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For a probability measure-preserving system, if $⟨U_T^{n}f,f⟩\to |\int fd\mu {|}^2$, then $U_T^{n}f$ converges weakly in $L^2$ to $(\int fd\mu )1$. Centre $v=f-(\int f)1$; its autocorrelations tend to zero. The isometry identity gives the same decay against each $U_T^{k}v$, hence by approximation against their closed forward span. Against its orthogonal complement the correlations are exactly zero. Orthogonal decomposition by a closed subspace finishes the proof without invertibility.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

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.

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.

For a counterexample to existence, take $K:{\ell }^2\to {\ell }^2$, $(Ku{)}_j=u_j/j$, and $f_j=1/j$. The range is dense, so the least residual infimum is zero; finite truncations realize residuals tending to zero. But an exact preimage would have every coordinate equal to one and would not belong to ${\ell }^2$. Thus no least-squares minimizer exists for this datum.

The fixed-point space $I=\ker (U-I)$ is closed. If $y\in I$ and $w=Ux-x\in W$, then unitarity givesso $I\perp W$.

Conversely, $z\in W^{\perp }$ exactly whenfor every $x\in H$. Thus $U^{\ast }z=z$, equivalently $Uz=z$, and $z\in I$. Hence $W^{\perp }=I$, so the double orthogonal complement identity givesThe orthogonal decomposition by a closed subspace now yields

Define the Cesaro averageFor $y\in I$, $A_{n}y=y$. For a generator $w=Uz-z$ of $W$, the sum telescopes:Moreover $\Vert A_n\Vert \le 1$, so approximation extends this convergence from $W$ to $\overline{W}$. Writing $x=Px+(I-P)x$ according to the orthogonal decomposition gives the Von Neumann mean ergodic theorem

Use the shape-rate convention for the Gamma distribution. DefineThe zero null space assumption makes $B$ a positive-definite matrix, and $u_{\ast }$ is the unique ordinary least squares solution. The Gaussian likelihood contributes ${\gamma }^{k/2}$, which must be retained because $\gamma$ is unknown. Multiplying it by the improper prior and the Gamma distribution gives the normal-gamma posterior with a flat prior:Here the constant density of the improper prior cancels on normalization. PutThe normal equation $A^{T}r=0$ makes the residual orthogonal to the range of $A$. The orthogonal decomposition by a closed subspace into that range and its orthogonal complement givesIntegration over $u$ by the Gaussian integral, followed by integration over $\gamma$ using the Gamma distribution, yieldsThus the posterior distribution is proper, despite the improper prior; here $k\ge d$ follows from the zero null space assumption.

Completing the square in $u$ and collecting the powers of $\gamma$ give the two conditional distributions:Explicitly their probability density functions areFor a joint maximum a posteriori estimate relative to $dud\gamma$, first fix $\gamma >0$. The unique maximizing $u$ is $u_{\ast }$. The remaining log density is $(a-1)\log \gamma -b_0\gamma$ plus a constant. Consequently, if $a>1$,The stated assumptions do not always imply $a>1$. If $a=1$, the supremum occurs as $\gamma \downarrow 0$ but is not attained on $\gamma >0$; if $a<1$, the density is unbounded there. In either case there is no joint maximum a posteriori estimate in the stated parameter space. Calling zero a boundary mode does not make it an admissible positive precision.

If “MAP estimators” instead means separate modes of the marginal distributions, integration givesThus the marginal maximum a posteriori estimate of $u$ is always $u_{\ast }$, and that of $\gamma$ is $(s-1)/b_0$ when $s>1$, with the same nonattainment at zero when $s\le 1$. Joint and marginal maximization are different operations; the usual joint interpretation gives the preceding boxed pair.

The real Riesz representation theorem states that for every bounded linear functional $f:H\to \mathbb{R}$ on a real Hilbert space there is a unique $y\in H$ such thatand $\Vert f\Vert =\Vert y\Vert$.

If $f=0$, take $y=0$. Otherwise $Y=\ker f$ is a closed proper subspace, so the supplied orthogonal decomposition by a closed subspace gives a nonzero $z\in Y^{\perp }$. Since $f(z)\ne 0$, for every $x\in H$,Taking its inner product with $z$ yieldsand thereforeThus $y=f(z)z/\Vert z{\Vert }^2$ represents $f$. If both $y$ and $y^{\prime }$ represent it, then $⟨x,y-y^{\prime }⟩=0$ for every $x$; taking $x=y-y^{\prime }$ proves uniqueness. Finally, Cauchy-Schwarz inequality gives $\Vert f\Vert \le \Vert y\Vert$, while evaluating at $x=y/\Vert y\Vert$ when $y\ne 0$ gives the reverse inequality. Hence

An orthonormal eigenbasis isThe starting vector has the orthogonal decompositionHence its component in the dominant eigenspace is exactly zero. Normalizing $A^k{\mathbf{x}}_0$ givesSince the two remaining eigenvectors are orthogonal,ThereforeThe limit is not the leading eigenvalue $9$ because the nonorthogonality assumption in part (a) fails. On the active spectrum of the power method, the leading eigenvalue is $6$ and the next is $2$; indeedin agreement with the squared spectral-ratio estimate.

For an open subdomain $U\subset D$, identify $H_0^1(U)$ with the closed subspace of $H_0^1(D)$ obtained by zero extension. Its orthogonal complement isIf $f\in {\mathcal{H}}_U$, then testing against $C_0^{\infty }(U)$ and integrating by parts gives $\Delta f=0$ in $U$ in the distributional derivative sense. The Weyl lemma therefore gives a representative harmonic on $U$.

The orthogonal decomposition by a closed subspace givesProject the isonormal process defining $h$ onto these two orthogonal subspaces. The projections are jointly Gaussian and uncorrelated, hence independent. The first projection is a zero-boundary Gaussian free field $h_U$ on $U$; the second is a random distribution $h^{\mathrm{har}}$ that is harmonic on $U$. Thuswith independent summands. This proves the Domain Markov property of the Gaussian free field.

For positive-semidefinite kernels $k_j$ with Reproducing-kernel Hilbert spaces ${\mathcal{H}}_j$, the Reproducing-kernel Hilbert space of $k={\sum }_j{k}_j$ consists of sums $f={\sum }_j{f}_j$, withThe minimum is attained uniquely. Indeed, in the Hilbert space direct sum $E={⨁}_j{\mathcal{H}}_j$, the tuples summing to zero form a closed subspace $N$: they are the intersection, over inputs $x$, of the kernels of the continuous maps $(f_j)\mapsto {\sum }_j{f}_j(x)$. Every affine fibre has a unique representative in $N^{\perp }$ by orthogonal decomposition by a closed subspace. The vector $(k_j(\cdot ,x){)}_j$ lies in $N^{\perp }$ and represents evaluation in the quotient; its evaluation at $y$ is ${\sum }_j{k}_j(y,x)$, identifying the positive-semidefinite kernel with $k$.