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A closed subspace invariant under a star-algebra is reducing, since it is invariant under every adjoint as well. Thus the orthogonal projections $P_1,P_2$ onto $H_1,H_2$ commute with $M$. Consider the bounded intertwinerIts kernel is $H_1\cap H_2$, so its initial polar space isIts closed range isFor the inclusion from left to right, $(I-P_2)h_1=h_1-P_2{h}_1$ lies in $H_1+H_2$ and is orthogonal to $H_2$. Conversely, approximate any $z$ in the displayed $K_2$ by $h_{1,n}+h_{2,n}$ and apply $I-P_2$. The images converge to $z$ and equal $T{h}_{1,n}$, proving the reverse inclusion after closure. The overline on the sum is present in the original PDF and is essential for potentially nonclosed sums.

Take the polar decomposition $T=V|T|$ between these Hilbert spaces. Because $T$ intertwines the two actions of $M$, its adjoint does also, so $T^{\ast }T$ commutes with the action on $H_1$. Its positive square root therefore commutes as well. The identity $V|T|\xi =T\xi$ shows on its dense initial range that $V$ intertwines, and continuity extends this to $K_1$. It is an isometry onto $K_2$. ConsequentlyThe unitary here is a unitary between the indicated Hilbert space modules, not an assertion that the unclosed algebraic sum is a Hilbert space.

The discrete Bochner–Herglotz theorem says that a sequence $r:\mathbb{Z}\to \mathbb{C}$ is a positive-definite function, meaningif and only if there is a unique finite positive Borel measure $\sigma$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ such thatThe mass is $\sigma (\mathbb{T})=r(0)$. The necessity of positive definiteness follows by integrating $|{\sum }_j{c}_j{e}^{2\pi ijt}{|}^2$.

For sufficiency, positive definiteness makes the relevant Gram matrices Hermitian positive semidefinite, so $r(-n)=\overline{r(n)}$ and $|r(n)|\le r(0)$. DefineTaking $c_j=e^{-2\pi ijt}$ in the positive-definiteness condition gives $p_N(t)\ge 0$, while ${\int }_{\mathbb{T}}{p}_N(t)dt=r(0)$. Moreover,Thus the integrals against every trigonometric polynomial converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by Fourier inversion for smooth functions gives this density. The uniform mass bound $r(0)$ then makes the integrals against every continuous $\phi$ converge as well. Their limit is a positive linear functional $L$ with $|L(\phi )|\le r(0)\Vert \phi {\Vert }_{\infty }$. The standard Riesz-Markov-Kakutani representation theorem gives a finite positive Borel measure $\sigma$ with $L(\phi )=\int \phi d\sigma$. The preceding coefficient computation proves its required Fourier coefficients. Finally, two such measures agree on all trigonometric polynomials, hence on all continuous functions by uniform density, and therefore are identical. This proves the Bochner–Herglotz theorem, including existence and uniqueness; if $r(0)=0$, the estimate $|r(n)|\le r(0)$ gives the zero measure directly.

Now let $U$ be the Koopman operator $Uf=f\circ T$ on the complex L2 space of the probability measure-preserving system. We use $⟨f,g⟩=\int f\overline{g}d\mu$, linear in the first entry. Since $U$ is an isometry, the sequence $r_f(n)=⟨U^{n}f,f⟩$ for $n\ge 0$, extended by $r_f(-n)=\overline{r_f(n)}$, is positive-definite: its matrix is the Gram matrix of $f,Uf,\dots$. The Bochner–Herglotz theorem gives its spectral measure of a Koopman observable ${\sigma }_f$.

We first prove exactly how an atom of a measure detects an eigenfunction. Fix $\zeta =e^{2\pi i{t}_0}$ and setThe mean ergodic theorem, applied to the isometry $V={\zeta }^{-1}U$, says these averages converge in L2 space to the orthogonal projection $P_{\zeta }f$ onto $\ker (U-\zeta I)$. Its usual Hilbert-space proof works without invertibility: the orthogonal complement of $\overline{\mathrm{ran}(I-V)}$ is $\ker (I-V^{\ast })=\ker (I-V)$, and the averages of $(I-V)h$ telescope to $(h-V^{N}h)/N\to 0$. The equality of the two kernels follows from $V$ being an isometry, by expanding $\Vert Vh-h{\Vert }^2$ when $V^{\ast }h=h$.

