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Interpret the angle as the directed subspace angle defined by the infimum of projected unit vectors; it is different from the smallest angle between two subspaces. WriteConsider the bounded linear operator $T:W⟶V^{\perp }$ given by $Tw=P_{V^{\perp }}w$. Its adjoint operator, between these two Hilbert spaces, is $T^{\ast }u=P_{W}u$: for $w\in W$ and $u\in V^{\perp }$, the orthogonal projections give $⟨Tw,u⟩=⟨w,u⟩=⟨w,P_{W}u⟩$. The two positive directed subspace angle cosines yieldThe first bound makes $T$ injective and gives a closed range. Explicitly, if $T{w}_j$ converges, then $\Vert w_j-w_k\Vert \le a^{-1}\Vert T{w}_j-T{w}_k\Vert$, so $(w_j)$ is a Cauchy sequence. The closed subspace of a Hilbert space $W$ is complete, and its limit maps to the proposed range limit. The second bound gives $\ker T^{\ast }=\{0\}$. A vector $u\in V^{\perp }$ orthogonal to the range has $T^{\ast }u=0$, so it must be zero. The range is therefore dense as well as closed in $V^{\perp }$, and $T$ is onto. This is the mechanism of invertibility from lower bounds on an operator and its adjoint.

For any $f\in H$, choose the unique $w\in W$ with $Tw=P_{V^{\perp }}f$. Then $P_{V^{\perp }}(f-w)=0$, so $f-w\in V$. Moreover, if $w\in W\cap V$, then $Tw=0$ and hence $w=0$. Every vector has a unique decomposition, andThe direct sum is a topological one as well: the component $w=T^{-1}{P}_{V^{\perp }}f$ depends boundedly on $f$, with operator norm at most $a^{-1}$.

For precision, the quoted equality of the norms of complementary oblique projections needs both summands nonzero. For example, with $H=\mathbb{R}$, $V=H$ and $W=\{0\}$, the oblique projection is $I$, so $\Vert I\Vert =1$ but $\Vert I-I\Vert =0$. The secant function has value one here, so the second equality in the quoted formula fails. With nonzero complementary summands its intended version is valid. The proof above does not use that formula. The angle itself is undefined on a zero source space because it has no unit vectors; expressing the hypotheses as the two lower bounds handles zero spaces without ambiguity.

Use $T=P_{V^{\perp }}{|}_U:U⟶V^{\perp }$. Positivity of the directed subspace angle cosine givesso $T$ is injective. The two vector spaces have the same finite dimension, $n$. By the rank-nullity theorem, $T$ is also surjective. There is no need to assume a second positive directed subspace angle cosine in this finite-dimensional case.

For any $f\in H$, find $u\in U$ with $Tu=P_{V^{\perp }}f$. Then $f-u\in V$. If $u\in U\cap V$, then $Tu=0$, and injectivity gives $u=0$. HenceThe direct sum is again bounded: $u=T^{-1}{P}_{V^{\perp }}f$. If $n=0$, then $U=\{0\}$ and $V^{\perp }=\{0\}$, so closedness of $V$ gives $V=H$ and the conclusion directly; no angle of an empty unit sphere is needed. Equal finite dimensions are essential to the surjectivity argument, whereas mere injectivity between infinite-dimensional Hilbert spaces is insufficient.

For an implicit Runge-Kutta method, algebraic stability means that $b_i\ge 0$ and the symmetric matrixis positive semidefinite. Here $b_1=b_2=1/2$ and $A+A^T=e{e}^T/2$, soThe method is algebraically stable. This also explains why the property matters for nonlinear problems. If a vector field satisfies $⟨f(v)-f(w),v-w⟩\le 0$, the Runge-Kutta contractivity identity for two solutions, with stage differences $D_i$ and vector-field differences $F_i$, isThe first sum is nonpositive and the second vanishes. Thus, whenever the implicit stage equations have the relevant solutions, their updates are contractive in the Hilbert space norm.

Work first in a real Hilbert space. A bounded linear operator $L$ is self-adjoint whenFor the variational interpretation, elliptic means uniformly coercive: some $\gamma >0$ satisfiesThis is ellipticity of a bounded Hilbert-space operator, expressed through its quadratic form, rather than the principal-symbol definition for an elliptic differential operator. In the nonsymmetric case this condition concerns the symmetric part of $L$ and does not itself imply self-adjointness.

A positive-definite operator is self-adjoint and strictly positive:Some variational conventions use “positive definite” for the stronger combination of self-adjointness and ellipticity; we distinguish strict from uniform positivity explicitly. A uniformly positive definite symmetric operator is coercive, whereas strict positivity alone in infinite dimension need not give a bounded inverse or solvability for every forcing. If positivity is instead defined only by the displayed real quadratic inequality, symmetry must be added separately for the next part. In a complex Hilbert space, use a self-adjoint operator, the real quadratic form, and $\mathrm{Re}⟨f,v⟩$ in the linear term.

