15.35 Exponential equations for an individual densityC3635E28D6B1

Proof. Since $f_ k\leq f_{k+1}$, lemma 14.19.28 gives $u_{k+1}\leq u_ k$.

We first show that every exponential subsolution $v$ lies below each $u_ k$. Let $t>0$. On the set $\{ u_ k<v-t\} $,

Since $u_ k$ is bounded, $P_{h/2}(u_ k)\geq \inf _ Xu_ k$ is not identically $-\infty $, and $v-t\preceq u_ k$. Strict domination, theorem 15.2.2, therefore gives $u_ k\geq v-t$. Letting $t\downarrow 0$ gives $u_ k\geq v$.

In particular $s\leq u_ k\leq u_1$ for every $k$, so the decreasing limit $u_\lambda \geq s$ is not identically $-\infty $ and belongs to $\mathrm{PSH}(X,h)$. By lemma 5.29.19, $u_ k\to u_\lambda $ in capacity. Choose a constant $A\geq \sup _ Xu_1$. The nonpositive functions $u_ k-A$ satisfy

and $\mathrm{e}^{\lambda A}f_ k\to \mathrm{e}^{\lambda A}f$ in $L^1(\mathrm{d}V)$. Theorem 15.34.2, applied with $\mu =\mathrm{d}V$ and constant exponents, shows that $\mathrm{MA}^{\mathrm{ext}}_ h(u_\lambda )$ is finite and that $u_\lambda $ solves (15.35.1). Finally, every exponential subsolution lies below all $u_ k$, hence below $u_\lambda $, which is itself a subsolution. $\square$

Proof. By theorem 15.28.4, a function $v$ as in the statement exists. Since $v\leq 0$, the function $s=v+\lambda ^{-1}\log b$ satisfies

so it is an exponential subsolution. Theorem 15.35.2 gives the greatest solution $u_\lambda $ and the bound $u_\lambda \geq s$; in particular, $c_\lambda \geq b$.

Set $g_\lambda =\mathrm{e}^{\lambda v_\lambda }f$. By lemma 5.44.5, applied to the finite measure $f\, \mathrm{d}V$, we have $g_\lambda \to f$ in $L^1$ as $\lambda \downarrow 0$. Each $g_\lambda $ is nonzero, since $v_\lambda $ is finite almost everywhere. The equation for $u_\lambda $ reads $\mathrm{MA}^{\mathrm{ext}}_ h(v_\lambda )=c_\lambda g_\lambda \, \mathrm{d}V$, so theorem 15.28.4, applied to the density $g_\lambda $, gives

The upper semicontinuity of the threshold (theorem 15.32.1) therefore gives

This proves $c_\lambda \to b$. Since $\mathrm{e}^{\lambda u_\lambda }f=c_\lambda g_\lambda $ and $g_\lambda \to f$ in $L^1$, it also proves $\mathrm{e}^{\lambda u_\lambda }f\to bf$ in $L^1$. $\square$

Proof. By lemma 5.26.2, normalized $h$-plurisubharmonic functions are uniformly bounded in every $L^ s$ with $s<+\infty $. By (15.32.8) with $r=2n$, their norms $\| v\| _{\mathcal D_ h}$ are therefore uniformly bounded; let $C_ h$ be $1/n$ times this bound. Let $v$ be normalized and $K>0$. On $\{ p\leq K\} $ we have

since $1-\mathrm{e}^{-x}\leq x$ for $x\geq 0$, while on $\{ p>K\} $ we use the bound $p(1-\mathrm{e}^{\lambda v/n})\leq p$. The monotonicity of $\| \cdot \| _{\mathcal D_ h}$ and the triangle inequality give

and taking the infimum over $K$ proves (15.35.6). By proposition 15.32.9, and since the $L^ n$ tails of $p$ tend to zero, $\| p\mathbf1_{\{ p>K\} }\| _{\mathcal D_ h}\to 0$ as $K\to \infty $. Choosing $K$ first and then letting $\lambda \downarrow 0$ shows that $\eta _ p(\lambda )\to 0$.

