11.34 Supporting probability measuresEE2EC5B3CDE0

Proof. The function $C_\varphi =-B_\varphi $ is convex and Lipschitz. Fix $F\in C(X)$. For a direction $h\in C(X)$, the limit

exists and is finite: convexity makes the difference quotients monotone in $t>0$, and their absolute values are at most $\| h\| _\infty $. Clearly $p$ is positively homogeneous. Given another direction $k\in C(X)$, apply convexity at the midpoint of $F+2th$ and $F+2tk$ and let $t\downarrow 0$; this gives $p(h+k)\leq p(h)+p(k)$. Thus $p$ is sublinear.

By the Hahn–Banach theorem, the zero functional on $\{ 0\} $ extends to a linear functional $\ell \leq p$. Since $p(h)\leq \| h\| _\infty $, applying this bound to $h$ and $-h$ shows that $\ell $ is continuous. Monotonicity of the convex difference quotients gives

which is (11.34.1) once $\ell $ is represented by a probability. Testing the translation rule with positive and negative constants gives $\ell (1)=1$. If $h\geq 0$, monotonicity of $B_\varphi $ gives $C_\varphi (F-th)\leq C_\varphi (F)$ for $t>0$, whereas the inequality just proved gives the lower bound $C_\varphi (F)-t\ell (h)$. Hence $\ell (h)\geq 0$. By the Riesz representation theorem, $\ell (h)=\int _ Xh\, \mathrm{d}\nu $ for a probability measure $\nu $, and $\nu \in \mathcal A_\varphi (F)$. Each inequality in (11.34.1) defines a weakly closed half-space in the compact space of probability measures, so their intersection is closed and convex, hence weakly compact.

It remains to prove absolute continuity. Let $K$ be compact with $\mathrm{d}V(K)=0$, and choose continuous functions $0\leq \chi _ j\leq 1$, equal to one on $K$, whose supports have volumes tending to zero. For any $\nu \in \mathcal A_\varphi (F)$, the supporting inequality with $G=F+\chi _ j$ gives

The right-hand side tends to zero by theorem 11.33.5, so $\nu (K)=0$. By inner regularity of finite Borel measures on $X$, the same holds for every Borel set of zero volume. The Radon–Nikodym theorem now gives $\mathrm{d}\nu =h_\nu \mathrm{d}V$ with $h_\nu \geq 0$ and $\int _ Xh_\nu \mathrm{d}V=1$. $\square$

Proof. Take a weakly convergent subsequence and fix $G\in C(X)$. In the inequality

the Lipschitz continuity of $B_\varphi $, the uniform convergence of $F_ j$ and the weak convergence of the probabilities allow passage to the limit. The limiting inequality holds for every $G$, so the limit measure belongs to $\mathcal A_\varphi (F)$. If that set is a singleton, every convergent subsequence has the same limit, and compactness of the space of probability measures proves the final assertion. $\square$

Proof. The one-sided limits exist and are finite by concavity and the Lipschitz bound. For $t>0$, every supporting probability $\nu $ at $F$ gives

Conversely, choose $\nu _ t\in \mathcal A_\varphi (F+tG)$ and apply its supporting inequality with the test function $F$; this bounds the same quotient below by $-\int _ XG\, \mathrm{d}\nu _ t$. Along a sequence $t\downarrow 0$, the measures $\nu _ t$ have a weak limit point, which belongs to $\mathcal A_\varphi (F)$ by proposition 11.34.3. The two bounds identify the right derivative with minus the maximum. Applying this identity to $-G$ gives the formula for the left derivative.

If $\mathcal A_\varphi (F)$ is a singleton $\{ \nu \} $, the two derivatives coincide for every $G$ and equal $-\int _ XG\, \mathrm{d}\nu $, a continuous linear functional of $G$. Conversely, if the two one-sided derivatives agree for every $G$, all supporting probabilities have the same integral against every continuous function, and the Riesz representation theorem shows that they coincide. $\square$

Proof. For $h\in C(X,\mathbb R)$ define the width

The extrema exist by weak compactness. We claim that the width is upper semicontinuous in $F$. Indeed, let $F_ j\to F$ and choose maximizing and minimizing probabilities at each $F_ j$. Along a subsequence realizing the upper limit of the widths, compactness of the space of probability measures gives weak limits of both choices. By proposition 11.34.3, both limits belong to $\mathcal A_\varphi (F)$, so the difference of their integrals against $h$ is at most $w_ h(F)$.

