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In congestion control, resource $j$ observes only its own aggregate load $y_j={\sum }_{r:j\in r}{x}_r$ and generates the nonnegative feedback price ${\mu }_j=p_j(y_j)$. User $r$ receives the sum $M_r={\sum }_{j\in r}{\mu }_j$ of prices along its route. Its desired increase $w_r$ competes with the congestion charge $x_r{M}_r$, while ${\kappa }_r$ sets the adjustment speed. This is local information: a resource needs the rates passing through it, and a user needs the feedback along its own route, rather than the complete network state.

There is a hypothesis to make explicit before claiming global convergence. If “increasing” means nondecreasing, the printed assumptions allow $p_j\equiv 0$, in which case ${\dot{x}}_r={\kappa }_r{w}_r>0$ and there is no equilibrium point. The intended theorem holds when every route contains at least one resource whose price is not identically zero, with nonempty routes and nonnegative, continuous, nondecreasing prices. Strictly increasing prices satisfy this condition automatically. We prove the theorem under this necessary qualification; without it the zero-price example disproves the literal claim.

Let $P_j(y)={\int }_0^y{p}_j(u)du$. Each $P_j$ is a nonnegative convex function. For positive rates define the primal congestion potentialThe weighted logarithmic utility is a strictly concave function, and subtracting the resource costs preserves strict concavity. Thus $U$ has at most one maximizer and at most one critical point on the positive orthant.

It also attains its maximum there. A nonzero nondecreasing price is bounded below by some positive constant at all sufficiently large loads; hence its primitive satisfies $P_j(y)\ge c_{j}y-K_j$ with $c_j>0$. Choose $c_j=0$ at zero-price resources. The route coverage assumption gives $c={\min }_r{\sum }_{j\in r}{c}_j>0$, soThis bounds all superlevel sets of $U$. Within a bounded set, approaching $x_r=0$ makes $U\to -\infty$. Consequently every nonempty superlevel set is a compact set contained in the positive orthant. A maximizing sequence stays in one such set, giving a unique interior maximizer $x^{\ast }$. Its critical-point equations areThe rate dynamics are the gradient flow with diagonal mobility ${\dot{x}}_r={\kappa }_r{x}_r{g}_r(x)$. Along any positive trajectory,with equality only at $x^{\ast }$. Thus $V(x)=U(x^{\ast })-U(x)$ is a strict Lyapunov function. The initial superlevel set traps the trajectory away from both infinity and the boundary, ensuring global continuation. Positive initial rates remain positive: variation of constants in ${\dot{x}}_r+{\kappa }_r{M}_r(t)x_r={\kappa }_r{w}_r$ makes this explicit. Starting with finite nonnegative rates also gives positive rates at every later positive time.

For completeness, mere continuity of $p_j$ suffices for convergence; one need not silently assume differentiable prices. On the trapping compact set, both the vector field and $D(x)={\sum }_r{\kappa }_r{x}_r{g}_r(x{)}^2$ are continuous and bounded. The trajectory satisfies a uniform Lipschitz condition in time, so $D(x(t))$ is uniformly continuous. Also ${\int }_0^{\infty }D(x(t))dt<\infty$, since $U$ increases to a finite limit. The decay of a nonnegative uniformly continuous integrable function applies: otherwise disjoint intervals of a fixed positive width would each contribute a fixed positive amount. Every accumulation point therefore has $D=0$ and is $x^{\ast }$. Compactness then proves the global convergenceThis is the dissipation argument behind the LaSalle invariance principle. If uniqueness of trajectories is desired under continuous prices, put $z_r=2\sqrt{x_r/{\kappa }_r}$. The transformed equation is $\dot{z}={\nabla }_{z}U(\kappa z^2/4)$. The transformed potential is concave: it is a sum of positive multiples of $\log z_r$, minus convex nondecreasing $P_j$ composed with nonnegative quadratic sums. Its gradient is monotone decreasing, so the squared distance between two solutions cannot increase. This proves uniqueness for positive initial data without a locally Lipschitz function hypothesis on $p_j$.

For the alternative price dynamics, restrict first to positive prices on resources used by at least one route, and discard unused resources. Define the dual congestion potentialIts domain has ${\mu }_j\ge 0$ and $M_r>0$ for every route. Strictly increasing $q_j$ make each $Q_j$ strictly convex, and $-\log M_r$ is convex. Thus $H$ is a strictly convex function. Each $q_j$ is eventually bounded below by a positive constant, so its integral grows at least linearly, dominating the negative logarithms as $\Vert \mu \Vert \to \infty$. If a route price tends to zero in a bounded set, $H\to \infty$. Therefore $H$ attains a unique minimum on its domain including the admissible boundary faces.

