Sharp integral fillings in CAT(0) spaces

Proof. Let T be the cycle in a proper \operatorname{CAT}(0) space X. If T=0, take the zero filling. Otherwise choose a closed ball Y=\overline B(o,R) whose interior contains \spt T. Properness makes Y compact. Triangle comparison implies that balls are convex, so the induced metric makes Y a compact \operatorname{CAT}(0) space.

For x\in X, let P(x) be the point of the segment [o,x] at distance \min\{R,d(o,x)\} from o. Geodesics in a \operatorname{CAT}(0) space are unique. Comparing two such radial points with their counterparts in the Euclidean comparison triangle, and then using nonexpansion of Euclidean radial projection onto a closed ball, gives

One can also identify P(x) as the nearest point of Y: the triangle inequality bounds the distance from x to Y below by \max\{0,d(o,x)-R\}, and the radial point attains this bound.

Lipschitz pushforward therefore gives an integral cycle P_\#T in Y with \M(P_\#T)\le\M(T). If \iota:Y\hookrightarrow X denotes inclusion, locality gives \iota_\#P_\#T=T, because \iota\circ P is the identity on a neighborhood of \spt T. Apply the compact-space inequality to P_\#T and include its filling into X. Its mass does not increase, its boundary is T, and its support is contained in the compact set Y.□

Proof. The pushforward and Borel-coefficient properties are part of the basic current calculus of [3]. The exact restriction identity also follows from the restriction mass bound and the decomposition C=C\restr E+C\restr(Y\setminus E), using minimality of the mass measure in both directions. The representation in (iii) is the integer-rectifiable representation [3, Lemma 4.1 and Theorem 4.5]; subdivision makes chart domains bounded and disjointification makes their images disjoint. The slicing assertion is [3, Theorems 5.6–5.7]. The compactness and closure theorems [3, Theorems 5.2 and 8.5] give (ii): completeness is automatic, the bounded normal masses are assumed, and compactness of Y supplies uniform tightness. The product statements, including the action convention, are [30, Definition 3.1 and Theorem 3.2]. They apply because an integral current is normal and its support in Y is bounded.

Here is the geometric verification of the filling hypotheses. Choose a point o in the support of a nonzero cycle B, and let H(t,x) be the point at fraction t along [o,x]. Triangle comparison gives

Thus H is jointly Lipschitz on [0,1]\times Y. On the support of B, its time speed is bounded by D=\diam(\spt B), and its spatial Lipschitz constant is at most one. In each of the k+1 terms of (2.5), a 1-Lipschitz scalar test after composition with H has time derivative at most D, and the remaining spatial derivatives have Lipschitz bounds at most one. The mass bound therefore gives

The boundary is B: the terminal map is the identity, the initial map is constant and hence has zero pushforward in positive dimension, and \partial B=0. The filling is supported on segments from o to \spt B, all lying within distance D of o. Consequently the space has local cone inequalities in every positive current dimension. Taking D\le\diam Y gives the second estimate in (2.6), for example with L_k=(k+1)\diam Y.

The complete space Y has the convex geodesic bicombing given by its unique geodesics. Wenger’s Euclidean-exponent isoperimetric theorem [29, Theorem 1.2 and Corollary 1.4], or its cone-inequality form applied successively in dimension, gives the first estimate in (2.6) for every k\ge1. No sharp value of K_k is used here.

For a fixed boundary B, the cone gives at least one filling. A minimizing sequence has bounded mass and fixed boundary, so (ii) gives an integral weak limit with that boundary. Lower semicontinuity proves attainment. Finally, Y is complete and geodesic, hence quasiconvex, and it has the local cone inequalities in dimensions 1,\ldots,k just verified. The cycle assertion in [30, Theorem 1.4] applies to the weakly convergent sequence B_j: its normal masses are bounded because its boundary masses are zero. It gives V(B_j-B)\to0 directly. Finiteness and subadditivity of filling volume imply

which proves the second limit.□

Proof. The linear bound in (2.6) implies

The right side has finite supremum, and positivity implies \M(B)<(L_n/c)^n. A positive maximizing sequence therefore has uniformly bounded masses and zero boundaries. By Theorem 2.2, a subsequence converges weakly to an integral cycle A. Filling continuity and lower semicontinuity of mass give

Thus A attains the positive supremum, and in particular \M(A)>0. Maximality and subadditivity now give

which is (2.9).□

Proof. At a density-one point z of E_i, every fixed finite configuration in \R^k can be approximated by points of (E_i-z)/t as t\downarrow0. Indeed, a point of such a configuration that stayed a positive distance from (E_i-z)/t along a sequence of scales would give a missing Euclidean ball of radius comparable to t at distance O(t) from z, contradicting density one. Applying this observation to a single direction and using the bi-Lipschitz lower bound for \phi_i shows that N_z(w)\ge\Lip(\phi_i^{-1})^{-1}|w|. The upper Lipschitz bound gives N_z(w)\le\Lip(\phi_i)|w|. Thus N_z is a norm.

For any four points a,b,c,d of a \operatorname{CAT}(0) space,

To verify this, let m be the midpoint of [a,c]. Midpoint comparison bounds d(b,m)^2 and d(d,m)^2 by the corresponding average squared distances to a,c, minus d(a,c)^2/4. Combine these bounds with d(b,d)^2\le2d(b,m)^2+2d(d,m)^2 to obtain (2.14).

Approximate the four Euclidean domain points 0,u,u+v,v by scaled points of E_i-z. Apply (2.14) to their images, divide by t^2, and use the two-point differential (2.11). It follows that

Apply the same inequality to (u+v)/2 and (u-v)/2, and use homogeneity, to obtain the reverse inequality. The parallelogram law therefore holds. Polarization produces a positive definite inner product and the matrix G_i(z) in (2.12). Its entries are measurable, since they are finite linear combinations of squared metric differentials in fixed rational directions. Setting w'=0 proves (2.13).□

Proof. Fix a tuple of d scalar 1-Lipschitz tests. Almost everywhere on each domain their compositions with \psi_i are differentiable in the sense of scalar Lipschitz extensions. If \ell is one of the resulting derivative rows, (2.17) gives

To see the bound in an arbitrary direction, approximate that direction by rescaled domain points at a density-one point, and then use differentiability and (2.17). If P_i is positive definite, transform it to the identity and apply Hadamard’s inequality to the rows. Their absolute determinant is at most \sqrt{\det P_i}. If P_i is singular, all rows vanish on its kernel; the full determinant and \det P_i both vanish, giving the same bound. Substitution in the representation yields (2.2) with the measure \eta.

The exceptional set is allowed to depend on the fixed test tuple. The measure \eta is independent of that tuple and controls every tuple separately; this is precisely what the definition of mass requires.□

Proof. Denote the right side of (2.19) by \eta, initially allowing \eta(Y)=\infty. We first prove \eta\le\|C\|; this will also establish finiteness before we apply the preceding lemma.

Fix 0<\varepsilon<1. Each chart domain, up to a Lebesgue-null set, has a countable Borel partition into pieces E on which there is a fixed positive definite matrix Q, depending on the piece, satisfying

Here is a construction. Approximate the measurable matrix G_i(z) by an element of a countable dense family of positive definite matrices, with a sufficiently small relative error. The centered estimate (2.13) then gives a radius 1/l, depending on the point, on which the two inequalities with this fixed matrix and the prescribed \varepsilon hold against every other point of the original chart domain. Group points according to the matrix and l, subdivide into sets of diameter less than 1/l, and disjointify. These sets can be chosen Borel. Indeed, for any fixed matrix and radius, the non-strict distance inequalities against all domain points with 0<|z-z'|<1/l can be tested on a fixed countable dense subset of the domain, using continuity in z'. All metric differentiability points are covered by this countable construction.