Expanding the averaged norm through the spectral measure of a Koopman observable givesThe integrand is at most one and converges to $1_{\{t_0\}}$. By the dominated convergence theorem,Thus a nonzero atom corresponds to a nonzero Koopman operator eigenfunction, with eigenvalue $\zeta$, in the cyclic spectral data of $f$.

We also need the Wiener Fourier coefficient lemma, which here has a short proof. For any finite positive circle measure $\sigma$ with coefficients $r(n)$,The bracket is bounded in absolute value by one and tends to the indicator of the diagonal $t=s$. Another application of the dominated convergence theorem givesIn particular the limit is zero for an atomless measure.

Suppose there are no nonconstant Koopman operator eigenfunctions. For every $f$ in the mean-zero L2 space, all its projections $P_{\zeta }f$ vanish: a nonzero projection would be a nonconstant eigenfunction, since it has mean zero. Hence ${\sigma }_f$ is atomless, and the Wiener Fourier coefficient lemma givesThe Cauchy-Schwarz inequality also gives vanishing averaged absolute autocorrelations. For two mean-zero observables $f,g$, the sesquilinear polarization identitythen gives vanishing averaged absolute cross-correlations. Taking $f=1_A-\mu (A)$ and $g=1_B-\mu (B)$ proveswhich is weak mixing.

Conversely, this set formulation of weak mixing extends to all pairs of mean-zero L2 space observables by finite-simple-function approximation and the uniform estimate $|⟨U^{n}f,g⟩|\le \Vert f{\Vert }_2\Vert g{\Vert }_2$. If $Uh=\zeta h$ and $h\ne 0$, the isometry property gives $|\zeta |=1$. If $\zeta \ne 1$, invariance of the integral gives $\int h=0$. If $\zeta =1$ and $h$ is nonconstant, replace it by $h-\int h$, a nonzero mean-zero eigenfunction. But in either casefor every $n$, contradicting weak mixing. We have proved weak mixing is equivalent to absence of nonconstant eigenfunctions, including for noninvertible transformations.

Finally, the doubling map $Tx=2x(\mathrm{mod}1)$ preserves normalized Lebesgue measure, as substitution on its two inverse branches shows. Let $f\in L^2(\mathbb{T})$ be an eigenfunction, $f(2x)=\zeta f(x)$, with Fourier coefficients $c_k$. The Fourier basis calculation givesSince $\zeta \ne 0$, all odd coefficients vanish. Every nonzero integer is $2^{r}m$ with $m$ odd, and repeated use of $\zeta c_{2^{r}m}=c_{2^{r-1}m}$ makes all nonzero coefficients vanish. Completeness of the Fourier basis then says $f$ is constant. Therefore the doubling map has no nonconstant eigenfunctions, and

Use metric signature $(+---)$ throughout this solution. Stress-energy conservation for the perfect fluid is ${\nabla }_b{T}^{ab}=0$. Expanding it and using metric compatibility givesThe normalization of the four-velocity implies $U_a{U}^b{\nabla }_b{U}^a=0$. Contracting with $U_a$ therefore gives the relativistic perfect-fluid energy equation,while subtracting the component parallel to $U$ gives the relativistic Euler equation,Equivalently, apply the orthogonal projection ${\delta }_a{}^c-U_a{U}^c$ to the conservation equation. This form makes clear that a pressure gradient orthogonal to the four-velocity supplies the fluid's four-acceleration.

