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Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Suppose $y$ in the capped simplex has a coordinate strictly between zero and one. Since ${\sum }_i{y}_i=k$ is an integer, it cannot have exactly one such coordinate: the other coordinates would contribute an integer sum. Hence there are distinct $i,j$ with $0<y_i,y_j<1$.

Choose $0<\epsilon <\min \{y_i,1-y_i,y_j,1-y_j\}$ and set $y^{\pm }=y\pm \epsilon (e_i-e_j)$, where $e_i,e_j$ are standard basis vectors. Both perturbed vectors satisfy the coordinate bounds and have the same coordinate sum. They are distinct and $y=(y^{+}+y^{-})/2$, so $y$ is not an extreme point.

Conversely, every zero-one vector in this convex polytope is an extreme point. If it were a nontrivial convex combination of two feasible vectors, each coordinate equal to zero would force both corresponding coordinates to be zero, and each coordinate equal to one would force both to be one. Thus the two vectors would equal the original one. The extreme points are therefore exactly the indicators of $k$-element subsets.

Each map $x\mapsto ⟨a_i,x⟩+b_i$ is an affine function. For $0\le t\le 1$, the pointwise maximum of convex functions satisfiesso $f$ is convex.

A vector $g$ is a subgradient of a convex function $f$ at $x$ whenfor every $y$. Choose any active index $j\in I(x):=\{i:f(x)=⟨a_i,x⟩+b_i\}$. Thenand hence $\boxed{a_j\in \partial f(x)}$. More generally, every convex combination of the active vectors is a subgradient, and in fact

Let the value function at time $n$ be the greatest conditional expected terminal utility attainable from wealth $x$. Independence of the innovations gives the Bellman equationfor $0\le n<N$. A policy obtained from the maximizing choices is admissible because each time-$n$ holding is a predictable process, hence is known before ${\xi }_n$ is observed.

We prove the claimed properties by backward induction. They hold at $n=N$ because $V(N,\cdot )=U$ is increasing and concave. Suppose they hold at time $n+1$. For fixed $\theta$, the mapis increasing because $1+r>0$, and taking a supremum preserves this inequality.

For concavity, let ${\theta }_x$ and ${\theta }_y$ be optimal at wealths $x$ and $y$. For $0\le t\le 1$, use the admissible portfolioThe next wealth is the same convex combination of the next wealths generated from $(x,{\theta }_x)$ and $(y,{\theta }_y)$. ThereforeThis completes the induction and proves monotonicity and concavity of a portfolio value function.

Put $z_n=x_n-x$, so $z_n\rightharpoonup 0$. We show that the norm closure of the convex hull of every tailcontains zero. If it did not, the Hahn-Banach separation theorem would give a bounded linear functional $\varphi$, a real number $c>0$, and, after multiplying $\varphi$ by a complex scalar if necessary,for every $z$ in that closed convex hull. In particular $\mathrm{Re}\varphi (z_n)\ge c$ for all $n\ge k$. By the Riesz representation theorem, $\varphi (z)=⟨z,y⟩$ for some $y\in H$, contradicting $z_n\rightharpoonup 0$.

Consequently, for every $k$ there is a finite convex combinationwith $\Vert w_k{\Vert }_H<1/k$. DefineThen every ${\tilde{x}}_k$ is a convex combination of terms of the original sequence andThis is Mazur lemma in the present Hilbert-space setting.

For particle positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrödinger equation isWrite $|a{⟩}_i$ for the localized wave packet ${\psi }_{ai}$ and $d_{ab}=|x_{a1}-x_{b2}|$. Neglecting packet spreading and branch overlap, the initial product state evolves branchwise asup to phases generated independently on the two particles. These branch-dependent phases generally cannot be separated into one phase depending only on $a$ and one depending only on $b$, so the Newtonian gravitational potential energy creates gravitationally induced entanglement.

If $d_{10}=|x_{11}-x_{02}|=d$ is much smaller than the other separations, remove their nearly common phase and retain onlyThe state is approximatelyIts concurrence is $|\sin (\varphi /2)|$, so it becomes maximally entangled first at $\varphi =\pi$. For $m_1=m_2=m$,Thus the near-maximal entanglement time is about $\boxed{10\mathrm{s}}$ within the stated approximation.

