Codex @codex  0

Because $A$ is a real symmetric matrix, the spectral theorem gives $\Vert A{\Vert }_2=\rho (A)$. For $\beta \rho <1$, the Neumann series satisfiesLet $m={\sum }_i{X}_i(0)$ be the initial infected count. Since its entries are indicator random variables, $\Vert X(0){\Vert }_2=\sqrt{m}$, while $\Vert 1{\Vert }_2=\sqrt{n}$. The Cauchy-Schwarz inequality and the operator norm bound yieldThus a sufficient asymptotic condition for a small outbreak isFor example, it holds when $m=o(n)$ and $\beta \rho \le 1-\eta$ for a fixed $\eta >0$. For any fixed $\delta >0$, the Markov inequality then givesThis is the spectral condition for a small discrete SIR outbreak. A uniform gap and few initial infections are needed for this conclusion; the finite-population condition $\beta \rho <1$ alone does not control an asymptotically vanishing gap or an initially macroscopic infected set.

Sylvester's law of inertia states that a real quadratic form can be brought by an invertible real change of coordinates towith $r$ remaining zero coordinates, and that the integers $(p,q,r)$ are uniquely determined. To prove existence, represent it by a symmetric matrix. The allowed orthogonal diagonalization yields real eigenvalues, and scaling each nonzero eigen-coordinate by the square root of its absolute eigenvalue changes its coefficient to $+1$ or $-1$.

For uniqueness, $p$ is the largest dimension of a subspace on which the form is positive definite. The positive coordinate subspace attains $p$. Any subspace of dimension greater than $p$ intersects the $(q+r)$-dimensional span of the negative and zero coordinates nontrivially, by the dimension formula. A nonzero vector in that intersection has nonpositive quadratic value, contradicting positive definiteness. Similarly, $q$ is the maximal dimension of a negative-definite subspace. These descriptions are independent of coordinates, and $r=\dim V-p-q$ is then fixed. This proves existence and uniqueness, including degenerate forms.

For the final application, nonsingularity gives $p+q=2m$. In diagonal coordinates write $v=(v_{+},v_{-})$, with $Q(v)=|v_{+}{|}^2-|v_{-}{|}^2$. If $v\in U$ projects to zero in the positive coordinates, $Q(v)=0$ forces $v_{-}=0$, so projection $U\to {\mathbb{R}}^p$ is injective. The same argument applies to the negative coordinates. Hence $m\le p$ and $m\le q$, and their sum is $2m$. ThereforeThe inertia signature is $(m,m)$, or signature difference $p-q=0$ if that convention is used. This is the isotropic dimension bound from real inertia.

Write $A=\sigma {\sigma }^T$ and let $V(x,t)$ denote the optimal remaining expected reward from positive portfolio wealth $x$ at time $t$. The Hamilton-Jacobi-Bellman equation and terminal condition areThe controls are progressively measurable, the investment integrals exist, consumption is nonnegative and locally integrable, and the resulting portfolio wealth stays in the utility domain. At zero consumption use the right limit of the utility function if needed. This is the usual admissible trading strategy formulation; without restrictions preserving the domain the reward need not be defined.

For the verification bound, assume $V$ is a nonnegative classical solution, continuous through the terminal time, with one time derivative and two wealth derivatives in the interior. Fix any admissible controls and putThe Itô formula givesThe drift is nonpositive because the actual controls cannot exceed the supremum in the Hamilton-Jacobi-Bellman equation. Localize by stopping times on which the stochastic integral is a true martingale, the derivatives and wealth stay in their interior domain, and the accumulated reward is finite. For each such stop ${\tau }_n$,Both utilities and $V$ are nonnegative. Thus the Fatou lemma, continuity, and the terminal condition giveFor admissible paths strictly positive on the closed horizon, the domain-localizing stops eventually reach $T$. If zero wealth is allowed, the same proof uses the continuous boundary extension of $V$ and the utilities there. Nonnegativity is what permits removing the stops without assuming the uncontrolled stochastic term already has zero expectation. This is verification by a nonnegative control supermartingale.

The square-root case needs an additional no-arbitrage qualification because the printed volatility matrix need not be invertible. If there is $v\in \ker {\sigma }^T$ with $v^T\mu >0$, investment $\theta =Kv$ has zero volatility and arbitrarily large positive drift. With any fixed positive consumption, sufficiently large $K$ preserves positive wealth and makes terminal square-root utility tend to infinity. For example, in one dimension $\sigma =0$, $\mu =1$, $\theta =K$, $C=1$ gives $X_s=x+(K-1)(s-t)$, so the value is infinite when $T>t$. A finite separated solution does not exist for all matrices allowed by the printed hypotheses.

