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The no-cloning theorem for two pure states states that no single unitary operator can clone two distinct nonorthogonal states. Suppose, to the contrary, thatPreservation of the inner product by $U$ givesWriting $z=⟨{\alpha }_0|{\alpha }_1⟩$, this requires $z=0$ or $z=1$. The first case is orthogonality; the second means that the normalized states represent the same ray. Both contradict the assumptions. Therefore distinct nonorthogonal quantum states cannot be cloned by a unitary process.

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.

The map $x\mapsto -x(\mathrm{mod}K)$ is a bijection and an involution, so its permutation matrix satisfiesIn particular $S^{†}S=I$, which proves that $S$ is a unitary operator. Equivalently, it permutes the computational basis and therefore preserves every inner product.

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.

For real ${\lambda }_1,{\lambda }_2$, computeThus $L$ is a non-normal matrix. For distinct eigenvalues, takeThese satisfy $L{v}_j={\lambda }_j{v}_j$ but have inner product $v_1^T{v}_2=1$. The eigenvectors are therefore nonorthogonal. When the eigenvalues coincide, only the direction $v_2$ remains as an eigenvector; $L$ is a defective Jordan block.

For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.

For an edge $ij$, put $\rho =⟨v_i,v_j⟩$ and $\theta =\arccos \rho$. Since $\rho \le -1/2$, we have $\theta \ge 2\pi /3$.

The random normal $a$ has the isotropic Gaussian distribution $N(0,I_p)$. Its distribution is invariant under orthogonal transformations; if $v_i,v_j$ are linearly independent, its projection onto their plane has a uniformly distributed direction. The signs of its inner products with $v_i,v_j$ differ in two sectors of total angle $2\theta$, out of $2\pi$. Thus random hyperplane rounding givesIf $v_j=-v_i$, the signs differ with probability one and the same formula holds with $\theta =\pi$. Zero inner products have probability zero, since each $v_i$ is a unit vector.

If the graph has an edge, its corresponding principal $2\times 2$ block of $U$ forces $t\ge 2$, so the division by $t-1$ used to obtain the Gram matrix is valid. An edgeless graph can instead be colored with one color directly.

A Runge-Kutta method with stage matrix $A$, weights $b$ and step size $h$ isIts order measures accuracy as $h\to 0$, while stability concerns how errors or stiff components propagate. Linear A-stability, strong stiff damping through L-stability, and nonlinear B-stability address different questions.

For the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is $\{z:|R(z)|\le 1\}$. The method is A-stable when it contains the closed left half-plane, and L-stable when it is also true that $R(z)\to 0$ as $|z|\to \infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.

The Forward Euler method has $R(z)=1+z$, so it fails A-stability, for example at $z=-3$. In fact no nontrivial explicit Runge-Kutta method is A-stable: its stability function is a nonconstant polynomial, which is unbounded along the negative real axis. The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z|\ge 1$, and the function tends to zero at infinity, proving both A-stability and L-stability. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability, but its limit $-1$ rules out L-stability.

For nonlinear equations, a dissipative vector field satisfies $⟨f(x)-f(y),x-y⟩\le 0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that inner product. A B-stable method preserves this contraction for every $h>0$, whenever its stages are well defined.

Let $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the update's squared norm, substituting $d=D_i-h{\sum }_j{a}_{ij}{F}_j$, and symmetrizing the double sum proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and the matrix $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the quadratic form of $M$ over each coordinate of the $F_i$. Thus algebraic stability implies B-stability. Applied to complex scalar linear equations with the real part of the Hermitian inner product, the same argument also implies A-stability, where the stages are defined.

The Backward Euler method has $b_1=a_{11}=1$, hence $M=(1)$, and is algebraically stable. One can prove its nonlinear contraction directly: $d_{n+1}=d+h(f(y_{n+1})-f({\tilde{y}}_{n+1}))$ gives $\Vert d_{n+1}{\Vert }^2\le ⟨d,d_{n+1}⟩\le \Vert d\Vert \Vert d_{n+1}\Vert$. The implicit midpoint rule has $a_{11}=1/2$, $b_1=1$ and $M=0$, so it is B-stable even though it is not L-stable.

Linear A-stability does not by itself imply nonlinear B-stability. The ODE trapezoidal rule has the same stability function as the implicit midpoint rule, but take the dissipative scalar field $f(y)=-\max (y,0)$ and $h=6$. For any positive starting value, its trapezoidal update is $y_{n+1}=y_n+3(f(y_n)+f(y_{n+1}))=-2y_n<0$. Thus two positive starting values have their distance doubled, violating B-stability. Its algebraic stability matrix also has a negative first diagonal entry. The field is globally Lipschitz continuous, so this is a well-defined nonlinear counterexample.