Consequently, for every $\epsilon >0$ the set $\{ F:w_ h(F)<\epsilon \} $ is open. It is also dense. If $h=0$ there is nothing to prove. Otherwise fix $F$ and an open ball about it. The function

is finite and concave on the real line, so it is differentiable outside an at most countable set: its one-sided derivatives are finite and monotone, and the nonempty open intervals between unequal one-sided derivatives at distinct points are disjoint, so each contains a different rational number. Choose a point of differentiability $t$ so close to zero that $F+th$ lies in the given ball. The derivative formulas of proposition 11.34.5 give $w_ h(F+th)=0$, which proves density.

The compact manifold $X$ is metrizable, so $C(X)$ is separable. Choose a countable uniformly dense family $(h_ j)$ in $C(X)$. By the Baire category theorem in the complete metric space $C(X)$, the set

is a dense $G_\delta $. At every point of this intersection, all supporting probabilities have the same integral against every $h_ j$. Since they have unit mass, uniform approximation gives equal integrals against every continuous function, and the Riesz representation theorem shows that the probabilities coincide. Conversely, a singleton has zero width in every direction, so the intersection is exactly $\mathscr U_\varphi $. The formula for the derivative follows from proposition 11.34.5, and the approximation assertion from proposition 11.34.3. $\square$

Proof. Suppose first that $n\geq 2$. By uniqueness of the constant, the normalization scalar of the smooth equation in theorem 14.8.1 is $B_0$, and that theorem gives, for every smooth real $G$,

Let $\nu \in \mathcal A_0(F)$. Testing the supporting inequality at $F+tG$ for both signs of $t$ and letting $t\to 0$ shows that $G$ has the same integral against $\nu $ and $\mu _ u$. Since smooth functions are uniformly dense in $C(X)$, it follows that $\nu =\mu _ u$, and the assertion follows from nonemptiness of $\mathcal A_0(F)$. In dimension one, Stokes' theorem gives $c_0(F)=\int _ X\omega /\int _ X\mathrm{e}^ F \mathrm{d}V$; differentiating its logarithm in the same way yields the stated measure, since $\omega +\mathrm{dd}^{\mathrm{c}}u=c_0(F)\mathrm{e}^ F\mathrm{d}V$. $\square$

Proof. The continuous trace of $\mathrm{dd}^{\mathrm{c}}u$ is the Euclidean Laplacian of $u$, up to the fixed normalization. Local constant-coefficient elliptic regularity gives $u\in W^{2,p}$ for every finite $p$. Indeed, subtract the Newtonian potential of a cutoff of this trace. Its second high-frequency derivatives are order-zero multipliers, and the low-frequency kernels are locally bounded; the remaining distribution is harmonic. Sobolev embedding gives a $C^{1,\alpha }$ representative for every $\alpha <1$, which agrees with the quasi-psh representative.

On a coordinate ball choose smooth $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \omega $. The radial regularizations of the bounded psh function $u+\rho $ decrease to it, and their complex Hessians converge uniformly on smaller balls. Continuity of bounded products and the finite smooth-background expansion identify $(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n$ with $\sigma ^ n$. For smooth real $h$, the same argument applies to $u+th$, which is admissible for all sufficiently small signed $t$ by uniform positivity. Define

Its equation has the same normalization scalar as the equation for $u$, so $B_0(F_ t)=B_0(F)$. The supporting inequality gives $\int _ X(F_ t-F)\, \mathrm d\nu \leq 0$ for each $\nu \in \mathcal A_0(F)$. Since $F_ t-F=tL_\sigma h+O(t^2)$ uniformly, using both signs yields stationarity.