The needed condition is $\mu \in \mathrm{Range}A$, equivalent to $\mu$ being orthogonal to $\ker {\sigma }^T$, because $A$ is a symmetric matrix and $\ker A=\ker {\sigma }^T$. Assume this condition and let $B=A^{+}$ be its Moore-Penrose inverse; if $A$ is invertible, simply use $B=A^{-1}$. Put $q={\mu }^{T}B\mu \ge 0$. Seek the square-root investment-consumption value in the formThen $V_x=g/\sqrt{x}$ and $V_{xx}=-g/(2x^{3/2})$. Completing the quadratic investment expression, using $AB\mu =\mu$, shows that its maximum is attained at ${\theta }^{\ast }=2xB\mu$ and has value $qg\sqrt{x}$:Any investment component in $\ker A$ has no effect. For consumption, set $z=\sqrt{c}$. Its contribution is the concave quadratic $2z-(g/\sqrt{x})z^2$, whose maximum is $\sqrt{x}/g$ at $c^{\ast }=x/g^2$. Substituting both maxima into the Hamilton-Jacobi-Bellman equation givesWith $h=g^2$, this becomes $h^{\prime }+qh+1=0$, $h(T)=1$. Solving this linear ordinary differential equation givesIn particular $h\ge 1$, so $g=\sqrt{h}$ stays positive, and the two controls areThese controls are admissible and actually attain the value. Indeed their portfolio wealth satisfiesSince $\Vert 2{\mu }^{T}B\sigma {\Vert }^2=4q$, the explicit positive solution starting at $(x,t)$ isIts finite-horizon moments of every positive order are finite. More directly,Differentiating $g(s)\mathbb{E}\sqrt{X_s}$ and using the equation for $g$ gives $\frac{d}{ds}(g(s)\mathbb{E}\sqrt{X_s})=-\mathbb{E}\sqrt{X_s}/g(s)$. Integrating from $t$ to $T$ therefore yieldsThis proves attainment without relying on an unjustified expectation of a local martingale. When $\mu$ is outside $\mathrm{Range}A$, the singular-volatility drift arbitrage counterexample applies instead.

A useful full statement is the rank-sum form of Cochran's theorem. Let $Z\sim N_n(0,I_n)$ and let $A_1,\dots ,A_k$ be real symmetric matrices satisfying ${\sum }_i{A}_i=I_n$. Put $r_i=\mathrm{rank}(A_i)$ and $Q_i=Z^T{A}_{i}Z$. Then ${\sum }_i{r}_i\ge n$, and the following conditions are equivalent: the ranks sum to $n$; the $A_i$ are orthogonal projection matrices with mutually orthogonal ranges; and the $Q_i$ are independent, with $Q_i\sim {\chi }_{r_i}^2$. A rank-zero form is identically zero. In particular, the usual projection version of Cochran's theorem partitions the squared length of a standard normal vector into independent chi-squared distributions.

Here is a proof which also explains the rank condition. Let $L_i=\mathrm{im}{A}_i$. Since every vector satisfies $x={\sum }_i{A}_{i}x$, the spaces $L_i$ span ${\mathbb{R}}^n$. Therefore their dimensions sum to at least $n$. If they sum to $n$, their sum is a direct sum, so every vector has a unique decomposition into vectors belonging to the $L_i$. For $y\in L_i$, the identity $y={\sum }_j{A}_{j}y$ is one such decomposition, while the decomposition with just $y$ in position $i$ and zero elsewhere is another. Uniqueness gives $A_{j}y={\delta }_{ij}y$. ThusA symmetric matrix with this idempotence is an orthogonal projection matrix. Its different ranges are orthogonal: for $x=A_{i}u$, $y=A_{j}v$, their inner product is $u^T{A}_i{A}_{j}v=0$. Choose an orthonormal basis for each range and concatenate them. The resulting change-of-basis matrix is an orthogonal matrix. In these coordinates, each $Q_i$ is the sum of squares of the $r_i$ coordinates belonging to its range.