At almost every density point of a piece E, differentiation of (2.20) in domain directions gives

On \phi_i(E), every scalar coordinate of Q^{1/2}\phi_i^{-1} is (1-\varepsilon)^{-1}-Lipschitz by the lower inequality in (2.20). By McShane’s theorem [23, Theorem 2.1], extend these coordinates separately to all of Y, preserving that Lipschitz constant, and call the resulting tuple \pi.

For any Borel subset D\subset E, choose the bounded Borel first coefficient supported on \phi_i(D) and equal there to \sgn(\theta_i)\circ\phi_i^{-1}. At density points of E, the extended coordinates have the same derivatives as Q^{1/2}z, since they agree with them on the piece. The images of all original charts are disjoint. Hence (2.3), applied to this coefficient and tuple, gives exactly

The mass bound is legitimate for this Borel coefficient by its standard extension from the first slot. Combining the last two estimates gives

For a Borel set B\subset Y, take D=E\cap\phi_i^{-1}(B) and sum over all pieces and all charts. Their images are disjoint, while discarded Lebesgue-null sets make no contribution to \eta. Thus

This proves that \eta is finite; letting \varepsilon\downarrow0 gives \eta\le\|C\|.

Conversely, (2.13) gives the quadratic majorant P_i=G_i for every chart. Its measure (2.18) is exactly the now finite measure \eta. Lemma 2.5 yields \|C\|\le\eta, which completes the proof of the equality.□

Proof. By Proposition 2.6, it suffices to take y=\phi_i(z), where z is a density-one point of E_i, the metric differential exists, z is a Lebesgue point of \rho_i=|\theta_i|J_i extended by zero off E_i, and |\theta_i(z)|\ge1. These conditions hold at almost every point carrying chart mass. Fix \varepsilon>0. For all sufficiently small \rho, the domain points in the ellipsoid

map into B(y,\rho), by (2.13). This ellipsoid has volume

Lebesgue differentiation on dilates of this fixed ellipsoid shows that the contribution of this chart to \rho^{-k}\|C\|(B(y,\rho)) has lower limit at least \omega_k|\theta_i(z)|/(1+\varepsilon)^k. Other charts contribute nonnegative mass. Letting \varepsilon\downarrow0 and using |\theta_i(z)|\ge1 proves (2.21).□

Proof. Fix y and g. Put R=\diam Y and M_g=\|g\|_\infty, and choose h>0 small enough that hM_g\leq 1/2. For 0\leq t\leq h, let

where [y,x]_a denotes the point at fraction a along the geodesic from y to x. In particular, F_0 is the identity and 1/2\leq\lambda\leq1. Equal-fraction CAT(0) comparison and the constant-speed parametrization of a geodesic give

Thus F is jointly Lipschitz. The currents

are integral by Theorem 2.2. Their boundaries are \partial B_h=0 and \partial D_h=B_h-A.

Comparison forms. We first establish precise differential upper bounds for these maps. For x,x'\in Y and two fractions \lambda,\lambda'\in[0,1], Euclidean comparison for the triangle with vertices y,x,x' gives

where r=d(y,x) and r'=d(y,x'). Indeed, the expression on the right is exactly the squared distance between the corresponding points in the Euclidean comparison triangle. This remains valid for degenerate triangles.

Take the disjoint charts \phi_i:E_i\to Y from Proposition 2.6, with metric matrices G_i, J_i=\sqrt{\det G_i}, and signed multiplicities \theta_i. At almost every density point z\in E_i, the metric differential (2.12) and the derivatives of r\circ\phi_i and g\circ\phi_i exist. At such a point write

and abbreviate r=r(\phi_i(z)), g=g(\phi_i(z)), and \lambda=1-tg. Thus \alpha and \gamma represent Dr and Dg in this chart. Equation (2.16) yields |\alpha|_{G_i^{-1}}\leq1 and |\gamma|_{G_i^{-1}}\leq\Lip(g).

Set F_i(t,z)=F(t,\phi_i(z)). The quadratic forms

are nonnegative. In the first expression, the first summand acts only on spatial vectors. For increments (s,w) for which (t+s,z+w)\in[0,h]\times E_i, they satisfy

To see this, the derivative of (1-tg)r is \lambda\alpha-tr\gamma-rg\,dt, while

Substitution into (3.2) proves (3.5); changing the factor \lambda\lambda' to \lambda^2 contributes only a higher-order error. Taking s=0 gives the corresponding upper bound with P_{\mathrm{sp}} for the fixed-time map. These statements are needed separately for each fixed y and g; no common exceptional set for all centers or all functions is required.

Determinants of the upper forms. We next compute the determinants. At the point under consideration choose spatial coordinates orthonormal for G_i. In these coordinates (3.4) becomes

The derivative at the identity of H\mapsto\sqrt{\det H} is H'\mapsto\tfrac12\operatorname{tr}H'. Returning to the original chart coordinates therefore gives

After decreasing h if necessary, the remainder is bounded in absolute value by Ct^2, where C depends only on n,R,M_g, and \Lip(g). In particular it is independent of the chart and the base point: all the matrix entries in the orthonormal coordinates are controlled by these bounds.

For the full form, put H=\lambda^2(I-\alpha\alpha^\top) in the same coordinates and u=\lambda\alpha-tr\gamma. Its matrix in time-first coordinates is

Its determinant is (rg)^2\det H. For positive definite H this follows directly by subtracting suitable multiples of the first row and column, or by a block determinant calculation; replacing H by H+\varepsilon I and letting \varepsilon\downarrow0 proves the identity for every nonnegative H. Since \det(I-\alpha\alpha^\top)=1-|\alpha|^2, we obtain the exact identity

Here r,g,\lambda are nonnegative. In particular, the formula also covers rg=0 and |Dr|=1.

Figure 1 separates the fixed-time current from the swept current and illustrates the unequal-fraction comparison. The forms above are upper bounds for the deformed metric; the exact identity is the determinant calculation for the comparison form.

From whole-current products to mass. The spatial determinant will bound the endpoint cycle mass; the full determinant will bound the swept filling mass. We now pass from the comparison forms to these two current bounds. The chart formula (2.3) gives, for a bounded Lipschitz coefficient b and Lipschitz functions \pi_1,\ldots,\pi_n on Y,

For the homotopy current, the corresponding formula is

To obtain this formula, apply the interval product to the whole current A. Write \psi_a(t,x)=\pi_a(F(t,x)) and \widetilde b(t,x)=b(F(t,x)). Inserting these functions into (2.5) expresses the left side as

The time derivatives admit bounded Borel versions and exist for dt\,d\mu-almost every (t,x). For a fixed test tuple they have uniform bounds, as do the spatial chart-covector norms of the \psi_a(t,\cdot). Inserting (2.3), every cofactor in the resulting expansion is bounded by a constant times J_i, by Euclidean determinant bounds. The sum of the absolute integrals is consequently bounded by C h\sum_i\int_{E_i}|\theta_i|J_i\dd z=C h m. Fubini’s theorem therefore permits both the insertion and the interchange of the sum and integrals. Expansion in the time column gives exactly (3.9). The separate derivatives agree almost everywhere with the joint derivatives of the Lipschitz scalar compositions on [0,h]\times E_i. This proves the formula without assigning a boundary to any individual chart piece.

