5.10 Limits, envelopes and local integrability5C7282E51B13

Proof. The limit $u$ is upper semicontinuous, since $\{ u<c\} =\bigcup _ i\{ u_ i<c\} $ is open for every real $c$. Fix a connected component $X_0$ of $X$; it has a countable basis.

Step 1: a decreasing sequence with the same limit. Let $i_0\in I$. We show that there are indices $i_0\leq i_1\leq i_2\leq \cdots $ such that $u_{i_ l}\downarrow u$ pointwise on $X_0$. Replacing $I$ by the cofinal directed set $\{ i\in I:i\geq i_0\} $, which does not change the limit, we only have to find indices $i_1\leq i_2\leq \cdots $. For each rational $c$, the open sets $\{ u_ i<c\} \cap X_0$ cover $\{ u<c\} \cap X_0$, and countably many of them suffice: for each basis element of $X_0$ contained in some set $\{ u_ i<c\} $, choose one such index $i$; every point of $\{ u<c\} \cap X_0$ lies in such a basis element. Let $k_1,k_2,\ldots $ be an enumeration, with repetitions allowed, of the indices chosen for all rational $c$. Since $I$ is directed, we can choose inductively $i_1\geq k_1$ and $i_{l+1}\geq i_ l,k_{l+1}$. The sequence $(u_{i_ l})$ is decreasing and bounded below by $u$. If $u(x)<c$ with $c$ rational and $x\in X_0$, then $u_{k_ l}(x)<c$ for some $l$, hence $u_{i_ l}(x)\leq u_{k_ l}(x)<c$. Therefore $u_{i_ l}\downarrow u$ on $X_0$.

Step 2: the limit. Let $V\subset X_0$ be a connected coordinate neighborhood, identified with a domain in $\mathbb {C}^ n$. On each connected component of each affine complex-line section of $V$, the restrictions of the $u_{i_ l}$ form a decreasing sequence of subharmonic functions, whose limit is subharmonic by theorem 3.4.9. Since $u$ is upper semicontinuous, it is either plurisubharmonic on $V$ or identically $-\infty $ on $V$, by definition 5.8.1. As $X_0$ is connected, lemma 5.9.4 shows that either $u$ is plurisubharmonic on $X_0$ or $u\equiv -\infty $ on $X_0$.

Step 3: convergence. Suppose first that $u$ is plurisubharmonic on $X_0$. Fix $i_0\in I$, take the sequence of Step 1, and let $K$ be a compact subset of a coordinate neighborhood in $X_0$. Then $0\leq u_{i_ l}-u\leq u_{i_0}-u$ almost everywhere, namely wherever $u$ is finite, and $u_{i_0}-u$ is integrable on $K$. By dominated convergence, $\| u_{i_ l}-u\| _{L^1(K)}\to 0$. Given $\varepsilon >0$, choose $l$ with $\| u_{i_ l}-u\| _{L^1(K)}<\varepsilon $; for every $i\geq i_ l$ we have $0\leq u_ i-u\leq u_{i_ l}-u$ almost everywhere, hence $\| u_ i-u\| _{L^1(K)}<\varepsilon $. Thus the net converges to $u$ in $L^1(K)$, and covering an arbitrary compact subset of $X_0$ by finitely many such sets $K$ gives convergence in $L^1_{\mathrm{loc}}(X_0)$.

Suppose now that $u\equiv -\infty $ on $X_0$, and let $K\subset X_0$ be compact and $A>0$. The open sets $\{ u_ i<-A\} $ cover $K$, so finitely many of them, say those with indices $j_1,\ldots ,j_ p$, already cover $K$. If $i\geq j_1,\ldots ,j_ p$, then $u_ i\leq u_{j_ q}$ for each $q$, and hence $u_ i<-A$ on $K$. This is locally uniform convergence to $-\infty $. $\square$

Proof. Since $F^*$ is defined by a local formula and plurisubharmonicity is a local property, it suffices to prove that $F^*$ is plurisubharmonic on every connected coordinate neighborhood $V$, which we identify with a domain in $\mathbb {C}^ n$. By theorem 3.4.23, $F^*$ is subharmonic on $V$, and $F$ is Lebesgue measurable and equal to $F^*$ almost everywhere on $V$. By Choquet's lemma, lemma 3.4.22, there is a countable set $J_ V\subset I$ with $(\sup _{i\in J_ V}u_ i)^*=F^*$ on $V$.

