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Hermiticity of the Hamiltonian operator gives $⟨{\psi }_m,H{\psi }_n⟩=⟨H{\psi }_m,{\psi }_n⟩$, soThe distinct eigenvalues force orthogonality for $m\ne n$, and the given normalization gives one for $m=n$. Therefore $⟨{\psi }_m,{\psi }_n⟩={\delta }_{mn}$.

Under the time-independent Schrödinger equation, each energy component acquires its own phase:The sum converges in the state Hilbert space because the squared coefficients are summable. Orthogonality makes its squared norm ${\sum }_{n\ge 0}{2}^{-(n+1)}=1$ at every time.

The Born rule for energy measurement givesBoth depend only on squared coefficient magnitudes, so both are time-independent. The expectation follows either by summing energy times measurement probability or by inserting the orthogonal expansion in the quadratic form of $H$.

Normalization alone does not guarantee a finite mean for an arbitrary spectrum. The displayed energy series is finite when $\sum |E_n|2^{-(n+1)}<\infty$. Since the spectrum is bounded below by $E_0$, a divergent positive tail gives an extended expectation $+\infty$, still independent of time; for example an abstract diagonal Hamiltonian with $E_n=4^n$ has this property. No growth assumption on $E_n$ is printed, so finiteness should not be asserted without this qualification.

Write $L(f)={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)={\sup }_z|f(z)|$. The quantity in the question is the bounded Lipschitz norm $\Vert f{\Vert }_{\mathrm{BL}}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a norm.

For the sequence bounded by one, both $L(f_i)\le 1$ and $B(f_i)\le 1$. Given real $x$, choose a nearby rational $q$. ThenFirst make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a Cauchy sequence, so $f_i$ converges pointwise to a real function $f$.

To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\Vert f{\Vert }_{\mathrm{BL}}\le 1}$, the pointwise closure of the bounded Lipschitz unit ball.

To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.

There is a qualification in the printed hypotheses: quadratic ellipticity does not bound a nonsymmetric coefficient matrix. The intended standard existence proof needs $a_{ij}\in L^{\infty }(\Omega )$, or more generally a bounded principal bilinear form. To see why this does not follow from the displayed inequalities, take $\Omega =(-1,1{)}^2$ andThen ${\zeta }^{T}A(x)\zeta =|\zeta {|}^2$ for every $x,\zeta$. However, choose $u_0,{\phi }_0\in C_c^{\infty }(\Omega )$ equal to $x_2,x_1$, respectively, on a small rectangle about the origin. There $(AD{u}_0)\cdot D{\phi }_0=k(x_1)$, whose positive part has infinite integral. Thus the standard principal pairing is not a finite Lebesgue integral even for these smooth test functions. This demonstrates the missing form hypothesis, rather than assuming that all unbounded skew coefficients prevent solvability.

Under the intended bounded-coefficient hypothesis the proof is as follows. On $H=H_0^1(\Omega )$ use the Hilbert space norm $\Vert Dw{\Vert }_2$, equivalent to the $H^1$ norm by the Poincaré inequality on a bounded domain. The form from part (i) is bounded, sinceIt is a coercive bilinear form even without symmetry:Here the nonpositive sign of $q$ is essential. With $u=\psi +w$, the required equation becomesThe Cauchy-Schwarz inequality, bounded coefficients and the Poincaré inequality show that $\ell$ is a bounded linear functional on $H$. The Lax-Milgram theorem states that a bounded coercive bilinear form on a real Hilbert space represents each bounded linear functional in this way with a unique $w\in H$. It therefore yields $u=\psi +w$ with the required boundary condition and weak equation. For completeness, if $u_1,u_2$ are solutions, $z=u_1-u_2\in H_0^1$ satisfies $B(z,z)=0$; coercivity and the Poincaré inequality imply $z=0$.

