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Suppose the underlying graph has a cycle through $k\ge 3$ distinct vertices, and let $v_1,\dots ,v_k$ be their unit vectors from admissibility. Adjacent vertices have $(v_i,v_j)\le -1/2$, and every other pair has nonpositive inner product. ThereforeBut linear independence makes the sum nonzero, whose squared norm in a Euclidean space must be positive. This contradiction proves that the underlying graph of an admissible Coxeter graph has no cycles. Double and triple lines record edge multiplicities; they are not cycles of length two. The argument also tolerates chords, which only make the bound smaller.

Write $N=2k+1$ and abbreviate the Dicke states by $|D_j^n⟩=|{\Psi }_j^n⟩$. Alice's shared resource has decompositionFollow the actual circuit order: a controlled-NOT gate from the input to $q_0$, a Hadamard gate on the input and a Pauli X gate on $q_0$, followed by computational basis measurements. Just before measurement, the joint vector isThe Dicke states in each branch are orthogonal, so each branch has squared norm $1/4$. Conditioning on outcome $00$ and normalizing therefore gives the state of the other $N$ parties:This is the encoding used in Dicke-resource telecloning. The extra Pauli X gate on Alice's resource qubit is essential to the displayed ordering of the amplitudes.

For target unitary operator $U_T$ and actual propagator $U_f(T)$ on an $n$-dimensional space, minimize the phase-insensitive unitary gate errorThe Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives a value in $[0,1]$, because both unitary matrices have squared norm $n$. The error is zero exactly when the matrices are proportional, namely $U_f(T)=e^{i\varphi }{U}_T$. It therefore tests the full process, not just its action on one state, and ignores a physically irrelevant common phase. If the gate phase itself must be fixed, instead minimize $\Vert U_f(T)-U_T{\Vert }_{\mathrm{HS}}^2/(2n)=1-\mathrm{Re}\mathrm{Tr}(U_T^{†}{U}_f(T))/n$, which vanishes only at exact equality.

The Korovkin theorem states that if positive linear operators on continuous functions $L_n:C[0,1]\to C[0,1]$ satisfythen $\Vert L_{n}f-f{\Vert }_{\infty }\to 0$ for every real continuous function $f$.

Let $r=k-1\ge 2$ and $h_n=|{\Delta }_n|$. For the B-splines in this question, $N_{i,n}\ge 0$ and their sum is one on the basic interval. Thus the Schoenberg spline operator $V_n$ is linear and positive. To verify the three test functions, suppress the index $n$ and write the Marsden identity with ${\omega }_i(x)={\prod }_{j=1}^r(x-t_{i+j})$. Comparing the coefficients of $x^r,x^{r-1},x^{r-2}$ gives the monomial B-spline coefficientswhereThe first coefficient is the partition of unity; the second is the Greville abscissa. Every internal knot and the sample point ${\tau }_i$ lie in $[t_i,t_{i+k}]$, of width at most $k{h}_n$. Consequently $|{\tau }_i-{\xi }_i|\le k{h}_n$. Since all these numbers lie in $[0,1]$, for any two internal knots $u,v$ we also haveAveraging gives $|{\tau }_i^2-q_i|\le 2k{h}_n$. Nonnegative weights of sum one now yieldHere the spline order $k$ is fixed, so all three errors tend to zero. The Korovkin theorem therefore provesNo special selection such as the Greville abscissae is required for the sample points. In fact the local-support error bound for a Schoenberg spline operator also gives $\Vert V_{n}f-f{\Vert }_{\infty }\le \omega (f,k{h}_n)$. Under the usual continuous-spline convention the operators have the codomain required by the stated theorem. If the permissive knot notation is taken to allow full interior multiplicities and hence discontinuous splines, the same conclusion in the uniform norm still follows from this direct bound, although each approximant need not then be continuous.

For a real continuous function on $[-1,1]$, the Chebyshev alternation theorem gives a unique best uniform approximation $p_n^{\ast }$ from the space of polynomials of degree at most $n$. If $f$ is not already in that space, a candidate $p$ is best if and only if there exist ordered points $x_0<\cdots <x_{n+1}$ and a sign $\sigma \in \{1,-1\}$ withThus a nonzero best error has at least $n+2$ alternating extrema of equal magnitude. If $f$ is already a polynomial of degree at most $n$, the unique best approximant is $f$ itself with zero error.

Existence does not require assuming that the infimum is attained: a minimizing sequence has uniformly bounded polynomial norms, and evaluation at $n+1$ fixed distinct nodes bounds its coefficients through an invertible interpolation matrix. A convergent coefficient subsequence then attains the infimum. The alternation criterion characterizes that minimum and gives uniqueness. In the following parts, $n\ge 1$ whenever $E_{n-1}$ occurs.