That minimum has every used-resource price positive. At a face ${\mu }_j=0$, all route prices are still positive and the inward derivative isHence the minimum cannot lie on that face. At the unique interior minimum,Since the price dynamics are ${\dot{\mu }}_j=-{\kappa }_j{\mu }_j{\partial }_{j}H$, their positive equilibrium points are exactly these critical points. There is one equilibrium in the positive price domain. An unused resource instead has its only nonnegative equilibrium at zero.

The positive-domain qualification matters here even with strictly increasing $q_j$. If zero prices are admitted, the multiplicative factor ${\mu }_j$ permits boundary equilibria of multiplicative price dynamics. Take one route using two resources, $w=1$, and $q_1(a)=q_2(a)=a$. Both $({\mu }_1,{\mu }_2)=(1,0)$ and $(0,1)$ are equilibria, with $x=1$, as is the positive equilibrium $(1/\sqrt{2},1/\sqrt{2})$, with $x=1/\sqrt{2}$. Thus the literal assertion of uniqueness on the whole nonnegative orthant is false. A boundary equilibrium need not satisfy load equals supplied capacity at its zero-price resources.

The relationship between the two intended equilibria is inverse resource response. The first system has ${\mu }_j=p_j(y_j)$; the positive equilibrium of the second has $y_j=q_j({\mu }_j)$. If $q_j=p_j^{-1}$ on the relevant ranges, these are the same resource equations, and both systems also impose $x_r{M}_r=w_r$. They therefore have the same rates and prices. For strictly increasing continuous $p_j$ with $p_j(0)=0$ and unbounded range, the ordinary inverses satisfy all the stated conditions on $q_j$. With arbitrary independently chosen $p_j$ and $q_j$, the equilibria need not agree. Under inverse response, $P_j$ and $Q_j$ are convex conjugates on the nonnegative half-line, and $H$ supplies the Lagrangian dual problem for maximizing the weighted logarithmic utility minus resource costs, up to constants independent of $\mu$.

The statistical Hamiltonian in this question is already in thermal units, as indicated by $e^{-H}$ without an additional ${\beta }_T$. Define $W[h]=\log Z[h]$ and $F[h]=-W[h]$. The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle $m_h$. It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.

The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, $h$ is chosen to produce $m$, and $\delta {\Gamma }_{\mathrm{L}}/\delta m=h$. Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in $F_{\mathrm{L}}$.

Differentiate the saddle equation with respect to $h(y)$. The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that $G$ is the inverse kernel of ${\delta }^2{\Gamma }_{\mathrm{L}}/\delta m(x)\delta m(y)$. The factor $u_0/2$ follows from twice differentiating the quartic term $u_0{m}^4/4!$; it is not $u_0/6$. This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.

For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so $m_h(x)=m_0$ and $G(x,y)=G(x-y)$. SetThe Fourier transform with the printed positive sign sends $-{\nabla }^2$ to $q^2$, while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is $G(x)=\int d^{D}q(2\pi {)}^{-D}{e}^{-iq\cdot x}\tilde{G}(q)$. For general inhomogeneous $h(x)$, $K$ has nonconstant coefficients and $G$ depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.

At zero source and for $u_0>0$, the homogeneous saddle is $m_0=0$ for $r_0>0$, and $m_0^2=-6r_0/u_0$ on either selected stable ordered branch for $r_0<0$. Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining $m_0=0$ below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning $r_0=a_0(T-T_c)$, $a_0>0$, both branches diverge as $|T-T_c{|}^{-1/2}$, so the correlation-length critical exponent isThe high-temperature amplitude is $\sqrt{2}$ times the low-temperature amplitude for the same $a_0$. This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.

When $u_0=0$, the field theory is a Gaussian field theory, so integrating out short-wavelength modes can be done exactly rather than merely at a saddle. Introduce an ultraviolet momentum cutoff $\Lambda$ and a scale factor $b>1$, then split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into $|q|<\Lambda /b$ and $\Lambda /b<|q|<\Lambda$. Orthogonality of Fourier modes eliminates cross terms in the quadratic Hamiltonian. A constant source couples only to the zero-momentum mode and therefore does not couple to ${\varphi }_{>}$.