With the missing bounded-form hypothesis, the weak solution exists and is unique. For nonsymmetric matrices the printed quadratic inequalities omit the hypothesis needed for this standard formulation and proof. No maximum principle is used. In part (iii), symmetry makes the quadratic upper bound a bound on the whole matrix, so no such qualification is needed there.

Let $d=\mathrm{dist}({\Omega }^{\prime },\partial \Omega )>0$. Choose a smooth cutoff function $\eta \in C_c^{\infty }(\Omega )$ with $0\le \eta \le 1$, $\eta =1$ on ${\Omega }^{\prime }$ and $|D\eta |\le C(n)/d$. Such cutoffs can be constructed by mollifying the indicator of a fixed small neighbourhood of $\overline{{\Omega }^{\prime }}$. Write $M=\Vert A{\Vert }_{\infty ,\mathrm{op}}$, $B=\Vert b{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$; these are controlled by the individual coefficient bounds and dimension.

The weak formulation, extended to compactly supported $H^1$ tests by density, isWe may use $\phi ={\eta }^{2}u$, since $u\in H_{\mathrm{loc}}^1$. Expanding its gradient and using uniform ellipticity givesApply Young inequality to the first two terms, allocating at most $\lambda /4$ of the gradient integral to each. Bound $|fu|\le (f^2+u^2)/2$ in the last term. Absorbing those gradient contributions yields the Caccioppoli inequality with bounded lower-order termsConsequently $\Vert Du{\Vert }_{L^2({\Omega }^{\prime })}\le C(\Vert u{\Vert }_{L^2(\Omega )}+\Vert f{\Vert }_{L^2(\Omega )})$. Adding the $L^2$ norm of $u$ proveswhere $C$ has precisely the stated dependence on dimension, ellipticity, coefficient bounds and $d$. The proof uses $|c|$, so no sign condition on the zeroth-order term is required.

First perform homogenization of Dirichlet boundary data: put $v=u-\psi \in H_0^1(B_1^{+})$ and $g=f-\Delta \psi \in L^2(B_1^{+})$. Then $\Delta v=g$ weakly. Extend both functions oddly across the flat face:The zero Sobolev trace of $v$ makes this odd reflection an $H^1(B_1)$ function. This can also be seen by approximating $v$ by smooth compactly supported functions in $B_1^{+}$ and reflecting the approximants. Its $H^1$ norm is $\sqrt{2}$ times that of $v$, and $\Vert \tilde{g}{\Vert }_{L^2(B_1)}=\sqrt{2}\Vert g{\Vert }_{L^2(B_1^{+})}$.

For $\varphi \in C_c^{\infty }(B_1)$, the function $\varphi (x^{\prime },t)-\varphi (x^{\prime },-t)$ on the upper half-ball has zero trace on the flat face and vanishes near the curved face, hence is an admissible $H_0^1$ test. Splitting the integrals over the two half-balls and changing $t$ to $-t$ showsThus $\Delta \tilde{v}=\tilde{g}$ weakly throughout the full ball; there is no extra distribution on the reflecting face.

Apply part (i) to $B_{1/2}⋐B_1$, restrict to the upper half-ball, and then add $\psi$. Since $\Vert \Delta \psi {\Vert }_2\le C(n)\Vert \psi {\Vert }_{H^2}$, this gives $u\in W^{2,2}(B_{1/2}^{+})$ and the second-derivative estimate at a flat Dirichlet boundaryThe fixed radii leave distance $1/2$ for the interior estimate, so the constant depends only on dimension.

Put $P=I-p^{-1}{1}_p{1}_p^T$, $Z=PX=X-\bar{X}{1}_p$, $Q=\Vert Z{\Vert }^2$ and $a=p-3$. Componentwise,Thus the James–Stein shrinkage toward the sample mean leaves the average unchanged and multiplies every contrast about it by the same data-dependent factor $1-a/Q$. If $Q>a$ the contrasts shrink without changing sign; if $Q<a$ their signs reverse, and if $Q<a/2$ their magnitudes increase. The event $Q=0$ has probability zero, and the estimator can be assigned $\bar{X}{1}_p$ there.