Now set $q_\lambda =p\, \mathrm{e}^{\lambda v_\lambda /n}$, so that $q_\lambda ^ n=g_\lambda =\mathrm{e}^{\lambda v_\lambda }f$, with $g_\lambda $ as in the proof of corollary 15.35.3. Since $0\leq q_\lambda \leq p$ and $\rho _ h(p-q_\lambda )\leq \eta _ p(\lambda )$ by (15.35.6), the triangle inequality and monotonicity give

The same proof gives $b\leq c_\lambda \leq b_ h(g_\lambda )=\rho _ h(q_\lambda )^{-n}$, which yields the scalar bound. For the measures, the factorization of $p^ n-q_\lambda ^ n$ gives $0\leq p^ n-q_\lambda ^ n\leq np^{n-1}(p-q_\lambda )$, and Hölder's inequality implies

Since $\mathrm{MA}^{\mathrm{ext}}_ h(v_\lambda )=c_\lambda g_\lambda \, \mathrm{d}V$, the identity $c_\lambda g_\lambda -bf=(c_\lambda -b)g_\lambda -b(f-g_\lambda )$, together with $\| g_\lambda \| _{L^1}\leq \| f\| _{L^1}=\| p\| _{L^ n}^ n$, proves the total-variation bound.

Finally, suppose $p\in L^ r$ with $n<r<+\infty $, and choose $n<s<r$. By (15.32.8) with exponent $s$, and since $\| \cdot \| _{L^ n}\leq \| \cdot \| _{L^ s}$, we have $\| \cdot \| _{\mathcal D_ h}\leq (1+C_ s)\| \cdot \| _{L^ s}$. Hölder's inequality and lemma 5.26.2 then give, for every normalized $v$,

Since $0\leq 1-\mathrm{e}^{\lambda v/n}\leq \lambda (-v)/n$, the left side of (15.35.6) is therefore at most $C_{h,r}\| p\| _{L^ r}\lambda $, and the two estimates above can be repeated with this bound in place of $\eta _ p(\lambda )$. For $\lambda $ so small that $C_{h,r}\| p\| _{L^ r}\lambda \leq R/2$, the function $t\mapsto (R-t)^{-n}$ has bounded derivative on $[0,R/2]$, so the scalar error $c_\lambda -b$, and hence the total-variation error, is $O(\lambda )$. $\square$

Proof. In the notation of the proof of lemma 15.30.1, $t=r/(1+r)$ and $s(r)=L$. Since the mass of $q_\alpha \kappa ^ k$ on $\{ r_1<r<r_2\} $ is $Q(r_2)^ k-Q(r_1)^ k$, this measure has radial distribution $-\mathrm{d}(Q^ k)/\mathrm{d}L=W(L)\, \mathrm{d}L$. The Fubini–Study measure $\kappa ^ k$ corresponds to $Q=t$, so its radial distribution is $k\mathrm e^{-k(L-1)}\mathrm{d}L$. Both measures are invariant under the unitary group of the chart, so $q_\alpha $ is the ratio of the two distributions, which is the stated formula. The lower bound $q_\alpha \geq b_0$ follows from $\kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha \geq (1-\alpha )\kappa $ off $o$. As $L\to +\infty $, $t$ tends to zero exponentially, so $Q\sim \alpha L^{\alpha -1}$ and $D\sim \alpha (1-\alpha )L^{\alpha -2}$; this gives the asymptotics of $W$. Consequently,

and the mass of $B_\tau \kappa ^ k$ on $\{ L>T\} $, which equals $\int _ T^\infty \mathrm e^{\tau (L^\alpha -1)}W(L)\, \mathrm{d}L$, is infinite for every $T$. $\square$