The two action formulas are determinant representations of the finite-mass currents B_h and D_h. Their maps have the respective quadratic majorants P_{\mathrm{sp}}(h,z) and P_{\mathrm{tot}}(t,z) from (3.5). Consider the measures obtained by pushing forward

under F_i(h,\cdot) and F_i, respectively, and summing over i. They are finite: (3.6) bounds the spatial density by a uniform multiple of |\theta_i|J_i, while (3.7) bounds the product density by RM_g|\theta_i|J_i. Their integrals are therefore controlled by m and h m, using (2.19). Lemma 2.5 now bounds the two current mass measures by these measures. That lemma includes singular quadratic forms and permits differentiation null sets to depend on the fixed test tuple.

Combining these bounds with (2.19), (3.6), and (3.7), and using \lambda^n\leq1, gives

The constant in the first remainder is finite because the pointwise remainder was uniform and \mu(Y)=m<\infty.

Using maximality. The spatial mass decrease and the swept filling mass are now controlled by the same radial test. We apply the extremizer inequality to compare them. Set

Choose a constant C so that \M(B_h)\leq U_h:=m-hI_g+Ch^2 for all sufficiently small h>0. Since m>0, also U_h>0 for such h. The current -D_h fills A-B_h, so (2.9) and the monotonicity of s\mapsto s^{q_0} give

Dividing by h and letting h\downarrow0 yields c q_0m^{1/n}I_g\leq H_g. The normalization c q_0m^{1/n}=1/n in (2.10) is precisely (3.1).□

Proof. If this inequality failed, Lemma 2.3 in dimension n=2 would produce a normalized nonzero cycle A as above with

Choose a center y for which Lemma 2.7 gives

Such a center exists because the density statement holds \mu-almost everywhere and m>0.

Let r=d(y,\cdot). For 0<\varepsilon<1 such that \mu\{r=\varepsilon\}=0, the function

is a nonnegative Lipschitz function on Y, so it is admissible in Proposition 3.1. On \{r<\varepsilon\} its gradient is zero and

On \{r>\varepsilon\} the chain rule gives Dg_\varepsilon=-2r^{-3}Dr. Writing b=r^{-1}\sqrt{1-|Dr|^2} there, the same two integrands are 2b^2 and 2b, respectively. The separating level has zero \mu-measure by our choice of \varepsilon. Therefore (3.1) implies

where the last inequality is the elementary bound 2b-2b^2=\tfrac12-2(b-\tfrac12)^2\leq\tfrac12.

Only countably many levels of a real function can carry positive mass under a finite measure, so eligible \varepsilon tend to zero. Taking the lower limit of (3.15) along such radii and using (3.13) yields 2\pi\leq m/2, or m\geq4\pi. This contradicts (3.12). The least filling mass is therefore at most c_2\M(T)^{3/2}, and attainment from Theorem 2.2 supplies the required integral filling. The zero cycle is filled by the zero current.□

Proof. Fix a level at which U_t is integral, and write a=a(t) and P=P(t). Its boundary is an integral (n-1)-cycle. By the induction hypothesis there is an integral n-current R in Y satisfying

Indeed, the coefficient on the right is the sharp filling coefficient in dimension n-1. The currents B=A-U_t+R and A-B=U_t-R are integral cycles. Restriction decomposes the mass measure exactly, so

No disjointness involving R is needed for these upper bounds.

Suppose first that d<a\le m/2. The extremizer inequality (2.9) and the coarse filling estimate in Theorem 2.2 give

The first inequality follows by integrating the derivative of s^{q_0} over [m-a+d,m]\subset[m/2,m]. Thus

If d\ge a, the same lower bound holds automatically. Choose 0<\delta<m/2 so small that

For 0<a\le\delta, combining (4.3) with the upper bound for d yields

The case a=0 is immediate. All constants used to choose \delta depend only on the fixed extremizer, the dimension, and \eta.□

Proof. Fix a point z_0\in Z and normalize each differentiated test by \pi_i(z_0)=0; adding constants does not change its current evaluation. The normalized 1-Lipschitz functions form a compact metric space in the uniform topology, by the Arzelà–Ascoli theorem. Choose a countable dense family of normalized k-tuples (\pi^j)_{j\ge1}.

Write \lambda=\|C\|. For each tuple \pi, the first-slot extension to bounded Borel functions gives a finite signed measure b\mapsto C(b,\pi) with density \rho_\pi relative to \lambda, where |\rho_\pi|\le1 almost everywhere. Choose measurable versions for the countable family and put

This satisfies 0\le\rho\le1. If a normalized tuple \pi is approximated uniformly by tuples in the family, current continuity, with their common Lipschitz bound, shows for every Lipschitz b that

Thus \rho\lambda is a mass-controlling measure for all normalized 1-Lipschitz tuples and hence, by rescaling the differentiated slots, for all tests. Minimality of the mass measure gives \lambda\le\rho\lambda. It follows that \rho=1 almost everywhere.

For any finite set of the tuples, select at each point the first index where the largest |\rho_{\pi^j}| is attained and select its sign. This produces bounded Borel functions b_j with \sum_j|b_j|\le1 and

These coefficients can be replaced, with an arbitrarily small error in the displayed sum, by Lipschitz coefficients satisfying the same constraint. To see this, regularity of the finite Borel measure \lambda gives Lipschitz approximations to each b_j in L^1(\lambda). For example, an indicator is approximated between a compact set and an open neighborhood by a Lipschitz function obtained from distances to those sets, and simple functions suffice for the general case. Apply to the vector of approximations the Lipschitz map

This retains L^1 convergence and enforces the constraint. Monotone convergence of the finite maxima to \rho=1 now proves that the supremum over all admissible finite Lipschitz lists equals \M(C); the reverse inequality follows directly from the mass bound.

It remains to make the lists countable without depending on C. For each fixed list length, the admissible lists of coefficients and normalized tuples form a separable metric space in the uniform topology: they are a subspace of a finite product of C(Z), which is separable because Z is compact metric. Choose a countable dense subset within that admissible space and take the union over list lengths. An admissible list can be approximated by these lists with its constraint and differentiated Lipschitz bounds preserved. The first-slot errors are controlled by their uniform norms times \M(C), and convergence in the differentiated slots follows from current continuity. Hence the countable collection has the same supremum. A common Lipschitz bound on the first coefficients is neither asserted nor needed.□

Proof. For a bounded open interval I, define the bounded Lipschitz function

Let b be Lipschitz and let \pi be an (n-1)-tuple of 1-Lipschitz functions. Restriction, the chart representation, and Fubini give

For the first equality, the boundary evaluation has the measurable representative U_t(1,b,\pi), obtained by integrating the chart determinants with the factor 1_{\{u>t\}}. It is integrable on bounded intervals. The second equality is the product rule applied to the cycle identity \partial A=0.

The chain rule in the charts gives

There is no ambiguity at the two endpoints of I: the differential of a Lipschitz scalar function vanishes almost everywhere on each fixed level set. This follows by applying Euclidean differentiation at density points of that level set in each chart. At points of the charts the other differentiated covectors have norm at most one, so the Euclidean determinant bound in Proposition 2.6 applies.

