the pairing $\int f\mathrm{div}\zeta$ is weakly continuous because $\mathrm{div}\zeta \in L^2$. Its supremum has weak lower semicontinuity. The squared Hilbert space norm also has weak lower semicontinuity. Thus the direct method in the calculus of variations gives a minimizer in $BV\cap L^2$.

The variation term is convex, and the squared fidelity term is strictly convex. Their sum is strictly convex on its effective domain, so the minimizer is unique.

Let $T(t)=\min (b,\max (a,t))$. The bounded-variation contraction under clipping gives $|D(T\circ f)|(\Omega )\le |Df|(\Omega )$. If $g(x)\in [a,b]$, then $|T(f(x))-g(x)|\le |f(x)-g(x)|$, strictly whenever $f(x)\notin [a,b]$. A positive-measure set outside the interval would therefore strictly decrease the fidelity integral, while not increasing variation, contradicting minimality. HenceNo compactness theorem for the $BV$ embedding is required for this weak-$L^2$ argument.

For a differentiable convex function $F$ with $L$-Lipschitz gradient and a proper lower-semicontinuous convex function $H$, the proximal gradient method isThe proximal operator is ${\mathrm{prox}}_{\tau H}(v)=\arg {\min }_z\{H(z)+\Vert z-v{\Vert }^2/(2\tau )\}$. To obtain an explicit iteration for discrete isotropic total variation, let $D=\nabla$ with the printed zero boundary differences andThe dual is minimization of $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $\mathcal{P}$, and the primal minimizer is $u_{\ast }=g-D^{\ast }{p}_{\ast }$. Here $D^{\ast }=-\mathrm{div}$ must be the exact adjoint operator of these differences. For example,with the analogous formula in $j$ for $(D_y^{+}{)}^{\ast }$; for $N=1$ both operators are zero. Equivalently use $p_{0j}^1=p_{Nj}^1=p_{i0}^2=p_{iN}^2=0$ in the backward-difference expression, ignoring unused edge components.

The gradient of the dual objective is $D(D^{\ast }p-g)$, with Lipschitz constant $\Vert D{\Vert }^2\le 8$. Taking $H$ to be the indicator functional of a constraint set makes its proximal operator the Euclidean projection onto a convex set. Thus a valid explicit algorithm isThe bound follows by summing $(a-b{)}^2\le 2a^2+2b^2$ over grid differences. Finite-dimensional proximal gradient method convergence gives a dual minimizer and convergence of the recovered $u^k$ to the unique primal minimizer. Projection is pointwise in the dual; applying pointwise soft thresholding directly to $Du$ would not compute the primal TV proximal operator.

Take $F(x)=\frac{1}{2}\Vert Ax-y{\Vert }_2^2$, with $\nabla F(x)=A^T(Ax-y)$ and Lipschitz constant $L=\Vert A{\Vert }_{2\to 2}^2$, and $H(x)=\Vert x{\Vert }_1$. The proximal operator separates over coordinates. Solving the scalar subgradient optimality condition $0\in \partial |z|+(z-v)/\tau$ gives soft thresholding $S_{\tau }(v)=\mathrm{sgn}(v)(|v|-\tau {)}_{+}$. Therefore the iterative soft-thresholding algorithm isHere soft thresholding is applied coordinatewise. The objective is convex and coercive because of its ${\ell }^1$ term, so minimizers exist and the proximal gradient method converges to a minimizer. Uniqueness need not hold if $A$ has a nontrivial null space. If $A=0$, the unique minimizer is zero and any positive step size is admissible.

Either essay option suffices; both are developed here to make the two mathematical constructions available.

Diffusion for images. Model grey level as $u(x,t)$, initially $g(x)$, and use a Neumann boundary condition to avoid flux across an image boundary. The linear heat equation $u_t=\Delta u$ smooths the data. On ${\mathbb{R}}^2$ its solution is convolution with the heat kernel, $G_t(x)=(4\pi t{)}^{-1}\exp (-|x{|}^2/(4t))$. Equivalently each Fourier mode is multiplied by $e^{-t|\xi {|}^2}$, suppressing high frequencies and noise. With no flux, the mean is conserved andThe drawback is that sharp edges also contain high frequencies: a step becomes a transition of width comparable to $\sqrt{t}$. Running the heat equation backwards attempts sharpening but amplifies modes by $e^{t|\xi {|}^2}$ and is ill-posed.

The Perona-Malik equation instead uses a decreasing diffusivity,Small gradients are smoothed strongly and large ones less strongly. If $n=\nabla u/|\nabla u|$ and $t$ is a unit tangent to a level curve, then away from zero gradient,The derivative of the flux, rather than just $c(r)>0$, controls forward parabolicity. Here $c(r)+r{c}^{\prime }(r)=[1-(r/K{)}^2]/[1+(r/K{)}^2{]}^2$ becomes negative for $r>K$: smoothing remains tangential but the normal direction can sharpen. This formal edge enhancement comes with forward-backward ill-posedness, so existence and stability of the unregularized continuum equation must not be presumed. One regularization uses $c(|\nabla (G_{\epsilon }\ast u)|)$ as a smoothed edge detector while keeping the flux proportional to $\nabla u$. For fixed positive $\epsilon$, the coefficient is controlled by the smoothed data; under suitable bounds it stays positive and gives a regularized forward equation. It preserves edges through reduced cross-edge transport, without the same local backward-diffusion calculation.

Diffusion from a gradient energy connects these equations with variational regularization. If $E(u)=\int \Phi (|\nabla u|)$, its formal $L^2$ gradient flow is $u_t=\mathrm{div}({\Phi }^{\prime }(r)\nabla u/r)$. Linear diffusion corresponds to $\Phi (r)=r^2/2$. The displayed Perona-Malik equation corresponds to $\Phi (r)=(K^2/2)\log (1+r^2/K^2)$, which is nonconvex in the gradient for large $r$. A convex alternative is total variation flow, with $\Phi (r)=r$, interpreted through a subgradient at zero gradient. Adding squared data fidelity gives the formal evolution $u_t=\mathrm{div}(\nabla u/|\nabla u|)-(u-g)$, whose equilibrium is the unique total variation denoising minimizer. Convex TV preserves sharp interfaces more effectively than the heat energy, but it may produce piecewise constant plateaux, known as staircasing in total variation denoising. A stopping time or fidelity weight controls the smoothing scale. Linear diffusion is stable but blurs edges; nonlinear diffusion must balance edge selectivity with parabolicity and regularization.

Wavelets on an interval. Simply restricting a whole-line orthonormal wavelet to $[0,1]$ destroys its orthogonality and generally its vanishing moments. Periodizing the basis restores orthogonality on the circle, but treats the two endpoints as neighbors. This is suitable for periodic data and can create an artificial endpoint jump for nonperiodic data.

A localized interval-adapted wavelet basis instead uses unchanged interior functions and finitely many special boundary wavelets at each endpoint. Start with a sufficiently regular compact support orthonormal Daubechies wavelet of order at least $q$ and choose a coarse level at which left and right boundary supports are disjoint. At each boundary, take appropriate finite combinations of the scaling functions that meet the endpoint, restricted to the interval. Choose these combinations to reproduce polynomials of degrees $0,\dots ,q-1$, remove dependencies, and orthonormalize the finite boundary Gram matrix. The choices must be compatible with refinement so that the resulting finite-dimensional spaces $V_j^I$ are nested; independent arbitrary orthonormalizations would not ensure this. Interior functions retain their whole-line filters, while the boundary functions use finite boundary refinement matrices.