Choose $p=4n+1$ and $q=p/(p-1)=1+1/(4n)$. By corollary 12.9.3, every supporting probability has an $L^ q$ density, since $q<n/(n-1)$. Smooth density in $W^{2,p}$ extends its stationarity to every $W^{2,p}$ test. By lemma 11.24.1, all these densities lie in a one-dimensional space. Their integrals equal one, so they coincide. Nonemptiness is theorem 11.34.2, and the derivative formula follows from proposition 11.34.5. $\square$

Proof. The Green solution of $\mathrm{dd}^{\mathrm{c}}u=(A-a)w$ on $\mathbb P^1$ has continuous complex Hessian and is bounded. Its pullback yields $\sigma =Aw+b\kappa >0$, and $\sigma ^{k+1}=(k+1)Ab^ kw\wedge \kappa ^ k$. Thus scalar uniqueness gives $c_0(F)=1$, and theorem 11.34.11 gives a unique support. For every smooth $h$ depending on all variables, the block decomposition

shows that $(A/a)w\otimes \nu _\kappa $ is stationary: the first term integrates to zero by Stokes on $\mathbb P^1$, and the second by stationarity on $Y$ and Fubini. This candidate has a bounded density with respect to $\mathrm dV$. The one-dimensional annihilator of lemma 11.24.1 and probability normalization therefore identify it with the supporting probability. $\square$

Proof. Take the Hopf surface $Y=(\mathbb C^2\setminus \{ 0\} )/\langle z\mapsto 2z\rangle $ and its unit-volume metric $\kappa =c|z|^{-2}\mathrm{dd}^{\mathrm{c}}|z|^2$, where $c>0$ is chosen so that $\int _ Y\kappa ^2=1$. Put $s=\log |z|^2$, $\alpha =\mathrm{dd}^{\mathrm{c}}s$ and $\beta =\mathrm ds\wedge \mathrm d^{\mathrm c}s$, so that $\kappa =c(\alpha +\beta )$. The differentiation in (23.1.3) applies in this dimension. Here $\alpha $ has rank one, so $\mathrm{dd}^{\mathrm{c}}\kappa =-c\alpha ^2=0$ and $\nu _\kappa =\kappa ^2$. Polar coordinates identify $Y$ with $S^3\times S^1$, hence $H^2(Y,\mathbb R)=0$. A closed positive form would be exact and have positive square integral, contradicting Stokes. Thus $Y$ is non-Kähler, and so is $\mathbb P^1\times Y$, by restriction.

Let $\tau \in [-1,1]$ be the height on $\mathbb P^1$, set $b=2+\tau $, and put

Since $0\leq f\leq 1$, this is a positive continuous function of mean $a$. Along a normal segment to the equator, $|\tau |$ is comparable to the distance $r$. The oscillation of $A$ is therefore bounded below by a positive multiple of $1/\log (e/r)$. Its Dini integral $\int _0^{r_0}\mathrm dr/(r\log (e/r))$ diverges, and this modulus is larger than every $r^\gamma $, $\gamma >0$, as $r\downarrow 0$. The logarithm of $A$ has the same lower bound, because $A$ is uniformly positive and bounded. The smooth term $2\log b$ contributes only $O(r)$, so $F=\log (3Ab^2)$ has neither regularity. The mixed component $\mathrm db\wedge \kappa $ shows that $\omega =aw+b\kappa $ is nonclosed. By proposition 11.34.13, its supporting probability is $(A/a)w\wedge \kappa ^2$. $\square$

Proof. The function $W$ is convex: both one-sided derivatives at zero equal one, and the derivative of $2\chi $ increases from one to two on $[0,\infty )$. Since $\omega _1=\mathrm{dd}^{\mathrm{c}}\chi (x)$ on the affine chart, $W-2\chi $ is $2\omega _1$-plurisubharmonic there. Since $W$ has slope one for $x\leq 0$ and agrees with $2\chi $ for $x\geq 0$, the function $W-2\chi $ extends to a global $2\omega _1$-plurisubharmonic function, with a logarithmic pole along $D$ and equal to zero where $|z|\geq 1$. The asserted bounds follow from $\psi =x-\chi $: the difference $\varphi -\psi $ equals $2\log 2-\chi $ for $x\leq 0$ and $\chi -x$ for $x\geq 0$.

