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For $S={\sum }_i{a}_i{ϵ}_i$ and $F=\Vert S\Vert$, the even-function spectral gap on a hypercube gives $2\mathrm{Var}(F)\le -\mathbb{E}[FLF]$. A norming functional supplied by the Hahn-Banach theorem and convexity of the norm gives $LF\ge -F$, because $LS=-S$. Hence the Dirichlet form is at most $\mathbb{E}{F}^2$, proving the inequality. Two equal scalar coefficients attain equality, so the constant is sharp.

Suppose $q$ is a sublinear functional on the bounded real sequences, dominated by the limit superior and nonpositive on every shift difference $x-Sx$. The Hahn-Banach theorem then yields a Banach limit. Indeed $0\le q(x)+q(-x)$ implies $q(x)\ge -q(-x)\ge \mathrm{liminf}{x}_n$, so $q$ agrees with the ordinary limit on convergent sequences. Extend this linear limit functional by Hahn-Banach to a linear $\psi \le q$. Domination on $x-Sx$ and on its negative forces $\psi (x)=\psi (Sx)$. The two-sided limit bounds give positivity, norm one and agreement with ordinary limits. In finite-shift averaging constructions, the needed bound on shift differences follows by telescoping consecutive averages to $(x_{n+1}-x_{n+k+1})/k$.

On $[0,1]$, the continuous functions $f_n(x)=\sin (nx)$ do not form a Cauchy sequence in the Lp space with $p=1$. Put $d_n=f_n-f_{2n}$. The identity ${\int }_0^1{d}_n^{2}dx=1-\sin (2n)/(4n)-\sin (4n)/(8n)-\sin n/n+\sin (3n)/(3n)$ tends to $1$. Since $|d_n|\le 2$, one has ${\int }_0^1|d_n|dx\ge \frac{1}{2}{\int }_0^1{d}_n^{2}dx$, which stays bounded away from zero. Thus arbitrarily late pairs are separated in the integral norm; in particular the sequence cannot converge in that norm.

Prepare a qubit in $|+⟩$, apply a phase gate $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$, and correlate its computational basis states with normalized environment states $|e_0⟩,|e_1⟩$. Write $⟨e_0|e_1⟩=v{e}^{i\alpha }$. After a final Hadamard gate, the environment amplitude accompanying output zero is $(|e_0⟩+e^{i\varphi }|e_1⟩)/2$. Its squared norm is $P_0=[1+v\cos (\varphi +\alpha )]/2$. Thus $v$ is the interference visibility, while $\alpha$ shifts the fringe. This quantum decoherence map commutes with the phase gate, so placing it immediately before that gate gives the same probabilities.

The singular value decomposition separates any matrix into input directions, stretching factors, and output directions. Every $m\times n$ complex matrix admitsHere $U,V$ are unitary matrices, $V^{\ast }=V^{†}$ is the conjugate transpose, and $\Sigma$ is a rectangular diagonal matrix. Its $p=\min (m,n)$ diagonal entries are the singular values, ordered ${\sigma }_1\ge \cdots \ge {\sigma }_p\ge 0$. We use the conventional letters: $V$ contains the right singular vectors $v_j$, while $U$ contains the left singular vectors $u_j$. Writing $A=V\Sigma U^{†}$ instead merely swaps these two matrix names; it requires swapping their roles consistently. For real matrices, choose orthogonal matrices and replace the conjugate transpose with the transpose.^[ref]

Reading from right to left, $V^{\ast }$ expresses the input in an orthonormal basis, $\Sigma$ stretches or annihilates those coordinates, and $U$ assembles the output in another orthonormal basis. Unlike matrix similarity, the input and output bases can differ, and even have different sizes. A $3\times 2$ matrix has full factors of sizes $3\times 3$, $3\times 2$, and $2\times 2$. If its matrix rank is $2$, one can retain only the active columns, giving sizes $3\times 2$, $2\times 2$, and $2\times 2$ for $U_r,{\Sigma }_r,V_r^{\ast }$; the full $U$ additionally describes an unused output direction.