Proof. (1) By theorem 15.30.3, $\mathrm{MA}^{\mathrm{ext}}_ H(u_\alpha )=CV_0q_\alpha \mu $. Off the pole fiber $X_0\times \{ o\} $ we have $\mathrm e^{\lambda u_\alpha }B_\lambda =q_\alpha $ and $f\, \mathrm{d}V=F\mu \leq AB_\lambda \mu $, hence

and both measures vanish on the pole fiber. Thus $s_\lambda $ is an exponential subsolution, and its mass is $CV_0$, since adding a constant does not change the Monge–Ampère measure. Theorem 15.35.2 now gives the greatest solution $u$ and the finiteness of its mass. The mass $\int _ X\mathrm{e}^{\lambda u}f\, \mathrm{d}V$ is positive because $u$ is finite almost everywhere and $f$ has positive integral.

(2) Fix $\lambda >0$. By hypothesis, there is $T$ such that $F/q_\alpha \leq \mathrm e^{(\lambda /2)L^\alpha }$ almost everywhere on $\{ L>T\} $. Since $\mathrm e^{(\lambda /2)L^\alpha }\leq \mathrm e^{\lambda }\mathrm e^{\lambda (L^\alpha -1)}$ for $L\geq 1$, we have $F\leq \mathrm e^{\lambda }B_\lambda $ there. On the complement $\{ L\leq T\} $, which is compact and disjoint from the pole fiber, $F$ is essentially bounded, while $B_\lambda \geq q_\alpha \geq b_0>0$. Increasing the constant, we obtain $F\leq AB_\lambda $ almost everywhere on $X$. $\square$

Proof. Put $w_ j=2^{-j}$ and $M_ j=\max _{X_0}F_ j$.

Step 1: a smooth baseline. Fix $C^ j$ norms $\| \cdot \| _{C^ j}$ on $X_0$, computed in a fixed finite atlas, so that they increase with $j$. Choose $0<\varepsilon _ j\leq 1/2$ with

Then $E=\sum _ jw_ j\varepsilon _ jF_ j$ converges in every $C^ l$ norm, so $E$ is smooth on $X_0$, and $0<E\leq b_0/4$.

Step 2: bumps escaping to the pole. Choose a smooth function $\chi :\mathbb R\to [0,1]$, supported in $[0,1]$, positive on $(0,1)$ and flat at $0$ and $1$. Starting with $T_0=0$, choose inductively $T_ j\geq 2$ with $T_ j>T_{j-1}+2$ and

This is possible because, by lemma 15.35.8 with $\tau =1/j$, the factor $\mathrm e^{(L^\alpha -1)/j}W(L)$ is uniformly large on sufficiently distant intervals. On the annulus $\{ T_ j\leq L\leq T_ j+1\} $ put $\beta _ j=\chi (L-T_ j)B_{1/j}/Z_ j$, extend $\beta _ j$ by zero, and set $g_ j=\varepsilon _ j+(1-\varepsilon _ j)\beta _ j$. Since $\chi $ is flat at the endpoints, $\beta _ j$ is smooth; it vanishes near $o$ and near the hyperplane $\{ L=1\} $. Its integral against $\kappa ^ k$ is one, so $g_ j$ is a smooth strictly positive probability density with respect to $\kappa ^ k$. Moreover,

Step 3: the scalars $c_ H(f_ j)$. By theorem 14.2.1 on $Y$, there is a smooth $\phi _ j$ with $\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ j>0$ and $(\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ j)^ k=g_ j\kappa ^ k$; the scalar is one because both sides have mass one. On $X_0$, the same theorem gives a smooth $v_ j$ with $h_0+\mathrm{dd}^{\mathrm{c}}v_ j>0$ and $(h_0+\mathrm{dd}^{\mathrm{c}}v_ j)^ m=c_{h_0}(F_ j)F_ j\, \mathrm dV_0$. Since only one term of the binomial expansion survives for dimensional reasons,

Uniqueness of the scalar gives $c_ H(f_ j)=C c_{h_0}(F_ j)\leq 2^{-2j}$.