Before rescaling, the Gaussian functional integral factorizes:where $H_{<}$ has the original $r_0$ and $h$, restricted to low momenta. Up to a field-independent normalization, the eliminated modes contributeThe determinant affects the additive free energy, but generates neither a mass correction nor a source correction, since no interaction mixes low and high modes. This remains true beyond the Landau approximation: it is an exact integration of a Gaussian field theory, not neglect of the high-mode fluctuations.

To restore the original cutoff, use $x=b{x}^{\prime }$, $q^{\prime }=bq$, and choose the field rescalingSubstitution into the low-mode Hamiltonian gives the Gaussian momentum-shell scalingThe gradient coefficient is unchanged: the volume factor is $b^D$, the two derivatives contribute $b^{-2}$, and the two fields contribute $b^{2-D}$. The mass term gains $b^2$, while the source term gains $b^{(D+2)/2}$. ThusFor a nonconstant source the corresponding formula is $h^{\prime }(x^{\prime })=b^{(D+2)/2}h(b{x}^{\prime })$, after restricting its coupling to retained modes. The constant-source case avoids that extra source filtering.

Writing $b=e^{\ell }$, the exact Gaussian renormalization-group flow hasBoth perturbations are relevant at the massless zero-source Gaussian fixed point, with $y_t=2$ and $y_h=(D+2)/2$. There is no anomalous field rescaling here, so $\eta =0$ and the correlation-length critical exponent is again $\nu =1/y_t=1/2$. The coefficient changes arise from rescaling, not from an interaction-induced shift in $r_0$ during shell integration.

A stable real Gaussian functional integral requires $r_0>0$, or an infrared prescription that treats the zero mode at the massless point. With $u_0=0$ and $r_0<0$, the field energy is unbounded below; the Gaussian field theory alone cannot describe a stable ordered phase. The scaling laws are therefore interpreted about the Gaussian fixed point from the stable side, with finite-volume zero-mode regularization when needed.

Put $h=u-\varphi -C$. Then $h=0$ on $S$, and the constant does not change a gradient, so $\nabla u=\nabla \varphi +\nabla h$. Expand the unnormalized Dirichlet energy:The cross term is zero by Green's first identity:Thus the useful stronger identity isEquality holds exactly when $h$ has zero gradient. It is then constant on every connected component and zero on that component’s boundary, hence $h=0$. Therefore equality holds precisely when $u=\varphi +C$ throughout $V$. This is the Dirichlet principle; an additive boundary constant does not alter the minimizing energy. The energy here omits the optional factor $1/2$, which does not affect minimizers or the inequality.

For the boundary constraint $g(x,y)=x^2+y^2-1=0$, the Lagrange multiplier equations for $z=xy$ areNeither coordinate can vanish at a stationary point. Substitution gives $4{\lambda }^2=1$. For $\lambda =1/2$, $x=y=\pm 1/\sqrt{2}$ and $z=1/2$; for $\lambda =-1/2$, $x=-y=\pm 1/\sqrt{2}$ and $z=-1/2$. The constraint circle is compact, its gradient is nowhere zero, and every extremum must occur among these stationary points. ThusThe paired signs in each expression are linked. The inequality $2|xy|\le x^2+y^2=1$ also confirms the global values.

Test the weak formulation with $u$ itself. Uniform ellipticity and the Cauchy-Schwarz inequality giveThe Poincare inequality for the bounded shell with zero trace gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$. If the gradient is nonzero, divide to obtain $\Vert \nabla u{\Vert }_2\le C_P{\lambda }^{-1}\Vert f{\Vert }_2$; if it is zero the estimate is immediate and the zero trace forces $u=0$. ThusThe constant depends on the shell and $g$, not on $u$ or $f$. Subtracting two solutions with the same forcing gives a zero-forcing solution and hence proves uniqueness. This is a coercive energy argument, not an appeal to real analytic Cauchy solvability.

For the spherical coordinates in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin \theta$. Their metric tensor is therefore $\mathrm{diag}(1,r^2,r^2{\sin }^2\theta )$ and the Jacobian determinant is $r^2\sin \theta$. Applying the divergence formula to the gradient givesFor example, the formula follows directly from $\Delta u=|\det h{|}^{-1/2}{\partial }_i(|\det h{|}^{1/2}{h}^{ij}{\partial }_{j}u)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the Laplacian is a smooth geometric operator there as well, described by other charts. The coordinate volume form integrates over $1/2<r<2$, $0<\theta <\pi$, $0\le \varphi <2\pi$, with the usual periodic identification.

