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For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

The convention used for planar Gaussian free field theory is the Dirichlet energy space:It is a real Hilbert space with the extended Dirichlet inner product. A conformal map from the unit disc onto $D$ identifies the two energy completions, so local representatives can be understood through the disc's zero-boundary space.

For an arbitrary unbounded domain this is the homogeneous energy completion; it should be distinguished from completion in the inhomogeneous norm $\Vert f{\Vert }_{L^2}+\Vert \nabla f{\Vert }_{L^2}$ used in another standard definition of the zero-boundary Sobolev space. On bounded domains with a Poincare inequality, the two norms are equivalent.

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.

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.

Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.

Fix a receiver position $\mathbf{\zeta }$ and define $g_{\mathbf{\zeta }}(\mathbf{z})=G_L(\mathbf{\zeta },\mathbf{z})$. Choose the two scalar fields in the Wigner distribution to be $u=v=g_{\mathbf{\zeta }}$. Fourier inversion in the separation variable givesIndeed, substituting the defining Wigner distribution makes the $\mathbf{p}$ integral a $(2\pi {)}^2$ Dirac delta function, canceling its normalization and setting the separation equal to $\mathbf{z}-{\mathbf{z}}^{\prime }$.

Average over the receiver plane, including the aperture:Then the receiver-averaged Wigner distribution represents the back-propagated field asThe fast decay of $f$ justifies its use as a test function. For oscillatory Green functions that are not integrable, the Wigner distribution and its inversion can be understood as tempered distributions, or derived with smooth cutoffs before taking their limits.

The printed $W_A$ expression can also be used literally, but its displayed ${\mathbf{z}}^{\prime }$ integral is over source-plane coordinates, not receiver positions. The missing receiver average and source weight can be encoded by choosing vector fields whose channels are the receivers. LetContract the receiver channels in the product of the vector fields, so thatWith exactly the printed definitionthe same Fourier inversion now yields the requested compact formFor a finite array, the receiver integral in the contraction is a weighted sum. For a continuous array, these are fields valued in the receiver Hilbert space. An uncontracted vector outer product instead gives a matrix-valued Wigner distribution, whose receiver trace must be taken. Thus the printed formula is usable with these choices and this contraction convention, but its phrase “over the plane of the receiver” does not describe the displayed integral.

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.

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.

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.

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrodinger equation example. Conformity, coercivity and approximation estimates together explain convergence.

The hypothesis says exactly that $X_n\rightharpoonup 0$ in the Hilbert space $L^2$. We give the constructive Weak Banach–Saks theorem in a Hilbert space argument. Choose $n_1<n_2<\cdots$ inductively so that, for $Y_k=X_{n_k}$,This is possible because each earlier $Y_j$ is an admissible test variable in the weak-convergence assumption. Since $\Vert Y_k{\Vert }_2\le 1$,Therefore

The Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives

The Lax-Milgram theorem states that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form, and there is an $\alpha >0$ such thatthen for every bounded linear functional $F\in H^{\prime }$ there is a unique $u\in H$ satisfyingMoreover, $\boxed{\Vert u{\Vert }_H\le {\alpha }^{-1}\Vert F{\Vert }_{H^{\prime }}}$. Symmetry of $B$ is not required.

On the product Hilbert space $H=H_0^1(U)\times H^1(U)$ defineandThe Cauchy-Schwarz inequality makes $B$ and $F$ bounded. On the diagonal,The Poincare inequality controls $\Vert u{\Vert }_2$ by $\Vert Du{\Vert }_2$, andThe displayed diagonal value therefore controls the full product $H^1$ norm, so $B$ is coercive. The Lax-Milgram theorem now gives exactly one pair $(u,w)\in H$ satisfying the weak identities. Hence a unique weak solution exists for every $f,g\in L^2(U)$.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

The Radon-Riesz property of a norm is the implicationIn a Hilbert space, expand the square using its inner product:Weak convergence gives $⟨u_k,u⟩\to \Vert u{\Vert }^2$, and convergence of norms gives $\Vert u_k{\Vert }^2\to \Vert u{\Vert }^2$. Hence the right side tends to zero. This proves the Hilbert-space Radon-Riesz theorem.

The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.