For fixed real $\kappa$, consider initial data $u_0\in L^2(0,1)$ and a solution continuous into $L^2$ at time zero, with homogeneous Dirichlet boundary conditions for positive times. Existence, uniqueness and continuous dependence in this norm give Hadamard well-posedness.

To construct a solution, put $v(x,t)=e^{\kappa x}u(x,t)$. Substitution givesLet $b_j=2{\int }_0^1{e}^{\kappa x}{u}_0(x)\sin (j\pi x)dx$. The Fourier sine series solution isMultiplication by $e^{\pm \kappa x}$ is bounded on $L^2(0,1)$. Completeness of the sine basis gives convergence to $u_0$ in $L^2$ at time zero, while the exponential decay gives smoothness for positive times and the required equation and boundary values. For a classical solution continuous up to the endpoints at time zero, the usual initial-boundary compatibility conditions are additionally required.

For the energy method, integration by parts givesBoth boundary terms vanish. The same identity applies to the difference of two solutions, proving uniqueness and continuous dependence. The Poincaré inequality further giveswith the identical estimate for differences of initial data. This Dirichlet convection-diffusion contraction holds for every real $\kappa$, and its energy constant is independent of the drift strength.

The stated hypotheses are precisely the algebraic criterion in part (a). To prove its stability content, compare two Runge-Kutta method steps for the same dissipative vector field. Let $d=y_n-z_n$ be the input difference, $D_i=Y_i-Z_i$ the stage differences, and $F_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,Z_i)$. ThenExpanding the squared norm of $d_{+}$ givesIn each term of the linear sum, replace $d$ by $D_i-h{\sum }_j{a}_{ij}{F}_j$. Symmetrizing the resulting double sum proves the Runge-Kutta contractivity identity:Dissipativity makes $⟨D_i,F_i⟩\le 0$, and the weights are nonnegative. For the second term, positive semidefiniteness permits a factorization $M=C^{T}C$, givingFor $h\ge 0$ both contributions in the distance-change identity are nonpositive. Therefore $\Vert y_{n+1}-z_{n+1}\Vert \le \Vert y_n-z_n\Vert$. This proves the contractivity assertion behind algebraic stability, usually called the Butcher contractivity theorem. It assumes the compared implicit stage equations have solutions; the matrix condition by itself should not be substituted for a separate stage-solvability argument.

Let $\Omega =(0,1{)}^2$ and $V=H_0^1(\Omega )$, the closure of smooth compactly supported functions in the Sobolev space $H^1$. Its elements have zero boundary trace. SetThe form is symmetric and continuous by Cauchy-Schwarz. For a smooth function with zero boundary values, $v(x,y)={\int }_0^x{v}_x(s,y)ds$ gives $\Vert v{\Vert }_{L^2}\le \Vert v_x{\Vert }_{L^2}\le \Vert \nabla v{\Vert }_{L^2}$ after integration. Density extends this Poincaré inequality to $H_0^1$. Therefore the gradient norm makes $V$ complete and equivalent to its usual $H^1$ norm, withAll the hypotheses of part (a) hold. Green's identity identifies the weak solution of the Dirichlet problem asFor smooth $u$ this is $-{\nabla }^{2}u=f$ with zero boundary values; in general it defines the unique weak solution without requiring pointwise second derivatives.

For the Ritz method, choose a finite-dimensional conforming space $V_h\subset H_0^1(\Omega )$ with basis ${\varphi }_1,\dots ,{\varphi }_N$ and put $u_h={\sum }_j{U}_j{\varphi }_j$. Minimize $I(u_h)$ in the coefficients. Its derivative in each basis direction gives the finite element method equationsThe stiffness matrix is symmetric positive definite: for a nonzero coefficient vector $U$, the corresponding basis combination is nonzero and $U^{T}KU=\Vert \nabla u_h{\Vert }_2^2>0$. Hence the Ritz system has a unique solution.

For example, triangulate the square and use continuous piecewise-affine nodal basis functions at the interior vertices, with boundary vertex values fixed to zero. On a triangle $T$ of area $A_T$, let ${\lambda }_i$ be its barycentric basis functions. If its counterclockwise vertices are $(x_i,y_i)$, define ${\beta }_i=y_j-y_k$ and ${\chi }_i=x_k-x_j$ cyclically. Then $\nabla {\lambda }_i=({\beta }_i,{\chi }_i)/(2A_T)$, giving the directly assemblable element matrices and loadsIf $f$ is constant on $T$, $F_i^T=f_T{A}_T/3$; otherwise integrate or quadrature the load. Assemble by adding element contributions with the same global vertex labels and eliminate prescribed boundary coefficients. Only overlapping basis supports interact, making $K$ sparse. This gives the Ritz-Galerkin equivalence for a symmetric coercive form in a practical finite element form.

