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Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.

Use a mostly-plus target Minkowski spacetime metric and restore the string tension by $T=(2\pi {\alpha }^{\prime }{)}^{-1}$. After Wick rotation, the Polyakov action on a Riemann surface isThe Polyakov path integral integrates over embeddings and worldsheet metrics, dividing by worldsheet diffeomorphisms and Weyl transformations. For a consistent flat bosonic string theory, take the critical dimension of the bosonic string $d=26$. Conformal gauge turns the embedding fields into free scalars; its Faddeev-Popov determinant is represented by the bc system. The remaining integrations are over worldsheet moduli and external insertion points.

A tachyon vertex operator isThe worldsheet Green function is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. The operator product expansion with the holomorphic stress-energy tensor gives conformal weights $({\alpha }^{\prime }{k}^2/4,{\alpha }^{\prime }{k}^2/4)$. An integrated string vertex operator must have weights $(1,1)$, so the mass-shell condition is $k^2=4/{\alpha }^{\prime }$, or tachyonic mass parameter $m^2=-4/{\alpha }^{\prime }$.

At lowest order in the string coupling, the worldsheet is a Riemann sphere. For four external states, three insertion points can be fixed by a Möbius transformation. In the worldsheet ghost field description these three vertices carry $c\bar{c}$, while the fourth is integrated. The three-point worldsheet ghost correlator gives the squared product of the three fixed-point separations, precisely the sphere gauge fixing for four string vertices factor. Thus fixing three points is a gauge choice, rather than deleting three integrations without their Jacobian.

The integral over the constant embedding worldsheet zero mode gives $(2\pi {)}^d{\delta }^{(d)}({\sum }_i{k}_i)$. The remaining Gaussian integral, using normal ordering to remove self-contractions, gives the Koba-Nielsen factorSet $(z_1,z_2,z_3,z_4)=(0,z,1,\infty )$; the vertex at infinity is defined with its conformal-weight factor. Let all momenta be incoming and defineThe mass-shell condition and momentum conservation imply $s+t+u=-16/{\alpha }^{\prime }$. Since ${\alpha }^{\prime }{k}_1\cdot k_2=-4-{\alpha }^{\prime }s/2$ and similarly for the pair $(2,3)$, the sphere tachyon position integral becomesHere $\mathcal{N}$ depends on the normalization of external string vertex operators; it does not affect the kinematic dependence.

To evaluate the remaining worldsheet modulus, put $a=-1-{\alpha }^{\prime }s/4$ and $b=-1-{\alpha }^{\prime }t/4$. The complex beta integral isFor example, this identity follows by writing the two powers as Schwinger-parameter Gamma integrals, doing the two-dimensional Gaussian integral, and changing the two positive parameters to their sum and ratio. The complex beta integral first holds for $\mathrm{Re}a,\mathrm{Re}b>0$ and $\mathrm{Re}(a+b)<1$; the scattering result is its analytic continuation. The third gamma-function argument is $1-a-b=-1-{\alpha }^{\prime }u/4$. Absorbing $\pi$ into $\mathcal{N}$, the result is the tree-level four-tachyon amplitude, the Virasoro–Shapiro amplitude:It is symmetric in the three Mandelstam variables and has generic poles at $q=4(N-1)/{\alpha }^{\prime }$, matching the bosonic string mass spectrum. The formula is the leading term of string perturbation theory. Higher orders are constructed by the same Polyakov path integral prescription on higher-genus worldsheets, with four marked points and the corresponding ghost and worldsheet moduli measure; genus $g$ has coupling power $g_s^{2g+2}$.

The ungauge-fixed Polyakov action has worldsheet diffeomorphism and Weyl transformation symmetries. They must survive quantization so that conformal gauge remains a valid gauge choice. In a curved target, the string nonlinear sigma model has a field-dependent coupling $g_{ab}(X)$. Its worldsheet Weyl anomaly is controlled by the renormalization of this coupling, the sigma-model beta function.