Use the standard nonnegative partition-normalized B-splines $N_j$; their sum is at most one, and is exactly one for the complete clamped basis on $[0,1]$. The printed question introduces $M_i$ only in its last matrix formula without defining it. The intended normalization is the unit-integral normalization of a B-spline:For supports wholly inside the interval, $c_i=(t_{i+k}-t_i)/k$. The positivity and partition convention, sometimes called $L_{\infty }$ normalization, should not be confused with individually rescaling every spline to have maximum exactly one.

The characterization in part 1(a) gives the normal equations for the orthogonal projection:Dividing each equation by $c_i$ yieldsThis is the mixed-normalization spline Gram matrix. It need not be symmetric, but $G=D^{-1}H$ with $D=\mathrm{diag}(c_i)$ and $H_{ij}=(N_i,N_j{)}_{L_2}$. The ordinary Gram matrix $H$ is positive definite because the splines are linearly independent, so $G$ is invertible and $a=G^{-1}d$.

Since each $M_i$ is nonnegative and has integral one,The induced matrix operator norm is the maximum absolute row sum, soFinally, nonnegativity and the subpartition of unity for B-splines giveTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the maximum-norm bound for spline projection:The formulas also establish linearity and boundedness of the projection on continuous inputs. Its range is exactly the finite-dimensional spline space $\mathcal{S}$, since it fixes every spline in that space. Thus, when the splines are continuous, it is an operator $C[0,1]\to C[0,1]$ with range $\mathcal{S}$; it is not onto all of $C[0,1]$ as the printed wording says. For example, $e^t$ cannot belong to a fixed finite-degree spline space because it is not a polynomial on any nonempty knot interval. The missing $M_i$ normalization and this range correction are needed for a precise reading of the result.

The ordinary least squares estimate, which is also the maximum-likelihood estimate in this normal linear model, gives the hat matrixIt is the orthogonal projection onto the column space of $X$, so $P^T=P$, $P^2=P$ and $\mathrm{rank}P=p$. Since $(I-P)X=0$, the vector of regression residuals is $\widehat{ϵ}=(I-P)ϵ$, with multivariate normal distributionFor $i\ne j$, its covariance is $-{\sigma }^2{p}_{ij}$, which is generally nonzero. Therefore the regression residuals are generally dependent, even though the original errors are independent random variables.

Fit the model in R, compute its internally studentized residuals with rstandard(fit), and examine a normal Q-Q plot with qqnorm(rstandard(fit)) and qqline(rstandard(fit)). A plot of these regression residuals against the fitted values helps detect a curved mean pattern or nonconstant variance. Approximate agreement with a straight line in the Q-Q plot, with no systematic residual pattern, supports the assumed error model.

Here $p\ll n$. The average leverage is $\mathrm{tr}P/n=p/n$, and ${\sum }_{ij}{p}_{ij}^2=\mathrm{tr}(P^2)=p$. Thus most diagonal corrections and most off-diagonal dependences are small when only a few directions have been removed from a large sample. Also ${\tilde{\sigma }}^2$ estimates ${\sigma }^2$ accurately when $n-p$ is large. These facts make the visual independent-normal approximation reasonable, though exceptional high-leverage observations still deserve attention.

The approximation is not an exact distributional statement: the distribution of an internally studentized Gaussian residual has ${\hat{\eta }}_i^2/(n-p)\sim \mathrm{Beta}(1/2,(n-p-1)/2)$ when $n-p\ge 2$ and $p_{ii}<1$. In particular these residuals are bounded, rather than exactly distributed according to Student's t-distribution. Formal goodness-of-fit calibration should account for fitting and dependence.

Let $d={\inf }_{v\in V}\Vert f-v\Vert$ and choose $v_n\in V$ with $\Vert f-v_n\Vert \to d$. The parallelogram identity givesCompleteness and closedness give $v_n\to v\in V$, attaining the distance. Put $w=f-v$. For $u\in V$ and real $t$, minimality gives $\Vert w-tu{\Vert }^2\ge \Vert w{\Vert }^2$, so $\mathrm{Re}⟨w,u⟩=0$. In a complex Hilbert space, also replace $u$ with $iu$ to get the imaginary part zero. ThereforeThe decomposition is unique since $V\cap V^{\perp }=\{0\}$. This proves the required orthogonal projection result rather than assuming it before the averaging argument.