Step 4: properties of $f$. By Tonelli's theorem, $\int _ Xf\, \mathrm dV=\sum _ jw_ j=1$. The density of $f\, \mathrm{d}V$ with respect to $\mu $ is

The annuli are disjoint and escape to the pole, so only finitely many bumps meet any compact subset of $Y\setminus \{ o\} $; hence $F$, and therefore $f$, is smooth off the pole fiber. On the pole fiber all bumps vanish and $F=E$, so $f$ is finite everywhere. Also $f\geq w_1f_1\geq \frac12\min _ Xf_1>0$. If $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq af\, \mathrm{d}V$ with $a>0$, then $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq aw_ jf_ j\, \mathrm{d}V$, and lemma 15.27.1 gives

This is impossible, so $b_ H(f)=0$ by theorem 15.28.4.

Step 5: exponential subsolutions. Fix $\lambda >0$, and choose $J$ with $1/j\leq \lambda $ for $j\geq J$. Since $L^\alpha -1\geq 0$, we have $B_{1/j}\leq B_\lambda $ for these $j$. As the bumps have disjoint supports, Step 2 bounds their total contribution $\sum _{j\geq J}w_ j(1-\varepsilon _ j)F_ j\beta _ j$ by $B_\lambda /2$. The baseline satisfies $E\leq b_0/4\leq B_\lambda /4$, because $B_\lambda \geq q_\alpha \geq b_0$. The remaining finitely many bumps are bounded by the constant

and $K_\lambda \leq (K_\lambda /b_0)B_\lambda $. Hence $F\leq A_\lambda B_\lambda $ with $A_\lambda =3/4+K_\lambda /b_0$, and theorem 15.35.9 gives the exponential subsolution $u_\alpha +\lambda ^{-1}\log (CV_0/A_\lambda )$ and the greatest solution. Finally, a bounded solution $u$ would satisfy $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq \mathrm{e}^{\lambda \inf _ Xu}f\, \mathrm{d}V$, a positive scaled subsolution, contradicting $b_ H(f)=0$. Hence every solution is unbounded. $\square$

Proof. Let $u$ be a solution at $\lambda _0$, and put $s=u-\max (0,\sup _ Xu)$. Then $s\leq \min (u,0)$, so for $\lambda \geq \lambda _0$,

Thus $s$ is an exponential subsolution at every $\lambda \geq \lambda _0$, and theorem 15.35.2 gives a solution. Hence $\Lambda _ h(f)$ contains $[\lambda _0,+\infty )$ together with each of its points $\lambda _0$. A nonempty subset of $(0,+\infty )$ with this property is determined by its infimum $a$ and by whether $a$ belongs to it, which gives the stated forms. If $b_ h(f)>0$, corollary 15.35.3 gives $\Lambda _ h(f)=(0,+\infty )$. $\square$

Proof. Set $g_ j=\mathrm e^{\lambda _ jv_ j}f$. By lemma 5.44.5, applied to the finite non-pluripolar measure $f\, \mathrm dV$, $g_ j\to f$ in $L^1$; each $g_ j$ is nonzero, since $v_ j$ is finite almost everywhere. Since the Monge–Ampère measure is unchanged by adding constants,

by theorem 15.28.4. The upper semicontinuity of the threshold, theorem 15.32.1, gives $\limsup _ jc_ j\leq b_ h(f)=0$. Since $0\leq g_ j\leq f$, the total masses are at most $c_ j\| f\| _{L^1}$ and tend to zero. Moreover $\lambda _ jM_ j=\log c_ j\to -\infty $, and $0\leq \mathrm e^{\lambda _ ju_ j}\leq \mathrm e^{\lambda _ jM_ j}=c_ j$. Finally, once $\lambda _ j<1$ and $\log c_ j<0$, we have $M_ j=\lambda _ j^{-1}\log c_ j\leq \log c_ j$, so $M_ j\to -\infty $. $\square$

