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Rearranging the outgoing radial null geodesics of the Vaidya metric equation along $r=r_0(v)$ givesFor the prescribed parametrization by $w$, the chain rule yieldsSince $n>1$, $v/w=1+2\beta w^{n-1}\to 1$ as $w\to 0^{+}$. The asymptotic equivalence $w\sim v$ therefore impliesMore precisely $w=v+O(v^n)$ and $M(v)=(\beta /2)v^n[1+O(v^{n-1})]$, using Big O notation. The positivity of $dv/dw$ justifies treating $w$ as a single-valued function of $v$ locally.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

Use the geometrical optics conventionBecause ${\partial }_t=\epsilon {\partial }_T$ and ${\partial }_x=\epsilon {\partial }_X$, differentiating $e^{i\theta /\epsilon }$ gives an order-one physical angular frequency and order-one physical wavenumbers. Absorb a constant leading amplitude phase into $\theta$ and take $A_0$ real; for a complex amplitude the corresponding quadratic density is $|A_0{|}^2$.

Substitution into the wave equation gives the eikonal equation for a variable-speed wave at leading order:Choose the positive-frequency branch $\omega =\Omega =cK$, $K=\sqrt{k^2+{\ell }^2}$; the negative branch can be treated in exactly the same way. The imaginary first-order terms giveMultiplying by $A_0$ puts this wave-action transport for a variable-speed wave in conservation form:Its density is $\omega A_0^2$ and its flux is that density times the group velocity. Independently, equality of mixed wave phase derivatives givesThis phase compatibility in geometrical optics is essential when converting the differentiated dispersion relation into equations along rays.

For completeness let $\omega =\Omega (\mathbf{k};\mathbf{X},T)$ be a general local dispersion relation. Define ray tracing by $d\mathbf{X}/dT={\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$. Using the chain rule in ${\nabla }_{\mathbf{X}}\omega$ and the symmetry ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ givesLikewise ${\omega }_T={\Omega }_T+{\mathbf{c}}_g\cdot {\mathbf{k}}_T$ cancels its advective term because ${\mathbf{k}}_T=-\nabla \omega$. Hence the Hamiltonian ray equations for a local dispersion relation areThe derivatives on the right hold the wavevector fixed. For this wave equation they reduce to

When $c=c(Y)$, the ray invariants are $k$ and $\omega$, while $d\ell /dT=-K{c}_Y<0$. For a monochromatic component with these fixed invariants write $\theta =kX-\omega T+S(Y)$, soUsing the stated independence of $A_0$ from $X,T$, the wave-action transport for a variable-speed wave becomes $(c^2\ell A_0^2{)}_Y=0$. ThereforeThis is the stationary wave amplitude in a stratified wavespeed; $\mu$ is the constant for that component or ray family.

At the proposed simple turning point of a variable-speed wave, take $c_s=c(Y_s)=\omega /k$ and $c_s^{\prime }=c_Y(Y_s)>0$. On the chosen positive branch this entails $k>0$. DefineAs the ray approaches from below, $\ell \sim [q(Y_s-Y){]}^{1/2}$ andThe amplitude diverges and the vertical group velocity $c^2\ell /\omega$ vanishes. The WKB approximation also fails its slow-variation condition $\epsilon |{\ell }_Y|/{\ell }^2\ll 1$ when $Y_s-Y=O({\epsilon }^{2/3})$. The turning-point enhancement of wave amplitude shows that this divergence is a failure of the outer ray approximation, rather than an actual infinite solution.

To derive the inner wavefield directly, separate the conserved horizontal wave phase and frequency in the original differential equation:Dividing by $c^2$ gives ${\epsilon }^2{\Psi }_{YY}+2{\epsilon }^2(c_Y/c){\Psi }_Y+Q(Y)\Psi =0$. Set $Y-Y_s=\delta \eta$. The second derivative is of size ${\epsilon }^2/{\delta }^2$, whereas $Q\sim -q\delta \eta$, so the Airy scaling at a variable-speed wave turning point isThe first-derivative term is smaller by $O(\delta )$, and higher coefficients in the expansion of $Q$ give the same relative order. Thus the general leading inner profile is a linear combination of the two Airy functions $\mathrm{Ai}(\eta )$ and $\mathrm{Bi}(\eta )$. For decaying Airy continuation and ray reflection, if the physical continuation is bounded and decays into $Y>Y_s$, its growing $\mathrm{Bi}$ coefficient is zero. In that usual evanescent continuation the wavefield has the formWithout a condition in the evanescent region the corresponding $\mathrm{Bi}$ term remains admissible locally; the given existence of a wavefield below the turning point alone does not select its coefficient.