For a fully computed nonsymmetric example, takeMultiplication gives $U\Sigma V^T=A$. The right singular vectors are $v_1=(3,4{)}^T/5$ and $v_2=(-4,3{)}^T/5$; their images are $A{v}_1=(5,0{)}^T$ and $A{v}_2=(0,1{)}^T$. Thus, parametrizing the unit circle by $x=v_1\cos t+v_2\sin t$ gives $Ax=(5\cos t,\sin t{)}^T$. Its image is the ellipse $y_1^2/25+y_2^2=1$. The two singular values are its semiaxis lengths, and the left singular vectors point along its axes. The singular value decomposition finds precisely the perpendicular input directions whose images stay perpendicular.

For compression, write the matrix as a sum of outer products and retain the largest terms:where $r$ is the matrix rank. Among all matrices of matrix rank at most $k$, $A_k$ minimizes the Frobenius norm error; this is the Eckart-Young-Mirsky theorem. The operator norm error is ${\sigma }_{k+1}$ for $k<r$. These statements concern specified norms, rather than guaranteeing that discarded details are visually or scientifically unimportant.^[ref]

Consider an $8\times 10$ brightness matrix $B=p{q}^T+a{b}^T/2$, withand $a_i=(-1{)}^{i-1}$, $b_j=(-1{)}^{j-1}$. The broad bright patch $p{q}^T$ and alternating checkerboard $a{b}^T/2$ have orthogonal row and column directions: $p^{T}a=q^{T}b=0$. Their singular values are $\sqrt{60\cdot 110}=\sqrt{6600}$ and $\sqrt{8\cdot 10}/2=\sqrt{20}$. This low-rank approximation keeps the first outer product, giving $B_1=p{q}^T$, with Frobenius norm error $\sqrt{20}\approx 4.472$ and relative error $\sqrt{20/6620}\approx 5.50\%$. For example, $B_{11}=1.5$ becomes $1$, and $B_{12}=1.5$ becomes $2$: every entry changes by $0.5$ in magnitude. The discarded pattern is small but entirely lost.

For this illustrative image compression, storing one normalized left singular vector, one normalized right singular vector, and one singular value requires $8+10+1=19$ scalars instead of $80$. This counts scalars at equal precision, before storage overhead. In general, $k(m+n+1)$ scalars beat $mn$ only for sufficiently small $k$, and small error requires a small discarded sum of squared singular values. The usefulness comes from measuring the tradeoff before throwing information away.^[ref]

Smooth convergence on a compact interval requires uniform convergence of the functions and of their derivatives of every order separately, as measured by the derivative seminorms. The need for several scales is visible inFor $n\ge 2$, $p_0(f_n)=1/n$ because $t=\pi /(2n)$ lies in $[0,1]$. The derivative has supremum norm one for every $n$, already at $t=0$. Thus $f_n$ converges uniformly to zero, but fails the test for smooth convergence to zero. Andreas Buttenschoen, Applied Functional Analysis, spaces of continuous and differentiable functions discusses this example and the family of smoothness measurements.

More generally, for any fixed $K\ge 0$, $h_n(t)=\sin (nt)/n^{K+1}$ has $p_K(h_n)\le 1/n$ but $p_{K+1}(h_n)=1$ for $n\ge 2$. If a norm $N$ induced the smooth topology, its continuity would give $N(f)\le C{p}_K(f)$ for some finite $K$, while continuity of $p_{K+1}$ for $N$ would give $p_{K+1}(f)\le C^{\prime }N(f)$. Substituting $h_n$ contradicts these two bounds. Smooth convergence therefore cannot be reduced to one norm. Its beauty is that every finite order can be controlled without fixing a final order in advance.

