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Set $K={\sum }_i\Vert A^i{\Vert }_{\mathrm{op}}^2$, where the norms are operator norms, and $D_m=y^{(m+1)}-y^{(m)}$. Independence of the Brownian coordinates and the vector Itô isometry giveInduction, as in part (b), yields the Factorial Picard estimate for a linear stochastic differential equationFor $t\le T$, the triangle inequality in $L^2$ therefore bounds any tail of the iterates by the corresponding tail ofThis numerical series converges by the ratio test: successive terms have ratio $\sqrt{KT/(m+2)}\to 0$. Thus the iterates are Cauchy in $L^2$, uniformly in the deterministic time parameter on each bounded interval. Completeness of Lp spaces gives a limit $y_t$. Once its continuous adapted version is chosen in part (d), the Itô isometry permits passage to the limit in the integral equation: the squared norm of each integral difference is bounded by $K{\int }_0^T\mathbb{E}|y_s^{(m)}-y_s{|}^{2}ds\to 0$. Consequently the Picard limit solves the stochastic differential equation.

The iteration matrix has eigenvalues $0,-1/2,1/2$: expanding its characteristic polynomial down the first column leaves the triangular two-by-two block with diagonal entries $-1/2,1/2$. The distinct eigenvalues ensure diagonalizability. Thus its spectral radius is $1/2<1$.

The previous bound gives $\Vert f_n\Vert \le 2^{-n}\Vert f_0\Vert \to 0$, so the Gauss-Seidel iteration converges for every initial guess under the stated diagonalization. Its limit is the exact solution $(1,1,-1{)}^T$. Since $e_n=S{f}_n$, a fixed matrix also sends this transformed error to zero; its operator norm can account for any amplification in the original coordinates.

Let $T(x)$ be the weighted average on the right, and define $F$ and $\Phi =\log F$ as in part (i). Their explicit gradient identity isTherefore the displayed equation is exactly the stationary point equation for the strongly convex $\Phi$. It has one solution, which is the unique minimizer from part (i). Reuse the BFGS iteration, withThe gradient is explicit, the weights can again be scaled by their largest exponent, and the usual full-memory BFGS assumptions give local superlinear convergence.

Simply setting $x_{n+1}=T(x_n)$ is not automatically a convergent fixed-point iteration. Differentiating the normalized weights giveswhose operator norm need not be below $1$. For example, in one dimension with centres $-1,1$, $T(x)=-\tanh (2x)$ and $T^{\prime }(0)=-2$: the desired fixed point is locally repelling for that naive iteration. If exact second derivatives are affordable, multivariable Newton iteration on $x-T(x)=0$ is another valid choice, with Jacobian matrix $I-J_T$ as given above and local quadratic convergence.

For $R(\theta )=\mathrm{diag}(1,e^{i\theta })$, the difference vanishes on $|0⟩$ and multiplies $|1⟩$ by $e^{i\theta }(e^{i\delta }-1)$. Its operator norm is therefore $|e^{i\delta }-1|=2|\sin (\delta /2)|$. Choosing the nearest angle on the grid $2\pi m/2^n$, with circular distance, gives $|\delta |\le \pi /2^n$ and a uniform bound of this size on the output-state norm error. This bound concerns state vectors, not an asserted equality with measurement-probability error.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.

For an $m$-quasirandom group, uniform expectations, and the normalized convolution on a finite group, a scalar mean-zero function $f$ satisfiesThe Fourier analysis on a finite group proof bounds each nontrivial matrix component of $f$ in operator norm using its weighted Hilbert-Schmidt norm. For subsets of subset density values $a,b,c$, the error in their normalized product count is at most $\sqrt{a(1-a)b(1-b)c(1-c)/m}$. In particular $abc>1/m$ guarantees a solution of $xy=z$ in the three subsets.

A Markov chain $Z$ with transition kernel $K$, sampled at an independent Poisson process $N$ of rate $r$, is a pseudo-Poisson process. Its infinitesimal generator is $r(K-I)$ and its transition semigroup is $e^{-rt}{\sum }_{n\ge 0}(rt{)}^n{K}^n/n!$. Conditioning on $N_t$ proves the latter formula, and differentiation at zero gives the former. Its resolvent operator is $(\lambda +r{)}^{-1}{\sum }_{n\ge 0}[r/(\lambda +r){]}^n{K}^n$ for $\lambda >0$, by integrating the nonnegative scalar Poisson weights. The series converge in operator norm on bounded Borel measurable functions.

Independent verifier circuits on $r$ witness registers induce a threshold acceptance effect built from tensor products of $M$ and $I-M$, where $M$ is the quantum verifier acceptance operator. In a product eigenbasis of $M$, the effect's eigenvalues are threshold probabilities for independent Bernoulli random variables with parameters given by the selected eigenvalues of $M$. Bounding every parameter on a NO instance bounds the effect's operator norm, even if the actual submitted witness is entangled. Independence is used only for the spectral calculation, not assumed for the witness.