In particular, for an admissible finite list \mathcal T as in Lemma 4.2, (4.6) yields

Let \gamma=u_\#(|Du|\mu). Lebesgue differentiation of (4.7), simultaneously for the countable collection of lists on Y, gives

Here we used (4.4) for each integral slice. Thus the passage from a signed integrated evaluation to total slice mass excludes only a countable union of null sets.

Set \lambda=u_\#\mu and \zeta=u_\#(|Du|^2\mu). For every interval I, Cauchy–Schwarz on u^{-1}(I) and the bound |Du|\le\Lip(u) imply

Divide by the appropriate powers of the interval length and let symmetric intervals shrink to an almost every point. The densities of \lambda_{\mathrm{ac}} and \zeta_{\mathrm{ac}} are respectively -a' and -e'. Combining the resulting inequalities with the bound for P proves (4.5).□

Proof. If u=0 almost everywhere, take u^*=0. Otherwise let L=\Lip(u) and U=\esssup_\mu u. Then U>0 and L>0: a function with Lipschitz constant zero is constant, and a positive constant would have positivity set of mass m>\delta. For 0\le t\le U, define

The function R is nonincreasing and right-continuous, R(t)>0 for t<U, and R(U)=0. For almost every t\in(0,U), Proposition 4.1 and Lemma 4.3 give

This estimate also controls finite differences, even if R has a singular part. Indeed, on every compact subinterval of (0,U), the nonnegative measure -dR has absolutely continuous density -R'\ge c_0 and a nonnegative singular part. Therefore

Define the generalized inverse for \rho\ge0 by

It is nonincreasing, h(0)=U, and h(\rho)=0 for \rho\ge R(0). It is also c_0^{-1}-Lipschitz. To check the latter claim, let \rho_1<\rho_2 with h(\rho_1)>h(\rho_2) and choose

The definition and monotonicity imply R(s)\le\rho_2 and R(t)>\rho_1, whence c_0(t-s)\le R(s)-R(t)\le\rho_2-\rho_1. Letting s and t approach the two inverse values proves the claim.

For 0<t<U, right continuity of R implies

In the reverse implication, if \rho<R(t), right continuity supplies some s>t with \rho<R(s); the other implication follows from monotonicity. Consequently u^*(x)=h(|x|) is Lipschitz and compactly supported, and its superlevel sets have Euclidean volume a(t). The layer-cake formula proves the asserted equality of all positive power integrals.

We give the exact energy substitution, including possible atoms and singular continuous parts of a. Strict decrease in (4.10) implies

Indeed, all levels s<t satisfy R(s)>R(t), while no s\ge t is eligible in (4.11) with \rho=R(t). If R is continuous at t, (4.12) and this continuity show that the entire fiber h^{-1}(\{t\}) is the singleton \{R(t)\}.

The monotone function R is continuous except at countably many points and differentiable with finite derivative almost everywhere. Its derivative, wherever it is relevant to (4.9), is nonzero. Let N be the null set where the Lipschitz function h is not differentiable. The set h(N) is null because Lipschitz functions map Lebesgue null sets to null sets. By (4.13), every level t with R(t)\in N lies in h(N). It follows that for almost every t\in(0,U) both derivatives needed below exist, and differentiating the inverse identity gives

For example, this follows directly by letting s\to t in h(R(s))-h(R(t))=s-t at a continuity and differentiability point of R.

Since h is absolutely continuous and nonincreasing on [0,R(0)], the measure |h'(\rho)|\dd\rho pushes forward under h to Lebesgue measure on (0,U). One may verify this first on an interval (a,b)\subset(0,U): integration of -h' over its inverse image equals b-a by the fundamental theorem of calculus; constant portions have zero derivative. Intervals determine the measure. At almost every target level the fiber consists of the single point R(t), so this substitution and (4.14) give

The denominators are positive and finite almost everywhere by (4.9). The preceding substitution also justifies the equality for a nonnegative integrand before its finiteness is known.

For clarity, a jump of R produces an interval on which h is constant. In particular, an atom at U produces a central constant part of u^*. Such intervals contribute no energy. A singular continuous decrease causes no additional term either: the exact substitution uses the absolutely continuous measure |h'|\dd\rho, and the inverse image of any null set of levels has zero measure for this measure. Thus no absolute-continuity assumption on a has entered (4.15).

Finally, (4.2) gives

Using (4.5) in (4.15) therefore yields

The last inequality only uses monotonicity of e and remains true if its distributional derivative has a singular part.□

Proof. The assertion is immediate for f=0. By homogeneity, assume \int f^p\dd x=1. Write f=f(r), r=|x|. Continuity and radial monotonicity give a finite r_0>0 such that f>0 on [0,r_0) and f=0 on [r_0,\infty). For D>0 define

where A_D>0 is chosen so that \int g_D^p\dd x=1. For 0<r<r_0 match the cumulative radial masses by

Both cumulative functions are C^1 with strictly positive derivative on their open radial intervals. Their inverse composition t is therefore C^1 on (0,r_0), strictly increasing, and satisfies t(0+)=0 and t(r_0-)=D. Differentiation gives

The radial map T(x)=t(|x|)x/|x| has positive eigenvalues t' and n-1 copies of t/r and transports f^p\dd x to g_D^p\dd x. This follows either from (4.17) by radial integration, or from the change of variables and (4.18). The jump of g_D at D is harmless: the calculation uses its positive continuous interior profile and never differentiates the cutoff.

Let

Change of variables, the arithmetic–geometric mean inequality for the positive eigenvalues of DT, and radial integration by parts give

The last equality needed for Cauchy–Schwarz is the transport identity \int f^pt^2\dd x=\int|x|^2g_D^p\dd x. To justify the integration by parts, first integrate over \varepsilon<r<R<r_0. The boundary term is n\omega_n[r^{n-1}tf^\ell]_{\varepsilon}^R. It tends to zero at r_0 because f(r_0)=0 and t\le D, and at zero because n>1 and t,f are bounded. The resulting derivative integral is absolutely integrable: f' is essentially bounded, t is bounded, and the source ball is bounded. The source profile is Lipschitz, so its radial integration by parts is justified by absolute continuity.

Now put

The normalizing constants A_D converge to A_\infty as D\to\infty. Moreover,

For both integrals the unnormalized radial integrand is of order r^{1-n} at infinity. Thus both are finite for every n>2, and the convergence follows by monotone convergence for the unnormalized integrals and convergence of A_D. In particular, no limit of the transport maps is required.

The ratio on the limiting left side of (4.19) can be evaluated from g itself. Direct differentiation shows that

If J=\int g^\ell\dd x and M_2=\int|x|^2g^p\dd x, radial integration by parts with the vector field x gives

Its boundary term r^ng(r)^\ell vanishes at zero and is O(r^{2-n}) at infinity. Therefore

Taking D\to\infty in (4.19) proves \int|\nabla f|^2\dd x\ge\int|\nabla g|^2\dd x.

The transport argument has reduced the lower energy bound to the single comparison profile g. It remains to compute this energy and thus fix the threshold used in the current’s critical minimization. The inverse stereographic map

pulls the sphere metric back to 4(1+|x|^2)^{-2} times the Euclidean metric. Its volume Jacobian is 2^n(1+|x|^2)^{-n}, so

For w(r)=(1+r^2)^{-(n-2)/2}, direct differentiation gives

Integration by parts consequently yields

Here the boundary term r^{n-1}g(r)g'(r) vanishes at zero and is O(r^{2-n}) at infinity; also 1-2/p=2/n. Multiplying the energy lower bound by 4/(n-2) proves (4.16) for the normalized f, and homogeneity proves the general statement. All moments and boundary limits used above remain valid when n=3 or n=4; no L^2 integrability of the full profile g is needed.