For $1\le p<\infty$, an inhomogeneous Sobolev space norm is $\Vert u{\Vert }_{W^{k,p}(D)}=({\sum }_{|\alpha |\le k}\Vert D^{\alpha }u{\Vert }_{L^p(D)}^p{)}^{1/p}$, using weak derivatives. In particular, $\Vert u{\Vert }_{H^1(D)}^2={\int }_D(|u{|}^2+|\nabla u{|}^2)$. The gradient seminorm alone defines a different completion, the Dirichlet energy space, unless a suitable Poincaré inequality makes the two norms equivalent on the chosen zero-boundary space.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

When the velocity distribution is independent of position and there are no forces, the free transport equation contributes zero and only collisions evolve $f$. For a bounded cutoff kernel, $\Vert Q(f,g){\Vert }_1\le C\Vert f{\Vert }_1\Vert g{\Vert }_1$ gives local existence and uniqueness by the Banach fixed-point theorem. Positivity and mass conservation bound the norm and permit global continuation in this particular model. Stronger kernels require moment or weighted estimates.

For a bounded Borel measurable function $f$ and the projection-valued measure $E$ of a normal operator, the displayed set is the spectrum of an element of $f(T)$. Outside it, a bounded reciprocal away from a spectral-null set supplies the inverse. Inside it, unit vectors supported by the indicated arbitrarily small spectral sets make $f(T)-\lambda$ fail to be bounded below. The kernel of the calculus consists of functions satisfying $E(|f|>\epsilon )=0$ for every $\epsilon >0$, and its norm is the corresponding essential supremum. Ordinary pointwise range is insufficient: for multiplication by $t$ on $L^2[0,1]$, $1_{\{0\}}(T)=0$ even though the function takes the value $1$.

The spectral projection $P_N$ retains the first $N$ Stokes eigenfunctions. On its finite-dimensional range, one derivative costs at most ${\lambda }_N^{1/2}$ and all Sobolev norms are equivalent.

Let $A$ be a nonzero unital C-star algebra, $x\in A$, and $\lambda \in {\sigma }_A(x)$. On the linear span of $1,x$, define $g(a1+bx)=a+b\lambda$. The spectral mapping theorem for affine polynomials gives $|a+b\lambda |\le \Vert a1+bx\Vert$, making $g$ well-defined with norm one and $g(1)=1$. A norm-preserving Hahn-Banach theorem extension to $A$ remains unital; the norm characterization of a positive functional on a C-star algebra makes it a state on a C-star algebra attaining $\lambda$.

The unit assumption is necessary for this formulation. The element $x(t)=e^{-t^2}$ of the nonunital algebra $C_0(\mathbb{R})$ has zero in its unitization spectrum, but every state integrates against a probability measure on $\mathbb{R}$ and takes a strictly positive value on $x$. The preceding extension proof can be performed in the C-star unitization; a state obtained there need not restrict to a norm-one functional on the original algebra.

For $B=\overline{\mathbb{C}[1,x]}$ inside $A$, the displayed set is the full spectrum of $x$ in $A$. On a bounded complementary component, the maximum modulus principle gives $|p(\lambda )|\le {\max }_{{\sigma }_A(x)}|p|\le \Vert p(x)\Vert$. Evaluation $p(x)\mapsto p(\lambda )$ therefore extends to a character of an algebra on $B$, showing $\lambda \in {\sigma }_B(x)$. Conversely the unbounded resolvent component survives by resolvent components in a closed unital subalgebra. In particular, if zero lies in that component, $x^{-1}$ belongs to the norm closure of polynomials in $x$.

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.

In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar{v}(s)=(4\pi {)}^{-1}{\int }_{S^2}v(s\omega )d\omega$. Outside $B_a$, the weak solution is a harmonic function. The weak formulation against radial test functions gives $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ for $a<s<R$. The zero Sobolev trace gives $\bar{v}(R)=0$; integration proves the displayed formula, without assuming the source is a radial function.