We next show that $\mathrm{dd}^{\mathrm{c}}F^*\geq 0$ on $V$. Fix $i_0\in I$, enumerate $J_ V\cup \{ i_0\} $ as $i_0,i_1,i_2,\ldots $, with repetitions if this set is finite, and put $v_ j=\max \{ u_{i_0},\ldots ,u_{i_ j}\} $. The functions $v_ j$ are plurisubharmonic by proposition 5.9.5, increase with $j$, are locally uniformly bounded above, and dominate the locally integrable function $u_{i_0}$. Their limit $G=\sup _ jv_ j$ lies between $\sup _{i\in J_ V}u_ i$ and $F$, so $G^*=F^*$, and hence $G=F^*$ almost everywhere by theorem 3.4.23 applied to the countable family. On a compact set $K\subset V$ on which every $u_ i$ is bounded above by $M$, we have $0\leq v_ j-u_{i_0}\leq M-u_{i_0}$, so dominated convergence gives $v_ j\to F^*$ in $L^1(K)$. Therefore $\mathrm{dd}^{\mathrm{c}}F^*\geq 0$, as a limit of positive currents. Since $F^*$ is subharmonic and dominates the locally integrable function $u_{i_0}$, it is locally integrable and hence the canonical representative of its class, and theorem 5.9.1 shows that $F^*$ is plurisubharmonic on $V$.

If $X$ has a countable basis, cover $X$ by countably many connected coordinate neighborhoods $V_1,V_2,\ldots $ and put $J=\bigcup _ mJ_{V_ m}$. On each $V_ m$ we have $F^*=(\sup _{i\in J_{V_ m}}u_ i)^*\leq (\sup _{i\in J}u_ i)^*\leq F^*$, and since the regularization is local, $(\sup _{i\in J}u_ i)^*=F^*$ on $X$.

Finally, in each chart the functions $u_ i$ are subharmonic, so the second assertion of theorem 4.1.26 shows that $\{ F<F^*\} $ is polar in the corresponding open subset of $\mathbb R^{2n}$, that is, real polar. $\square$

Proof. By the Cauchy–Schwarz inequality, with equality for $\lambda $ proportional to $(\overline{g_1(z)},\ldots ,\overline{g_ r(z)})$, we have

for every $z\in \Omega $, both sides being $-\infty $ at a common zero. The supremum does not change if we omit the $\lambda $ for which $\sum _ j\lambda _ jg_ j$ vanishes identically: at a point $z$ that is not a common zero, the maximizing $\lambda $ gives $\sum _ j\lambda _ jg_ j(z)\neq 0$, and at a common zero all terms equal $-\infty $. Some $g_ j$ is not identically zero, so the remaining family is nonempty. Its members are plurisubharmonic by proposition 5.5.3, theorem 5.9.1, and they are locally uniformly bounded above by the left-hand side. Their supremum $\log \sum _ j|g_ j|^2$ is already upper semicontinuous: it is continuous as a map to $[-\infty ,\infty )$, because $\sum _ j|g_ j|^2$ is continuous and its logarithm tends to $-\infty $ at a common zero. Hence it equals its regularization, which is plurisubharmonic by theorem 5.10.2. The particular case is the tuple $(z_1-a_1,\ldots ,z_ n-a_ n)$. $\square$

Proof. Plurisubharmonic functions are subharmonic by theorem 5.9.1. Hence, by the compactness theorem (theorem 3.4.19), a sequence as in the first assertion either tends to $-\infty $ locally uniformly, or has a subsequence converging in $L^1_{\mathrm{loc}}(\Omega )$ to a locally integrable subharmonic function $v$. In the second case $\mathrm{dd}^{\mathrm{c}}v\geq 0$, as a distributional limit of positive currents, and $v$ is the canonical representative of its class; so $v$ is plurisubharmonic by theorem 5.9.1.