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.

Let $a=\mu (A)$, $b=\mu (B)$ and $c=\mu (A+B)$. The sum is also symmetric negative definite. Part (a) gives the exponential norm bounds, while the triangle inequality and submultiplicativity giveApplying these estimates to the exponential product defect identity yieldsIf $c<a+b$, evaluate the scalar integral to obtain the decaying bound for symmetric negative matrix splitting:The sign condition is natural: for every unit vector $v$, $v^T(A+B)v\le a+b$, so the Rayleigh quotient gives $c\le a+b$.

When $c=a+b$, the integrand equals $e^{tc}$ independently of $x$, givingThis also follows by taking the continuous limit of the quotient as $a+b-c\to 0$. There is no division-by-zero exception in the original integral argument. All exponents are negative, so these bounds decay for large $t$; they are coarse bounds and need not reproduce the sharper quadratic small-$t$ error.

The element $\omega$ satisfies ${\omega }^2-\omega +6=0$, so is an algebraic integer. Conversely, write an algebraic integer as $(a+b\sqrt{-23})/2$ with $a,b$ rational. Its trace gives $a\in \mathbb{Z}$, and its field norm implies $23b^2\in \mathbb{Z}$. If $b=p/q$ in lowest terms, $q^2$ divides 23, forcing $q=1$. The integral field norm condition modulo four then makes $a,b$ have the same parity. Therefore it equals $(a-b)/2+b\omega$, provingBoth reductions of $x^2-x+6$ modulo two and three split into the distinct roots zero and one. The index is one, so ideal factorization givesTheir norms are two and three respectively. The product ${\mathfrak{p}}_2{\mathfrak{p}}_3$ is generated by $6,2\omega ,3\omega ,{\omega }^2$. All are multiples of $\omega$, using $6=\omega \overline{\omega }$; and $\omega =3\omega -2\omega$ belongs to the product. HenceThe Minkowski bound is $(2/\pi )\sqrt{23}<4$, in fact about $3.05$. Every class therefore has a representative of ideal norm one, two or three. The displayed primes and their conjugates exhaust those possibilities. Their classes are all $1,c,c^{-1}$, with $c=[{\mathfrak{p}}_2]$, because $[{\mathfrak{p}}_3]=c^{-1}$ and conjugation inverts a quadratic ideal class. There are at most three classes.

There is no element of field norm two: $N(a+b\omega )=a^2+ab+6b^2=(a+b/2{)}^2+23b^2/4$, which exceeds two for $b\ne 0$ and is a square for $b=0$. Thus ${\mathfrak{p}}_2$ is not principal. Moreover $N(\omega -2)=8$, and $\omega -2$ belongs to ${\mathfrak{p}}_2$ but not its conjugate, so $(\omega -2)={\mathfrak{p}}_2^3$. Consequently $c$ has order three, and

Concavity of $\log$ gives $\log (a/p+b/q)\ge p^{-1}\log a+q^{-1}\log b$ because $1/p+1/q=1$. Exponentiating proves Young inequalitywith equality precisely when $a=b$. Equivalently it follows from convexity of the exponential.

Define ${\ell }^p=\{x=(x_j):{\sum }_j|x_j{|}^p<\infty \}$ with $\Vert x{\Vert }_p=({\sum }_j|x_j{|}^p{)}^{1/p}$. Positivity, definiteness and homogeneity are immediate. Summing Young inequality for normalized sequences gives Hölder's inequality $\sum |x_j{y}_j|\le \Vert x{\Vert }_p\Vert y{\Vert }_q$. For a finite truncation $z=x+y$, it yieldsDividing and passing to larger truncations proves Minkowski inequality and shows $x+y\in {\ell }^p$, so this is a normed vector space.

Let $x^{(m)}$ be a norm-Cauchy sequence. Each coordinate is Cauchy because $|x_j^{(m)}-x_j^{(l)}|\le \Vert x^{(m)}-x^{(l)}{\Vert }_p$, and let its limit be $x_j$. The sequence norms are bounded by some $C$. For every finite $N$, passage to coordinate limits gives ${\sum }_{j\le N}|x_j{|}^p\le C^p$, so $x\in {\ell }^p$. Given $\epsilon >0$, choose $M$ with $\Vert x^{(m)}-x^{(l)}{\Vert }_p<\epsilon$ for $m,l\ge M$. Letting $l\to \infty$ in finite truncations and then $N\to \infty$ gives $\Vert x^{(m)}-x{\Vert }_p\le \epsilon$ for $m\ge M$. Thus ${\ell }^p$ is complete, hence a Banach space.