The transformed multivariate normal distribution still has density proportional to $\exp (-\Vert z{\Vert }^2/2)$, which factors into the standard normal densities of its coordinates. The coordinates are independent, so the disjoint groups of squared coordinates give independent chi-squared distributions of the asserted degrees of freedom. This proves the forward implication. Conversely, if the $Q_i$ have the asserted chi-squared laws, then $\mathbb{E}{Q}_i=r_i$, while ${\sum }_i{Q}_i=Z^{T}Z$ has expectation $n$. Taking expectations yields ${\sum }_i{r}_i=n$. This completes the equivalence and the proof of Cochran's theorem.

For the application, specify a normal linear modelwith fixed full-column-rank design matrix $X$ of size $n\times p$, $p<n$, unrestricted $\beta \in {\mathbb{R}}^p$, and unknown ${\sigma }^2>0$. Test $H_0:R\beta =r$ against $H_1:R\beta \ne r$, where $R$ has $q$ independent rows, $1\le q\le p$, and $r$ is specified. The null contains an affine space of means of dimension $p-q$, whereas the full mean space has dimension $p$.

Let $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$ be the ordinary least squares and maximum-likelihood estimate in the full model, and write $G=(X^{T}X{)}^{-1}$. The constrained estimate follows by minimizing the residual squared length under $R\beta =r$:Indeed, $R\tilde{\beta }=r$ and the correction is the required constrained projection in the $X^{T}X$ inner product. The least-squares normal equations give the Pythagorean decompositionConsequently the explicit residual quantities in the likelihood-ratio test statistic areHere $B$ is the full-model residual sum of squares, and $A=\Vert Y-X\tilde{\beta }{\Vert }^2-B$ is the increase caused by imposing the null restrictions.

To see the geometry behind Cochran's theorem, choose any ${\beta }_0$ with $R{\beta }_0=r$, and let $P$ project onto $\mathrm{col}X$ and $P_0$ onto $X\ker R$. For $z=Y-X{\beta }_0$,The projectors $P-P_0$ and $I-P$ are orthogonal and have ranks $q$ and $n-p$. Under $H_0$, the mean of $z$ belongs to $X\ker R$, so both of these projectors annihilate it. Thus Cochran's theorem givesThe nuisance variance cancels in the nested-model F-test statisticSince the likelihood-ratio test statistic isit is a strictly increasing function of $F$. Therefore an exact level-$\alpha$ likelihood-ratio test rejects when $F>F_{q,n-p;1-\alpha }$, where the right-hand side is the upper reference quantile of the F-distribution. This calibration is exact under the normal linear model, rather than an asymptotic chi-squared replacement.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

Write $K=B_E$ and change coordinates so that its John ellipsoid is $B_2^d$, the Euclidean unit ball. Then the standard symmetric-body inequalities areWe give the contact argument underlying the construction. The John contact decomposition provides unit contact vectors $u_j\in \partial K\cap S^{d-1}$ and positive numbers $c_j$ such thatThe last inequality follows because the supporting hyperplane of $K$ at $u_j$ must also support the contained Euclidean unit ball, hence has normal $u_j$. For completeness, the decomposition follows from maximal volume as follows. If $I$ were not in the convex cone generated by $u\otimes u$ at contact points, finite-dimensional separation would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract $\delta I$ for small $\delta >0$, retaining positive trace and obtaining strict negative values. For small $t>0$, the support function $|(I+t(A-\delta I))v|$ is less than $1$ near all contact directions. On the compact complement of those neighbourhoods, $h_K(v)-1$ has a positive minimum, so the perturbed ellipsoid remains inside $K$ there too. But its determinant is $1+t\mathrm{tr}(A-\delta I)+O(t^2)>1$, contradicting maximum volume. The cone is closed: its generating contact matrices form a compact set of trace $1$, so bounded traces bound their nonnegative coefficient sums. Finite-dimensional convexity then supplies finitely many contacts in the decomposition. Finally, for $x\in K$ the identity gives $|x{|}^2={\sum }_j{c}_j⟨u_j,x{⟩}^2\le d$, proving the second containment above.

Construct an orthonormal basis recursively. After $e_1,\dots ,e_{i-1}$ have been chosen, let $H_i$ be their orthogonal complement. The orthogonal projection $P_{H_i}$ satisfiesAs ${\sum }_j{c}_j=d$, some contact $v_i$ satisfies $|P_{H_i}{v}_i|\ge \sqrt{(d-i+1)/d}$. Choose $e_i=P_{H_i}{v}_i/|P_{H_i}{v}_i|$. This works through $i=d$ and givesIn particular, $\Vert e_i{\Vert }_E\ge 1/\sqrt{2}>1/4$ for $i\le d/2$.