Now let $(u_ j)$ converge in $L^1_{\mathrm{loc}}(\Omega )$ to $T$. As above, $\mathrm{dd}^{\mathrm{c}}T\geq 0$, so $T$ has a plurisubharmonic representative $u$, namely its canonical subharmonic representative. The $u_ j$ are locally integrable subharmonic functions converging in $L^1_{\mathrm{loc}}(\Omega )$, and $u$ is the canonical representative of their limit, so the remaining assertions are Hartogs' lemma for subharmonic functions, theorem 3.4.20, which requires no upper bound on the sequence; the last one is its first assertion with the constant sequence $x_ j=x$. $\square$

Proof. In this proof $C$ denotes constants that depend only on two cutoff functions $\chi $ and $\eta $ fixed below, and $C'$ constants that depend in addition on $p$. Since $v(0)>-\infty $, the function $v$ is not identically $-\infty $, and hence it is locally integrable by proposition 3.4.8. For $0<r<1$, the ball submean inequality (theorem 3.4.5) and $v\leq 0$ give

Let $\mu =\Delta v$ be the Riesz measure of $v$, and choose a smooth cutoff $\chi \geq 0$ supported in $D(0,19/20)$ and equal to one on $D(0,9/10)$. By the definition of the distributional Laplacian and (5.10.7),

Next choose a smooth cutoff $0\leq \eta \leq 1$ supported in $D(0,9/10)$ and equal to one on a neighborhood of $\overline{D(0,4/5)}$. The fundamental solution in the plane is $(2\pi )^{-1}\log |z|$, so theorem 3.4.14, applied with the measure $\nu =\eta \mu $, which is finite and supported in $\operatorname{Supp}\eta \subset D(0,9/10)$, gives

for every $z\in D(0,4/5)$, with $h$ harmonic on $D(0,4/5)$.

We estimate $P$. For $|\zeta |\leq 9/10$, the translated disc $D(0,4/5)-\zeta $ lies in $D(0,2)$, so

the integral $L_ p$ is finite because its part over $D$ equals $2\pi \int _0^1r|\log r|^ p\, \mathrm dr$, whose integrand is bounded on $(0,1]$, while $|\log |z||\leq \log 2$ on the rest of $D(0,2)$. Let $m=\nu (D)=\int \eta \, \mathrm d\mu $, so that $m\leq \mu (D(0,9/10))\leq C|v(0)|$. If $m=0$, then $P=0$. Otherwise, $|P(z)|\leq \frac m{2\pi }\int |\log |z-\zeta ||\, \mathrm d\nu (\zeta )/m$, and Hölder's inequality for the probability measure $\nu /m$ gives

and integrating over $z\in D(0,4/5)$ with Tonelli's theorem yields $\| P\| _{L^ p(D(0,4/5))}\leq L_ p^{1/p}m/(2\pi )\leq C'|v(0)|$.

We now estimate $h$. The case $p=1$ of the bound for $P$, in which $L_1$ is an absolute constant, and (5.10.7) give $\| h\| _{L^1(D(0,4/5))}\leq C|v(0)|$. For $|z|\leq 1/2$, the closed disc $\overline{D(z,1/4)}$ lies in $D(0,4/5)$, and the ball mean-value property of the harmonic function $h$ (proposition 3.1.5) gives $|h(z)|\leq (16/\pi )\| h\| _{L^1(D(0,4/5))}$. We have thus found a constant $C_0\geq 1$, depending only on the cutoff functions $\chi $ and $\eta $, such that $\nu (D)=m\leq C_0|v(0)|$ and $\sup _{\overline{D(0,1/2)}}|h|\leq C_0|v(0)|$; this proves (5.10.6). Adding the estimates for $P$ and $h$ on $D(0,1/2)$, we obtain

which proves the last assertion. $\square$

Proof. We prove the first estimate by induction on $n$. For $n=1$ it is lemma 5.10.5. Let $n\geq 2$, write $z=(z_1,z')$ with $z'\in \mathbb C^{n-1}$, and let $v$ be as in the statement with $v(0)>-\infty $. The function $w(z')=v(0,z')$ is the composition of $v$ with the holomorphic map $z'\mapsto (0,z')$, so by proposition 5.9.3 it is plurisubharmonic or identically $-\infty $ on $D^{n-1}$. Since $w(0)=v(0)>-\infty $, it is plurisubharmonic, hence locally integrable and finite outside a Lebesgue null set $E\subset D^{n-1}$. For $z'\in D^{n-1}\setminus E$, the function $z_1\mapsto v(z_1,z')$ is subharmonic and nonpositive on $D$, with the finite value $w(z')$ at the center, so lemma 5.10.5 gives

We integrate over $z'\in (\tfrac 12D)^{n-1}$ using Tonelli's theorem; the null set $E$, on which the inner integral may be infinite, does not contribute. The induction hypothesis applied to $w$ then gives

which is the first estimate with $C_{n,p}=C_ pC_{n-1,p}$.

