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Let $r=dim(U\cap W)$. Choose a basis $e_1,\dots ,e_r$ of the intersection of vector subspaces, extend it to a basis $e_1,\dots ,e_r,u_{r+1},\dots ,u_m$ of $U$, and independently extend it to a basis $e_1,\dots ,e_r,w_{r+1},\dots ,w_n$ of $W$. The combined listspans the sum of vector subspaces $U+W$. It is also linearly independent: a vanishing linear combination says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the dimension formula for a sum of subspaces

Parametrize the two linear subspaces asPutting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thusand the dimension formula gives $\dim (U+W)=2+2-1=3$.

The linear functionalvanishes on each displayed generator of $U$ and $W$. Its kernel is three-dimensional because $\ell \ne 0$, so the inclusion $U+W\subseteq \ker \ell$ between two three-dimensional subspaces is equality.

Write an element of the diagonal Cartan subalgebra asand define the linear functionals ${\epsilon }_i(t)=t_i$. The root-space decomposition of the symplectic Lie algebra ${\mathfrak{sp}}_6(\mathbb{C})$ then has the C3 root system

The mapis a nonzero linear functional: otherwise $v$ would lie in the radical of a bilinear form, contradicting nondegeneracy. Its kernel is $v^{\perp }$, so the rank-nullity theorem gives $\dim v^{\perp }=n-1$. Antisymmetry gives $⟨v,v⟩=-⟨v,v⟩=0$, hence $v\in v^{\perp }$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^{\perp }$, it is also orthogonal to $v$, and therefore toFor a nondegenerate bilinear form, $(v^{\perp }{)}^{\perp }=\mathbb{R}v$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap \mathbb{R}v=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional symplectic vector space has even dimension.

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

Using multi-index notation, the Schwartz space isA sequence ${\phi }_m$ converges to $\phi$ in this Fréchet space when $p_{\alpha ,\beta }({\phi }_m-\phi )\to 0$ for every $\alpha ,\beta$. The space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ of tempered distributions is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$, and $u_m\to u$ there means weak convergence of distributions, namely $⟨u_m,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{S}$.

Continuity of a linear functional immediately implies that ${\phi }_m\to 0$ entails $⟨u,{\phi }_m⟩\to 0$. Conversely, enumerate the Schwartz seminorms as $p_1,p_2,\dots$. If $u$ were not continuous, then for each $m$ one could choose ${\phi }_m$ such thatEvery fixed seminorm tends to zero along this sequence, so ${\phi }_m\to 0$ in $\mathcal{S}$, contradicting the assumed sequential property. This is the sequential continuity criterion for a linear map on a metrizable topological vector space.

Since ${\delta }_{1/t}\phi (x)=\phi (x/t)$ maps the Schwartz space continuously to itself, the formuladefines a continuous linear functional on $\mathcal{S}$, hence a tempered distribution. For a locally integrable function $u$, the change of variables formula $y=tx$ givesso this dilation of a distribution agrees with ordinary function dilation.

Let $A$ be a rational matrix with columns $a_1,\dots ,a_n$. It is a partition regular matrix when every finite coloring of the positive integers has a monochromatic vector $x$ with $Ax=0$. It has the columns property when the column indices have an ordered partition $B_1,\dots ,B_r$ such thatand, for every $s>1$,Rado's theorem states that $A$ is partition regular if and only if it has the columns property.

First suppose $A$ is partition regular, and clear its denominators. For a prime number $p$, apply the last nonzero digit coloring in base $p$: if $x=p^{v_p(x)}u$ with $p\nmid u$, its color is $u\mathrm{mod}p$. Choose a monochromatic solution and group its coordinates into blocks $B_1,\dots ,B_r$ of equal $p$-adic valuation, in increasing order of valuation. Only finitely many ordered partitions are possible, so one partition occurs for infinitely many primes.

For any such prime, divide ${\sum }_i{x}_i{a}_i=0$ by the lowest power of $p$ and reduce modulo $p$. All coordinates in $B_1$ have the same nonzero leading digit $d$, while later blocks vanish, soThe integer vector ${\sum }_{i\in B_1}{a}_i$ is divisible by infinitely many primes and therefore equals zero.

Now fix $s>1$, and let $t$ be the common valuation on $B_s$. Reduction modulo $p^{t+1}$ givesIf the first sum of columns were outside the rational linear span of the earlier columns, an integer linear functional would vanish on every earlier column but not on that sum. Applying it to the congruence would say that infinitely many primes divide one fixed nonzero integer, a contradiction. Thus the displayed partition has the columns property.

Conversely, suppose $A$ has the columns property. For $s>1$, choose rational numbers $q_{is}$ such thatLet $d_{is}$ equal $1$ for $i\in B_s$, equal $-q_{is}$ in the earlier blocks, and equal zero in the later blocks. For $s=1$, let $d_{i1}$ be the indicator of $B_1$. Thenfor every $s$. Choose a positive integer $c$ clearing all denominators and a positive integer $p$ with $|c{d}_{is}|\le p$ whenever $s$ is later than the block containing $i$.

