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The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

Write $N_2=\Vert M_{\infty }{\Vert }_2$ and $N_{\ast }=\Vert {\sup }_{t\ge 0}|M_t|{\Vert }_2$. The L2 martingale convergence theorem ensures the existence of $M_{\infty }$ and yields $|M_{\infty }|\le {\sup }_t|M_t|$ almost surely. Hence $N_2\le N_{\ast }$.

For a finite horizon $T$, the Doob L2 maximal inequality for the submartingale $|M|$ givesThe inequality does not require $M_0=0$. Send $T$ to infinity: the left side increases to $N_{\ast }^2$ by the monotone convergence theorem, and the right side tends to $4N_2^2$ by convergence in the Lebesgue space $L^2$. Thus the equivalent terminal and maximal norms for L2-bounded continuous martingales satisfyBoth are genuine norms on L2-bounded continuous martingales modulo indistinguishability of stochastic processes: a zero terminal norm gives $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]=0$, first at rational times and then at all times by continuity. This proves that the two norms are equivalent norms.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

Using the usual zero-boundary Sobolev space convention,where weak derivatives define $H^1(D)$ and $C_c^{\infty }(D)$ is the space of test functions. In the Gaussian free field convention, the same notation often denotes the Dirichlet energy space, the completion in the gradient norm alone. On bounded domains the Poincaré inequality makes the two definitions equivalent; on unbounded domains one must distinguish them. The following gradient-pairing argument applies in either setting whenever the energy completion is realized as weak functions.

Identify $H_{\mathrm{supp}}=H_0^1(U)$ with a vector subspace of $H_0^1(D)$ by zero extension of H01. Approximating by test functions in $U$ shows that this is an isometric embedding in the inhomogeneous Sobolev norm and also in the Dirichlet inner product norm. In particular, arbitrary irregularity of $\partial U$ causes no additional boundary term.

DefineThese are weakly harmonic Sobolev functions. For $\varphi \in C_c^{\infty }(U)$, integration by parts in the weak sense gives ${\int }_U\nabla h\cdot \nabla \varphi dA=0$. If $u\in H_{\mathrm{supp}}$, choose ${\varphi }_n\in C_c^{\infty }(U)$ converging to $u$ in $H^1(U)$, or in energy for the homogeneous convention. The Cauchy-Schwarz inequality givesHence the orthogonality of supported and harmonic Dirichlet functions isBoth are linear vector subspaces. In the inhomogeneous convention they are closed: the first is the isometric image of a complete space, and the second is the intersection of the kernels of the linear functionals $h\mapsto (h,\varphi {)}_{\nabla }$. No spanning assertion is needed. The PDF contains this orthogonality statement; the TeX has badly corrupted it into an assertion about openness.

A real positive-definite kernel is a symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ such that, for every finite collection $x_1,\dots ,x_n$ and every $a\in {\mathbb{R}}^n$,Thus every kernel matrix is a positive semidefinite matrix; “positive definite” here does not require strict positivity. A Reproducing kernel Hilbert space is a Hilbert space $\mathcal{H}$ of functions on $\mathcal{X}$ in which every point evaluation is a continuous linear functional. Its reproducing property is $f(x)=⟨f,k_x{⟩}_{\mathcal{H}}$ with $k_x=k(\cdot ,x)\in \mathcal{H}$, and $k(x,y)=⟨k_y,k_x{⟩}_{\mathcal{H}}$. For a sum of positive-definite kernels, the quadratic form of its kernel matrix is a sum of nonnegative quadratic forms. Symmetry also survives summation. Consequently $\boxed{k=\sum _j{k}_j\text{ is a positive-definite kernel}.}$

For the representer theorem, put $V=\mathrm{span}\{k_{x_1},\dots ,k_{x_n}\}$. This is a closed finite-dimensional vector space. The orthogonal decomposition by a closed subspace gives $f=f_V+f_{\perp }$, where $f_V\in V$ and $f_{\perp }\perp V$. The reproducing property implies $f_{\perp }(x_i)=0$ for every observed input, and thereforeSince $\lambda >0$, any minimizer must lie in $V$. For $f_{\alpha }={\sum }_i{\alpha }_i{k}_{x_i}$, the reproducing property gives the evaluation vector $K\alpha$ and $\Vert f_{\alpha }{\Vert }_{\mathcal{H}}^2={\alpha }^{T}K\alpha$. Hence $Q_1(f_{\alpha })=M(\alpha )$. If $\hat{f}$ minimizes $Q_1$, its representation yields a minimizing $\hat{\alpha }$. Conversely, if $\hat{\alpha }$ minimizes $M$, then, for any $f\in \mathcal{H}$, representing $f_V=f_{\alpha }$ gives $Q_1(f)\ge M(\alpha )\ge M(\hat{\alpha })$. ThusThis is an equivalence of minimizers, not an existence or uniqueness assertion: the loss function is arbitrary. A singular kernel matrix also allows nonunique coefficients, since differences in $\ker K$ represent the zero function. Strict convexity of the loss was not used.