Apply the weak maximum principle for elliptic operators to $q=|Dw{|}^2$. The product rule and the equation for the harmonic function giveThe function $q$ is twice differentiable inside and continuous on the closure, by the prescribed regularity. Hence ${\sup }_{\Omega }q\le {\sup }_{\partial \Omega }q$. Conversely, each boundary value of $q$ is a limit of values at interior points, so the reverse inequality holds for the suprema. Taking square roots proves the gradient maximum principle for harmonic functions:This argument controls the length of the entire gradient, without comparing its separate components.

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

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

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

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.

Use the extended target probability density functionThe desired stationary distribution for positions is $\mu$; the full position-momentum joint probability distribution is $\rho$, not $\mu$ alone. Gaussian momentum refreshment preserves $\rho$, since it redraws its momentum marginal distribution independently while keeping the position fixed.

Let $R(x,p)=(x,-p)$ and $T=T_{\epsilon ,L}$ be the surrogate leapfrog integration map. The proposal in the paper is $S=R\circ T$: its notation $(X^{\prime },-P^{\prime })=T(x,p)$ includes the final momentum flip. Reversibility gives $RTR=T^{-1}$, so $S^2=I$. Both $R$ and $T$ preserve volume. Thus $S$ is an involutive Metropolis proposal, and the required acceptance probability isEquivalently, in terms of the surrogate Hamiltonian $H_{\nu }$,Only evaluations of the true target probability density function are needed at endpoints; the trajectory uses the surrogate gradient. Target and surrogate normalizing constants cancel.

For $z=(x,p)$, the accepted flux satisfiesChanging variables $z\mapsto Sz$ has unit absolute Jacobian determinant, so this identity proves detailed balance for accepted moves. The rejection mass stays at the same point and is also reversible. The accept/reject step therefore preserves $\rho$. Since momentum refreshment also preserves $\rho$, their composition and either phase of the alternating process preserve $\rho$. Projecting onto positions proves the position stationary distribution is exactly the original target. Stationarity alone does not establish uniqueness or convergence from every start; those require additional irreducible Markov chain and aperiodic Markov chain hypotheses.

There is a genuine defect in the printed smoothing claim. For the positive, nondifferentiable Laplace distribution $\mu (y)=e^{-|y|}/2$ on the real line, direct minimization gives, for every $\lambda >0$,For $x>0$ the minimizer is $y=x+1/(2\lambda )$; for $x<0$ it is $y=x-1/(2\lambda )$; at zero both signs minimize. Hence normalization leaves $\nu =\mu$, still nondifferentiable at zero. The displayed construction does not in general produce the asserted smooth surrogate. For example, the also nondifferentiable target $\mu (y)\propto e^{-y^4-|y|}$ makes that infimum $-\infty$ for every $x$, since the negative quartic term dominates the quadratic penalty. The invariance proof above is valid conditional on actually having a usable smooth surrogate, as the subsequent algorithm assumes.

A corrected sufficient construction is Moreau smoothing of a negative log-density. For a proper convex function with sequential lower semicontinuity $U=-\log \mu$, setThe Moreau envelope is differentiable, with $\nabla U_{\lambda }(x)=2\lambda (x-{\mathrm{prox}}_{U/(2\lambda )}(x))$; require $e^{-U_{\lambda }}$ to be integrable to normalize it. The signs differ from the printed formula. For the Laplace distribution, this givesa genuinely differentiable potential with integrable exponential tails. Using that surrogate with the boxed acceptance probability still targets the original Laplace distribution.

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.

Compute the coupling flow at zero magnetic field, or assume the source has support only in retained Fourier modes, so that the shell Gaussian measure is centered. First justify the shell propagator. For a positive quadratic kernel on the shell, the Gaussian functional integral gives covariance equal to the inverse kernel,with both momenta restricted to $\Lambda /b<|\mathbf{p}|\le \Lambda$. Inserting the Fourier transforms and using the Dirac delta function yields the Gaussian shell covarianceChanging $\mathbf{p}$ to $-\mathbf{p}$ gives the negative-exponent convention as well. The dependence is on the separation, as required by translation invariance. The printed numerator uses $\mathbf{x}$ alone: it is correct only if that symbol means the separation or if $\mathbf{y}=0$. For example at $\mathbf{x}=\mathbf{y}\ne 0$, the covariance must equal the constant coincident value $G_{>}(0)$, while the literal printed integral generally depends on $\mathbf{x}$. Thus the missing separation is a real source qualification, not a change of Fourier-sign convention.