Restore the Euclidean normalization $(4\pi {\alpha }^{\prime }{)}^{-1}\int g_{ab}(X)\partial X^a\partial X^b$. Use a covariant background field expansion of a string sigma model, with geodesic fluctuations about a slowly varying embedding. The quadratic fluctuation operator contains the target Riemann curvature tensor coupled to two background embedding derivatives. The ultraviolet coincident propagator is $⟨{\xi }^a{\xi }^b{⟩}_{\mathrm{UV}}=\alpha \prime g^{ab}\log (\Lambda /\mu )$: the kinetic operator supplies $2\pi \alpha \prime$, and the two-dimensional momentum integral supplies $(2\pi {)}^{-1}\log (\Lambda /\mu )$. Contracting the curvature vertex with this propagator gives a logarithmic metric counterterm proportional to $\alpha \prime R_{ab}$. The resulting one-loop metric counterterm givesThe trace of the worldsheet stress tensor contains this coefficient multiplying $\partial X^a\partial X^b$, together with the curvature anomaly controlled by total central charge. Thus, with zero antisymmetric background, constant dilaton, and the critical matter/ghost system,The central-charge condition also requires $d=26$ if there is no extra internal conformal theory. This is a leading-order equation: higher-curvature terms enter the beta function at higher orders in ${\alpha }^{\prime }$, so a Ricci-flat metric alone is not a general all-orders quantum consistency criterion.

To derive the local T-duality, work where the spacelike Killing vector has nonzero norm $V>0$. Use dimensionless coordinates and ${\alpha }^{\prime }=1$ units for the duality formulas. The isometry makes the action depend on $z$ only through its derivatives. Replace those derivatives by an independent worldsheet one-form $A_{\mu }$, and enforce its flatness with a Lagrange multiplier $\tilde{z}$. With ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$ and ${ϵ}^{\tau \sigma }=1$, the relevant first-order worldsheet duality action isVarying $\tilde{z}$ sets $dA=0$. Locally $A=dz$, recovering the original theory. For the other elimination, integrate the multiplier term by parts. Varying $A_{\nu }$ gives $V{A}^{\nu }={ϵ}^{\mu \nu }{\partial }_{\mu }\tilde{z}$, orSubstituting into the entire first-order worldsheet duality action, including the multiplier term, givesThe multiplier contribution is necessary for the sign of the dual kinetic term. This is the first-order action for Abelian worldsheet duality, with Buscher rulesThe coordinate map is ${\partial }_{\tau }\tilde{z}=V{\partial }_{\sigma }z$, ${\partial }_{\sigma }\tilde{z}=V{\partial }_{\tau }z$. The original $z$ equation is precisely the integrability condition for $\tilde{z}$; conversely the original identity $d(dz)=0$ becomes the dual equation. Thus this transformation exchanges an equation of motion with the Bianchi identity for an Abelian p-form. It reverses one chiral derivative and preserves the other.

For full closed-string equivalence, the local derivation must include the global data. For a free compact circle isometry, the period of $\tilde{z}$ and the allowed gauge-field holonomies are chosen so that the Polyakov path integral exchanges momentum and winding modes. A circle of radius $R$ becomes a circle of radius ${\alpha }^{\prime }/R$ when dimensions are restored. A spacelike Killing field alone does not fix these periods or ensure a free global circle action; the global qualifications of Abelian T-duality are additional to the local field transformation.

At the quantum level integrating out $A_{\mu }$ also gives a regulated functional determinant. The Buscher dilaton shift isin the stated units and conventional Euclidean dilaton normalization. It generates the dilaton curvature coupling even though the original dilaton vanished. The dilaton curvature coupling is defined before fixing conformal gauge; its metric variation improves the worldsheet stress tensor even when the reference worldsheet is flat. As a normalization check, $e^{-2\tilde{\Phi }}\sqrt{|\tilde{g}|}=e^{-2\Phi }\sqrt{|g|}$ for this block-diagonal background. The local dual metric by itself is therefore only part of the equivalent quantum background.

With a nonconstant dilaton and no antisymmetric field, the leading metric-dilaton Weyl condition istogether with the scalar dilaton anomaly condition. It does not demand ${\tilde{R}}_{ab}=0$ separately. The new Ricci tensor can be nonzero while the Hessian of the shifted dilaton compensates it.