For $C>0$ write $v_ C=P_\theta (\min \{ \psi +C,0\} )$. Rotations of the first factor and the transitive unitary action on the second preserve the whole competitor class, so the envelope is radial and independent of the second factor. The function $V_ C=2\chi +v_ C$ is then convex in $x$ and satisfies $V_ C\leq \chi +x+C$. This upper bound as $x\to -\infty $ forces the slope of $V_ C$ at $-\infty $ to be at least one, and by convexity all its slopes are at least one. For $x<0$,

For $x\geq 0$, the zero obstacle gives $V_ C\leq 2\chi =W$. Consequently $v_ C\leq \varphi $. If $C\geq 2\log 2$, the bounds already proved make $\varphi $ a competitor, so $v_ C=\varphi $. Letting $C\to \infty $ proves $P_\theta [\psi ]=\varphi $. Since $\varphi \sim \psi $ and the ordinary envelope depends only on the singularity type, also $P_\theta [\varphi ]=\varphi $.

Finally

Since $\psi \leq \varphi $, the function $\psi $ is a competitor for $P_{(3/4)\theta }(\varphi )$, which is therefore not identically $-\infty $. $\square$

Proof. On $X\setminus D$, the function $w=v-\psi $ is bounded, because $v\sim \varphi \sim \psi $, and satisfies $\beta +\mathrm{dd}^{\mathrm{c}}w\geq 0$. After adding local potentials of the closed form $\beta $, lemma 5.24.7 extends $w$ across $D$; the extensions agree on overlaps and give a bounded $\beta $-plurisubharmonic function on $X$. Conversely, for every bounded $w\in \mathrm{PSH}(X,\beta )$, the function $\psi +w$ is $\theta $-plurisubharmonic with current $\beta +\mathrm{dd}^{\mathrm{c}}w+[D]$, and it has the ordinary type of $\varphi $. The decomposition is unique, since two such functions $w$ agree off $D$ and hence everywhere.

Put $T=\beta +\mathrm{dd}^{\mathrm{c}}w$. The bounded-product construction gives

and Stokes' theorem against the constant function yields $\int _ XT^2=4$. The product gives no mass to $D$ by proposition 5.28.2; we also give a direct cutoff estimate, which will be used again below. Shift $w$ so that $0\leq w\leq M$, and set $f_ t=\max \{ \psi /t,-1\} $ for $t>0$. Choose smooth convex functions $\sigma _\epsilon $ with derivative in $[0,1]$, equal to $-1$ below $-1-\epsilon $ and to the identity above $-1+\epsilon $, converging uniformly to $\max \{ \cdot ,-1\} $. Then $f_{t,\epsilon }=\sigma _\epsilon (\psi /t)$ is globally smooth, constant near $D$, and

Testing the distributional identity against $f_{t,\epsilon }$, using positivity of $T+\beta $, and letting $\epsilon \to 0$, we obtain

Here the last integral equals four by Stokes' theorem. The right-hand side tends to zero as $t\to \infty $, since $f_ t\to 0$ off $D$ and $\beta ^2(D)=0$; as $-f_ t=1$ on $D$, this gives $T^2(D)=0$. Off $D$ the currents $\theta +\mathrm{dd}^{\mathrm{c}}v$ and $T$ agree, so locality and the fact that the extended measure puts no mass on the pole set prove (11.34.19).

For the Hermitian background $\beta $, theorem 14.19.2 with density $\mathrm e^ F$ gives a bounded solution $w$ and a constant $c>0$, and the correspondence above turns it into a solution $\psi +w$ in the ordinary type of $\varphi $. Every potential of that type has mass four, so integration forces $c=4/\int _ X\mathrm e^ F\mathrm dV$, which is the formula for $B_\varphi $. Jensen's inequality for the probability $\nu _ F=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$ gives

Thus $\nu _ F$ is a supporting probability. For any supporting probability, testing its inequality at $G=F\pm th$ and differentiating at $t=0$ shows that its integral against every $h\in C(X)$ equals that of $\nu _ F$.