Integrating out the shell gives $H_{\mathrm{eff}}[{\varphi }_{<}]=H_0[{\varphi }_{<}]-\log ⟨e^{-V[{\varphi }_{<}+{\varphi }_{>}]}{⟩}_0^{\mathrm{shell}}$, up to a field-independent constant. At first order in $u_0$, the cumulant expansion of a coarse-grained free energy retains $⟨V⟩$. Odd moments of the centered Gaussian measure vanish and Wick theorem gives $⟨{\varphi }_{>}^4⟩=3G_{>}(0{)}^2$. Writing $I_1=G_{>}(0)$,The last term affects only the constant; the second is a momentum-independent tadpole diagram correction. Before rescaling it changes $r_0$ to $r_0+u_0{I}_1/2$, leaves the quartic coefficient at $u_0$, and produces no gradient correction. Combining with part (a) gives the leading coupling recursionThis is one-loop shell mass renormalization in scalar quartic theory. Here $I_1={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}\alpha /(p^2+\alpha r_0)$. The shell Gaussian is well defined if $p^2+\alpha r_0>0$ there; near $r_0=0$ this follows from $\alpha >0$. A negative slow-mode mass may be stabilized by the retained quartic term and does not require pretending that the full unconstrained quadratic measure at negative mass is normalizable.

Linearizing about $(r_0,u_0)=(0,0)$ givesIts mass and quartic eigenvalues are $b^2$ and $b^{4-D}$. The off-diagonal term is an additive critical-mass shift: setting the bare $r_0=0$ is not generally the critical tuning when $u_0>0$. The mass remains relevant for every $D$; at $D=2$ the two linear eigenvalues coincide and the matrix can have a Jordan block, but both perturbations are still relevant.

For $D>4$, the quartic interaction is an irrelevant operator, and the Gaussian fixed point attracts weak quartic perturbations after the mass and magnetic field have been tuned. The quartic can nevertheless be a dangerously irrelevant coupling: its positive value stabilizes the ordered phase. For $D<4$, it is a relevant operator, so the Gaussian description is unstable toward interactions. At $D=4$ it is a marginal operator; first-order perturbation theory alone does not decide its fate.

To settle that borderline case, retain the leading quartic contribution from the second cumulant, at zero external momentum in the local expansion. The two-fast-field part of $V$ is $(u_0/4)\int {\varphi }_{<}^2{\varphi }_{>}^2$. Since $⟨{\varphi }_{>}^2(\mathbf{x}){\varphi }_{>}^2(\mathbf{y}){⟩}_c=2G_{>}(\mathbf{x}-\mathbf{y}{)}^2$, its connected second cumulant contributesKeeping the local quartic term and using Parseval identity gives $\int d^{D}y{G}_{>}(\mathbf{x}-\mathbf{y}{)}^2=I_2$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}[\alpha /(p^2+\alpha r_0){]}^2$. Thus the one-loop shell quartic renormalization isThe coefficient follows from multiplying the quartic density correction $-u_0^2{I}_2/16$ by $4!=24$. This local, zero-external-momentum coupling extraction is not a claim that the exact finite-shell effective action retains only its original polynomial; further operators are generated.

More explicitly, for a thin shell $b=e^{d\ell }$, let $c_D=\mathrm{area}(S^{D-1})/(2\pi {)}^D$, $R=\alpha r_0/{\Lambda }^2$ and $g=c_D{\alpha }^2{\Lambda }^{D-4}{u}_0$. With increasing length scale $\ell$, the leading flow isThis convention has the opposite scale direction to a renormalization-group beta function defined using increasing momentum. On the tuned critical surface at $D=4$, $R=O(g)$ and $dg/d\ell =-3g^2/2+O(g^3)$, so weak positive $g$ is marginally irrelevant, tending to zero with logarithmic corrections rather than remaining an exactly marginal parameter. Just below four dimensions, $D=4-\epsilon$ with $0<\epsilon \ll 1$, the same calculation yields the Wilson-Fisher fixed point $g_{\ast }=2\epsilon /3+O({\epsilon }^2)$, $R_{\ast }=-\epsilon /6+O({\epsilon }^2)$. It does not justify extrapolating a small-$\epsilon$ expansion to every lower dimension. The ordinary scalar quartic upper critical dimension is therefore