A single prescribed classical gravitational potential gives a Hamiltonian of the form $H_1[\Phi ]\otimes I+I\otimes H_2[\Phi ]$. Its evolution factorizes as $U_1\otimes U_2$ and preserves every initial product state, so it cannot generate this entanglement. A semiclassical mean field sourced only by expectation values likewise gives each particle a local one-body potential and does not provide a quantum mediator carrying branch correlations.

An entanglement witness is a Hermitian operator $W$ whose expectation is nonnegative on every separable state but negative on at least one entangled state. At $\varphi =\pi$, defineThe largest Schmidt coefficient of $|{\Psi }_{\ast }⟩$ is $1/\sqrt{2}$, so every product state $|u⟩|v⟩$ in the four-dimensional branch subspace satisfies $|⟨{\Psi }_{\ast }|u,v⟩{|}^2\le 1/2$. By closure under convex combinations, $\mathrm{Tr}(W{\rho }_{\mathrm{sep}})\ge 0$ for every separable mixture, whereasA negative measured value therefore certifies entanglement. Under the assumptions that the masses began unentangled and interacted only through gravity, such certification would show that the mediator can transmit quantum coherence; it would be evidence against a purely classical gravitational channel and for the quantum nature of gravity.

The Quantum de Finetti theorem says that the fixed-$k$ reduced density matrix of an exchangeable $N$-particle state approachesas $N\to \infty$. Finite versions bound the trace norm error by a constant of order $d_{\mathrm{loc}}^{2}k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixtureThis is the mean-field ansatz from the quantum de Finetti theorem.

The bond energy is a linear functional of ${\rho }_{AB}$. A convex combination cannot have energy below its lowest product component, so it remains only to minimizeThe Cauchy-Schwarz inequality gives ${\mathbf{r}}_A\cdot {\mathbf{r}}_B\ge -|{\mathbf{r}}_A||{\mathbf{r}}_B|\ge -1$, with equality for pure antiparallel vectors. This reproduces the Néel state and $-J/4$ per bond found in part (a).

Yes. The limiting state saturates the de Finetti mean-field lower bound: the minimizing product state belongs to the allowed de Finetti mixture, and the finite-de-Finetti error tends to zero as $z\to \infty$. At finite $z$ the theorem gives only an approximation; entanglement and correlated fluctuations can lower the energy below the product-state value by corrections that vanish in the infinite-coordination limit.

A bipartite density operator is a separable quantum state when it has a convex combination decompositioninto product states. A state for which no such decomposition exists is an entangled state.

At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yieldsBoth $\sigma$ and $D/d$ are convex combinations of product states, so ${\rho }_{p_0}$ is a separable quantum state. For $0\le p\le p_0$, the state ${\rho }_p$ is a convex combination of ${\rho }_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the partial transpose of the maximally entangled projector is $F/d$, where $F$ is the swap operator. ThereforeOn the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of ${\rho }_p^{T_B}$ iswhich is negative exactly when $p>1/(d+1)$. The positive partial transpose criterion then proves that ${\rho }_p$ is entangled. Thus

If $\epsilon =0$, then ${\rho }_{AB}={\sigma }_{AB}$ and the desired continuity bound is immediate, so assume $\epsilon >0$. Apply the positive-negative decomposition of a Hermitian operator toBecause $\mathrm{Tr}X=0$ and $\Vert X{\Vert }_1=2\epsilon$, one hasThus ${\Delta }_{AB}=X_{+}/\epsilon$ is positive with trace one, hence is a density operator. DefineIt is a convex combination of states. The equation $\epsilon {\Delta }_{AB}^{\prime }=(1+\epsilon ){\omega }_{AB}-{\rho }_{AB}$ giveswhich is likewise positive and has trace one.

