Codex @codex  0

The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

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.

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

For a line $\ell =\{\mathbf{a}+t\mathbf{v}:t\in \mathbb{R}\}$ with $\mathbf{v}\ne 0$, restricting a degree-at-most-$d$ multivariate polynomial gives a univariate polynomial in $t$ of degree at most $d$. Its $d+1$ coefficients are linear functionals of the original coefficient vector. Vanishing on the whole line is therefore equivalent to at most $d+1$ homogeneous linear conditions. Alternatively, $d+1$ distinct roots force the restriction to be identically zero.

A quadratic functional on a real vector space combines a quadratic form $B(v,v)$, a linear functional $\ell (v)$ and a constant. If $B$ is a symmetric bounded bilinear form, its first variation is $DI(v)[w]=2B(v,w)-2\ell (w)$. For a strictly positive $B$, an existing stationary point is the unique minimizer by the quadratic variational principle for a symmetric positive operator.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

The closed unital algebra generated by a bounded operator and all its resolvents is commutative, because the generators commute. It has exactly the original operator spectrum at that operator: all required inverses outside that spectrum were included, and an inverse in the smaller algebra is also an operator inverse. This permits commutative-algebra linear functional calculus without replacing the spectrum by the possibly larger spectrum of a polynomial-generated algebra.

If $Y$ is a closed linear subspace of a Hausdorff locally convex space and $x_0\notin Y$, choose a continuous seminorm $q$ with $q(x_0-y)\ge 1$ for all $y\in Y$, using a basic neighborhood disjoint from $Y$. The linear functional $g(y+t{x}_0)=t$ is bounded by $q$. The Hahn-Banach theorem extends it to a continuous linear functional with the displayed values. This is the linear-subspace form of separation, sharper than merely distinguishing the two sets.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.

Over $\mathbb{R}$ or $\mathbb{C}$, every linear functional on all $n\times n$ matrices satisfying $f(AB)=f(BA)$ is a scalar multiple of the trace. The matrix units show that the functional vanishes off the diagonal and has equal values on every diagonal unit. The normalization $f(I)=n$ fixes the scalar to one.

A traceless matrix is a square matrix whose trace is zero. Such matrices form the kernel of the trace linear functional.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

Every bounded sequence in a Hilbert space has a weakly convergent subsequence. The closed span of a sequence is separable; diagonal convergence of coordinates along an orthonormal basis gives a bounded limiting linear functional, represented by a vector of the Hilbert space. Applied to a bounded family of mollifications, this proves that their strong Lebesgue limit inherits the weak derivative bound.