The sample mean is a linear normal statistic with mean $\bar{\theta }$ and variance $1/p$, soThe error splits into orthogonal mean and contrast components:Since the second term is perpendicular to $1_p$, its cross product with the first is zero for every sample. Taking squared norms and expectations, $p\mathbb{E}(\bar{X}-\bar{\theta }{)}^2=1$, which yields the risk function

To compute the contrast risk, choose an orthonormal basis matrix $U$ for the range of $P$, with $U^{T}U=I_m$, $U{U}^T=P$ and $m=p-1\ge 3$. Then $W=U^{T}X\sim N_m(\eta ,I)$ with $\eta =U^T\theta$, $Q=\Vert W{\Vert }^2$, and the contrast norm is the norm of $(1-a/Q)W-\eta$. Expand it:For $g_j(w)=w_j/\Vert w{\Vert }^2$, the Gaussian Stein identity givesThis application at a singularity is valid: near zero the reciprocal squared radius is integrable in dimension $m>2$, since its radial integral is ${\int }_0^1{r}^{m-3}dr$. Apply Gaussian integration by parts outside a ball of radius $\epsilon$; the inner boundary contribution is bounded by a constant times ${\epsilon }^{m-2}$ and vanishes. The Gaussian tail eliminates the boundary at infinity. Therefore all required expectations are finite and the displayed identity holds.

Since $a=m-2=p-3$, the Stein risk identity now gives contrast risk $m-a^2\mathbb{E}(1/Q)$. Adding the unit mean risk proves

Finally $Q$ has the noncentral chi-squared distribution ${\chi }_m^2(\lambda )$ with $\lambda =\Vert P\theta {\Vert }^2$. To establish where its reciprocal expectation is largest, complete the square in the normal representation $Q=\Vert W{\Vert }^2$ to obtain its Laplace transform:Since $1/q={\int }_0^{\infty }{e}^{-tq}dt$, the Tonelli theorem gives the reciprocal moment of a noncentral chi-squared variableIts integrand is positive and strictly decreasing in $\lambda$ for every $t>0$. The expectation is therefore strictly largest at $\lambda =0$, whereBecause this expectation is subtracted in the risk, the minimizing set is exactly $P\theta =0$, that is, all coordinates of $\theta$ equal. On this line the risk is $p-(p-3)=3$:

Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is $-u^{\prime \prime }+xu$, so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find $u\in V=H_0^1(0,1)$ such thatHere $V$ is the zero-boundary Sobolev space and can be equipped with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The interval Poincaré inequality states $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$, so this norm makes $V$ a Hilbert space and is equivalent to its usual $H^1$ Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, $u^{\prime \prime }=xu+1\in L^2$ gives $u\in H^2$ on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.

Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at $u$. This proves the minimization equivalence, including the sign of the linear term.

Restricting to a conforming finite element space $V_h\subset V$ gives a unique discrete solution and Galerkin orthogonality, $a(u-u_h,v_h)=0$ for every $v_h\in V_h$. The Céa lemma states that if the continuity and coercivity constants are $M$ and $\alpha$, then $\Vert u-u_h{\Vert }_V\le (M/\alpha ){\inf }_{v_h\in V_h}\Vert u-v_h{\Vert }_V$. For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.

The Butcher contractivity theorem states that an algebraically stable Runge-Kutta method is B-stable: for a dissipative vector field and any positive step for which the stage problems are solved, its step does not increase the distance between two numerical solutions. The required condition on the dissipative vector field isfor all compared states at each common time. The inner product can be real Euclidean or complex Hermitian. The stage solvability of an implicit Runge-Kutta method is assumed in this comparison, not a conclusion from a distance estimate alone.