Finally, the outer amplitude at distance $\delta$ is $O({\delta }^{-1/4})=O({\epsilon }^{-1/6})$ for $\mu =O(1)$ and fixed nonzero $q$. The oscillatory Airy turning-point connection formula has exactly the same $(-\eta {)}^{-1/4}$ envelope on the allowed side. Therefore the inner wavefield has magnitude $O({\epsilon }^{-1/6})$. The decaying Airy continuation matches an incident and reflected pair of rays, so a single upward $\ell >0$ branch must be supplemented by its reflected branch near the turning point; no formal coefficient matching is needed to identify this magnitude.

For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.

Assume $B$ is continuously differentiable. Apply the chain rule to each component with $z_j=t{x}_j$:Contracting the first identity with $x_j$ provesHere $\nabla$ differentiates with respect to $x$ at fixed $t$, while the time derivative holds $x$ fixed. The equality follows directly even at $t=0$; no division by $t$ is needed. It holds componentwise for the vector field, so it is not restricted to scalar functions.

The chain rule for Frechet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

Write $X(t,a)=X(t;0,a)$ and $A(t,x)=X(0;t,x)$. If $u_0\in C^1({\mathbb{R}}^d)$, the unique global classical solution isIt is $C^1$, and the chain rule makes its value constant along every characteristic curve. The same calculation proves classical uniqueness. No boundedness of $u_0$ is needed for this classical statement.

For $u_0\in L^{\infty }$, a bounded weak solution on the positive half-line is defined byfor every compactly supported $C^1$ test function on $[0,\infty )\times {\mathbb{R}}^d$. The analogous backward-time identity defines the solution on all real time. The $u\mathrm{div}F$ term is essential: this is transport, not the conservative equation $u_t+\mathrm{div}(Fu)=0$.

The same inverse-flow formula defines the unique weak solution, with $\Vert u(t){\Vert }_{\infty }=\Vert u_0{\Vert }_{\infty }$, equality almost everywhere, and a representative continuous into $L^1$ on each compact spatial set. For a direct uniqueness argument, change variables $x=X(t,a)$ in the weak identity and put $w(t,a)=u(t,X(t,a))$. Using $J_t=(\mathrm{div}F)(t,X)J$ givesinitially for $C^1$ compactly supported $\zeta$. Since $J>0$ and $J,J_t$ are continuous on compact sets, approximation in the spatial parameter permits $\zeta =\eta /J$: both $\zeta$ and ${\zeta }_t$ are uniformly approximable there, which are the only derivatives left in this identity. Hence $\int w{\eta }_t+\int u_0\eta (0)=0$ and $w=u_0$ almost everywhere. This also verifies existence by reversing the argument, and avoids assuming second derivatives of the $C^1$ field. Local $L^1$ continuity follows by approximating $u_0$ by continuous functions on compact flow tubes.

This is the weak transport solution under characteristic coordinates. The weak-strong uniqueness principle says that a weak solution in the specified class agrees with a classical solution having the same data for as long as that classical solution exists. Here it holds globally in the bounded class whenever the classical initial data are also bounded.

For a bounded $C^1$ classical solution and a smooth convex entropy $\eta$, let its entropy flux for a scalar conservation law satisfy ${\psi }^{\prime }=f^{\prime }{\eta }^{\prime }$. The chain rule givesIntegration against a nonnegative test function, including its initial boundary contribution, gives equality in the entropy inequality. Approximate a convex piecewise $C^1$ entropy by smooth convex functions on the bounded state range. Their values and flux antiderivatives converge uniformly, so the integrated equality passes to the limit. Therefore every bounded classical solution is an entropy solution. The bounded qualifier is required by the definition's $L^{\infty }$ membership; an arbitrary unbounded $C^1$ solution would not satisfy that membership condition.

