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Use the kinetic-energy inner product $⟨v,u⟩=\int v^{\ast }\cdot udV$, with perturbation energy $\Vert u{\Vert }^2/2$. On divergence-free fields the primal generator, before the pressure projection, isThe base state is also divergence-free. The periodic boundary terms in $x,z$ cancel, and decay at $|y|\to \infty$ removes the remaining boundary terms. Integration by parts changes $-\mathbf{U}\cdot \nabla$ to $+\mathbf{U}\cdot \nabla$; the Laplacian is self-adjoint under these conditions. Writing $J_{ij}={\partial }_j{U}_i$, the shear term $-Ju$ has adjoint operator $-J^{T}v$. A pressure gradient pairs to zero with a divergence-free field. Hencewhere the last term enforces the divergence-free projection.

To conserve $⟨v(t),u_p(t)⟩$ over the finite interval, the physical-time adjoint obeys $-{\partial }_{t}v=A^{†}(t)v$. Put $\tau =-t$ and $u_d(\tau )=v(-\tau )$; then ${\partial }_{\tau }{u}_d=A^{†}(-\tau )u_d$. This reverses the interval from $t=T$ towards $t=0$, or $\tau =-T$ towards $\tau =0$.

For $\mathbf{\Omega }=\nabla \times \mathbf{U}$, the vorticity cross-product identity and the curl of a cross producthold when both fields are divergence-free. Their difference equals the transport and shear terms in the adjoint operator. Therefore the adjoint linearized Navier-Stokes evolution isThe Leray-Helmholtz projection determines the adjoint pressure. The time-dependent base state is evaluated at $-\tau$; no extra time derivative of that base state appears in the spatial adjoint operator.

Write $\mathbf{r}=R\mathbf{\theta }$, $|\mathbf{\theta }|=1$. Uniformly for ${\mathbf{r}}^{\prime }$ in the source ball,For fixed $k$ and $r_0$, the outgoing Green function for the three-dimensional Helmholtz equation therefore has the far-field approximation for an outgoing sourceSubstitution into the source integral yieldsThus the source-to-far-field operator at fixed frequency isIt samples the Fourier transform of the source on the sphere of radius $k$. Its square-integrable kernel makes it a Hilbert-Schmidt operator, hence a compact operator. In factUse complex inner products linear in the first argument: $(f,g)=\int f\overline{g}$. Interchanging the volume and surface integrals givesHence the adjoint source-to-far-field operator isThis is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

Use the standard bounded linear operator setting for $K:U\to V$, so the adjoint operator $K^{\ast }:V\to U^{\ast }$ in the updates is defined on every datum. Put $q=K^{\ast }K{u}_{\lambda }$. The generalized singular vector relation is $\lambda q\in \partial J(u_{\lambda })$ with $\Vert K{u}_{\lambda }\Vert =1$ and $\lambda =J(u_{\lambda })\ge 0$. For an absolutely one-homogeneous functional, a subgradient $p$ at $u$ is characterized by $⟨p,u⟩=J(u)$ and $⟨p,v⟩\le J(v)$ for all $v$. Consequently $\lambda q$ is also a subgradient at every positive multiple of $u_{\lambda }$, and $bq\in \partial J(0)$ for every $0\le b\le \lambda$.

Use the initialization $u^0=p^0=0$ and start the first update at $k=0$; the PDF lists $k=1,2,\dots$ despite specifying only index-zero initial data. Suppose $p^k=b_{k}q$ with $0\le b_k\le \lambda$. DefineIf it is positive, $u^{k+1}=c_{k+1}{u}_{\lambda }$ and $p^{k+1}=\lambda q$ satisfy the optimality equation and the dual update. If it is zero, take $u^{k+1}=0$ and $p^{k+1}=(b_k+\gamma /\alpha )q\in \partial J(0)$. Thus this compatible branch hasInduction proves the finite termination of Bregman iteration on a singular vector formulasOnce the dual coefficient reaches $\lambda$, the next reconstruction is $\gamma u_{\lambda }$. Therefore a guaranteed stopping index for this branch isThe formulas also cover $\lambda =0$, where exact reconstruction occurs at the first update. The subgradient conditions are sufficient for global minimization because each objective is convex.

There is a genuine uniqueness issue in the printed claim about arbitrary iterates. Strict convexity of data fidelity in $Ku$ ensures that any two minimizers at a given step have the same image under $K$. Induction therefore makes every minimizer branch have the same dual variables and attain exact data at the stated finite index. It does not make their states equal to the specified $\gamma u_{\lambda }$ if $K$ has a null space.

For a counterexample take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $K(x,y)=x$, $J(x,y)=|x|$, $u_{\lambda }=(1,0)$, and $\lambda =1$. Every step leaves $y$ completely unconstrained. Choosing the minimizing state's second coordinate to be $k$ gives valid updates and subgradients but never reaches $(\gamma ,0)$ and does not converge as a state sequence. Thus finite exact-data recovery holds for all minimizer branches; exact recovery of the stated vector needs uniqueness or the aligned branch. If $K$ is injective, equal reconstructed data identify the state, and the literal vector conclusion follows.