Let $d$ be the difference of starting values, $D_i$ the differences of corresponding stages and $F_i$ the differences of their vector-field values. ThenExpanding the squared norm of the update givesSubstitute $d=D_i-h{\sum }_j{a}_{ij}{F}_j$ in the linear term. Symmetrizing the double sum proves the Runge-Kutta contractivity identityEach inner product in the first sum is nonpositive by the defining inequality for a dissipative vector field, and its weight is nonnegative. The final quadratic form is nonnegative: resolve the vectors $F_i$ into components and apply positive semidefiniteness of $M$ to each component vector. Hence $\boxed{\Vert d_{+}\Vert \le \Vert d\Vert }$, which is exactly B-stability. No linearization of the vector field has been used.

For the Dahlquist test equation $y^{\prime }=\lambda y$ with $\mathrm{Re}\lambda \le 0$, this also bounds the amplification factor by one wherever the stages are well defined. With strictly positive weights, no stage pole can occur in the open left half-plane. Indeed, if $(I-zA)v=0$ for nonzero $v$, thenwhen $\mathrm{Re}z<0$, contradicting $M⪰0$. This proves the usual A-stability corollary for the scalar stability function; boundedness from the open half-plane makes any boundary pole in that rational function removable. Zero-weight redundant stages require their own solvability treatment. The nonlinear B-stability statement is stronger than A-stability, and the two notions should not be conflated.

The eigenvalue stability analysis of a finite difference method reduces a linear evolution discretization to its modal behavior, but it must retain the norm and the dependence on the mesh. For a homogeneous system obtained by the method of lines, $U^{\prime }=A_{h}U$, the solution is $e^{t{A}_h}U(0)$. A one-step time discretization gives $U^{n+1}=G_{h,k}{U}^n$; for a Runge-Kutta method, where its stage matrices are invertible, $G_{h,k}=R(k{A}_h)$. With a multistep method one instead uses an amplification matrix on the augmented vector of several time levels.

The relevant finite-time stability estimate iswith $C_T$ independent of the refining spatial mesh and allowed time steps. Similarly, semidiscrete stability requires $\Vert e^{t{A}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. The norm should represent the continuum problem, such as the discrete L2 norm $\Vert U{\Vert }_{2,h}^2=h{\sum }_j|U_j{|}^2$ in one dimension. A fixed finite matrix having decaying solutions as $t\to \infty$ is not by itself a mesh-uniform result for the partial differential equation.

If $G=X\Lambda X^{-1}$, thenThus eigenvalues of modulus at most one give a uniform bound when the diagonalizing bases have uniformly bounded condition numbers. For a normal matrix in the chosen inner product, the spectral theorem for normal operators supplies an orthonormal basis, so the condition number is one and the norm of $G^n$ is exactly ${\max }_j|{\lambda }_j{|}^n$. For a self-adjoint or skew-adjoint spatial matrix, applying a scalar stability function preserves this favorable modal structure. This is the setting where eigenvalue calculations provide especially clean, reliable step restrictions.

A precise fixed-matrix statement is also useful: a matrix is power bounded for all nonnegative integers $n$ if and only if all its eigenvalues lie in the closed unit disk and every Jordan block at a unit-modulus eigenvalue has size one. Indeed, powers of an interior Jordan block contain a polynomial in $n$ times a decaying geometric factor and are bounded; a larger block on the unit circle produces unbounded polynomial growth. For a family of matrices, the resulting bounds must still be uniform. Moreover, finite-time stability versus power boundedness distinguishes this all-time condition from the physical-time estimate above. For example $G_k=I+kN$, $N^2=0$, has $G_k^n=I+nkN$, bounded for $nk\le T$ despite a nontrivial Jordan block at one. Even the scalar update $1+k$ is stable on fixed intervals, since $(1+k{)}^n\le e^T$, although it represents a growing equation rather than a contraction.