For the second estimate, let $Q=(\tfrac 1{10}D)^ n$ and $|Q|$ its volume. If $\| v\| _{L^1(D^ n)}=0$, then $v=0$ almost everywhere, hence $v\equiv 0$ by theorem 5.9.1, and there is nothing to prove; we may also assume that $\| v\| _{L^1(D^ n)}<\infty $. Put $\lambda =2|Q|^{-1}\| v\| _{L^1(D^ n)}>0$. By Chebyshev's inequality, the set of points of $Q$ with $|v|>\lambda $ has measure at most $\lambda ^{-1}\int _ Q|v|\, \mathrm dV\leq |Q|/2$, so its complement in $Q$ has positive measure. Since $v$ is finite almost everywhere, there is a point $a\in Q$ with $v(a)>-\infty $ and

The polydisc of radius $4/5$ centered at $a$ lies in $D^ n$, and the polydisc of radius $2/5$ centered at $a$ contains $(\tfrac 14D)^ n$, since $\tfrac 14+\tfrac 1{10}<\tfrac 25$. We apply the first estimate to the plurisubharmonic function $\zeta \mapsto v(a+\tfrac 45\zeta )\leq 0$ on $D^ n$ and change variables:

This is (5.10.9). $\square$

Proof. If $X$ has dimension zero, every compact subset of $X$ is finite, plurisubharmonic functions are real-valued, and convergence in $L^ p_{\mathrm{loc}}$ is pointwise convergence; so both assertions hold. Let $X$ have dimension $n\geq 1$. Every compact subset of $X$ is covered by finitely many coordinate polydiscs, so it suffices to work near a point $x\in X$ in a chart in which $x=0$ and the closed polydisc $\overline{(RD)^ n}$ lies in the chart, for some $R>0$.

Let $v$ be plurisubharmonic. Being upper semicontinuous, $v$ is bounded above on $\overline{(RD)^ n}$ by some constant $A$. The function $\zeta \mapsto v(R\zeta )-A$ is plurisubharmonic and nonpositive on $D^ n$, and its $L^1$ norm on $D^ n$ is finite because $v$ is locally integrable. By (5.10.9), $v$ belongs to $L^ p$ on the polydisc $(\tfrac R4D)^ n$. This proves the first assertion.

Now let $(u_ j)$ be a sequence of plurisubharmonic functions converging in $L^1_{\mathrm{loc}}(X)$. For $p=1$ there is nothing to prove, so let $p>1$. By theorem 5.10.4 applied in charts, together with the uniqueness in theorem 5.9.1, the limit is represented by a plurisubharmonic function $u$. The norms $\| u_ j\| _{L^1((RD)^ n)}$ are bounded because the sequence converges. If $y\in (\tfrac R2D)^ n$, the ball $B(y,R/2)$ lies in $(RD)^ n$, so the ball submean inequality gives $u_ j(y)\leq |B(y,R/2)|^{-1}\| u_ j\| _{L^1((RD)^ n)}$. Hence there is a constant $M$, independent of $j$, with $u_ j\leq M$ and $u\leq M$ on $(\tfrac R2D)^ n$. Applying (5.10.9) to $\zeta \mapsto u_ j(\tfrac R2\zeta )-M$ and to $\zeta \mapsto u(\tfrac R2\zeta )-M$, whose $L^1$ norms on $D^ n$ are bounded independently of $j$, we obtain a bound for the $L^{2p}$ norms of $u_ j$ and $u$ on $(\tfrac R8D)^ n$ that is uniform in $j$. By a finite covering, for every compact set $K\subset X$ the norms $\| u_ j\| _{L^{2p}(K)}$ and $\| u\| _{L^{2p}(K)}$ are bounded independently of $j$.

We interpolate between $L^1$ and $L^{2p}$. For a function $f$ on $K$, write $|f|^ p=|f|^{p/(2p-1)}\cdot |f|^{p(2p-2)/(2p-1)}$ and apply Hölder's inequality with the conjugate exponents $(2p-1)/p$ and $(2p-1)/(p-1)$; since $\frac{p(2p-2)}{2p-1}\cdot \frac{2p-1}{p-1}=2p$, this gives

that is, $\| f\| _{L^ p(K)}\leq \| f\| _{L^1(K)}^{1/(2p-1)}\| f\| _{L^{2p}(K)}^{(2p-2)/(2p-1)}$. With $f=u_ j-u$, whose $L^{2p}(K)$ norms are bounded independently of $j$, we obtain