Write $D=\nabla$ and use the real pixelwise inner products. Let $D^{\ast }$ be the adjoint operator, defined by $⟨Du,p⟩=⟨u,D^{\ast }p⟩$. For each nontrivial grid direction its one-dimensional formula iswith the analogous formula in $j$ for $D_y^{+\ast }{p}^2$. Their sum is $D^{\ast }p$; unused components $p_{N,j}^1$ and $p_{i,N}^2$ do not contribute. For $N=1$, $D=D^{\ast }=0$. If discrete divergence is defined, its sign convention is $\mathrm{div}=-D^{\ast }$, as in the adjoint of a discrete forward gradient.

Define the compact convex set $P_{\lambda }=\{p\in X^2:|p_{i,j}{|}_2\le \lambda \}$ and its image $C=D^{\ast }{P}_{\lambda }$. The image is compact and convex, hence closed, and contains zero. The finite-dimensional Euclidean duality formula $\lambda |a{|}_2={\max }_{|p{|}_2\le \lambda }a\cdot p$, applied independently at every pixel, givesThus the penalty is the support function ${\sigma }_C$. Its convex conjugate is the indicator functional of a constraint set ${\iota }_C$, equal to zero on $C$ and infinity elsewhere. For $w\in C$, the conjugate supremum is zero; for $w\notin C$, the separation theorem supplies a direction $u$ with $⟨u,w⟩>{\sigma }_C(u)$, and scaling that direction makes the supremum infinite.

The proximal operator is ${\mathrm{prox}}_{{\sigma }_C}(g)=\arg {\min }_u\{{\sigma }_C(u)+\frac{1}{2}\Vert u-g{\Vert }_2^2\}$. It exists uniquely because the objective is continuous, coercive and strictly convex. The Moreau decomposition for a proper closed convex function gives ${\mathrm{prox}}_{{\sigma }_C}(g)+{\mathrm{prox}}_{{\iota }_C}(g)=g$. The second term is the Euclidean projection onto a convex set, so the projection residual for discrete total variation isAlternatively, the variational characterization of convex projection says $w=P_{C}g$ exactly when $⟨g-w,z-w⟩\le 0$ for every $z\in C$. Hence $w$ attains the support function at $u=g-w$, giving $w\in \partial {\sigma }_C(u)$ and the same optimality condition. This supplies a direct verification of the Moreau decomposition step in this instance.

The printed wording needs “the residual after a projection”, rather than “the projection itself”. In general this denoising map is not a projection onto any fixed closed convex set. A metric projection onto a closed convex set is idempotent, while denoising a sufficiently large jump twice shrinks it twice. Concretely, for $N=2$ and data with rows $(0,0)$ and $(6\lambda ,6\lambda )$, the minimizer has rows $(\lambda ,\lambda )$ and $(5\lambda ,5\lambda )$. Reapplying the map gives rows $(2\lambda ,2\lambda )$ and $(4\lambda ,4\lambda )$, so it is not idempotent. These formulas have a global certificate: take $p_{1,j}^1=\lambda$, all other dual entries zero, giving $D^{\ast }p$ with rows $(-\lambda ,-\lambda )$ and $(\lambda ,\lambda )$; it saturates every positive row difference and yields the primal optimality condition. The qualification is therefore not merely a consequence of a restricted two-level ansatz.

Computing $P_{C}g$ is equivalent to minimizing $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $P_{\lambda }$, since $D^{\ast }p$ ranges over precisely $C$. Differentiating with the adjoint operator gives $\nabla F(p)=D(D^{\ast }p-g)$, whose Lipschitz constant is $L=\Vert D{D}^{\ast }\Vert =\Vert D{\Vert }^2$. Starting with any $p^0\in P_{\lambda }$, for example zero, the projected-gradient dual total variation algorithm isThe block formula is the Euclidean projection onto a convex set $P_{\lambda }$, since it clips each two-dimensional block to a Euclidean ball of radius $\lambda$. Both components must be clipped together for the isotropic block mixed norm.

For the unscaled forward differences, $\Vert D_x^{+}u{\Vert }_2^2\le 4\Vert u{\Vert }_2^2$ and likewise in the other direction, so $L\le 8$. More exactly, for $N>1$ the one-dimensional path difference has eigenvalues $4{\sin }^2(k\pi /(2N))$, $0\le k<N$, and the square-grid operator sums two such spectra. Thus $L=8{\cos }^2(\pi /(2N))$. Finite-dimensional projected gradient descent for a convex objective with an $L$-Lipschitz gradient converges from every feasible starting point to a dual minimizer for fixed $0<\tau <2/L$. The theorem assumes a nonempty closed convex set, a convex differentiable objective with globally Lipschitz gradient, and a nonempty minimizer set. All hold here: $P_{\lambda }$ is compact and convex, and $F$ is a continuous convex quadratic. Finite-dimensional convergence is of the full sequence, rather than only its objective values.