For a constant-coefficient periodic difference operator, Fourier modes diagonalize the spatial matrix. The discrete Fourier transform is a unitary operator, so Parseval's identity identifies its modal maximum with the operator norm. The eigenvalue stability analysis of a finite difference method then becomes von Neumann stability analysis. For the Dirichlet discrete Laplacian, a discrete sine transform plays the same role. These facts explain its advantages: a large matrix calculation becomes a scalar symbol or a known spectral interval, and the extremal eigenvalue directly determines a safe time step.

As a first example, centered space discretization of the heat equation $u_t=u_{xx}$ has eigenvaluesForward Euler method gives $R(k\lambda )=1-4r{\sin }^2(\theta /2)$ with $r=k/h^2$. Requiring all factors in $[-1,1]$ yields $0\le r\le 1/2$ for all periodic meshes. In two dimensions the corresponding uniform bound is $r\le 1/4$. Because these are normal matrices, these scalar bounds prove contraction rather than merely suggest it.

Backward Euler method has $R(z)=1/(1-z)$, so the centered discretization of the heat equation is contractive for every $k>0$. The Crank-Nicolson method has $R(z)=(1+z/2)/(1-z/2)$ and is likewise contractive for these negative real modes. However, its factor tends to $-1$ for very stiff modes, leaving high-frequency oscillations weakly damped; Backward Euler method tends to zero. The eigenvalues therefore expose the presence of stiff differential equations and damping as well as binary stability. Unconditional stability does not remove accuracy requirements.

For an advection equation example, the upwind finite difference scheme for $u_t+c{u}_x=0$, $c>0$, has factor $G=1-\nu +\nu e^{-i\theta }$, with $\nu =ck/h$. Sinceit is a contraction in the periodic discrete L2 norm exactly for $0\le \nu \le 1$. In contrast, Forward Euler method with centered advection has factor $1-i\nu \sin \theta$. Its modulus exceeds one at nonzero active frequencies, and repeated amplification makes it unstable under fixed nonzero Courant refinement. A much smaller scaling $k=O(h^2)$ can bound its finite-time growth, demonstrating again that exact contraction and the most general stability estimate are different claims.

The major limitation is the possible presence of a nonnormal matrix. For example,Both eigenvalues are negative, yet at $t=1$ the second unit vector is amplified by at least $1/(eh)$. This proves that negative spectra do not imply uniform semidiscrete stability. Ill-conditioned eigenvectors, Jordan normal form and transient amplification can invalidate a spectral-only argument. Direct energy estimates or estimates for the resolvent can supply the missing norm control.

Boundary conditions are another limitation. A periodic symbol proves a periodic or whole-line result, not an arbitrary initial-boundary-value problem. Boundary closure of a difference scheme can introduce growing modes or mesh-dependent amplification. A spatial operator with variable coefficients can still be studied through its assembled matrix, but then whether it is a normal matrix, its condition number and the physical inner product must be checked rather than borrowed from a Fourier stability analysis. For the finite element method the natural mass matrix inner product often turns a generalized eigenvalue problem into a self-adjoint one.

For time-dependent or split updates, separate spectra are insufficient to control products. For instance, $G_1=\left(\begin{array}{cc}0& K\\ 0& 0\end{array}\right)$ and $G_2=\left(\begin{array}{cc}0& 0\\ K& 0\end{array}\right)$ each have only zero eigenvalues, while $G_2{G}_1$ has eigenvalue $K^2$. For $K>1$, alternating the two is unstable. A common contractive norm would control the products, but their individual eigenvalues do not.

Finally, stability and consistency of a numerical method have different roles. The Lax equivalence theorem states that, for a well-posed linear initial-value problem and a consistent finite difference method on compatible grid spaces, stability on every fixed time interval is equivalent to convergence of a numerical method as the admissible meshes refine. Bounded grid representation and the correct initial and boundary setup are part of that framework. An eigenvalue estimate can establish its stability hypothesis in the favorable settings described above; it neither proves consistency of a numerical method nor bypasses its uniformity requirements. The method is therefore a powerful modal tool, provided its spectral calculation is connected to a genuine norm estimate.