The coefficient is also optimal within the class in the statement. Indeed, choose a nonnegative nonincreasing Lipschitz cutoff \chi equal to one on [0,1] and zero on [2,\infty), and set g_R(x)=g(x)\chi(|x|/R). Then g_R\to g in L^p, and \nabla g_R\to\nabla g in L^2: the tail of \nabla g tends to zero, while the extra cutoff energy is bounded by

Each g_R is admissible, and its Sobolev quotient tends to the computed value for g.□

Proof. Fix temporarily \eta\in(0,1) and its corresponding \delta. Combining Lemmas 4.4 and 4.5 gives

This also holds when z=0 almost everywhere.

For a general Lipschitz u with \|u\|_2>0, set

Chebyshev’s inequality gives \mu\{z>0\}\le\delta. Scalar Lipschitz composition in the charts gives E(z)\le E(u); at the finitely many corner levels the gradient vanishes almost everywhere on the corresponding level sets, so the usual almost-everywhere chain rule applies. In addition, 0\le|u|-z\le\tau, whence

For every \xi>0, the triangle inequality followed by (b+d)^2\le(1+\xi)b^2+(1+\xi^{-1})d^2 now yields

Choose \eta>0 and \xi>0 sufficiently small that S_*(1-\eta)^2/(1+\xi)\ge S_*-\epsilon, and then fix the associated \delta. The last display proves the assertion with, for example,

If \|u\|_2=0, then \|u\|_p=0 and the asserted inequality is immediate.□

Proof. Let \mathcal L^2 denote the Hilbert space of chart covector fields L with \int|L|^2\,d\mu<\infty. The Lipschitz level-set property from Section 2, applied to the difference of two representatives, shows that their gradients agree whenever the functions agree \mu-almost everywhere. We first prove that the graph of the Lipschitz gradient in L^2(\mu)\times\mathcal L^2 is closable. Suppose that u_j are Lipschitz,

For a Lipschitz scalar function b and a Lipschitz (n-1)-tuple \pi, the identity \partial A=0 gives

The right-hand side tends to zero by the mass bound and \|u_j\|_1\le m^{1/2}\|u_j\|_2. For the fixed tuple \pi, the limit of the left-hand side is integration of b against the signed measure

Here L_i is the row of coordinate components of L in the ith chart. The determinant estimate of Proposition 2.6 shows that this is a finite signed measure and that

The same estimate applied to Du_j-L justifies the asserted limit. Thus \int b\,d\nu_\pi=0 for every Lipschitz b on Y. Lipschitz functions are uniformly dense in C(Y), so they determine finite signed measures on the compact metric space Y. It follows that \nu_\pi=0 as a measure.

Fix a chart i. Extend each coordinate of \phi_i^{-1}:Z_i\to\R^n to a Lipschitz scalar function F^a on Y. At almost every density point of E_i,

Take for \pi each of the n tuples obtained by omitting one member of (F^1,\ldots,F^n). Restricting the corresponding zero measures \nu_\pi to Z_i shows, component by component, that \theta_i L_i=0 almost everywhere on E_i. The multiplicity is nonzero, so L_i=0. There are only countably many charts and finitely many tuples per chart; hence L=0 in \mathcal L^2. Notice the order of this argument: we first obtained zero signed measures by testing on the whole cycle, and then localized those measures. No boundary or normality property of a chart restriction is used.

The closure of the graph is therefore the graph of an operator D, and, with its graph norm, it is the asserted Hilbert space \Hcal. In particular, functions that agree almost everywhere have the same gradient. Applying (4.22) to differences of Lipschitz approximants shows that a graph-norm Cauchy sequence is Cauchy in L^p. Its L^p limit agrees with its L^2 limit, and passing to the limit proves both the continuous inclusion and the extension of (4.22).

To prove the first calculus assertion, choose Lipschitz approximants u_{a,j}\to u_a in \Hcal, and pass to a common subsequence converging almost everywhere. Write U_j=(u_{1,j},\ldots,u_{k,j}) and U=(u_1,\ldots,u_k). Bounded first derivatives give \Phi(U_j)\to\Phi(U) in L^2. The constant \Phi(0) is harmless because \mu is finite. The chart chain rule gives

For each summand, subtract the proposed limit by writing

The first term tends to zero in \mathcal L^2 by the derivative bound; the second does so by dominated convergence. Closedness proves assertion (i).

For assertion (ii), fix a level t and choose \eta\in C_c^\infty((-1,1)) with 0\le\eta\le1 and \eta(0)=1. The functions

converge uniformly to zero, their derivatives are bounded by one, and \Phi_\epsilon'(s)\to\mathbf1_{\{t\}}(s). Assertion (i) and dominated convergence give

Closedness forces the latter limit to be zero. A Lipschitz, piecewise C^1 scalar function with finitely many corners can now be smoothed by convolution. The smoothed functions converge uniformly, their derivatives have a common bound, and the derivatives converge at every point other than the corners. The level-set conclusion removes the contribution of those corners, so dominated convergence and closedness give the claimed chain rule.

In particular, D(u_+)=\mathbf1_{\{u>0\}}Du. If u_j\to u in \Hcal, the positive parts converge in L^2, and their gradient difference is

Along any subsequence with almost-everywhere convergence, the second term tends to zero in \mathcal L^2 by the level-set conclusion at zero and dominated convergence. The first term is bounded in norm by \|Du_j-Du\|_2. Applying this observation to subsequences proves continuity for the original sequence. Taking positive parts of Lipschitz approximants proves assertion (iii).

Finally fix y\in Y. The distance r=d(y,\cdot) is bounded, |Dr|\le1 almost everywhere, and \mu is finite. Each term of (3.1) is therefore continuous as a linear functional of g in the graph norm. Approximate a nonnegative g\in\Hcal by the nonnegative Lipschitz functions just constructed and pass to the limit. This proves assertion (iv).□

Proof. Projecting the whole weighted current. First consider Lipschitz functions u_j with \sup_j\|u_j\|_{\Hcal}\le M_0<\infty. Fix a chart i, extend its inverse coordinates to a Lipschitz map F=(F^1,\ldots,F^n):Y\to\R^n, and fix a Lipschitz function \chi on Y. Define finite signed measures \lambda_{j,\chi} on \R^n by

for bounded Borel h. The chart representation defines these measures directly. They are supported in the fixed compact set F(Y), and the current mass bound gives

For h\in C_c^\infty(\R^n), expanding the row D(h\circ F)=\sum_a(\partial_a h\circ F)DF^a in a determinant shows that

All terms except a=l in the row expansion vanish because they have a repeated row. Apply the cycle identity (5.1), with a permutation of rows if necessary, to move h\circ F to the first coefficient. The intrinsic determinant bound then gives

uniformly in j and l. Indeed,

Constants here may depend on the fixed chart, \chi, and M_0.