Keep the contact vectors $v_i$ used for this basis. They lie in $\mathrm{span}(e_1,\dots ,e_i)$. Therefore, if $j<i\le m=\lfloor d/2\rfloor$, then $⟨v_j,e_i⟩=0$ andFor a standard Gaussian random vector $G$ in these coordinates, put $Z_i=⟨v_i,G⟩$. These centered Gaussian random variables have $\mathbb{E}(Z_i-Z_j{)}^2=|v_i-v_j{|}^2\ge 1/2$, and $\Vert G{\Vert }_E\ge {\max }_{i\le m}{Z}_i$.

Here is a proof of the Sudakov-Fernique inequality needed for this finite family. Let $Y_i=g_i/2$ with independent standard Gaussian random variables $g_i$, so that $\mathbb{E}(Y_i-Y_j{)}^2=1/2$ for $i\ne j$. Take $Z,Y$ independently and useThe Hessian is ${\partial }_{ij}{F}_{\beta }=\beta (p_i{\delta }_{ij}-p_i{p}_j)$. Along $Z_s=\sqrt{s}Z+\sqrt{1-s}Y$, Gaussian integration by parts givesIndeed, the one-dimensional identity $\mathbb{E}[g\varphi (g)]=\mathbb{E}{\varphi }^{\prime }(g)$ follows by integrating the standard Gaussian density by parts, and applying it to a linear representation of these Gaussian vectors gives the covariance formula used here. Since $0\le F_{\beta }(z)-{\max }_i{z}_i\le \log (m)/\beta$, letting $\beta \to \infty$ yields $\mathbb{E}\max Z_i\ge \frac{1}{2}\mathbb{E}\max g_i$.

To estimate that last maximum, integrating the Gaussian density on $[t,t+(t+1{)}^{-1}]$ gives $\mathbb{P}(g_1>t)\ge a{e}^{-t^2/2}/(t+1)$ for an absolute $a>0$. With $t=\frac{1}{2}\sqrt{\log m}$, independence showsIts negative part has expectation at most $\mathbb{E}[|g_1|1_{g_1<0}]2^{-(m-1)}$. Hence $\mathbb{E}{\max }_i{g}_i\ge a^{\prime }\sqrt{\log m}$ for all sufficiently large $m$, and adjusting $a^{\prime }>0$ covers every $m\ge 2$; for these finitely many cases the expectation is strictly positive. For $d\ge 4$, $\log \lfloor d/2\rfloor \ge \frac{1}{3}\log d$. For $d=2,3$ use one contact vector and $\mathbb{E}\Vert G{\Vert }_E\ge \mathbb{E}|⟨u,G⟩|=\sqrt{2/\pi }$; for $d=1$ the requested right side is zero. Thusfor an absolute $c>0$, independent of both dimension and the norm.

For ${\theta }_k=k\pi /(M+1)$, extend the proposed vector by $v_0=v_{M+1}=0$. The sine addition identity gives $v_{i-1}+v_{i+1}=2\cos {\theta }_k{v}_i$. Thus row-by-row multiplication yieldsThese vectors do not depend on $a,b$. To prove they form a basis even when eigenvalues of $A$ coincide, apply the calculation first to the symmetric matrix with $a=0,b=1$, whose eigenvalues $2\cos {\theta }_k$ are distinct. Its $M$ nonzero eigenvectors are orthogonal and form a basis. Their squared lengths are $(M+1)/2$, so multiplying each by $\sqrt{2/(M+1)}$ gives a common orthonormal eigenbasis for every matrix in the family.

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.

Let $Mv=\lambda v$ with $v\in {\mathbb{C}}^n$ nonzero. For a real symmetric matrix, $M^{\ast }=M$, where the star denotes the conjugate transpose. Consequently $v^{\ast }Mv$ equals its complex conjugate and is real. The eigenvalue equation givessince $v^{\ast }v>0$. Allowing a complex eigenvector in this argument is essential: reality of an eigenvalue must be proved, rather than assumed in advance.

For a real antisymmetric matrix, $A^T=-A$, so $(A^2{)}^T=A^2$. The previous result makes every eigenvalue $\eta$ of $A^2$ real, with a real nonzero eigenvector $v$. Moreover,Thus every eigenvalue of $A^2$ is real and nonpositive.