Proof. Set $q=\min (f,1)$, a nonzero bounded density. By theorem 14.19.3, there is a continuous $w\in \mathrm{PSH}(X,h)$ with $\mathrm{MA}_ h(w)=\mathrm e^{\tau w}q\, \mathrm dV$. Put $t_\tau =w+\max (0,-\inf _ Xw)$. Then $t_\tau \geq 0$ and $t_\tau \geq w$, so for $\lambda \geq \tau $,

Let $u$ be an exponential subsolution at an exponent $\lambda \geq \tau $, and let $\epsilon >0$. On $\{ t_\tau <u-\epsilon \} $,

The bounded function $t_\tau $ satisfies $P_{h/2}(t_\tau )\not\equiv -\infty $, and $u-\epsilon \preceq t_\tau $. Thus theorem 15.2.2 gives $t_\tau \geq u-\epsilon $, and letting $\epsilon \downarrow 0$ gives $t_\tau \geq u$.

For the order estimate, put $\delta =K_\tau (\lambda _2-\lambda _1)/\lambda _2$. By the first part, $U_{\lambda _1}\leq t_\tau \leq K_\tau $, so $(\lambda _2-\lambda _1)U_{\lambda _1}\leq \lambda _2\delta $, that is, $\lambda _1U_{\lambda _1}\geq \lambda _2(U_{\lambda _1}-\delta )$. Since the Monge–Ampère measure is unchanged by adding constants,

Thus $U_{\lambda _1}-\delta $ is an exponential subsolution at $\lambda _2$, and the maximality of $U_{\lambda _2}$ gives $U_{\lambda _2}\geq U_{\lambda _1}-\delta $. Finally, $1-x^{-1}\leq \log x$ for $x\geq 1$; with $x=\lambda _2/\lambda _1$ this gives $\delta \leq K_\tau \log (\lambda _2/\lambda _1)$, which proves the monotonicity. $\square$

Proof. Suppose first that $a\in \Lambda _ h(f)$, let $u_ a$ be a solution at $a$, and put $s=u_ a-\max (0,\sup _ Xu_ a)$. As in the proof of proposition 15.35.12, $s$ is an exponential subsolution at every $\lambda \geq a$, so $U_\lambda \geq s$. Hence the greatest solutions at any exponents $\lambda _ j\downarrow a$ have suprema at least $\sup _ Xs>-\infty $.

Conversely, let $\lambda _ j$ and $u_ j$ be as in the statement. The lemma with $\tau =a$ gives the common upper bound $u_ j\leq K_ a$. Write $M_ j=\sup _ Xu_ j$. After extraction, $M_ j\to M\in \mathbb R$ and, by theorem 5.25.2, $u_ j-M_ j\to v$ in $L^1$ for some $v\in \mathrm{PSH}(X,h)$. Then $u_ j\to u=v+M$ in $L^1$ and, after a further extraction, almost everywhere. Since $a\leq \lambda _ j\leq \lambda _1$, $u_ j\leq K_ a$ and $K_ a\geq 0$, the densities $\mathrm e^{\lambda _ ju_ j}f$ are bounded by $\mathrm e^{\lambda _1K_ a}f$, and dominated convergence gives

Lemma 15.28.3 now gives $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq \mathrm e^{au}f\, \mathrm dV$. Thus $u$ is an exponential subsolution at $a$, and theorem 15.35.2 gives a solution at $a$.

Finally, suppose $a\notin \Lambda _ h(f)$, and let $u_ j$ be solutions at exponents $\lambda _ j\downarrow a$. If $\sup _ Xu_ j$ did not tend to $-\infty $, a subsequence would satisfy the condition just shown to imply $a\in \Lambda _ h(f)$. Hence $M_ j\to -\infty $. Once $M_ j<0$, since $\lambda _ j\geq a>0$,

and the total mass $\int _ X\mathrm e^{\lambda _ ju_ j}f\, \mathrm dV$ is at most $\mathrm e^{aM_ j}\| f\| _{L^1}$. $\square$