By the monochromatic m-p-c set theorem, the coloring contains a monochromatic $(r,p,c)$-set with generators $z_1,\dots ,z_r$. DefineIf $i\in B_t$, thenso every $y_i$ lies in that one monochromatic M-p-c set. Moreover,Thus $A$ is partition regular, proving Rado's theorem.

For the equation $4x+3y-6z=0$, a monochromatic solution in the last nonzero base-$p$ digit coloring would, at the least $p$-adic valuation among $x,y,z$, force a nonempty subset sum of $4,3,-6$ to vanish modulo $p$. The seven possible sums areNone is divisible by $5$, so base $5$ gives no monochromatic solution. The smaller primes do admit solutions: $(x,y,z)=(3,2,3)$ works for base $2$, whose coloring has one color, and $(3,4,4)$ is monochromatic in base $3$. Hence the smallest prime is

For $x+2y-7z=0$, use the last nonzero base-$11$ digit itself. The nonempty subset sums of $1,2,-7$ arenone zero modulo $11$. This is the required $10$-coloring.

For a $5$-coloring, identify each nonzero residue $d$ modulo $11$ with $-d$, giving the five colorsIf a monochromatic solution existed, reduction at the least $11$-adic valuation would give a signed nonempty subset sum of $1,2,-7$ equal to zero modulo $11$. Singles have absolute residues $1,2,7$; pairs have absolute residues $1,3,5,6,8,9$; and triples have absolute residues $4,6,8,10$. None is zero modulo $11$, which proves that this $5$-coloring works.

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

Define the linear functional $L:F^n\to F$ by $L(x)={\sum }_{i=1}^n{a}_i{x}_i$. The given hyperplane is $H=\ker L$. Choose $j$ with $a_j\ne 0$. For any $t\in F$, the vector whose only nonzero coordinate is $x_j=t/a_j$ satisfies $L(x)=t$. Hence $L$ is surjective and has rank of a linear map equal to ${\dim }_{F}F=1$.

Applying the rank-nullity theorem to this particular linear map givesThe nonzero-coefficient hypothesis is essential: if all $a_i$ were zero, the kernel of a linear map would instead be all of $F^n$.

Every $v\in V$ has a unique basis expansion $v={\sum }_{i=1}^n{a}_i{v}_i$. Define the linear functionalUniqueness of the coordinates makes this definition well defined. The coordinates of a sum or a scalar multiple are the corresponding sums or scalar multiples of the coordinates, so this formula defines a linear map. It has $T(v_i)={\mu }_i$. Conversely, any linear map with those values must satisfy the same formula by linearity, proving uniqueness.

If the finite field $\mathbb{F}$ has $q$ elements, each of the $n$ values ${\mu }_i$ can be chosen independently in $q$ ways. The number of linear maps $V\to \mathbb{F}$ is therefore $\boxed{q^n}$, including the zero linear map.

If $T^{\prime }=\lambda T$ with $\lambda \ne 0$, then $T^{\prime }(x)=0$ if and only if $T(x)=0$, so their kernels agree. Conversely, suppose both nonzero linear functionals have kernel $K$. Choose $u$ with $T(u)=1$, by scaling a vector on which $T$ is nonzero. For each $x\in V$,Therefore $T^{\prime }(x)=T(x)T^{\prime }(u)$. Moreover $T^{\prime }(u)\ne 0$, since $u\notin K$. Thus $\boxed{T^{\prime }=T^{\prime }(u)T}$, with a nonzero scalar.

Every nonzero linear functional $V\to {\mathbb{F}}_p$ is surjective, so its kernel has dimension $n-1$ by the rank-nullity theorem. Conversely, every hyperplane $K$ is such a kernel: extend a basis of $K$ by one vector $u$, and assign value zero on that basis and value one on $u$. Hence the hyperplanes containing $v$ are exactly the kernels of the $p^{n-1}-1$ nonzero linear functionals counted above. Each kernel occurs for exactly $p-1$ of them, its distinct nonzero scalar multiples. The required number is

Use basis extension to choose a basis $(v,v_2,\dots ,v_n)$ of the vector space $V$ over the finite field ${\mathbb{F}}_p={\mathbb{Z}}_p$. The condition $T(v)=0$ fixes the first value of a linear functional; its remaining $n-1$ values are arbitrary. There are $p^{n-1}$ such linear maps, exactly one of which is zero. Hence the required number isFor $n=1$ this is zero, as it should be: vanishing at a basis vector then forces the entire linear functional to vanish.

The real Lax-Milgram theorem states: let $H$ be a Hilbert space and let $a:H\times H\to \mathbb{R}$ be a bounded bilinear form such that $a(v,v)\ge \alpha \Vert v{\Vert }_H^2$ for some $\alpha >0$ and every $v\in H$. For every bounded linear functional $\ell \in H^{\ast }$ there exists a unique $u\in H$ satisfyingSymmetry of $a$ is not required. The complex version replaces bilinearity by sesquilinearity and the lower bound by $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_H^2$.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

For a scalar-valued function, the gradient is the column vectorWhen $f$ is differentiable, its total derivative is the linear functionalThus the Jacobian is the transpose of the gradient in standard Euclidean coordinates.