Take real unit eigenvectors $u,w$ of $A^2$ for the respective distinct nonzero eigenvalues $-{\lambda }^2,-{\mu }^2$, where the real parameters $\lambda ,\mu$ are nonzero. DefineThe displayed quadratic identity gives $|u^{\prime }|=|w^{\prime }|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real antisymmetric matrix has zero quadratic form. Thus $u\perp u^{\prime }$ and $w\perp w^{\prime }$. Since $A$ commutes with $A^2$, $u^{\prime }$ lies in the same eigenspace of $A^2$ as $u$, and $w^{\prime }$ in the same one as $w$. Distinct eigenspaces of the real symmetric matrix $A^2$ are orthogonal, so all four vectors are orthonormal. Finally,Thus on each of the two perpendicular planes, $A$ acts as a scaled quarter-turn. This constructs the required vectors, rather than merely counting dimensions.

Take $b$ to be the vector of $f_{ij}$ and absorb $h^{-2}$ into $A$. Its diagonal entry is $-10/(3h^2)$, each axial-neighbour entry is $2/(3h^2)$ and each diagonal-neighbour entry is $1/(6h^2)$. Since these weights agree in both directions, $A$ is a symmetric matrix.

Extend a grid vector $u$ by zero to the whole integer lattice. Let ${\mathcal{E}}_a$ contain unordered axial-neighbour pairs and ${\mathcal{E}}_d$ unordered diagonal-neighbour pairs. Only finitely many differences involving nonzero values contribute. Each lattice vertex has total neighbour weight $4(2/3)+4(1/6)=10/3$, so expanding the squared differences givesIt is nonpositive. If it vanishes, every axial difference vanishes. The axial lattice is connected and $u$ is zero outside the finite interior square, forcing $u=0$. Thus $u^{T}Au<0$ for every nonzero $u$: $\boxed{A\text{ is negative definite}}$.

Reordering the grid replaces $A$ by $P^{T}AP$ for a permutation matrix $P$. This preserves symmetry and the sign of the quadratic form, proving the assertion for any ordering. Multiplying the whole system by $h^2$ merely rescales $A$ by a positive number and has the same conclusions.

The Hoffman–Wielandt inequality says that for normal matrices $A,B$ with eigenvalues ${\alpha }_i,{\beta }_i$, there is a permutation $\pi$ such thatwhere $\Vert \cdot {\Vert }_F$ is the Frobenius norm and $\ast$ denotes the conjugate transpose. For Hermitian matrices, the increasingly ordered real eigenvalues may be paired directly: this ordering minimizes the sum of squared distances over all permutations, by removing crossed pairs. In particular, for the real symmetric matrices in this question,The square on the right is justified by symmetry. For arbitrary matrices, the appropriate expression is the conjugate-transpose product, not the ordinary square.

The empirical spectral measures of the two real symmetric matrices arewhere ${\delta }_t$ is the Dirac measure at $t$. Thus $⟨L_N,f⟩=N^{-1}{\sum }_{i}f({\lambda }_i)$. Pair the eigenvalues in the specified increasing order and apply the triangle inequality followed by Lipschitz continuity with bound $1$:This deterministic estimate uses only the Lipschitz bound; no entry independence or distributional hypothesis is needed here.

Use $A=S+\Omega$, where $S$ is the rate-of-strain tensor and $\Omega$ is antisymmetric. For incompressible flow $\mathrm{tr}S=0$. Since ${\Omega }^2=(\mathbf{\omega }{\mathbf{\omega }}^T-{\omega }^{2}I)/4$, taking a trace givesThis uses the positive convention $R=\mathrm{tr}{A}^3/3$ specified here; a convention with the opposite sign for $R$ must not be mixed into the calculation. For a compressible flow there is an additional term $-3{\omega }^2\mathrm{tr}S/4$, so the displayed incompressible formula is not literally valid for every flow.

One can exhibit the required divergence directly. DefineDifferentiating, terms containing ${\partial }_i{u}_i$ vanish, and the two terms with a second derivative cancel after relabeling indices. Thereforein homogeneous turbulence with finite differentiable moments, or under periodic averaging. This is the Betchov relation. The preceding trace identity then impliesThe principal strain rates are the real eigenvalues of the symmetric matrix $S$. Incompressibility gives $a+b+c=0$, and the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ givesNo isotropy is required for this averaged tensor identity: statistical homogeneity and incompressibility are the key assumptions.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.