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The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:

Conversely, supposeand defineFor every $N$, Bessel inequality and passage to the limit giveHence ${\sum }_i|a_i{|}^2\le C^2$. The partial sums ${\sum }_{i=1}^N{a}_i{e}_i$ are therefore Cauchy; completeness of the Hilbert space gives an element

Let $P_{N}h={\sum }_{i=1}^N⟨h,e_i⟩e_i$. For fixed $N$, coordinate convergence givesThe Hilbertian basis property gives $P_{N}h\to h$, andFirst choose $N$ large and then $n$ large. This provesfor every $h\in H$, so $x_n\rightharpoonup x_{\infty }$. This establishes the coordinate criterion for weak convergence in a separable Hilbert space.

The Lax-Milgram theorem says that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form satisfyingfor some $\alpha >0$, and $F\in H^{\ast }$, then there is a unique $u\in H$ such that $B(u,v)=F(v)$ for every $v\in H$.

For this problem take $H=H_0^1(U)$ andThe Cauchy-Schwarz inequality and the continuous embedding $H_0^1(U)\hookrightarrow L^2(U)$ make both maps bounded. For smooth zero-boundary functions, integration by parts givesdensity extends this identity to $H_0^1(U)$. Consequentlyso $B$ is coercive. Lax–Milgram supplies the unique weak solution.

Choose a test function $\chi$ that equals one on the unit ball and vanishes outside the ball of radius two. Sinceintegration by parts givesThe first term on the right is supported where $1\le |x|\le 2$, where $u^2\le |x{|}^{\alpha }{u}^2$. The Cauchy-Schwarz inequality and Young inequality bound the second term byAfter absorbing the first integral,

This estimate also proves completeness. Indeed, a Cauchy sequence in $\Sigma$ is Cauchy in $H^1$ locally, while its gradients and the functions $|x{|}^{\alpha /2}{u}_n$ converge in $L^2$. The limits agree locally with a function $u$, so $u\in \Sigma$ and $u_n\to u$ in the energy norm. Thus $\Sigma$ is a Hilbert space; it is the confining-potential energy space for $V(x)=|x{|}^{\alpha }$.

For a bounded operator $A:X\to Y$ between Hilbert spaces, the Moore--Penrose inverse of a Hilbert-space operator is defined onand has range $\mathcal{R}(A^{†})=\mathcal{N}(A{)}^{\perp }$. Its Penrose equations areEquivalently, $A^{†}A=P_{\mathcal{N}(A{)}^{\perp }}$ and $A{A}^{†}$ is the restriction of $P_{\overline{\mathcal{R}(A)}}$ to $\mathcal{D}(A^{†})$.

Subtracting $T_{\alpha }{u}_n=z_n$ and $T_{\alpha }{u}_m=z_m$ and applying the coercive estimate from part i yieldsThus $(z_n)$ Cauchy implies $(u_n)$ Cauchy. Completeness of the Hilbert space $X$ gives $u_n\to u\in X$, and boundedness of $T_{\alpha }$ gives $z_n=T_{\alpha }{u}_n\to T_{\alpha }u$.

For distinguishable particles with one-particle Hilbert spaces ${\mathcal{H}}_1,\dots ,{\mathcal{H}}_N$, the multiparticle quantum state belongs toA product of one-particle states gives a product vector, and arbitrary linear combinations give entangled states. A noninteracting Hamiltonian is the sum of one-particle Hamiltonians acting on their respective tensor factors, so product-state energies add.

For two identical particles, the particle exchange operator $P_{12}$ swaps every degree of freedom. Since $P_{12}^2=I$, its eigenvalues are $\pm 1$. Particles whose allowed total states satisfyare bosons, while those satisfyingare fermions. Thus bosonic states are symmetric and fermionic states antisymmetric under interchange. The same condition applies to every transposition in a system of more than two identical particles.

No. Take the Hilbert space $H={\ell }^2$ and its standard orthonormal sequence $(e_n)$. Letthe unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$; equivalently, this is a weakly null orthonormal sequence. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.

The Sobolev trace theorem makes evaluation at zero a continuous linear map $T_0:H^1(0,1)\to \mathbb{R}$. Henceis a closed vector subspace of the Hilbert space $H^1(0,1)$. Every closed vector subspace of a Hilbert space is complete with the restricted inner product, so $H$ is a Hilbert space with the standard $H^1$ inner product.

A symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ is a positive-semidefinite kernel when, for every $m$, every $x_1,\dots ,x_m\in \mathcal{X}$, and every $c\in {\mathbb{R}}^m$,Equivalently, every associated kernel matrix is a positive semidefinite matrix.

A Reproducing-kernel Hilbert space $\mathcal{H}$ on $\mathcal{X}$ is a Hilbert space of real-valued functions such that every point-evaluation map $f\mapsto f(x)$ is continuous. The Riesz representation theorem then gives a function $k(x,\cdot )\in \mathcal{H}$ satisfying the reproducing propertyIts reproducing kernel is $k(x,y)=⟨k(x,\cdot ),k(y,\cdot ){⟩}_{\mathcal{H}}$.

The identity operator is positive and self-adjoint, but on the infinite-dimensional Hilbert space $L^2[0,1]$ it has eigenvalue one with infinite multiplicity. ConsequentlyA square-integrable Hilbert-space random variable has a trace-class covariance operator with trace $\mathbb{E}\Vert X{\Vert }^2<\infty$. The identity therefore cannot be such a covariance operator.

A shift-invariant basis is a common eigenbasis of all cyclic shift operators $U(\gamma )$. For the displayed Fourier states, orthonormality follows from the root-of-unity filter:There are $N$ vectors in this orthonormal set in the $N$-dimensional Hilbert space, so they form a basis. Reindexing the finite sum givesThus every $|{\xi }_{\alpha }⟩$ is simultaneously an eigenvector of every shift, with eigenvalue ${\omega }^{\alpha \gamma }$ for $U(\gamma )$.

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.

The restricted trace map is continuous, and $V$ is its kernel, so $V$ is a closed vector subspace of the Hilbert space $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient normis equivalent to that norm, so it is complete as well. It comes from the inner productConsequently $(V,\Vert \cdot {\Vert }_V)$ is a Hilbert space.

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 density operator $\rho$ on a finite-dimensional Hilbert space, the Von Neumann entropy iswith $0\log 0=0$. Its concavity says that, for $0\le p\le 1$,

A symmetric operator $L$ on a real Hilbert space is strictly positive definite when $⟨Lv,v⟩>0$ for every nonzero $v$ in its operator domain. Strict positivity proves uniqueness of a solution of $Lu=f$, but in infinite dimension it need not prove existence for every $f$. A uniformly positive definite symmetric operator additionally has a positive lower bound independent of $v$. Symmetry must be stated separately when positivity is defined only by a real quadratic expression: adding a skew-symmetric matrix changes the operator without changing that expression.