For a smooth real $h$ and sufficiently small $|t|$, we have $\beta +t\mathrm{dd}^{\mathrm{c}}h>0$, so $\psi +th$ stays in the same ordinary type and its residual measure is $(\beta +t\mathrm{dd}^{\mathrm{c}}h)^2$, with derivative $2\mathrm{dd}^{\mathrm{c}}h\wedge \beta $ at $t=0$. Since $\beta $ is closed, the constant weight $\rho =1/4$ annihilates this derivative after integration, and $Q=\rho \beta $ gives the asserted probability. This normalization uses the residual mass: since $\int _ X\theta \wedge Q=3/2$, the cofactor $Q$ is not a normalized Gauduchon current for $\theta $ in the sense of section 11.7. $\square$

Proof. Smoothness and monotone convergence follow directly from the defining formula, also on $D$, where $\psi =-\infty $. Off $D$, set

The chain rule gives $\Omega _{\delta ,g}=A_\delta +R_\delta $, where

Thus $A_\delta \geq \omega _1+\omega _2$ and $R_\delta \geq 0$, and the inequality $\Omega _{\delta ,g}\geq \omega $ extends across $D$ by continuity. Closedness of $\theta $ and Stokes' theorem give the mass eight. Since $\Omega _{\delta ,g}$ is closed, the Gauduchon condition $\mathrm{dd}^{\mathrm{c}}(\rho \Omega _{\delta ,g})=0$ reduces to the elliptic equation $\mathrm{dd}^{\mathrm{c}}\rho \wedge \Omega _{\delta ,g}=0$ on the connected surface, whose smooth solutions are constant by the maximum principle. Normalization against $\Omega _{\delta ,g}^2$ gives $\rho =1/8$ and the stated adjoint cofactors and probabilities.

The rank-one form $R_\delta $ has square zero, hence $\Omega _{\delta ,g}^2=A_\delta ^2+2A_\delta \wedge R_\delta $. The bounded coefficients of $A_\delta $ converge off $D$ to those of $\beta $, which gives $A_\delta ^2\to 4\mathrm dV$. Write $N=\mathrm d\psi \wedge \mathrm d^{\mathrm c}\psi $. In $A_\delta \wedge R_\delta $, the part in which $R_\delta $ contributes only derivatives along the first factor (the normal part) is

Terms mixing a derivative along the first factor with one along the second vanish for degree reasons. The remaining tangential term is $\lambda (1-\lambda )(2-\lambda )\omega _1\wedge \mathrm dg\wedge \mathrm d^{\mathrm c}g$, which tends to zero by dominated convergence.

The normal coefficients can be computed in polar coordinates. Put $s=\delta \mathrm e^{g}$ and $t=|z|^2$. Then

After angular integration the radial measure of $\lambda (1-\lambda )N$ is

It is bounded by $\mathrm dy/(1+y)^2$ and tends to that measure, whose integral is one. With an additional factor $1-\lambda $, the rescaled measure becomes $\mathrm dy/(1+(1+s)y)^3$, which is bounded by $\mathrm dy/(1+y)^3$ and has limiting integral $1/2$. The mass of both measures outside every fixed tube around $D$ tends to zero, uniformly in the point of the second factor because $g$ is bounded. Testing against continuous functions therefore gives

Combining the normal and the remaining terms proves

Dividing by eight gives the second line of (11.34.22). The limit is positive since $4\omega _2+\gamma \geq 3\omega _2$, and its mass on $D$ is $4/8=1/2$ by Stokes' theorem on the second factor. Taking $g=0$ and a small nonconstant smooth $g$ gives different limits, since $\mathrm{dd}^{\mathrm{c}}g$ vanishes identically only for constant $g$.

The difference of the potentials satisfies

off $D$. The right-hand side tends to zero and is dominated by an integrable logarithmic pole, which proves $L^1$ convergence. Continuity of distributional differentiation then gives the limit of the cofactors. A locally integrable function times the smooth form $\beta $ has absolutely continuous coefficients, so it cannot produce the singular coefficients of $[D]$.