Translation compactness. We give the compactness consequence of (5.3) in detail. This is the usual translation and mollification proof of Euclidean BV compactness [25, Chapter 2, Theorem 2.6], expressed directly for the projected signed measures. Let \tau_w(x)=x+w. For a smooth compactly supported test h, the fundamental theorem of calculus along x+tw and (5.3) imply

Smooth compactly supported tests recover the total variation of finite signed Radon measures, by regularity and approximation of continuous tests. After changing the constant, we obtain

Choose a nonnegative smooth convolution kernel \rho_\epsilon of integral one supported in the ball of radius \epsilon. Averaging (5.4) gives

For fixed \epsilon, the densities \rho_\epsilon*\lambda_{j,\chi} have a common compact support and uniform bounds on their values and first derivatives. The Arzelà–Ascoli theorem makes them precompact in the uniform norm on that support and hence in L^1(\R^n). Approximation by these families, first fixing \epsilon and then letting it decrease to zero, proves that \{\lambda_{j,\chi}:j\ge1\} is totally bounded in total variation. The space of finite signed measures is complete in that norm, so this family is precompact.

Isolating a chart in total variation. We now isolate the chart. Formula (5.2) also defines \lambda_{j,\mathbf1_{Z_i}}, using the bounded Borel coefficient \mathbf1_{Z_i}. By regularity of \mu and distance cutoffs on the compact metric space Y, choose Lipschitz \chi_k with

The mass bound and Cauchy–Schwarz give, uniformly in j,

For each fixed k the family on the left with coefficient \chi_k is precompact by the preceding argument. Uniform approximation (5.5) therefore makes \{\lambda_{j,\mathbf1_{Z_i}}:j\ge1\} precompact as well. There is no need for the Lipschitz constants of \chi_k to remain bounded.

Since F\circ\phi_i is the identity on E_i, the density of the isolated measure is exactly

Consequently, for any j,k,

Here J_i\le\Lip(\phi_i)^n. This calculation allows arbitrary signed, unbounded integer multiplicities: the fixed \theta_i appears as |\theta_i| in the total variation of a difference. Likewise, (5.5) already uses the full mass measure |\theta_i|J_i\,dz, so it requires no bound or regularity of the multiplicities. All integrations by parts above concerned the whole cycle A with a Lipschitz coefficient. None concerned a restriction to one chart.

Assembling the charts. We may now take a diagonal subsequence that is Cauchy in L^1(Z_i,\mu) for every i. The omitted charts have uniformly small contributions, because

uniformly in j. The chart images cover \mu up to a null set, and \mu is finite. The subsequence is thus Cauchy in L^1(\mu). Proposition 5.1 supplies a uniform L^p bound. Since p>2, interpolation with \alpha=(p-2)/(2(p-1))>0 gives

Finally, approximate the jth member of an arbitrary bounded sequence in \Hcal by a Lipschitz function with graph-norm error at most 1/j. The result just proved for those approximants proves compactness for the original sequence.□

Proof. Taking \epsilon=S_*/2 in (4.22) shows that

Thus \Lambda>0. The constant function m^{-1/p} has L^p norm one, so

Let u_j be an L^p-normalized minimizing sequence. It is bounded in \Hcal, since Q is a positive definite quadratic form equivalent to the squared graph norm. By weak compactness in a Hilbert space and Proposition 5.2, after taking a subsequence we have

Put z_j=u_j-u and t=\int|u|^p\,d\mu. Fatou’s lemma gives 0\le t\le1. We give the Brézis–Lieb splitting [8, Theorem 1] in the form needed here:

For every \epsilon>0, the mean-value theorem and Young’s inequality give a finite C_\epsilon such that, for real a,b,

Apply this with a=z_j and b=u, and set R_j=|u_j|^p-|z_j|^p-|u|^p. The functions

converge to zero almost everywhere and are bounded by (C_\epsilon+1)|u|^p. Dominated convergence, followed by \epsilon\downarrow0, shows that R_j\to0 in L^1, since the L^p norms of z_j are bounded. This proves (5.10).

Weak convergence in \Hcal also gives the quadratic splitting

The definition of \Lambda gives Q(u)\ge\Lambda t^{2/p}, with the same conclusion if u=0. For fixed \epsilon>0, apply (4.22) to z_j and use \|z_j\|_2\to0 to obtain

Letting \epsilon\downarrow0 therefore yields

If t<1, the inequalities a^{2/p}\ge a for 0\le a\le1 and S_*>\Lambda make the right-hand side strictly larger than \Lambda: indeed it is at least \Lambda t+S_*(1-t)>\Lambda. Thus t=1. Weak lower semicontinuity now gives Q(u)\le\Lambda, so u is a minimizer with \|u\|_p=1. Replacing it by |u| preserves both its L^p norm and its energy, by Proposition 5.1. We take u\ge0.

For each \psi\in\Hcal, differentiate the quotient Q(u+s\psi)/\|u+s\psi\|_p^2 at s=0. Its denominator is nonzero for sufficiently small s, and Hölder’s inequality justifies the derivative of the L^p integral: for |s|\le1 the derivative integrand is bounded by p(|u|+|\psi|)^{p-1}|\psi|\in L^1. Since u is a minimizer and \|u\|_p=1, this gives

Set

Homogeneity preserves the quotient minimum, and \Lambda(\Lambda/n)^{-1}=n gives (5.8). Since p/(p-2)=n/2, its pth moment is W=(\Lambda/n)^{n/2}<s_n by (5.9). Testing (5.8) with \psi=v gives Q(v)=nW, completing (5.7).□

Proof. We first prove the power assertion under the additional assumption v\in L^{2\gamma}. It is immediate for \gamma=1, so suppose \gamma>1. For N\ge1 define

The scalar functions in these definitions, extended by zero for negative arguments, are Lipschitz and piecewise C^1. Thus both compositions belong to \Hcal. Their derivatives show that

Indeed there is equality below N, while above N the ratio of the coefficients is one and 0<d_\gamma\le1. The gradient vanishes on the level set \{v=N\}, so no value at the corner matters. Testing (5.8) with \psi_N is legitimate; for example, v^{p-1}\psi_N\le N^{2\gamma-2}v^p is integrable. Dropping the nonnegative term n\int v\psi_N\,d\mu gives

Apply (4.22) to w_N with \epsilon=S_*/2 and multiply by d_\gamma. By (5.13),

Hölder’s inequality with conjugate exponents p/(p-2) and p/2 gives

Since v\in L^p, choose K\ge1, depending on \gamma but not N, so that the coefficient on the right is at most d_\gamma S_*/4. The contribution of \{v\le K\} is at most nK^{p-2}\|w_N\|_2^2. Absorbing the tail yields

The assumed L^{2\gamma} moment bounds the right-hand side uniformly in N, because w_N\le v^\gamma. Thus \|w_N\|_p is uniformly bounded. Equation (5.13), with another application of Hölder, then uniformly bounds E(w_N).

On \{v<N\} the gradient of w_N is \gamma v^{\gamma-1}Dv. Monotone convergence in these sets shows that \gamma v^{\gamma-1}Dv belongs to \mathcal L^2. At every truncation level the formulas and the level-set property give

The values and gradients converge almost everywhere to v^\gamma and \gamma v^{\gamma-1}Dv, respectively. The displayed bounds give strong L^2 convergence of both by dominated convergence. Closedness proves v^\gamma\in\Hcal with the asserted gradient, and the continuous inclusion into L^p gives v^\gamma\in L^p.

