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Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence

The expectation value is the mean outcome over many identically prepared measurements of the observable. For real $\lambda$, positivity of the norm in the Hilbert space givesThis real quadratic has nonpositive quadratic discriminant, soApply this to the centered observables $Q-⟨Q⟩$ and $P-⟨P⟩$ to obtain the Robertson uncertainty principle

For the quantum harmonic oscillator, $[x,p]=i\hslash$ andThe arithmetic-geometric mean inequality and $\Delta x\Delta p\ge \hslash /2$ now give

Put $I={\int }_0^{1}f(t)dt$. If $I\ne 0$, take the inner product with the unit vector $u=I/\Vert I{\Vert }_2$. Thenby the Cauchy-Schwarz inequality; the case $I=0$ is immediate. This is the triangle inequality for a vector-valued integral.

Equality in the Euclidean argument requires $u\cdot f(t)=\Vert f(t){\Vert }_2$ for every $t$: the nonnegative continuous difference has integral zero. Equality in Cauchy-Schwarz inequality then says $f(t)=\lambda (t)u$ with $\lambda (t)\ge 0$. Conversely, if $f(t)=\lambda (t)v$ for one fixed $v\in {\mathbb{R}}^m$ and a continuous $\lambda \ge 0$, then for every norm,Thus these nonnegative-ray functions, including $f\equiv 0$, are exactly the required functions.

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

Use the Hilbert space interpretation of a linear inverse problem, with a bounded linear operator $K:\mathcal{U}\to \mathcal{V}$. A least-squares solution minimizes $\Vert Ku-f{\Vert }_{\mathcal{V}}$ over $u\in \mathcal{U}$. A minimum-norm least-squares solution additionally has the smallest $\mathcal{U}$-norm among all least-squares solutions. When the equation is consistent this is the smallest-norm exact solution.

The least-squares solutions form a nonempty closed affine subspace precisely whenThe closest point theorem in a Hilbert space then gives a unique nearest point to zero, hence a unique minimum-norm least-squares solution. Equivalently, it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If $\mathcal{R}(K)$ is closed, this exists for every datum; a nonclosed operator range can leave some data without any least-squares solution.

For example, on the l2 sequence space, take $(Ku{)}_n=u_n/n$ and $f_n=1/n$. The equation would require $u_n=1$ for every $n$, which is not square summable. However, setting the first $N$ coordinates of $u$ to one and the rest to zero givesThus the least-squares infimum is zero but is not attained, and no minimum-norm least-squares solution exists for these data.

The Moore–Penrose inverse of an operator is the linear mapthat assigns each admissible datum its minimum-norm least-squares solution. If $f=y+z$ with $y\in \mathcal{R}(K)$ and $z\in \mathcal{R}(K{)}^{\perp }$, then $K^{†}f$ is the unique $u\in \mathcal{N}(K{)}^{\perp }$ satisfying $Ku=y$. In particular it vanishes on $\mathcal{R}(K{)}^{\perp }$, andThe continuity criterion, with the inherited data norm on its domain, isFor a closed operator range, the restriction of $K$ from $\mathcal{N}(K{)}^{\perp }$ to $\mathcal{R}(K)$ is a bounded bijection between Banach spaces, so its inverse is bounded by the bounded inverse theorem. Conversely, if $K^{†}$ is bounded and $K{u}_n\to y$, replace each $u_n$ by $P_{\mathcal{N}(K{)}^{\perp }}{u}_n=K^{†}K{u}_n$. This is a Cauchy sequence, hence converges to $u$, and $Ku=y$. Thus every limit of range elements remains in the operator range.

Let $\eta =f^{\delta }-f$, so $\Vert \eta \Vert \le \delta$. By linearity, $u_{\delta }^{(k)}-u^{(k)}=R_{1/k}\eta$. The preceding operator norm estimate givesMoreover, $K{R}_{1/k}$ has factors $1-(1+\tau {\sigma }_j^2{)}^{-k}$ on the data singular vectors, and is zero on $\mathcal{R}(K{)}^{\perp }$. Every factor is in $[0,1]$, henceThus, for any fixed $k$, one can take $C=\sqrt{k\tau }$ and $\gamma =1$.

The first constant cannot generally be chosen independently of the iteration count. On the l2 sequence space, take $(Ku{)}_n=u_n/n$, $\tau =1$, and noise $\eta =\delta e_n$. At $k=n^2$ the solution perturbation has normThis disproves a uniform-in-$k$ interpretation of the stated first estimate. For a closed operator range, boundedness of the Moore–Penrose inverse of an operator instead supplies the uniform constant $\Vert K^{†}\Vert$.

Assume $K{u}^{†}=f$, as in the exact-solution setting used for the preceding source condition for quadratic regularization. Let $r=r_{k+1}$, $\eta =f^{\delta }-f$, and $s=u_{\delta }^{(k+1)}-u_{\delta }^{(k)}=-\tau K^{\ast }r$. Then $K(u_{\delta }^{(k+1)}-u^{†})=r+\eta$, and expanding the two squared errors givesThe Cauchy-Schwarz inequality supplies the last line. Thus while the residual remains at least the noise bound,Compatibility of the exact data is essential. If only $f\in \mathcal{D}(K^{†})$ is retained, let $Kx=(x,0)$, $f=(0,1)$, $f^{\delta }=(1/2,1)$, $\delta =1/2$ and $\tau =1$. The minimum-norm least-squares solution is zero, but the first iterate is $1/4$ and its residual norm is $\sqrt{17}/4>\delta$. Its error has increased from zero. Thus the original assertion uses the implicit exact-data assumption from part (c), and is false without it.

The Iterated Tikhonov regularization update is the proximal operator stepThe objective has coercivity and is weakly sequentially lower semicontinuous and strongly convex, so the direct method in the calculus of variations and uniqueness of a minimizer of a strictly convex function give existence and uniqueness. Its gradient equation iswhich is exactly $(I+\tau K^{\ast }K)u=u_{\delta }^{(k)}+\tau K^{\ast }{f}^{\delta }$. The second term penalizes the change from the previous iterate, rather than the norm of $u$ relative to zero.

Comparison with zero and nonnegativity of the two terms givesThereforeFor a concrete variational regularization example, take a Hilbert space and $J(u)=\Vert u{\Vert }^2$. The bound forces $u_{\alpha }\to 0$ in norm. Since $K$ is a bounded linear operator, $K{u}_{\alpha }\to 0$, and continuity of the squared norm givesThe same conclusion is not forced by $J(u_{\alpha })\to 0$ for a regulariser with a nontrivial zero set.

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.

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.

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 $p<q$.

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.

Let $T$ be the $M\times M$ tridiagonal matrix with diagonal two and adjacent entries minus one, incorporating zero Dirichlet boundary conditions. The actual PDF scheme isAn orthonormal basis of eigenvectors of $T$ has components $\sqrt{2/(M+1)}\sin (jm\pi /(M+1))$, and substitution givesThey form an orthonormal basis by the spectral theorem for real symmetric matrices. Hence $I+\mu T$ is invertible for every $\mu >0$, and the amplification eigenvalues are $g_j=(1+\mu {\lambda }_j{)}^{-1}\in (0,1)$. For the discrete L2 norm $\Vert U{\Vert }_h^2=h{\sum }_m|U_m{|}^2$,This bound has constant one independent of time step and mesh, which is the required stability of a numerical method. Therefore every positive Courant number is stable. If a forcing or local-defect sequence is added, the discrete variation-of-constants formula gives a bound by the initial norm plus the sum of those perturbation norms; this also justifies the global error conclusion in part (a).