Now use the stated minimum-norm description of a sum of reproducing-kernel Hilbert spaces. Let $\hat{f}={\sum }_j{\hat{f}}_j$. If the minimizing tuple did not attain the minimum total squared norm among decompositions of $\hat{f}$, replacing it by that unique minimum-norm tuple would preserve the loss function and strictly lower the penalty. This contradicts optimality. ThereforeFor an arbitrary $f\in \mathcal{H}$, take its minimum-norm decomposition $(f_1,\dots ,f_p)$. Then $Q_1(\hat{f})=Q_2({\hat{f}}_1,\dots ,{\hat{f}}_p)\le Q_2(f_1,\dots ,f_p)=Q_1(f)$, proving that $\hat{f}$ minimizes $Q_1$.

Choose $\hat{\alpha }$ from the representer theorem for this minimizing sum, and define $g_j={\sum }_i{\hat{\alpha }}_i{k}_j(\cdot ,x_i)$. Write $K_j$ for the component kernel matrices. ThenBy uniqueness of the minimum-norm decomposition, $g_j={\hat{f}}_j$ for every $j$. This proves the common representer coefficients for a sum of kernels conclusion:The same coefficient vector works simultaneously. Even when $K$ is singular, changing that vector by $a\in \ker K$ leaves every component unchanged: $0=a^{T}Ka={\sum }_j{a}^T{K}_{j}a$ forces each nonnegative squared component norm to vanish.

Put $\hat{\Sigma }=X^{T}X/n$. The Debiased Lasso correction isFor the Nodewise Lasso, let $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions areso $z_{jk}=\mathrm{sgn}({\hat{\gamma }}_k^{(j)})$ at a nonzero coefficient and $|z_{jk}|\le 1$ at a zero coefficient. Multiplying by ${\hat{\gamma }}^{(j)}$ gives the nodewise-Lasso residual identityWe require nonzero design columns, a necessary qualification missing from the unrestricted printed formulation. Because ${\lambda }_j>0$, ${\hat{\tau }}_j^2=0$ would force both $r_j=0$ and ${\hat{\gamma }}^{(j)}=0$, and hence $X_j=0$. Thus nonzero columns suffice for ${\hat{\tau }}_j^2>0$ even when $p>n$ or the full Gram matrix is singular. A zero column instead gives ${\hat{\tau }}_j^2=0$ and its coefficient is unidentifiable from $Y$; the proposed normalization and confidence interval do not exist there.

For each $j$, define $c_j\in {\mathbb{R}}^p$ by $(c_j{)}_j=1$ and $(c_j{)}_{-j}=-{\hat{\gamma }}^{(j)}$. Set row $j$ of $\hat{\Theta }$ to $c_j^T/{\hat{\tau }}_j^2$. Since $X{c}_j=r_j$,This row construction is the nodewise normalization used by van de Geer and coauthors. The matrix $\hat{\Theta }$ need not be symmetric or be an actual inverse.

Substitute the linear model into the correction. The exact decomposition isIn the fixed-design normal linear model, or with $\epsilon \mid X\sim N_n(0,{\sigma }^{2}I)$ and nodewise estimates and tuning chosen from $X$ alone, a linear transformation of a multivariate normal distribution gives $W\mid X\sim N_p(0,{\sigma }^2\hat{\Omega })$. Marginal normality of $\epsilon$ without this conditional assumption would not suffice for a response-dependent random design. No independence between $W$ and $\Delta$ is asserted. The diagonal cancellation and the Holder inequality give the debiased-Lasso remainder bound

For a coordinate variance, ${\hat{\Theta }}_j{X}^T=r_j^T/{\hat{\tau }}_j^2$, whenceWriting $z_{1-\alpha /2}$ for the corresponding standard normal quantile, an approximate confidence interval isEquivalently its half-width is $z_{1-\alpha /2}\sigma \Vert X_j-X_{-j}{\hat{\gamma }}^{(j)}{\Vert }_2/(n{\hat{\tau }}_j^2)$. We assume $0<\alpha <1$ and $\sigma >0$. For asymptotic coverage, the standardized remainder ${\Delta }_j/(\sigma \sqrt{{\hat{\Omega }}_{jj}})$ must have convergence in probability to zero and the coordinate variance must be nondegenerate. Exact conditional normality of $W$ alone does not give finite-sample exact coverage for $\hat{b}$.