Proof. Let $K=K_ a$ be the constant of the lemma with $\tau =a$, and let $\lambda _ j>a$ decrease to $a$. By the lemma, the functions

decrease and satisfy $V_ j\geq U_ a$. Their limit $V$ therefore belongs to $\mathrm{PSH}(X,h)$ and satisfies $V\geq U_ a$. Dominated convergence between the integrable bounds $U_ a$ and $V_1$ gives $V_ j\to V$ in $L^1$, and lemma 5.29.19 gives convergence in capacity. Since the added constants tend to zero, $U_{\lambda _ j}\to V$ in both senses.

The nonpositive functions $U_{\lambda _ j}-K$ satisfy

and $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. Theorem 15.34.2 therefore gives $\mathrm{MA}^{\mathrm{ext}}_ h(V)=\mathrm e^{aV}f\, \mathrm dV$, together with total-variation convergence of the measures $\mathrm{MA}^{\mathrm{ext}}_ h(U_{\lambda _ j})$. Hence $V$ is a solution at $a$, so $V\leq U_ a$ by maximality, while $V\geq U_ a$ by construction. Thus $V=U_ a$. Since every sequence of exponents converging to $a$ from above has a decreasing subsequence, this proves the assertions. $\square$

Proof. Choose $\tau \in \Lambda _ h(f)$ with $0<\tau <a$; the upper-interval property gives such a choice. Put $K=K_\tau $ from lemma 15.35.15 and

This family is increasing. For a sequence $\lambda _ j\uparrow a$ with $\tau \leq \lambda _ j<a$, its members lie between the integrable potentials $V_\tau $ and $V_ a$. The upper-envelope construction in section 5.40 gives

Indeed cofinality and the order give the same pointwise supremum for every such sequence. Regularization changes it only on a pluripolar set. Dominated convergence between the integrable bounds gives $L^1$ convergence to $V^-_ a$, and lemma 5.29.19 gives convergence in capacity. Since the added constants converge, set

Then $U_{\lambda _ j}\to L_ a$ in both topologies and $L_ a\leq U_ a$; the latter inequality survives regularization because $V_ a$ is upper semicontinuous.

The potentials $U_{\lambda _ j}-K$ and their limit $L_ a-K$ are nonpositive. Apply theorem 15.34.2 with exponents tending to $a>0$ and densities $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. It proves the exact equation and total-variation convergence.

Every sequence approaching $a$ from the left has a cofinal increasing subsequence. If potential convergence in either topology, or total-variation convergence of the equation measures, failed, a sequence staying outside a fixed neighbourhood of the proposed limit would therefore have an increasing subsequence, contrary to the preceding argument. This proves the full left limit and measure convergence. Repeating the construction with another $\tau $ gives the same $L^1$ limit; canonical quasi-plurisubharmonic representatives that agree almost everywhere agree everywhere. Thus the auxiliary choice does not change $L_ a$. $\square$

Proof. Fix $a_0<\alpha <\beta $, choose an admitted $0<\tau <\alpha $, and use the corrected family $V_\lambda $ from the preceding proof on $[\alpha ,\beta ]$. Its integrals

form a finite bounded increasing real function. For $a\in (\alpha ,\beta ]$, the left-limit proposition gives

If $L_ a\ne U_ a$, this quantity is positive: the integrable nonnegative difference cannot vanish almost everywhere without equality of the canonical potentials. For any finite ordered collection of such points, the sum of their jumps is at most $I(\beta )-I(\alpha )$. Consequently there are only finitely many jumps at least $1/m$ for each positive integer $m$, and at most countably many positive jumps. A countable cover by intervals with rational endpoints inside $(a_0,+\infty )$ proves the assertion about $D$.

If $a\notin D$, the left limit is $U_ a$, and proposition 15.35.17 supplies the same right limit in both topologies and for the equation measures. Splitting an arbitrary approaching sequence into its two sides proves the stated two-sided convergence. $\square$