Starting with v\in L^p, set \ell_j=p(p/2)^j for j\ge0. The conditional assertion just proved, with \gamma=\ell_j/2, shows

Since p/2>1, these exponents tend to infinity. Finiteness of \mu then gives every finite positive moment of v. The conditional assertion therefore applies to every \gamma\ge1. Finally, for q\ge0 take \gamma=1+q/2 to obtain

which is (5.12).□

Proof. Let t_N=\min(v,N). The scalar calculus in Proposition 5.1 gives t_N\in\Hcal and Dt_N=\mathbf1_{\{v<N\}}Dv, where the level-set property handles v=N. For each N, choose a smooth, compactly supported cutoff in the given open neighborhood and equal to one on a neighborhood of the compact rectangle [0,R]\times[0,N]. Multiply F by this cutoff and extend by zero to obtain a global C^1 function with bounded first derivatives. It agrees with F and its first derivatives on the rectangle. The bounded-derivative calculus consequently gives F(r,t_N)\in\Hcal and

Writing L=\|\,|Dr|\,\|_\infty, the growth assumption gives bounds independent of N,

The squares of these bounds are integrable by Lemma 5.4: they are bounded by a constant times (1+v^{2M})(1+|Dv|^2). Since v is finite almost everywhere, the values and the displayed gradient formulas converge almost everywhere to the values and right-hand side in (5.14). Dominated convergence gives strong convergence in the two L^2 spaces, and closedness proves the result. A distance function has bounded values and |Dr|\le1, so it satisfies the hypotheses.□

Proof. We verify the hypotheses of Lemma 5.5 for the actual functions of (r,v) that occur here. This is necessary when \beta<1, since the scalar function v^\beta itself is not C^1 at zero.

For j=1,2 and z>0, define

On this half-plane,

As z\downarrow0, uniformly for r in bounded intervals, w(t)=1+O(t^2) and w'(t)=O(t). Thus

Extension by F_j(r,z)=z on z\le0 is therefore jointly C^1. The functions and their first derivatives have polynomial growth in z\ge0 on bounded r intervals.

For the other tests, write

Their partial derivatives are

Their values and all these derivatives tend to zero as z\downarrow0, uniformly for bounded r. The smallest power of z in the derivative formulas is 2\beta>0. Consequently, extension by zero on z\le0 makes both functions jointly C^1, even if f(0) or f'(0) is nonzero. Their values and first derivatives again have polynomial growth on the range of r.

These C^1 extensions and polynomial bounds satisfy all the hypotheses of Lemma 5.5. That lemma gives membership in \Hcal and the chain rules without a boundedness assumption on v. On \{v>0\}, differentiation gives (6.6). On \{v=0\}, Proposition 5.1 gives Dv=0 almost everywhere, while (6.5) gives \xi=0. Thus these formulas hold \mu-almost everywhere on all of Y.□

Proof. The extension of (3.1) to these tests is part of Proposition 5.1. Pointwise, with b=\sqrt{1-\abs{Dr}^2},

Multiply by ng\ge0 and combine this inequality with (3.1). All terms are integrable by boundedness of r, \abs{Dr}\le1, finite \mu, and g\in\Hcal.□

Proof. Fix the center and suppress its subscript on J. For a>0 let w(t)=(1+at^2)^{-(n-2)/2} as in Lemma 6.1. Since p(n-2)/2=n,

Testing (5.8) with vw(s)^2 and using the two product formulas (6.6) gives

Indeed the terms w^2\abs{Dv}^2+2ww'\ip{Dv}{\xi} in \abs{D(vw)}^2 equal \ip{Dv}{D(vw^2)}. The quotient minimality of v gives, on the other hand,

The test vw(s) is nonzero because w is strictly positive and W>0.

We next control the last integral in (6.10). Let f be smooth, bounded and nonnegative with bounded derivative, and use g,\psi from (6.7). Add (6.8) to (5.8) divided by 4\beta, with test \psi. The identity v\psi=\beta r^2g cancels the two undifferentiated terms and yields

Here we used v^{p-1}\psi=\beta s^2f(s)v^p. For completeness, after removing the common factor \kappa^2, the coefficients of \abs{Dr}^2, \ip{Dr}{Dv}, and \abs{Dv}^2 in the left integrand are respectively

The scalar functions f,f' in this display are evaluated at s. Since

these are precisely the coefficients of (nf(s)+sf'(s))\abs{\xi}^2/\kappa^2. Thus

To bound the energy term in (6.10), we want nf(t)+tf'(t)=w'(t)^2 in (6.14). Integrating this first-order equation leads to the choice

This is smooth and nonnegative. Both w' and w'' are bounded on \R, so f and f'(t)=2\int_0^1\tau^nw'(\tau t)w''(\tau t)\dd\tau are bounded. Integrating the derivative of \tau^nw'(\tau t)^2 gives

The boundary term at \tau=0 is zero. Combining (6.10), (6.11), (6.14), and (6.16), and dividing by nW, we obtain

The current and variational estimates have now reduced the comparison to (6.17). We evaluate its scalar right side and solve the resulting differential inequality for the transform. Put b=as^2\ge0 and q=(n-2)/2>0. Since

the substitution u=\tau^2 in (6.15) gives

The exact derivative identity

has an integrable right side on (0,1); this includes n=3, when q-1=-1/2. Integrating it shows that

In particular this expression lies in (0,1]. Tonelli’s theorem therefore turns (6.17) into

We have G(0)=1 and G(a)>0. For a>0, changing variable t=au gives

Continuity of J suffices for this differentiation. The finite second moment in (6.3) also allows differentiation at zero from the right, because

Consequently

Set h=G^{-1/q}=G^{-2/(n-2)}. For a>0,

where the last step follows from (6.19). Thus

By (6.20), the quotient on the left has limit 2L_y at zero. It follows that h(a)\ge1+2aL_y. Finally (6.19) gives

At a=0 both sides equal one. If L_y=0, then s=0 \nu-almost everywhere, so J=G=h=1 and the same conclusion holds. No higher chord moment is needed in the scalar comparison.□

Proof. Use the disjoint chart representation of Proposition 2.6. For one chart \phi_i:E_i\to Z_i\subset Y, write

and extend u_i and \rho_i by zero to \R^n. The bi-Lipschitz bounds give a positive lower bound for J_i on this chart almost everywhere. Since \abs{\theta_i}\ge1, v\in L^p(\mu) and \mu(Y)<\infty, u_i is Lebesgue integrable. The function \rho_i is integrable as well, by the mass representation. Thus ordinary Lebesgue differentiation applies to both functions and to 1_{E_i}.

For \nu-almost every y=\phi_i(z), we may therefore require all of the following: z is a density point of E_i; the centered metric differential is the Euclidean norm \abs{\cdot}_{G_0}, with G_0=G_i(z)>0; and u_i,\rho_i have Lebesgue values

Indeed the exceptional coordinate sets are Lebesgue-null and therefore have zero chart mass, while \{v=0\} has zero \nu-mass. There are only countably many charts. The representative of v agrees with its Lebesgue value at almost every such point. Fix a center with these properties.

Let \ell=\liminf_{a\to\infty}a^{n/2}J_y(a). If \ell=\infty, the desired lower bound is automatic. Otherwise, choose a_j\to\infty realizing this lower limit and put \varepsilon_j=a_j^{-1/2}. On every bounded set of h\in\R^n, the functions

converge in measure to v_0,\rho_0,1, respectively. For example, this follows by changing variables in the defining mean convergence at the Lebesgue point z. A diagonal subsequence over bounded balls makes all three convergences hold almost everywhere in \R^n. The subsequence still realizes \ell.

At almost every h, the points z+\varepsilon_jh consequently belong to E_i for all sufficiently large j. The centered metric differential then gives

When the chart point is absent, define the integrand below to be zero. Restricting the nonnegative integral defining J_y to this chart and changing variables gives, for every fixed bounded ball D_R\subset\R^n centered at zero,

The factor a_j^{n/2} has canceled the Jacobian \varepsilon_j^n. Fatou’s lemma, (6.23), and (6.24) now imply

This step uses no uniform integrability of the rescaled densities. Letting R\to\infty and applying monotone convergence gives the same bound with D_R replaced by \R^n.