Using the complex Euclidean norm and the normal matrix identity,Both sides are nonnegative. Therefore $\Vert Cx\Vert =0$ if and only if $\Vert C^{†}x\Vert =0$. This proves the stronger reusable fact that normal matrices have equal adjoint norms.

On a common domain where all products are defined, expansion of the operator commutator givesUse the usual quantum position and momentum operators with $p_j=-i\hslash {\partial }_j$, the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, and wavefunctions with the decay/domain conditions needed for adjoints. The cross-coordinate operators commute, soFor an eigenstate $L\psi =\lambda \psi$, hermiticity gives $\lambda \Vert \psi {\Vert }^2=\overline{\lambda }\Vert \psi {\Vert }^2$, hence its eigenvalue is real.

The product identity gives $[L,x]=i\hslash y$, $[L,y]=-i\hslash x$, $[L,p_x]=i\hslash p_y$, $[L,p_y]=-i\hslash p_x$. Thus the commutators with $x^2,y^2$ cancel, as do those with $p_x^2,p_y^2$, proving $[L,H]=0$. Each oscillator energy eigenspace is finite-dimensional and preserved by $L$; the Hermitian restriction of $L$ can be diagonalized within it. Therefore an energy eigenbasis can be chosen to diagonalize angular momentum as well.

For ${\varphi }_0=e^{-(x^2+y^2)/(2\hslash )}$, differentiation gives $\Delta {\varphi }_0=[(x^2+y^2)/{\hslash }^2-2/\hslash ]{\varphi }_0$, so $H{\varphi }_0=\hslash {\varphi }_0$. Also $[H,x]=-i\hslash p_x$ and $p_x{\varphi }_0=ix{\varphi }_0$, giving $H(x{\varphi }_0)=2\hslash x{\varphi }_0$; the same calculation holds for $y{\varphi }_0$. Directly $L=-i\hslash (x{\partial }_y-y{\partial }_x)$, henceThe simultaneous normalized states areTheir squared unnormalized norm is $\int r^2{e}^{-r^2/\hslash }{d}^{2}x=\pi {\hslash }^2$. Their inner product vanishes because the integral of $(x-iy{)}^2{e}^{-r^2/\hslash }$ is zero: the $xy$ term is odd and the $x^2,y^2$ terms cancel by rotational symmetry.

Put $a=e\mathcal{E}$. The new Hamiltonian hasFor a nonzero applied coupling $a$, a common normalizable eigenstate would be killed by the commutator, so $y\psi =0$. It would have to be supported on the line $y=0$, a set of zero planar measure, and would be the zero vector in $L^2$. Thus there are no nonzero common eigenstates in this case. Merely having a nonzero commutator would not prove that conclusion for arbitrary operators; the multiplication operator here has trivial kernel. If $a=0$, the original simultaneous eigenstates remain.

For the shifted two-dimensional oscillator in a uniform electric field, complete the square:Therefore ${\psi }_1(x^{\prime },y^{\prime })$ and ${\psi }_2(x^{\prime },y^{\prime })$ have the same shifted energyThe modified angular momentum isIt is angular momentum about the displaced centre, commutes with $H^{\prime }$, and has eigenvalues $\pm \hslash$ on the two translated states.

A norm is a function $N:{\mathbb{R}}^n\to [0,\infty )$ such that $N(x)=0$ exactly when $x=0$, $N(tx)=|t|N(x)$, and $N(x+y)\le N(x)+N(y)$. Both specified Lp norms are nonnegative, vanish only at zero and satisfy absolute homogeneity immediately from their formulas.

For the $1$-norm, the scalar triangle inequality gives ${\sum }_i|x_i+y_i|\le {\sum }_i|x_i|+{\sum }_i|y_i|$. For the $2$-norm, Cauchy-Schwarz inequality givesTaking nonnegative square roots proves its triangle inequality. Hence both are norms.

For the sharp comparison of l1 and l2 norms, Cauchy-Schwarz yieldsThe vector $(1,\dots ,1)$ attains equality, so no smaller constant works. Conversely, $({\sum }_i|x_i|{)}^2\ge {\sum }_i|x_i{|}^2$, giving $\Vert x{\Vert }_2\le \Vert x{\Vert }_1$. A standard basis vector attains equality. Thus, for $n\ge 1$,