Write $q=P(B\mid A)$, $s=P(B\mid A^c)$ and $p=P(A)$. The law of total probability gives $P(B)=pq+(1-p)s$, a convex combination of the two fixed conditional values. HenceThese are the closed sharp bounds in the usual mixture-model interpretation, obtained by allowing $p$ to range from zero to one. The joint masses $pq,p(1-q),(1-p)s,(1-p)(1-s)$ realize every such mixture.

There is an endpoint qualification if both conditionals must literally be defined by ratios: then $0<p<1$. If $q\ne s$, the marginal lies strictly between them, so the stated endpoints are an infimum and supremum rather than achieved extrema. If $q=s$, every allowed mixture has exactly that common value. This is the second case of sharp bounds from binary conditional probabilities.

Set $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. Since these vectors form a basis, write $\mathbf{r}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. The reciprocal basis givesA point is in the closed tetrahedron precisely when it is a convex combination of the four vertices, with weights $1-\alpha -\beta -\gamma ,\alpha ,\beta ,\gamma$. Thus the necessary and sufficient conditions areThese four half-spaces are the four inward sides of its faces. This is tetrahedron membership via scalar triple products; making all inequalities strict describes the interior.

Use the normalized eigenfunctions with the stated weighted inner product. Orthogonality gives $G[\hat{y}]={\sum }_n|c_n{|}^2$. The corrected Sturm-Liouville problem and the vanishing boundary term giveFor a complex trial the numerator means $\int (p|{\hat{y}}^{\prime }{|}^2+q|\hat{y}{|}^2)dx$, agreeing with the given formula for real trials. ThereforeIt is a convex combination of the eigenvalues, with equality only for a trial proportional to the lowest eigenfunction. The Rayleigh-Ritz variational principle consequently estimates ${\lambda }_0$ by choosing a nonzero admissible trial family and minimizing this quotient over its parameters:The result is an upper bound. A more flexible trial space can lower it, while equality requires that the exact lowest eigenfunction belong to the trial space. One need not know the expansion coefficients to calculate the quotient directly from the trial function.

List the roots with their multiplicities. By the factor theorem, $Q(z)=a{\prod }_{k=1}^n(z-{\alpha }_k)$ with $a\ne 0$. The product rule, followed by division where $Q(z)\ne 0$, gives its logarithmic derivative:Let $Q^{\prime }(\beta )=0$. If $Q(\beta )=0$ too, then $\beta$ is already one of the roots, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all roots have positive imaginary parts but $\mathrm{Im}\beta \le 0$, thenfor every $k$. Their sum cannot be zero. Thus $\mathrm{Im}\beta >0$.

For the unit disk conclusion, the same sum gives an especially transparent convex combination. Taking complex conjugates and setting $w_k=|\beta -{\alpha }_k{|}^{-2}>0$ givesTherefore, if every $|{\alpha }_k|<1$,The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the Gauss-Lucas theorem: every critical point of a nonconstant polynomial lies in the convex hull of its roots. The common-root case, handled before division, is essential when multiplicities occur.

The bold black curve is the pointwise minimum of the two opposing-column expected payoff functions. It is the row player's security level payoff at the specified mixed strategy:The lines meet where $3p-1=1-4p$, hence at $p=2/7$. The first branch increases with slope $3$ and the second decreases with slope $-4$, so their intersection is the unique global maximum of the lower envelope:The endpoint values are $g(0)=-1$ and $g(1)=-3$, both smaller. Since an opposing mixed strategy forms a convex combination of the two pure-column expected payoffs, it cannot make the row player's expected payoff lower than this envelope.

The bold black curve is now the pointwise maximum of the two pure-row expected payoff functions:It measures the largest expected payoff that the row player can achieve against column mixed strategy $q$. The column player minimizes this quantity because its own payoff has the opposite sign. Equality of the two lines gives $5q-3=1-2q$, or $q=4/7$. The envelope decreases before that point and increases afterwards, givingThe endpoint values $h(0)=1$ and $h(1)=2$ are both larger. A row mixed strategy cannot exceed the greatest pure-row expected payoff, because its expected payoff is a convex combination of those two values.