Let $B$ be a real square matrix and suppose $x^{T}Bx\ge c|x{|}^2$ for every real vector $x$, with $c>0$. The Cauchy-Schwarz inequality gives $x^{T}Bx\le |x||Bx|$, hence $|Bx|\ge c|x|$ for all $x$. In particular the kernel is zero, so $B$ is an invertible matrix. Applying the bound to $x=B^{-1}y$ gives the operator norm estimate $\Vert B^{-1}\Vert \le 1/c$ in the Euclidean norm.

Symmetry of $B$ is unnecessary: $x^{T}Bx=x^{T}Hx$ for its symmetric part of a matrix $H=(B+B^T)/2$. Thus the hypothesis is exactly a positive lower bound on that symmetric matrix, even when the antisymmetric part is nonzero.

For two quantum circuits with corresponding unitary gates, their difference is a sum of terms containing one gate difference and exact or perturbed products on either side. Unitary invariance and the triangle inequality for the operator norm therefore bound the circuit difference by the sum of individual gate errors. Tensoring a gate with an identity leaves that norm unchanged. On a normalized input, the same bound controls the output vector distance with gate phases retained. This is a worst-case coherent bound; cancellation can make a particular circuit less sensitive, but cannot be assumed without further information.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

Let $\hat{P}$ project onto a leading unit eigenvector of $H+W$, with $H$ as in the Davis-Kahan curvature lemma. Its maximizing Rayleigh quotient gives $\mathrm{tr}(H(P-\hat{P}))\le \mathrm{tr}(W(\hat{P}-P))$. Duality of the operator norm and trace norm, together with the rank-two bound $\Vert \hat{P}-P{\Vert }_{\ast }\le \sqrt{2}\Vert \hat{P}-P{\Vert }_F$, proves the estimate. No gap condition on the perturbed matrix is needed.

Let $S$ be a bounded multiplicatively closed subset of a unital Banach algebra. The displayed auxiliary norm $p$ is equivalent to the original norm, and each left multiplication $L_s$ is a contraction for $s\in S$. Taking the operator norm of the faithful left regular representation produces an equivalent algebra norm with $\Vert 1{\Vert }_S=1$ and $\Vert s{\Vert }_S\le 1$. If $M=\max (\Vert 1\Vert ,{\sup }_S\Vert s\Vert )$, then $\Vert a\Vert /M\le \Vert a{\Vert }_S\le M\Vert a\Vert$. The auxiliary norm is itself submultiplicative, since $p(ab)\le p(a)\Vert b\Vert \le p(a)p(b)$, but $p(1)=M$ can exceed one. The operator norm enforces the required unital normalization.

If an associative unital algebra has a complete norm and separately continuous multiplication, the Uniform boundedness principle applied to $L_x(y)=xy$, for $\Vert x\Vert \le 1$, yields $\Vert xy\Vert \le C\Vert x\Vert \Vert y\Vert$. The operator norm of $L_x$ then defines an equivalent algebra norm: $\Vert x\Vert /\Vert e\Vert \le \Vert L_x\Vert \le C\Vert x\Vert$, and $L_{xy}=L_x{L}_y$ proves submultiplicativity. It normalizes the identity to norm one.

The linear-system residual $r^{(k)}=b-A{x}^{(k)}$ is computable without knowing $x_{\ast }$. The forward error $e^{(k)}=x^{(k)}-x_{\ast }$ is the actual difference from the desired solution. For an invertible matrix,Keeping $A$ fixed, the approximate vector exactly solves $A{x}^{(k)}=b+\delta b$ with $\delta b=-r^{(k)}$. Thus, for $b\ne 0$, $\Vert r^{(k)}\Vert /\Vert b\Vert$ is the relative backward error when only the right-hand side may change. For a vector norm and its induced operator norm,A small relative linear-system residual ensures a small relative forward error when the condition number is moderate; an ill-conditioned matrix can amplify it. The bound follows directly from $e=-A^{-1}r$ and $\Vert b\Vert =\Vert A{x}_{\ast }\Vert \le \Vert A\Vert \Vert x_{\ast }\Vert$.^[ref]

For the Jacobi iteration for a three-variable tridiagonal system, the first linear-system residual is $(1,3,1{)}^T$, while the first forward error is $(-1/2,-1,-1/2{)}^T$. In the Euclidean norm, the relative forward error is $\sqrt{3/28}\approx 0.3273$ and the relative linear-system residual is $\sqrt{11/120}\approx 0.3028$. Its spectral condition number of a positive-definite matrix is $(4+\sqrt{2})/(4-\sqrt{2})\approx 2.094$, so the bound is about $0.6340$. The bound is a guarantee, not an equality. In floating-point arithmetic, recompute the linear-system residual from the current vector when checking termination; a small update alone need not certify a small forward error.

When $\beta \rho (A)<1$, summing the adjacency-matrix bound for a discrete SIR epidemic gives the displayed Neumann series bound on the expected total infected count $Z$. For an undirected graph, the adjacency matrix of a graph is a symmetric matrix, so its Euclidean operator norm equals its spectral radius. The Cauchy-Schwarz inequality then yields $\mathbb{E}Z\le \sqrt{nm}/(1-\beta \rho )$ for $m$ initially infected vertices.