Conversely, the affine entropies ${\eta }_{+}(z)=z$ and ${\eta }_{-}(z)=-z$ are both convex, with fluxes $f$ and $-f$. Their inequalities force the same linear weak integral to be both nonnegative and nonpositive. It is therefore zero for every nonnegative test, hence for every test by subtraction of nonnegative compactly supported tests. Thus $\boxed{\text{entropy solution}⟹\text{weak solution}.}$ No assumption that the flux itself is convex is used.

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

Write $D=(2\pi i{)}^{-1}d/dz=qd/dq$ and use the Serre derivative $D_{k}f=Df-(k/12)E_{2}f$. Translation invariance of $f$ and the Eisenstein series of weight two gives $D_{k}f(z+1)=D_{k}f(z)$.

Differentiate the weight-$k$ transformation $f(-1/z)=z^{k}f(z)$. The chain rule yieldsThe assumed transformation of the Eisenstein series of weight two givesSubtracting cancels the extra term. Hence $D_{k}f(-1/z)=z^{k+2}{D}_{k}f(z)$. Since $S$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ generate the modular group, these two transformations prove the weight-$k+2$ law for all its elements. Holomorphy on the complex upper half-plane follows from the formula.

For the Fourier expansion of a modular form $f={\sum }_{n\ge 0}{a}_n{q}^n$, differentiation gives $Df={\sum }_{n\ge 1}n{a}_n{q}^n$. The product of the convergent series for $E_2$ and $f$ likewise has no negative powers. Thus $D_{k}f$ is holomorphic at a cusp at infinity, and therefore at every cusp of the modular group. Its constant coefficient is $-k{a}_0/12$. Since $k>0$, it vanishes exactly when $a_0=0$. Consequently

For real smooth functions of compact support in a planar domain $D$, use the Dirichlet inner productChanging the positive normalization factor does not change orthogonality. For complex functions, insert complex conjugation in the second factor to obtain the corresponding Hermitian form.

Let $f:D\to \tilde{D}$ be a conformal bijection, and let $\tilde{u},\tilde{v}$ be smooth functions of compact support on $\tilde{D}$. Its real Jacobian matrix is $Df=|f^{\prime }|R$, where $R$ is a rotation. By the chain rule,The change of variables formula has Jacobian determinant $|f^{\prime }{|}^2$, so this factor cancels:This proves conformal invariance of the planar Dirichlet inner product. By completion it is also an isometry between the corresponding Dirichlet energy spaces. It asserts invariance of the energy form, not of the inhomogeneous Sobolev norm, whose $L^2$ term has a different transformation rule.

Let ${\psi }_A=g_A-g_A(0)$, and take another admissible compact H-hull $B$ in the image complex upper half-plane. Form the hull $C=A\cup {\psi }_A^{-1}(B)$, including its relative closure and filling bounded complementary components if needed. Its mapping-out function of a compact H-hull isIndeed, this composition maps the correct remaining domain onto $\mathbb{H}$, and its constant term at infinity is zero. Since ${\psi }_A(0)=0$, the chain rule givesBoth derivatives are positive by local Schwarz reflection principle and boundary orientation. Therefore the assumed avoidance formula givesThe conditioning event has positive probability because $g_A^{\prime }(0)>0$.

It remains to justify that these equalities determine the law. For a proper simple curve from $0$ to infinity, its complement has two sides, each connected to an interval of the real boundary. Any point off the curve lies in a thin polygonal tube attached to that boundary which avoids both the curve and zero; its closure, with an end cap, is an admissible compact H-hull. A countable collection of tubes with rational geometry therefore reconstructs the complement as the union of the tubes avoided by the curve.

The joint distribution of their avoidance indicators is also determined. Avoiding finitely many tubes is equivalent to avoiding the filled union when that union is admissible. If the union separates $0$ from infinity, avoidance is impossible and the joint probability is zero. Otherwise a simple curve cannot enter the bounded complementary pockets without crossing the union, so filling does not change the avoidance event. Finite joint avoidance probabilities consequently come from the same single-hull formulas; inclusion-exclusion principle supplies every finite indicator pattern. These patterns determine the random complement and hence the unparameterized simple trace.