Assume the relative permittivity is real and the field decays sufficiently at transverse infinity, or impose boundary conditions with zero transverse flux. Then $L(x)$ is self-adjoint on its transverse domain by integration by parts. The scalar parabolic wave equation givesThis is norm conservation for the scalar parabolic wave equation. It remains true when $L$ depends explicitly on $x$; its instantaneous self-adjointness suffices. Equivalently, the local balance isSince $|E|=|U|$, the invariant can also be writtenFor a unit transverse polarization in a fixed reference medium with impedance $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$, the leading forward paraxial Poynting vector has $S_x\simeq |E{|}^2/(2Z_0)$. Thus the scalar model conserves its leading reference-normalized power $P_0=\mathcal{N}/(2Z_0)$. This is the intended consequence of the Hermitian-operator hint. Complex permittivity would add absorption or gain and invalidate this norm invariant.

Two qualifications are necessary for the printed physical interpretation. First, $(1,0,1)$ is neither a unit vector nor transverse to nearly axial propagation: its longitudinal electric component is as large as its transverse one. In a homogeneous source-free medium, the example $\mathbf{E}=(1,0,1)e^{ikx}$ satisfies the componentwise Helmholtz equation but violates $\nabla \cdot \mathbf{E}=0$. It therefore is not the stated near-axial electromagnetic plane wave. The same scalar calculation formally conserves $\int |\mathbf{U}{|}^2=2\mathcal{N}$ for that fixed vector, but does not make it a physical Maxwell field. A unit transverse polarization, for example $(0,0,1)$, is the consistent paraxial choice, with only small longitudinal corrections.

Second, for varying local refractive index $n(x)=\sqrt{{ϵ}_{\mathrm{rel}}(x)}$, the quoted local plane-wave impedance relation gives $Z(x)=Z_0/n(x)$ and hence a flux proportional to $n(x)|E{|}^2$, not merely $|E{|}^2$. The scalar norm proof alone cannot make this exactly constant for arbitrary $n(x)$. For example a transversely uniform envelope obeying the printed parabolic equation hasso its intensity is constant while the locally inferred plane-wave flux changes with $n$. This supplies a direct obstruction to the literal exact-flux claim within the supplied scalar assumptions.

The consistent physical statement follows from Poynting theorem: for a lossless Maxwell solution and no lateral energy leakage,For a slowly varying transverse plane wave, the corresponding forward WKB field has amplitude proportional to $n^{-1/2}$, making $n|E{|}^2$ constant. In one dimension this is also encoded in the exact conserved flux $(2{\mu }_0\omega {)}^{-1}\mathrm{Im}(E^{\ast }{E}_x)$. Thus the requested physical conservation holds for the corrected lossless Maxwell description, while the given scalar equation proves the reference-normalized paraxial invariant above. The supplied fixed longitudinal polarization and arbitrary local-impedance inference cannot be used to claim more.

The restriction of a representation $\rho :G\to \mathrm{GL}(V)$ to $H$ is the same vector space and operators with the domain limited to $H$; its character is ${\chi }_H(h)=\chi (h)$. Averaging a positive Hermitian inner product over $H$ makes it invariant, so invariant subspaces have invariant orthogonal complements. Repeating gives a direct sum of irreducible $H$-spaces and hence ${\chi }_H={\sum }_i{d}_i{\psi }_i$ with nonnegative integer multiplicities.

By character inner product orthogonality,since $\chi$ is irreducible and has normalized squared norm one on $G$. Equality holds precisely when the omitted nonnegative terms vanish, namely $\boxed{\chi (g)=0\text{ for every }g\notin H}$. This is the restriction multiplicity bound.