The linear substitution H=v_0^{2\beta}G_0^{1/2}h has Jacobian v_0^{2n\beta}\sqrt{\det G_0}. Since 2n\beta=p, we obtain

The middle equality is the stereographic integral (4.21). The final inequality uses the nonzero integer multiplicity \abs{\theta_i(z)}\ge1. This proves (6.21) without a bound on the multiplicity.

If L_y>0, Proposition 6.3 and (6.21) give

Rearranging yields (6.22). If L_y=0, that upper bound is immediate. Its strict inequality follows from W<s_n.□

Proof. Lemma 5.4 gives all finite moments of v, so B_* is finite. It is positive because W>0. Write D_Y=\diam Y. Then 0\le I(y)\le D_Y^2B_*, and the triangle inequality gives

This proves the stated continuity.

Fix any y\in Y and put r=d(y,\cdot). By Lemma 5.4, the nonnegative function g=v^p belongs to \Hcal, and Dg=pv^{p-1}Dv. The extension of the radial inequality (3.1) furnished by Proposition 5.1 therefore applies to this g. Dividing by n and using p/n=2\beta gives

At almost every point of each current chart, take the direct sum of its cotangent inner-product space with \R. The vector

has norm one. Moreover, the identities

give the factorization

All powers in this formula are nonnegative powers of v, and the identity also holds on \{v=0\}. Applying the integral Cauchy–Schwarz inequality to (7.4) now yields

Both factors are finite: the first was bounded above, and the second is finite by the moment estimates and the Sobolev power v^{1+\beta} in Lemma 5.4.

To identify the last integral, use \psi=v^{1+2\beta}\in\Hcal in the Euler equation (5.8). Its gradient is D\psi=(1+2\beta)v^{2\beta}Dv. Thus

Since 1+2\beta=n\beta, division by n proves

Equations (7.5) and (7.6) prove (7.2). The argument fixed an arbitrary center before taking chart derivatives. Hence its conclusion holds for every center, although the null set on which Dr is undefined may depend on that center.

We have proved the all-center estimate W^2\le B_*I(y). It remains to combine it with the CAT(0) variance inequality [26, Proposition 4.4], which doubles this lower contribution when we average over centers. We recall its proof for the finite measure v^{p+2\beta}\mu. Compactness of Y and continuity of I provide a minimizer o\in Y. This is a barycenter of the finite measure v^{p+2\beta}\mu. For y\in Y, let o_t be the point at fraction t\in(0,1) on the segment from o to y. The CAT(0) squared-distance inequality, integrated against v(x)^{p+2\beta}\dd\mu(x), gives

Subtracting (1-t)I(o), dividing by t, and letting t\downarrow0 proves the variance inequality

This uses only the segment from o to y.

The chord kernel is a nonnegative Borel function on Y\times Y. Its integral is finite, since its expression below is bounded above by D_Y^2B_*^2/W^2. Tonelli’s theorem and (7.1) give

Here the first inequality is (7.7), and the last is (7.2) at y=o. The averaged expressions contain only distances and values of the Borel representative of v; the proof requires no joint choice of the derivatives of distance functions. This proves (7.3).□

Proof of Theorem 1.1. We first prove the statement in every compact CAT(0) space by induction on the cycle dimension n\ge2. For such a space Y, let V(T) denote the infimum of the masses of integral fillings of an integral n-cycle T in Y. Theorem 2.2 ensures that this infimum is finite and attained. It therefore suffices in each dimension to prove

The zero cycle has the zero current as a filling.

In dimension n=2, a violation of (7.9) would, by Lemma 2.3, produce a nonzero extremizing cycle that can be rescaled to satisfy (2.10). Proposition 3.2 excludes this cycle: its radial inequality and Euclidean lower density force \M(A)\ge s_2, whereas the normalization requires \M(A)<s_2. Thus (7.9) holds in dimension two, and attainment gives an integral filling with that mass bound.

Now let n>2, and assume the asserted existence of sharp integral fillings in dimension n-1 in every compact CAT(0) space. Suppose that (7.9) fails for a cycle T in a compact CAT(0) space Y. Since T\ne0, we may choose

Lemma 2.3 produces a nonzero cycle A maximizing V(B)-c\M(B)^{(n+1)/n} over integral n-cycles. After a constant rescaling of the metric it satisfies (2.9) and (2.10), in particular

The rescaled space remains compact and CAT(0), so the induction hypothesis continues to apply there.

Proposition 3.1 gives (3.1) for this cycle. The sharp filling statement in dimension n-1 is exactly the induction input to Proposition 4.1. Consequently Section 4, and in particular the almost sharp Sobolev inequality of Proposition 4.6, applies. Propositions 5.1 and 5.2 supply the completed calculus and compactness used in Proposition 5.3. We therefore obtain a nonnegative minimizer v satisfying (5.8) and (5.7), with

Lemmas 5.4 and 5.5 justify the powers and nonlinear tests used in Section 6 and in Lemma 7.1.

For the probability measure \nu=v^p\mu/W, the chord upper bound (6.22) holds for \nu-almost every center. Integrating that bound and using Lemma 7.1 gives

a contradiction. Hence no violating cycle T exists, and (7.9) holds in dimension n. By attainment in Theorem 2.2, every integral n-cycle has an integral filling realizing V(T), with the required mass bound. Its support is compact because it is a closed subset of Y. This completes the induction for every n\ge2.

Lemma 2.1 now transfers this bound to every compactly supported integral cycle T\in\I_n(X) in a proper CAT(0) space. It supplies a compactly supported integral filling S with

The zero cycle has the zero filling. This proves the theorem.□

Proof. In a relatively compact oriented coordinate ball, the top-dimensional representation of an integral current is integration against an integer-valued locally integrable coefficient \theta. Testing \partial C=0 with compactly supported smooth (d-1)-forms shows that every distributional partial derivative of \theta is zero. Thus \theta is constant almost everywhere on that ball. For completeness, convolution with a smooth kernel gives functions with zero ordinary gradient on each smaller ball; passing to the local L^1 limit proves the assertion. Compatibility on overlaps and connectedness give one constant on M. Some nonempty open set lies outside the compact support, so this constant is zero.□

Proof of Corollary 1.2. By the Cartan–Hadamard theorem and Hopf–Rinow, M is a proper CAT(0) space and is diffeomorphic to \R^d; in particular it is oriented and noncompact. See [9, Sections I.3, II.1A and II.4] for the metric comparison and completeness statements. Let [[\Omega]] be its integration current. Its mass is \vol(\Omega), and the mass of T=\partial[[\Omega]] is \operatorname{area}(\partial\Omega). Theorem 1.1, with n=d-1, gives a compactly supported S\in\I_d(M) with \partial S=T and

Lemma 8.1 applies to S-[[\Omega]], so S=[[\Omega]]. Rearranging its mass bound proves (1.3).

For a relatively compact set \Omega of finite perimeter, the standard representation of its integration current gives [[\Omega]]\in\I_d(M), with mass \vol(\Omega) and boundary mass equal to its perimeter; see [25, Chapter 3, Section 4, Theorem 4.3; Chapter 6, Section 3, Remark 3.15]. The same argument applies verbatim.□