Apply the proved projection result to $V=\overline{X}$ and write $f=v+w$ with $w\perp V$. By part (iii), $A_{n}v\to 0$, while part (iv) gives $U^{k}w=w$ and hence $A_{n}w=w$. ThusThe limit is the orthogonal projection onto the fixed vectors of $U$. This is norm convergence, not just weak convergence.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.

Take $\mathbf{n}$ to be the unit normal, so $P(v)=v-⟨v,\mathbf{n}⟩\mathbf{n}$ is orthogonal projection onto $TM$. With $D$ the Euclidean Levi-Civita connection, the proposed derivative isIt is a smooth tangent field, and independence of the extensions is given. Linearity follows from linearity of $D$ and $P$. Moreover, for $f\in C^{\infty }(M)$, take local extensions of $f,X,Y$ to obtainsince $PY=Y$. Thus it is an affine connection on $M$.

For tangent fields $Y,Z$, the normal component of $DY$ pairs to zero with $Z$. Differentiating the ambient inner product along $X$ givesHence it is a metric connection for the induced Riemannian metric. The ambient torsion vanishes, soHere the restriction of the ambient bracket is the intrinsic bracket and is tangent to $M$: derivations of functions restricted to $M$ give the same bracket, or this follows in submanifold coordinates. Thus the projected derivative is also a torsion-free connection. Uniqueness now provesThis is the projected ambient connection. The projection formula requires a unit normal. If a nonunit normal is used, its normal term must instead be $⟨DY,\mathbf{n}⟩\mathbf{n}/⟨\mathbf{n},\mathbf{n}⟩$. For example, on the unit circle, taking $\mathbf{n}=2p$ and differentiating its unit tangent along itself gives $DY=-p$; the unnormalized printed expression would be $3p$, which is not tangent. Normalizing the normal removes that ambiguity.

Let $\varphi$ be the scaling function and $\psi$ an orthonormal wavelet, with ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. For $f\in L^2(\mathbb{R})$, the wavelet series at a coarse level $j_0$ isThe series converges in the L2 norm. Its coarse part captures broad variation and successive wavelet levels add finer detail. Truncating at levels $j<J$ produces the orthogonal projection onto $V_J$; Parseval identity makes its squared approximation error the sum of the omitted squared coefficients. Merely square-integrable functions need not have pointwise convergence.

For the Haar wavelet, use the cell convention $I_{j,k}=(k{2}^{-j},(k+1)2^{-j}]$ and let $I_j(x)$ be the unique such interval containing $x$. The Haar scaling functions are $2^{j/2}{1}_{I_{j,k}}$, so the Haar projection isFor a merely locally integrable function this cell-average formula remains defined, even if a global $L^2$ orthogonal projection is unavailable. If $\Vert f^{\prime }{\Vert }_{\infty }=L$, the mean value theorem gives Lipschitz continuity and, since $|t-x|\le 2^{-j}$ on the cell,Thus the requested bound holds with $\boxed{c=\Vert f^{\prime }{\Vert }_{\infty }}$, by the Haar projection error for a Lipschitz function.

To estimate a density, replace each scaling or wavelet coefficient $\int f{\varphi }_{j,k}$ or $\int f{\psi }_{j,k}$ by the sample average of that basis function. The finite-resolution Haar density estimator is thereforeIt is the histogram on the dyadic cells, is nonnegative, and integrates to one. Its expectation is $K_{j}f(x)$. A density with bounded derivative is bounded: the Lipschitz density height bound gives $f(x{)}^2\le L$ by integrating the triangular lower envelope $(f(x)-L|t-x|{)}_{+}$. The case $L=0$ cannot be a probability density on the whole real line. Hence $B=\Vert f{\Vert }_{\infty }<\infty$.