For convergence in capacity, use the capacity $\operatorname {Cap}_\omega $ of section 5.29; after subtracting one, its test potentials range over $v\in \mathrm{PSH}(X,\omega )$ with $-1\leq v\leq 0$. For such $v$, apply the cutoff argument of (11.34.20) with $\beta $ replaced by $\omega $ and $w=v+1$. Stokes' theorem gives $\int _ X\omega _1\wedge (2\omega +\mathrm{dd}^{\mathrm{c}}v)=2$, so

The same bound shows that $D$ has capacity zero. For $\eta >0$, the set of points off $D$ where $u_{\delta ,g}-\psi >\eta $ is contained in

This is a sublevel set of $\psi $ whose level tends to $-\infty $ as $\delta \downarrow 0$. The capacity bound for such sublevel sets, together with $\operatorname {Cap}_\omega (D)=0$, proves convergence in capacity.

Finally, $\Omega _{\delta ,g}\geq \omega $ gives $F_{\delta ,g}\geq -\log 2$. In the affine chart $|\mathrm d\psi |_\omega ^2\leq C/r$, while $g$ and its first two derivatives are bounded. The decomposition above yields $\Omega _{\delta ,g}\leq C_ g(1+r^{-1})\omega $, hence

Every finite power of this logarithmic bound is integrable. Smooth convergence off $D$ gives $F_{\delta ,g}\to 0$ there, and dominated convergence proves convergence in $L^ p$. $\square$

Proof. The capacity is the one of section 5.29, and $\beta \leq 2\omega $ gives $\operatorname {Cap}_\beta \leq 4\operatorname {Cap}_\omega $. Set $a_ j=(w_ j+M)/(2M)$ and $S_ j=\beta +\mathrm{dd}^{\mathrm{c}}a_ j$. Then $0\leq a_ j\leq 1$ and

Positivity of bounded mixed products justifies these inequalities. Thus $\int _ E\beta \wedge T_ j\leq 2M\operatorname {Cap}_\beta (E)$ for every Borel set $E$, and the same bound holds for $T$. By closedness and Stokes' theorem, $\int _ X\beta \wedge (T_ j+T)=8$. Now test the identity

against $f$. On $X\setminus E_{j,\eta }$ the potential difference has absolute value at most $\eta $, and on $E_{j,\eta }$ at most $2M$. Consequently

Dividing by four proves (11.34.25). Letting first $j\to \infty $ with $\eta $ fixed and then $\eta \downarrow 0$ shows that $\int _ Xf\, \mathrm d\nu _ j\to \int _ Xf\, \mathrm d\nu $. All $\nu _ j$ and $\nu $ have mass one by proposition 11.34.18, and smooth functions are uniformly dense in $C(X)$, so $\nu _ j\rightharpoonup \nu $.

The constant test potential in the definition of capacity gives $4\mathrm dV(E)\leq \operatorname {Cap}_\beta (E)$. Splitting the potential difference over the same sets yields

Taking the limits in the same order proves $L^1$ convergence, and hence convergence of the currents $Q_ j$ by continuity of distributional differentiation. The limiting cofactor satisfies $\mathrm{dd}^{\mathrm{c}}Q=0$, $\int _ X\beta \wedge Q=1$, and $T\wedge Q=\nu $ as a bounded product. For smooth real $h$, the signed measure $2\mathrm{dd}^{\mathrm{c}}h\wedge T$ is the formal linearization of the residual operator at $w$ in the direction $h$, and its integral against the constant $1/4$ vanishes by Stokes' theorem. It is the derivative of the residual measure along $w+th$ whenever these functions remain $\beta $-plurisubharmonic.

If $T_ j$ is smooth and strictly positive, the maximum principle on the connected surface makes its Gauduchon multiplier constant, and the mass four normalizes it to $1/4$. Finally, if $\psi +w$ solves the equation at $F$, then $\nu =T^2/4=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$, which is the unique supporting probability by proposition 11.34.18. $\square$