One sufficient random-design regime is the Gaussian identity design for debiased Lasso: take independent rows $X_i\sim N_p(0,I_p)$, independently of $\epsilon$, fixed $\sigma >0$, $p=p_n\to \infty$, $s=s_n\ge 1$, andChoose $\lambda =C\sigma \sqrt{\log p/n}$ and all ${\lambda }_j=D\sqrt{\log p/n}$ for sufficiently large fixed $C,D$. Here is why this regime has the requested remainder bound. Uniformly over columns, their squared norms divided by $n$ tend to one, while all off-diagonal sample cross-correlations have magnitude at most a constant times $\sqrt{\log p/n}$ with probability tending to one. Taking $D$ larger makes the zero vector satisfy all nodewise Karush-Kuhn-Tucker conditions strictly, so ${\hat{\gamma }}^{(j)}=0$ for every $j$ on that event, and ${\hat{\tau }}_j^2=\Vert X_j{\Vert }_2^2/n\ge 1/2$ eventually. This also gives ${\hat{\Omega }}_{jj}=1/{\hat{\tau }}_j^2\to 1$ uniformly.

In this regime, the entrywise entrywise concentration of a Gaussian Gram matrix and Lasso cone condition argument in Gaussian identity design for debiased Lasso gives a Compatibility condition for the Lasso bounded away from zero with probability tending to one. Together with the noise-score event and the Basic inequality for the Lasso, this yields, for a fixed $C_1$,For example, the cone inequality and compatibility bound give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le 12\lambda s/{\varphi }^2$ on their joint event. Therefore, with a fixed $A\ge 2D{C}_1$,The stated sparsity condition makes this remainder vanish and supplies the preceding confidence-interval requirement. Both dimensions may grow: $p=n^2$ and $s=\lfloor n^{1/4}\rfloor$ satisfy these assumptions. The concentration and Gaussian-design compatibility results are sufficient standard conditions here; the question does not require their proofs. The displayed rate is a high-probability fixed-constant bound, rather than only a claim of boundedness in probability.

Assume a complete relativistically normalized basis of scalar momentum eigenstates transforms asThis states both reversal of spatial momentum and preservation of the normalization of basis quantum states. Antilinearity alone would not suffice: multiplying a conjugation operator by two gives an antilinear operator that multiplies squared norms by four.

Expand $|\varphi ⟩=\int d{\Pi }_p\varphi (\mathbf{p})|\mathbf{p}⟩$ and similarly for $|\psi ⟩$, with $d{\Pi }_p=d^{3}p/((2\pi {)}^{3}2E_p)$. This Lorentz-invariant phase-space measure is unchanged under $\mathbf{p}\mapsto -\mathbf{p}$. The antilinearity of the quantum time-reversal operator gives conjugated expansion coefficients. Orthogonality of the momentum eigenstates and cancellation of their unit-modulus phases yieldSince momentum reversal is a bijection of the complete basis, $\hat{T}$ is also onto. ThusThis proves antiunitarity, with the normalization hypothesis explicitly included. For scalar multiparticle quantum states the same argument uses the complete occupation-state basis and reverses all momenta; it is not restricted to a single-particle wave packet.

Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.

Let $V$ denote the coherent exact quantum phase estimation unitary operator from the preceding part, with the data quantum register written first. For every eigenvector of $A$,because $e^{2\pi iA}|u_j⟩=e^{2\pi i{\lambda }_j}|u_j⟩$ and ${\lambda }_j=c_j/2^n$ lies in $[0,1)$, with no phase-aliasing ambiguity. Apply $V$ coherently to the input quantum state and a clean quantum register; do not measure the eigenvalue label. By linearity it produces ${\sum }_j{\beta }_j|u_j⟩|c_j⟩$.

Append a quantum ancilla in $|0⟩$, and use the provided quantum variable rotation with ${\theta }_{c_j}=\arcsin {\lambda }_j$, choosing the nonnegative cosine. The resulting normalized quantum state isThe sum covers both terms; this fixes the potentially ambiguous sum placement in the abbreviated display. The orthonormal basis of eigenvectors and the unit length of each rotated quantum ancilla show that its norm is one. This is quantum spectral filtering by multiplication, not the reciprocal rotation used in the HHL algorithm.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, a least-squares solution of data $f$ is a minimizer of the residual norm:The minimum-norm least-squares solution is the one of smallest $U$-norm among these minimizers. Equivalently it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If the equation is consistent, this is the smallest-norm exact solution.

Let $P$ be the orthogonal projection onto $\overline{\mathcal{R}(K)}$. The orthogonal decomposition by a closed subspace givesThe first term has infimum zero by the definition of closure; that infimum is attained exactly if $Pf\in \mathcal{R}(K)$. This proves the least-squares existence criterion:They exist for every datum precisely when the range is closed. Once a solution exists, all solutions differ by elements of the closed null space, and projecting onto its orthogonal complement gives the unique minimum-norm solution.