Cauchy-Schwarz inequality on spheres gives $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding balls cannot remain bounded in the full $H^1({\mathbb{R}}^3)$ norm after zero extension of H01.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

A spin ladder operator changes the magnetic quantum number within a fixed spin multiplet. For simultaneous eigenstates of $S^2$ and $S_z$, the labels satisfy $S^2|s,m⟩={\hslash }^{2}s(s+1)|s,m⟩$ and $S_z|s,m⟩=\hslash m|s,m⟩$.

The operators $S_{\pm }=S_x\pm i{S}_y$ satisfy

The angular momentum commutation relations give $S_{\mp }{S}_{\pm }=S^2-S_z^2\mp \hslash S_z$. Its expectation value is the squared norm of the raised or lowered state, fixing the square-root factors by normalization. For spin one half, $S_{+}|\frac{1}{2},-\frac{1}{2}⟩=\hslash |\frac{1}{2},\frac{1}{2}⟩$. At $m=s$ the spin raising operator gives zero. A spin ladder operator is therefore not a unitary operator or, by itself, a deterministic physical rotation. ^[ref]

The Square-root Lasso minimizesIts residual norm is unsquared. For errors $\epsilon =\sigma Z\ne 0$, the magnitude of the loss gradient at the true regression coefficients is governed byThe unknown noise standard deviation $\sigma$ cancels. This makes a penalty calibrated to that score function pivotal with respect to noise scale.^[ref]

For normalized columns and Gaussian noise, a typical theoretical penalty has order $\sqrt{\log (p)/n}$ without an extra factor of $\sigma$. Prediction guarantees still require assumptions on the number of nonzero regression coefficients, the design matrix and the noise. Pivotal tuning does not remove those assumptions. At an exact zero residual the Euclidean norm is nondifferentiable, so its subdifferential must replace the displayed nonzero-residual gradient.

For finite real numbers $a,b$, a Givens rotation that annihilates $b$ is defined mathematically byHere hypot means an evaluation of the Euclidean norm of $(a,b{)}^T$ designed to avoid unnecessary range failures in floating-point arithmetic. The zero case selects the identity matrix, avoiding division by zero. If $a=0,b\ne 0$, then $c=0,s=\mathrm{sign}(b)$; if $b=0,a\ne 0$, then $c=\mathrm{sign}(a),s=0$. Thus the retained component is nonnegative, including when $a<0$. In exact arithmetic, $ca+sb=r$ and $-sa+cb=0$.

Computing the square root of a*a + b*b directly can fail even when the final norm is representable. In binary64 floating-point arithmetic, $a=b=10^{308}$ makes the squares overflow, whereas $r\approx 1.4142\times 10^{308}$ is finite. With $a=b=10^{-300}$ the squares round to zero, although $r\approx 1.4142\times 10^{-300}$ is representable. A range-safe hypot avoids these intermediate failures.

The simple divisions $a/r,b/r$ still need care when $r$ is a subnormal number: rounding $r$ can cause large relative errors. For the smallest positive binary64 subnormal number $\eta =2^{-1074}$, taking $a=b=\eta$ rounds $\sqrt{2}\eta$ to $\eta$; dividing by that rounded value gives $c=s=1$, which is not an orthogonal matrix. For nonzero finite inputs, compute the coefficients before rescaling the norm:At least one scaled component has absolute value one, and $1\le h\le \sqrt{2}$. This construction keeps the coefficients meaningful even when rounding $r$ is severe. It cannot make an unrepresentably large $r$ finite, preserve an arbitrarily tiny $s$ against underflow, or repair inputs already rounded to zero. The row updates themselves can also overflow if their exact results exceed the available range.

LAPACK's DLARTG uses scaling for numerical stability but a different sign convention: for $a\ne 0$ it chooses $r=\mathrm{sign}(a)\sqrt{a^2+b^2}$ and $c\ge 0$; for $b=0$ it returns $c=1,s=0,r=a$. Its block still has the lower-left minus sign. Do not combine its returned coefficients with an assumed nonnegative $r$.^[ref]

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.