The cell probability is $p_j(x)={\int }_{I_j(x)}f\le B{2}^{-j}$. The sample count has a binomial distribution, soThe Cauchy-Schwarz inequality bounds the mean centered absolute deviation by the square root of this variance. Adding the projection bias givesChoose $j_n=\lfloor ({\log }_{2}n)/3\rfloor$, so $2^{j_n}\asymp n^{1/3}$. The two terms have the same order, andThis is a bias-variance tradeoff: refining the cells decreases the averaging bias but increases the sampling variance.

The Haar projection averages this normal distribution density on each cell; the finer cells follow its variation more closely. A Haar density estimator replaces these exact cell probabilities by empirical frequencies and therefore also introduces sampling noise.

The log-likelihood in the normal linear model, up to an additive constant, isFull column rank makes $X^{T}X$ positive-definite, so minimizing the residual sum of squares gives the normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Maximizing over ${\sigma }^2$ afterward gives the maximum-likelihood estimatorsThe residual sum of squares is positive almost surely because $n-p>0$ and ${\sigma }^2>0$, so this maximization has an interior variance solution almost surely. The first estimator is a linear Gaussian statistic with mean $\beta$ and covariance ${\sigma }^2(X^{T}X{)}^{-1}$, giving

Let $P_X=X(X^{T}X{)}^{-1}{X}^T$ be the orthogonal projection onto the design column space, and let $A=I-P_X$. This is a symmetric idempotent matrix of rank $n-p$. Choose an orthonormal basis for its range and collect its columns in $L$, so $A=L{L}^T$, $L^{T}L=I_{n-p}$ and $L^{T}X=0$. ThenThe definition of the chi-squared distribution therefore givesThe maximum-likelihood denominator is $n$; the unbiased residual-variance estimator uses $n-p$.

Finally put $B=(X^{T}X{)}^{-1}{X}^T$. It has rank $p$ and $BA=0$, because $X^T(I-P_X)=0$. Part (a) implies that $BY=\hat{\beta }$ is independent of $Y^{T}AY=n{\hat{\sigma }}^2$. Multiplication by a fixed constant preserves independence, so the normal linear model maximum-likelihood sampling distributions satisfy

Let $P$ be the orthogonal projection onto the code, and let $\{E_a\}$ be the specified error operators, including their linear span. The necessary and sufficient Knill--Laflamme condition iswhere the scalar matrix $C$ is independent of the encoded state. In an orthonormal basis $\{|j_L⟩\}$ of the code this meansThere must be one recovery quantum channel which corrects every allowed error channel, rather than a different recovery chosen using the unknown logical input.

The scalar condition means that any information carried away by the errors depends on their labels but not on the encoded amplitudes. To see how recovery works, diagonalize the positive semidefinite matrix $C$ and take corresponding linear combinations $F_{\ell }$ of the errors. Then $P{F}_{\ell }^{†}{F}_{m}P={\lambda }_{\ell }{\delta }_{\ell m}P$. For each positive ${\lambda }_{\ell }$, the map $F_{\ell }P/\sqrt{{\lambda }_{\ell }}$ is an linear isometry into an error-image subspace, and these subspaces are mutually orthogonal. Measuring their label and reversing the associated linear isometry recovers the logical state without measuring its amplitudes. Zero eigenvalues correspond to errors that annihilate the code. This also explains why the condition corrects arbitrary coherent combinations of the listed errors.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

Take the rank-one orthogonal projections $P_i=|{\varphi }_i⟩⟨{\varphi }_i|$. Their pinching map givesThis has exactly the specified diagonal entries and no off-diagonal entries. Applying the entropy increase under nonselective projective measurement proved in part (c) therefore givesWriting $q_i=⟨{\varphi }_i|\rho |{\varphi }_i⟩$, the output Von Neumann entropy is the Shannon entropy $S({\rho }_d)=-{\sum }_i{q}_i\log q_i$. Equality holds precisely when $\rho$ was already diagonal in this basis.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

Let $u:X\to U_{\mathcal{K}}$ be continuous. The family $\{u(x)-I:x\in X\}$ is norm compact in the compact-operator ideal. For any $\epsilon >0$ there is one finite-rank orthogonal projection $P$ withTo prove it, take a finite $\epsilon$-net of compact operators, approximate each by a finite-rank operator, and include both their ranges and adjoint ranges in $P\mathcal{H}$. Contractivity of compression transfers the estimate to the whole family.