The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincare inequality makes this a Hilbert space norm equivalent to the usual $H^1$ norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous $p,q$ on the closed interval with $p>0$ and $q\ge 0$, setIf $p_{\ast }=\min p>0$, then Cauchy-Schwarz inequality and the Poincare inequality giveBy the Riesz representation theorem, a unique bounded linear operator $T:H\to H$ satisfies $(Tu,v{)}_H=a(u,v)$. Symmetry of $a$ makes $T$ self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by $a$: for smooth functions, integration by parts giveswith the boundary term zero. A forcing $f\in L^2$ becomes the bounded functional $v\mapsto \int fv$, or its Riesz representative in $H$.

Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on $L^2(-1,1)$ with domain $H^2\cap H_0^1$. Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator $T$ used above.

The PDF states sign conditions without coefficient regularity. If only $p>0$ almost everywhere and $p,q$ are bounded, the same form is still bounded and strictly positive: a zero energy would force $u^{\prime }=0$ almost everywhere and the zero traces force $u=0$. Uniform ellipticity, however, requires a positive lower bound. For instance $p(x)=x^2$ for $x\ne 0$, $p(0)=1$, and $q=0$ satisfy literal pointwise positivity. Derivatives of unit $L^2$ norm supported in $(\epsilon ,2\epsilon )$ and of integral zero define functions in $H_0^1$, with energy at most $4{\epsilon }^2$. Thus pointwise positivity without regularity does not imply coercivity in this $H$ norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.

The Holstein–Primakoff transformation acts on the physical Fock states $|n⟩$, $0\le n\le 2S$, with the specified operator ordering. Its matrix elements areThe unavailable endpoint states have zero coefficient. These formulas directly give $[S^z,S^{\pm }]|n⟩=\pm S^{\pm }|n⟩$. MoreoverThus the spin commutation relations hold in units $\hslash =1$. The Holstein–Primakoff occupation constraint is essential: the spin Hilbert space has dimension $2S+1$, so unrestricted boson Fock space is not itself this finite-spin representation. The square root and oscillator operators must retain their order for the displayed matrix elements and endpoint conditions to agree.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

Fix a classical globally hyperbolic spacetime with metric signature $(-+++)$; in this solution take $\hslash =1$. A free real Klein-Gordon field can be specified bywhere $\mu$ is its mass, $\xi$ its curvature coupling and $\mathcal{R}$ the scalar curvature. Global hyperbolicity ensures a well-posed initial-value problem on a Cauchy hypersurface $\Sigma$ and the existence of retarded and advanced propagators. Thus compactly supported field and normal-derivative data determine a classical solution. This fixes the dynamics, but not a Fock vacuum.

For real solutions with suitable support or falloff, the symplectic form on scalar-field solutions iswith $n$ the future unit normal. The field equation makes the current conserved, so this symplectic form is independent of $\Sigma$ when boundary flux vanishes. Quantize the initial data by the canonical commutation relation: with $\pi =n^a{\nabla }_a\varphi$ and the delta function defined relative to $d\Sigma$, $[\hat{\varphi }(x),\hat{\pi }(y)]=i{\delta }_{\Sigma }(x,y)$ and the two equal-field commutators vanish. Equivalently, construct the field algebra using the causal propagator. A state on that algebra is additional input.

To construct a particle representation, complexify the classical solution space. Its conserved Klein-Gordon inner product isIt is indefinite on all complex solutions. Choose a complete positive-norm subspace and an orthonormal mode basis $u_i$ with $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Such a choice is encoded by a compatible complex structure on the Klein-Gordon solution space. Its positive subspace gives the one-particle Hilbert space, and the associated bosonic Fock space contains symmetrized many-particle states. The field expansion isThe creation operator $a_i^{†}$ adds a particle in mode $i$, the annihilation operator removes one, and the number operator is $N_i=a_i^{†}{a}_i$. For continuous mode labels the sums and Kronecker deltas become integrals and delta functions, or one can work with normalized wave packets.

The ambiguity is precisely that the field equation and global hyperbolicity do not select that positive subspace. A different normalized basis may mix $u_j$ and $u_j^{\ast }$ by a Bogoliubov transformation, and then its annihilation operators mix $a_j$ and $a_j^{†}$. Its Fock vacuum and number operators differ. The Hadamard condition constrains physically acceptable short-distance singularities and allows local renormalization, but it still leaves many states. Hence there is generally no observer-independent particle count on an arbitrary dynamical geometry.