For a direct example, take flat polar coordinates away from the origin, with $d{s}^2=d{r}^2+r^{2}d{z}^2$ and flat spectator directions. The polar-coordinate T-dual background isHereBoth metric equations vanish. In critical dimension the scalar equation also holds: $4|\tilde{\nabla }\tilde{\Phi }{|}^2-4\tilde{\square }\tilde{\Phi }-\tilde{R}=4/r^2-8/r^2+4/r^2=0$. The example is local on $r>0$; the circle degenerates at the excluded origin.

Nonzero dual Ricci curvature is consistent because quantum T-duality transforms the dilaton as well as the metric. The displayed curvature equations and classical Buscher rules are interpreted at their stated leading derivative order; higher-order renormalized descriptions require the corresponding corrections and field-redefinition conventions. A constant $V$ is a special case in which the dual metric may also remain Ricci-flat.

Let $u_i=⟨v_i⟩$, ${\sigma }_{ij}=⟨(v_i-u_i)(v_j-u_j)⟩$, and $P_{ij}=\rho {\sigma }_{ij}$. The Jeans equation obtained by taking the first velocity moment of the Collisionless Boltzmann equation isFor an equilibrium population multiply by $x_i$ and integrate. Integration by parts givesAssume the integrals exist and the boundary flux vanishes, as for an isolated finite system with adequate decay. With the stellar kinetic-energy tensor, stellar pressure-energy tensor and stellar potential-energy tensor defined bywe obtain the equilibrium tensor virial theoremThe stellar kinetic-energy tensor describes ordered motion, whereas the stellar pressure-energy tensor describes random motion. In particular, $T_{ij}=0$ does not mean the stars have no kinetic energy. If a surface flux is retained, the right side of the displayed equilibrium identity is that surface tensor; it cannot be discarded for an arbitrary finite aperture. For an isolated self-gravitating population, $W_{ij}$ is symmetric by pairwise interchange in the Newtonian interaction integral, and the time-dependent identity is $\frac{1}{2}{\ddot{I}}_{ij}=2T_{ij}+{\Pi }_{ij}+W_{ij}$ with $I_{ij}=\int \rho x_i{x}_j{d}^{3}x$.

For the spherical Newtonian gravitational potential, ${\partial }_j\Phi ={\Phi }^{\prime }(r)x_j/r$. In the pressure-supported case the tensor virial theorem gives ${\Pi }_{ii}=\int \rho {\Phi }^{\prime }(r)x_i^2/r{d}^{3}x$. The notation ${\Pi }_{RR}$ used in this question is the planar virial trace ${\Pi }_{xx}+{\Pi }_{yy}$, not just the integral of the local cylindrical radial velocity dispersion. In cylindrical coordinates this trace includes both $v_R^2$ and $v_{\varphi }^2$. Summing the two Cartesian components therefore givesThe distinction explains the factor of two in the spherical limit.

Use ordinary spherical coordinates, $R=r\sin \theta$, $z=r\cos \theta$. The mass density separates as ${\rho }_0{r}^{-\gamma }h(\theta )$, whereBoth potential-energy integrals have the same radial factor $2\pi {\rho }_0{\int }_0^{\infty }{r}^{3-\gamma }{\Phi }^{\prime }(r)dr$. When this factor is finite and positive it cancels, leaving the scale-free tracer virial ratioThe angular factor in the printed intermediate formula has its sine and cosine interchanged. Keeping its sine/cosine weights while using that printed factor would instead produce the opposite sign in the final small-flattening correction. Merely measuring $\theta$ from the equatorial plane cannot fix this: the measure and both moment weights would have to change as well.

The angular ratio is finite and positive for every fixed $q>0$ and finite real $\gamma$, because $h$ is positive and bounded above and below. However, the claim that the unrestricted global virial integrals are always well-defined needs a hypothesis. In a point-mass Newtonian gravitational potential, the common radial factor is proportional to ${\int }_0^{\infty }{r}^{1-\gamma }dr$: convergence at zero requires $\gamma <2$, while convergence at infinity requires $\gamma >2$. There is no such $\gamma$. If the global energies diverge, the boxed angular ratio is still the ratio of the two potential-energy integrals with identical spherical inner and outer cutoffs, independent of the cutoffs. It is not a quotient of two finite global stellar pressure-energy tensors; applying a finite-aperture tensor virial theorem additionally requires its surface terms. These convergence and boundary qualifications are separate from the printed angular typo.