Take $\epsilon <1$ and set $g(x)=I+P(u(x)-I)P$. It is invertible because it is within distance one of the unitary $u(x)$. It has the form $g_V(x)\oplus I$ on $V\oplus V^{\perp }$, where $V=P\mathcal{H}$ is finite-dimensional and $g_V:X\to GL(V)$ is continuous. The straight segment from $u$ to $g$ stays invertible; taking its unitary polar factors supplies a homotopy in $U_{\mathcal{K}}$, since inverse square root depends continuously on the positive invertible factors and preserves identity modulo compacts.

Conversely a map $g:X\to GL(V)$ defines such a class by the continuous deformation $g(g^{\ast }g{)}^{-t/2}$ to its unitary polar factor, then extension by the identity on $V^{\perp }$. Apply the same uniform finite-rank approximation to a homotopy on the compact space $X\times [0,1]$. Choosing $P$ to include the original endpoint supports keeps those endpoints unchanged under compression. This proves that two finite-dimensional maps give the same class exactly when, after adding identity blocks, they are homotopic through a common finite-dimensional general linear group. Hencewith stabilization $g\mapsto \mathrm{diag}(g,1)$.

Let $M$ be the linear subspace of $L^2(P)$ consisting of classes admitting $\mathcal{G}$-measurable representatives. It is closed. Indeed, if $Y_n\in M$ converges in $L^2$ to $Y$, choose a subsequence converging almost surely: take $\Vert Y_{n_j}-Y{\Vert }_2^2\le 2^{-3j}$ and apply Markov inequality and the Borel-Cantelli lemma to errors exceeding $2^{-j}$. The pointwise limit of its $\mathcal{G}$-measurable representatives, set to zero wherever the finite limit does not exist, is $\mathcal{G}$-measurable and equals $Y$ almost surely. This also works when $\mathcal{G}$ is not complete.

Here is a proof of the orthogonal projection construction using only completeness. Put $d={\inf }_{W\in M}\Vert X-W{\Vert }_2$ and take $W_n$ approaching this infimum. The parallelogram law givesCompleteness and closedness give a minimizer $Y\in M$. For every $Z\in M$, expand $\Vert X-Y-tZ{\Vert }_2^2$ for real $t$. Its minimum at zero impliesThis $Y$ is unique almost surely, because the difference of two possible choices belongs to $M$ and is orthogonal to itself. It is the conditional expectation $E(X\mid \mathcal{G})$, as also follows by testing indicators of events in $\mathcal{G}$.

For the jointly Gaussian pair, if $u>0$, setThe residual $R=X-Y$ has mean zero and covariance $\mathrm{Cov}(R,G)=v-(v/u)u=0$. Since $(R,G)$ is jointly Gaussian, its characteristic function factors when this covariance is zero, so $R$ and $G$ are independent. Hence $E(RZ)=E(R)E(Z)=0$ for every square-integrable function $Z$ of $G$; independence first proves this for bounded functions, and truncation plus Cauchy-Schwarz inequality gives the general case. This proves the Gaussian conditional expectation formula, including a zero residual variance. If $u=0$, $G=\mu$ almost surely and the covariance inequality gives $v=0$; $Y=\nu$ is the answer.

Here $F=1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-sigma-algebra is the orthogonal projection in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the probability measure.

The increasing coordinate sigma-algebras ${\mathcal{E}}_n$ generate the whole product sigma-algebra $\mathcal{F}$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal{F})$: their cylinder indicators generate $\mathcal{F}$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is ${\mathcal{E}}_N$-measurable, then for $n\ge N$ the $L^2$ projection contraction givesApproximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.