Proof. Since $v(0)\geq -1>-\infty $, lemma 5.10.5 gives a finite positive measure $\nu $ with support in $D(0,9/10)$ and a harmonic function $h$ on $D(0,4/5)$ such that

at every point of $D(0,4/5)$. Since $|v(0)|\leq 1$, the bounds (5.10.6) give $M=\nu (D)\leq C_0$ and $\sup _{D(0,1/2)}|h|\leq C_0$, where $C_0\geq 1$ is the universal constant of that lemma. Fix $\beta \in (0,1]$ with $\beta C_0\leq 2\pi $.

We show that $\int _{D(0,1/2)}\mathrm e^{-\beta P}\, \mathrm dV\leq 8\pi $. If $M=0$, then $P=0$, and this integral is $\pi /4$. Assume $M>0$ and put $b=\beta M/(2\pi )\in (0,1]$. Fix $z\in D(0,1/2)$. For $N\geq 1$ the Borel function $f_ N(\zeta )=-b\max \{ \log |z-\zeta |,-N\} $ is bounded on $D(0,9/10)$, so Jensen's inequality (lemma 2.2.4), applied to the convex function $\exp $ on $\mathbb R$ and the probability measure $\nu /M$, gives

As $N\to \infty $, the functions $f_ N$ increase to $-b\log |z-\zeta |$, which may take the value $+\infty $, and they are bounded below by the bounded function $f_1$. By monotone convergence, $\int f_ N\, \mathrm d\nu /M$ increases to $\int -b\log |z-\zeta |\, \mathrm d\nu (\zeta )/M=-\beta P(z)$, and therefore

for every $z\in D(0,1/2)$, with the convention $\mathrm e^{\infty }=\infty $. For every $\zeta \in D(0,9/10)$ the translated disc $D(0,1/2)-\zeta $ lies in $D(0,2)$, and $r^{-b}\leq 1+r^{-1}$ for $r>0$ because $0<b\leq 1$; hence

Integrating the bound for $\mathrm e^{-\beta P(z)}$ over $z\in D(0,1/2)$ with Tonelli's theorem, we obtain $\int _{D(0,1/2)}\mathrm e^{-\beta P}\, \mathrm dV\leq 8\pi $. Finally, since $\mathrm e^{-\beta v}=\mathrm e^{-\beta P}\mathrm e^{-\beta h}\leq \mathrm e^{\beta C_0}\mathrm e^{-\beta P}$ on $D(0,1/2)$, the lemma follows with $C=8\pi \mathrm e^{C_0}$. $\square$

Proof. Put $A_ n=\max \{ 1,2/|\frac18B|\} $. The mean of $|v|$ over $\frac18B$ is at most $1/|\frac18B|<A_ n$. Hence the set of points of $\frac18B$ with $|v|\leq A_ n$ has positive measure, since otherwise $|v|>A_ n$ almost everywhere on $\frac18B$ and the mean would be at least $A_ n$; so there is a point $a\in \frac18B$ with $|v(a)|\leq A_ n$. For each unit vector $e\in \mathbb C^ n$, the disc $\{ a+\tfrac 34\zeta e:|\zeta |<1\} $ lies in $B$, so the function

is subharmonic and nonpositive on $D$, with value at least $-1$ at $0$. Set $\alpha _ n=\beta /A_ n$, with $\beta $ as in lemma 5.10.11. That lemma and the change of variables $\zeta \mapsto \frac34\zeta $ give a bound, independent of $e$ and $v$, for

It remains to pass from these integrals over discs in complex lines to an integral over a ball. Let $\tau $ and $s_{2n-1}$ be as in (5.8.3). By that identity, applied with center $a$, and since $|\zeta |^{2n-2}\leq r^{2n-2}$ for $|\zeta |<r$, every Borel function $F\geq 0$, possibly taking the value $+\infty $, satisfies

for every $r>0$. We apply this with $F=\mathrm e^{-\alpha _ n v}$ and $r=3/8$. Since $\frac14B\subset B(a,3/8)$, the uniform bound for the line integrals proves (5.10.13). Finally, if $\| v\| _{L^1(B)}\leq M$ with $M>0$, then $v/M$ is a nonpositive plurisubharmonic function with $\| v/M\| _{L^1(B)}\leq 1$, and (5.10.13) for $v/M$ is the last assertion. $\square$