In a stable strictly stationary spacetime, a globally future timelike Killing vector field $K$ gives a preferred time translation. Fix its normalization and suitable boundary conditions, and choose positive-frequency solutions satisfying $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. The corresponding positive spectral subspace gives the preferred vacuum state in a stationary spacetime. Unitary changes of basis within it leave the vacuum and the particle notion unchanged. This construction assumes a well-defined positive stationary generator; stationarity by itself is insufficient if $K$ becomes spacelike, as in a Kerr ergoregion, or if unstable or zero modes obstruct the ground-state construction. It selects a preferred ground state under the stated assumptions, not a unique state among all thermal and excited states.

If the geometry is suitably stationary in the asymptotic past and future, choose those preferred mode spaces separately, giving the in-vacuum and out-vacuum. Propagate the past modes through the intervening region using the field equation and compare them to the future modes using the conserved Klein-Gordon inner product. Adopt the conventionThe canonical identities for a bosonic Bogoliubov transformation read $\alpha {\alpha }^{†}-\beta {\beta }^{†}=I$ and $\alpha {\beta }^T=\beta {\alpha }^T$. Extracting the future annihilation operator with the same inner product givesIn the in-vacuum, only $⟨a_j{a}_k^{†}⟩={\delta }_{jk}$ contributes to $⟨b_i^{†}{b}_i⟩$. Therefore the particle number from Bogoliubov coefficients isNonzero $\beta$ is the production of future particles from the past vacuum. Summing over future modes gives the total expected particle number when that sum is finite. For infinitely many modes, a Hilbert-Schmidt operator $\beta$ is the condition for unitary implementability between these pure bosonic Fock space representations; finite-volume or wave-packet calculations must respect the relevant measures and convergence. Particle production is determined by the negative-frequency mixing, rather than by identifying a single instantaneous vacuum throughout the time-dependent region.

Embed the given purifications into a common reference Hilbert space $R$, enlarging it if necessary, and add a flag register $X$ with orthonormal basis $\{|x⟩\}$. A flagged purification of a quantum ensemble isOrthogonality of the flags gives $⟨\Psi |\Psi ⟩={\sum }_x{p}_x=1$. Moreover,Thus it is a purification of a density operator. The extra flag is essential: simply superposing the purifications without orthogonal labels would generally leave unwanted cross terms.

Uhlmann's theorem identifies quantum fidelity with the largest absolute overlap of purifications of a density operator. Choose a common reference Hilbert space $R$ of dimension at least that of the original system. Thenwhere $|u⟩,|v⟩\in {\mathcal{H}}_A\otimes {\mathcal{H}}_R$ have reduced density operators ${\rho }_A,{\sigma }_A$. One purification may be fixed in advance: the maximum is over the other, with the freedom to apply a unitary on the reference system. This is the unitary freedom of purification. Enlarging the reference by unused dimensions does not change the maximum.

Take maximizing purifications $|u⟩,|v⟩$ of ${\rho }_{AB},{\sigma }_{AB}$ on $ABR$. Their overlap has magnitude $F({\rho }_{AB},{\sigma }_{AB})$. Regard these same vectors as purifications of ${\rho }_A,{\sigma }_A$ with reference system $BR$. They are candidates in the larger optimization, so Uhlmann's theorem givesThis is monotonicity of quantum fidelity under partial trace: discarding a system cannot make two states more distinguishable according to their quantum fidelity.

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore

Set $K_y=|y⟩⟨y|$. The map is in Kraus representation:For any ancillary Hilbert space $R$ and positive operator $M$ on $R\otimes \mathcal{H}$,This verifies complete positivity directly, not merely positivity on unextended states. Cyclicity of the trace gives $\mathrm{Tr}\Lambda (X)=\mathrm{Tr}[X{\sum }_y{K}_y^{†}{K}_y]=\mathrm{Tr}X$. Thus $\Lambda$ is a CPTP map, the completely dephasing channel in the given basis.

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

Assume $|\xi ⟩$ is normalized and the Hilbert space has dimension $d$. The rank-one orthogonal projection $P=|\xi ⟩⟨\xi |$ satisfies $P^{†}=P$ and $P^2=P$.

For $d>1$, $P$ has eigenvalue zero on the orthogonal complement of $|\xi ⟩$, so it cannot be a unitary operator. The complementary orthogonal projection $I-P$ kills $|\xi ⟩$ and is not unitary in any positive dimension. In contrast, the Householder reflectionis Hermitian and obeys $R^{†}R=R^2=I-4P+4P^2=I$. Its eigenvalues are $-1$ along $|\xi ⟩$ and $+1$ on the orthogonal complement.

In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.