Write $x(S)={\sum }_{i\in S}{x}_i$. The core of a cooperative game isTake any $x,y\in C(v)$ and $\alpha \in [0,1]$, and form the convex combination $z=\alpha x+(1-\alpha )y$. Its total payoff is $z(N)=\alpha v(N)+(1-\alpha )v(N)=v(N)$. For every coalition $S$,Hence $z$ remains in the core. The core is a convex set, even when the transferable utility game is not a convex cooperative game. An empty core is a convex set vacuously; the argument does not claim nonemptiness. Geometrically, the same conclusion follows because the constraints intersect an affine hyperplane with closed half-spaces.

Assume the standard empty-coalition normalization $v(\varnothing )=0$. A convex cooperative game has increasing marginal contributions: whenever $A\subseteq B$ and $i\notin B$,This follows from the supermodular inequality applied to $A\cup \{i\}$ and $B$. Fix a permutation $\pi$ of all players, let $P_i$ be the players preceding $i$, and form its marginal contribution vectorSumming along the full permutation is a telescoping sum, so ${\sum }_{i\in N}{m}_i^{\pi }=v(N)$. For an arbitrary coalition $S$, list its members in their induced order $i_1,\dots ,i_k$, and put $S_j=\{i_1,\dots ,i_j\}$. Since $S_{j-1}\subseteq P_{i_j}$, increasing marginal contributions giveThus every marginal contribution vector belongs to the core. The vector of Shapley values is exactly their uniform average over the $n!$ permutations, hence a convex combination. By the preceding part, that average is in the core. This proves that the Shapley value allocation belongs to the core of a convex cooperative game, including every coalition constraint. It is the result Shapley value belongs to the core of a convex game.

Write $M_0={\sup }_x{u}_0(x)<\infty$ and assume $\lambda (1+p{M}_0^{p-1})\le 1$. Suprema are understood where the printed maxima on the infinite domain are not attained. We prove the invariant interval $0\le U_j^m\le M_0$ simultaneously with the stated speed bound.

Suppose the interval property holds at time $m$. By the mean value theorem, the divided differenceis between $1$ and $1+p{M}_0^{p-1}$; when the two states coincide define it as $f^{\prime }(U_j^m)$. The update is the convex combinationBoth coefficients are nonnegative and sum to one, so the same interval property holds at time $m+1$. It is true initially and hence for all times by induction. ThereforeThe propagated nonnegativity is essential; it keeps the flux increasing on the entire range of the numerical solution.

Use the fine matrix $A_h$ and right-hand side from part (b), with diagonal $D_h$. Starting from an iterate $v$, take $\nu$ weighted Jacobi smoothing stepsForm the residual $r_h=b_h-A_{h}v$. For coarse interior indices $j=1,\dots ,m/2-1$, use full-weighting restrictionThe prolongation operator matrix $P$ is linear interpolation of a coarse correction: $(Pe{)}_{2j}=e_j$ and $(Pe{)}_{2j+1}=(e_j+e_{j+1})/2$, with zero endpoint corrections. Thus $R=P^T/2$. Choose the Galerkin coarse-grid operator $A_c=R{A}_{h}P$, solve $A_c{e}_c=R{r}_h$ exactly, and correct $v\leftarrow v+P{e}_c$. The entries of $A_h$, $R$ and $P$ above specify every matrix in the cycle. A rediscretized coarse operator is another standard choice if used consistently.

For a nonsingular coarse operator, the fine-grid iteration matrix of this pre-smoothed two-grid cycle isThe purpose of under-relaxation is to damp fine-grid oscillations, leaving smoother error for coarse-grid correction. For the diffusion-dominated one-dimensional operator, $\omega =2/3$ is the usual effective choice; the high-frequency symbol $1-\omega (1-\cos \xi )$ then has modulus at most $1/3$ for $\pi /2\le |\xi |\le \pi$. Coefficients and convection can change the best parameter, so this is a motivated choice rather than a universal optimum. Under the diagonal-dominance condition, the unweighted Jacobi matrix already has norm less than one in the maximum norm, and its convex combination with the identity remains convergent for $0<\omega <1$.