These examples distinguish accuracy, scalar stiff-mode stability, nonlinear contraction, and rapid stiff-mode damping. For implicit methods, solvability of the stage equations remains an additional requirement; a formal stability function alone does not establish it for arbitrary nonlinear problems.

Suppose $m<n$. The Rodrigues formula and $n$ integrations by parts givebecause $p_m^{(n)}=0$. Every boundary term vanishes: the exponential controls infinity, while the remaining power of $x$ has positive exponent at zero. Symmetry of the inner product gives orthogonality whenever $m\ne n$.

For the norm, monicity gives $p_n^{(n)}=n!$, so the same calculation yields

An almost complex structure $J$ on $(M,{\omega }_M)$ is a compatible almost complex structure when $J^2=-I$, ${\omega }_M(Ju,Jv)={\omega }_M(u,v)$, andis a positive-definite inner product. To prove existence, choose any Riemannian metric $h$ and define $A$ by ${\omega }_M(u,v)=h(Au,v)$. The metric construction of a compatible almost complex structureis smooth and compatible, so the space is nonempty.

Identify the compact symplectic manifold $N$ with its image under the symplectic embedding. Along $N$ there is a symplectic splittingChoose compatible almost complex structures on both summands and take their direct sum. Its associated metric makes the two summands orthogonal. Extend this metric from the closed submanifold $N$ to all of $M$ using a partition of unity, and apply the metric construction again. Along $N$ it recovers the prescribed direct sum, so the resulting global compatible $J$ satisfies $J(TN)=TN$. This is the relative extension of a compatible almost complex structure.

For the two compact complex curves, use the supplied holomorphic coordinates at their transverse intersection. There $C_1\cup C_2$ is $\{xy=0\}$. Replace it in a small ball by the complex annulus $\{xy=ϵ\}$ and use a cutoff in a surrounding annulus to rejoin the unchanged curves. For sufficiently small nonzero $ϵ$, the result is an embedded symplectic surface $\Sigma$. This local replacement is a bordism between the old and new cycles, soThis is the symplectic smoothing of a positive transverse node.

For every $h\in L^2[0,1]$, the continuous linear functional induced by the inner product givesThis is a normal random variable because it is a linear combination of independent normal random variables. Hence the law of $U$ is a Gaussian measure. Its mean is zero, and independence together with $\mathrm{Var}({\xi }_i)=1/2$ givesThus its covariance operator of a Gaussian measure is

The no-cloning theorem for two pure states states that, for distinct nonorthogonal states $|c_0⟩$ and $|c_1⟩$, there is no unitary operator $U$ and fixed blank state $|0⟩$ such that

To derive this from state discrimination, letand suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The inner product of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the Helstrom-Holevo bound would beAlready for $N=2$,because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy Helstrom measurement for two pure states would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.

Equivalently, clone-assisted asymptotic state discrimination would give $P_N\to 1$, although two nonorthogonal states are not perfectly distinguishable. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.

Assume (ii). Weak convergence givesUsing the inner product identity,Hence (ii) implies (i); this is the Radon-Riesz theorem in a Hilbert space.

For each $p\in S$, the first fundamental form is defined without coordinates bythe restriction of the Euclidean inner product to the tangent space of $S$.

If $\varphi (u,v)$ is a local parametrization andthenTaking their dot product giveswhere

Bilinearity and symmetry are immediate. Since $r\ge 0$, $⟨u,u⟩\ge 0$. If this is zero, continuity of $u^{2}r$ makes it identically zero. The points where $r>0$ are dense because only finitely many are excluded, so $u$ vanishes on a dense set and hence everywhere. Thus the form is an inner product.

For $A=(A_1,A_2,A_3)$ and $B=(B_1,B_2,B_3)$, the scalar product is the inner productThe cross product isIt is perpendicular to both vectors and has magnitude $|A||B|\sin \theta$.

Let $Av=\lambda v$ for a nonzero eigenvector $v$ of the Hermitian matrix $A$. Hermitian symmetry of the inner product givesSince $lVertv{\Vert }^2>0$, $lambda=\overline{\lambda }$, so every eigenvalue is real.

Use the normalized inner productThe test-function seminorm generated by $\mathcal{F}$ and its dual test-function norm areThe first quantity may only be a seminorm if $\mathcal{F}$ does not separate all functions in ${\mathcal{H}}_N$; correspondingly, the second may be infinite outside the linear span detected by $\mathcal{F}$.