Consequently any fixed step with $0<\tau <1/4$ is safe on this grid; the more conservative $0<\tau \le 1/8$ is also safe. The dual minimizer need not be unique because $D^{\ast }$ has a null space, but $D^{\ast }{p}^k\to P_{C}g$ and $u^k\to u$ uniquely. For $N=1$, the gradient penalty vanishes and $u=g$, so no positive-$L$ step restriction is needed.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

To first order in $n_{ϵ}$, replace $q$ by $2n_{ϵ}$ and use the known incident field in the outgoing integral. The Born approximation for scalar wave scattering givesBoth the $n_{ϵ}^2$ contrast term and the internal-field correction are second order for fixed geometry in the perturbative regime established in part i.

Specify the data space before taking an adjoint operator. For example, let $X=L^2(B)$ and $Y=L^2(\Gamma )$, where $\Gamma$ is a bounded measurement surface separated from $B$. Define the linear operatorFor bounded incident field this is a Hilbert-Schmidt operator, hence a compact operator. The equation is $A{n}_{ϵ}=y$, with $y={\psi }_s{|}_{\Gamma }$. Other sampling geometries give corresponding data spaces and weights; the paper does not specify one. Using the usual complex $L^2$ inner products, its adjoint operator isThe complex conjugations are required by the adjoint operator identity, not by wave reciprocity alone.

The Landweber iteration starts from $f_0=0$ and applies gradient descent to $\frac{1}{2}\Vert Af-y{\Vert }^2$:Each step back-propagates the data residual. The Landweber relaxation parameter controls stability, and early stopping of Landweber iteration prevents small singular values from amplifying measurement errors. For an explicitly real-valued index contrast, use the real Hilbert space structure and replace $A^{\ast }$ by $\mathrm{Re}{A}^{\ast }$ in this update. Additional sign or support constraints require corresponding projections; none are assumed here.

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.

For $u\in U$, define $\varphi (u)\in U^{\ast }$ by $\varphi (u)(u^{\prime })=⟨u,u^{\prime }⟩$. If $\varphi (u)=0$, then in particularso $u=0$. Thus $\varphi$ is injective. Since a finite-dimensional vector space and its dual space have the same dimension, $\varphi$ is also surjective and hence an isomorphism. This is the finite-dimensional real case of the Riesz representation theorem.

The adjoint operator of $\alpha :V\to W$ is the unique linear map ${\alpha }^{\ast }:W\to V$ satisfyingfor every $v\in V$ and $w\in W$. In the stated orthonormal bases, if $v$ and $w$ denote coordinate columns, thenTherefore the matrix of ${\alpha }^{\ast }$ is the matrix transpose $A^T$.

For $w\in W$,Hence $\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\perp }$. Taking orthogonal complements in the finite-dimensional space $W$ proves the image-kernel orthogonality for an adjoint

Put $r=\alpha (v_0)-w_0$. For any $h\in V$,This is minimized at $h=0$ exactly when $r\perp \mathrm{im}\alpha$, equivalently when ${\alpha }^{\ast }r=0$. Thus the normal equation for a linear inverse problem is

For the given linear least-squares problem,The normal equations $2x+y=3$ and $x+2y=5$ have the unique solution

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a bounded linear functional. The Riesz representation theorem supplies a unique vector $Lv$ such thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. The invertibility from lower bounds on an operator and its adjoint therefore shows that $L$ is invertible and $L^{-1}$ is bounded.

Set $v_f=L^{-1}f$. Then $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus the unique vector is $v_f=L^{-1}f$.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

For fixed $y$, the map $x\mapsto ⟨Tx,y⟩$ is a bounded functional. The Riesz representation theorem gives a unique vector $T^{\ast }y$ representing it. Linearity, uniqueness, and $\Vert T^{\ast }y\Vert \le \Vert T\Vert \Vert y\Vert$ make $T^{\ast }$ the unique bounded adjoint operator.

The adjoint operator ${\alpha }^{\ast }$ is defined bySince $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an eigenvector $v$ with eigenvalue $\lambda$. Normality givesfor every $x$, by expanding both sides and using $\alpha {\alpha }^{\ast }={\alpha }^{\ast }\alpha$. Taking $x=v$ yields

For the Volterra integration operatorthe adjoint operator isSince $1/{\sigma }_n=(2n-1)\pi /2$, direct integration givesSimilarly,because $\cos ((2n-1)\pi /2)=0$. The half-integer sine and cosine families are orthonormal bases of $L^2([0,1])$, so $({\sigma }_n,u_n,v_n)$ is a singular system of a compact operator.

The Tikhonov solution for noisy data $g^{(\delta )}$ is thereforeWriting out the inner product and the singular functions makes this explicit:The factors ${\sigma }_n/({\sigma }_n^2+\alpha )$ suppress the unstable reciprocal growth $1/{\sigma }_n$ at high index.

Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.