Set $ϵ=q^{-2}-1$. For modest flattening, the Taylor expansion is $h=1-(\gamma /2)ϵ{\cos }^2\theta +O({ϵ}^2)$. The required elementary angular integrals areThese follow directly by $t=\cos \theta$, or from the supplied gamma function identity. Hence the numerator is $(4/3)[1-\gamma ϵ/10+O({ϵ}^2)]$ and the denominator is $(2/3)[1-3\gamma ϵ/10+O({ϵ}^2)]$. The flattening–anisotropy virial relation becomesFor a decreasing tracer mass density ($\gamma >0$), oblate flattening ($q<1$) requires more random kinetic support per in-plane direction than vertically. In a spherical Newtonian gravitational field, directional orbital anisotropy supplies the flattening rather than an anisotropic force law. A steeper radial mass density amplifies the required velocity dispersion anisotropy. A prolate tracer reverses the leading inequality. The relation constrains integrated velocity dispersions, not a unique local galactic distribution function.

Choose $R_0$ with $K\subset \{(y,w):|y|\le R_0,|w|\le R_0\}$ and let $A_0={\max }_{|w|\le R_0}|a(w)|$, which is finite by continuity. Every point transported from $K$ satisfiesHence anycontains the support of a function in the required phase space ball throughout $[0,T]$. Equivalently, the image of the compact set $[0,T]\times K$ under the characteristic flow map is compact. This uniform support control puts the solution in ${\mathcal{C}}_T$ and justifies the following differentiations under the integral.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

The planar vorticity velocity kernel has magnitude $(2\pi |z|{)}^{-1}$. Split its defining integral into $|x-y|<R$ and $|x-y|\ge R$ for any $R>0$. The singular part is absolutely integrable in two dimensions:In particular $R=1$ provesWhen both norms are nonzero, optimization in $R$ also gives $\Vert U[\Omega ]{\Vert }_{\infty }\le \sqrt{2/\pi }\Vert \Omega {\Vert }_1^{1/2}\Vert \Omega {\Vert }_{\infty }^{1/2}$. If either norm is zero, $\Omega =0$ almost everywhere. Absolute convergence supplies a well-defined velocity at every $x$, with these uniform bounds.

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.

The exact normalizing sum can be expensive on a large loss network. The Erlang fixed point approximation replaces joint resource acceptance by a product of marginal acceptances. Here consider unit requirements $A_{jr}\in \{0,1\}$, positive integer capacities and finitely many fixed routes. Let $B_j$ be resource blocking and $q_j=1-B_j$. A route-$r$ call contributes to the offered traffic at resource $j$ after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is ${\prod }_{j\in r}{q}_j$. This is an independence approximation, not an alternative exact factorization of the stationary law in part (i).

Uniqueness follows from a convex potential for the Erlang fixed point. For a single resource, the carried load of an Erlang loss resource is $m_C(v)=v[1-E(C,v)]$. It increases strictly from zero to $C$ as $v$ increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to $\log v$ gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.

Use $p_j=-\log q_j\ge 0$. Let $v_C(p)$ be the unique offered load with $1-E(C,v_C(p))=e^{-p}$, and set $g_C(p)=v_C(p)e^{-p}=m_C(v_C(p))$, with $g_C(0)=0$. This function is strictly increasing and tends to $C$. DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore $F$ is a strictly convex function. It is also a coercive function on the nonnegative orthant: an unbounded coordinate makes its integral grow asymptotically linearly with positive slope $C_j$. A unique minimizer exists.

If resource $j$ carries some positive offered route, its inward derivative at $p_j=0$ is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by $q_j$ gives precisely $v_j={\sum }_{r:j\in r}{\rho }_r{\prod }_{k\in r\setminus \{j\}}{q}_k$. A resource with no positive offered route uniquely has $p_j=0$, hence $B_j=0$. Thus the minimizer and the fixed point coincide, proving existence and uniqueness of the Erlang fixed point for fixed unit-resource routing. This does not by itself guarantee convergence of every simultaneous substitution algorithm; the uniqueness claim concerns the solution of the equations.