Thus avoidance probabilities determine a simple chordal curve law, and the conditional image law above equals the original law. This proves the derivative avoidance criterion for chordal restriction, giving the chordal restriction property. Simplicity is essential to this reconstruction argument; it should not be replaced by a claim that arbitrary random closed sets are determined by these hull-avoidance tests.

Parameterize the initial product normal distribution by $\eta =(m,\rho )$, usingRun the given recursion with independent random variables with uniform distributions to obtain $X_{\eta }=F_t(X_{\eta }^{(0)},Z_{1:t})$. Because the noise laws do not depend on $\eta$, the chain rule and automatic differentiation give reparameterization gradients through the entire simulated trajectory. The variational objective is $J(\eta )=\mathrm{KL}(q_{\eta }\Vert \pi )$, where $q_{\eta }=K^t{\nu }_{\eta }$ and $\pi =\mu (\cdot \mid y)$.

One must account for the output differential entropy, not just optimize $\mathbb{E}[\log \pi (X_{\eta })]$. Simulation and differentiability of $f$ do not themselves supply an evaluable $q_{\eta }$. If $K$ retains the reversible Markov chain assumption from part (b), a concrete algorithm avoids a transition-density oracle using reversible density-ratio propagation. Write $\pi =h/Z_{\pi }$, with $h$ an evaluable unnormalized posterior density, and defineDetailed balance impliesFor example, integrate ${\nu }_{\eta }(b)K^t(b,dx)$ and replace $h(b)K^t(b,dx)$ by $h(x)K^t(x,db)$ to obtain the first identity. This also proves absolute continuity of measures of $K^t{\nu }_{\eta }$ with respect to $\pi$ when ${\nu }_{\eta }\ll \pi$. An everywhere finite differentiable log-posterior has positive density, so an ordinary product normal distribution has this domination.

For each outer sample $X_{\eta }$, independently run $M$ trajectories of length $t$ starting at $X_{\eta }$, with endpoints $B_{\eta ,j}=F_t(X_{\eta },Z_{1:t}^{(j)})$. These are reverse trajectories because of detailed balance; the same recursion simulates them. Estimate $A_{\eta }(X_{\eta })$ byAverage $\log \hat{A}$ over an outer batch. Compute it stably by a logarithmic sum of exponentials. Differentiate through both outer and reverse trajectories and through the explicit product-normal probability density function ${\nu }_{\eta }$. Only the given state gradients of $f$ and $\log h$, together with elementary normal-density derivatives, are needed. Apply stochastic gradient descent to $m$ and $\rho$, refreshing the independent noise batches. The logarithmic scale parameters keep all variances positive.

This is nested Monte Carlo optimization of the exact objective in the limit of adequately increasing inner and outer sample sizes. For fixed $M$, $\log \hat{A}$ is generally biased, even though $\hat{A}$ is an unbiased estimator. Finite logarithmic moments, suitable uniform integrability, and justified differentiation under the integral sign are needed for consistency and gradient interchange. Under those conditions increasing $M$ and controlling the optimization error yields a justified numerical approximation; nonconvexity prevents a generic guarantee of a global optimum.

There is a real assumption issue: part (c) redefines $K$ without explicitly restating reversibility. The algorithm above uses the reversible interpretation inherited from part (b). For an arbitrary differentiable recursion, one additionally needs access to output densities or a simulatable reverse kernel and regularity of the objective. Differentiability alone is insufficient: $f(x,z)=0$ is differentiable and, with a standard-normal target, makes $K^t\nu$ a point mass for $t\ge 1$, so every candidate has infinite Kullback-Leibler divergence. Thus a universal finite-objective algorithm cannot be justified solely from the literal differentiability assumptions. An invertible recursion with computable density transformations offers a different sufficient route, but invertibility is not stated here.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

In the smooth-radiation, subhorizon approximation of part (c), the cosmological Poisson equation gives