PomeLemma 5.27.28.55862640C106 Let $\Omega \subset \mathbb C^ n$ be a domain. Suppose that $u_ j\in \mathrm{PSH}(\Omega )\cap L^\infty _{\mathrm{loc}}(\Omega )$ tend to zero in $L^1_{\mathrm{loc}}(\Omega )$ and satisfy $\sup _ j\| u_ j\| _{L^\infty (K)}<\infty $ for every compact set $K\subset \Omega $. Then $(\mathrm{dd}^{\mathrm{c}}u_ j)^ n(K)\to 0$ for every such $K$.
5.27.27 Measure concentration under integrable convergence4FE21E50E832
Decreasing approximation controls Bedford–Taylor products, but convergence of the potentials in $L^1$ alone does not. If plurisubharmonic functions tend to zero in $L^1_{\mathrm{loc}}$ and are locally uniformly bounded, the range estimate (5.27.3) shows that their Monge–Ampère measures tend to zero. Without a common bound, smooth bounded potentials that tend to zero, even in capacity, can have Monge–Ampère measures converging to a point mass.
Proof. Fix concentric balls $B'\Subset B\Subset B_1\Subset \Omega $, and put $\beta =\mathrm{dd}^{\mathrm{c}}|z|^2$. If $\psi $ is a nonnegative smooth cutoff function supported in $B_1$ and equal to one near $\overline B$, integration by parts gives
\[ 0\leq \int _ B\mathrm{dd}^{\mathrm{c}}u_ j\wedge \beta ^{n-1} \leq \int _{B_1}u_ j\mathrm{dd}^{\mathrm{c}}\psi \wedge \beta ^{n-1} \leq C\| u_ j\| _{L^1(B_1)}\longrightarrow 0. \]
Choose $M>0$ with $|u_ j|\leq M$ on $B$ for all $j$, and apply the first estimate in (5.27.3) to $u_ j+M$, whose values lie in $[0,2M]$. It bounds the top-degree mass of $B'$ by $C(2M)^{n-1}$ times the trace integral over $B$, which tends to zero; for $n=1$ the trace estimate itself suffices. Covering $K$ by finitely many such balls proves the lemma. $\square$
The construction uses a regularized maximum of a small multiple of a logarithm in one variable and a potential in the remaining variables. Its Monge–Ampère measure can be computed exactly by integration over circles in the first variable, and this will keep its top-degree mass fixed while the logarithmic coefficient tends to zero.
Lemma 5.27.29.E2307F7792A2 Let $n\geq 2$, let $U\subset \mathbb C^{n-1}$ be a domain, and let $v$ be a smooth plurisubharmonic function on $U$. Fix $\varepsilon ,R,\eta >0$ and $A\in \mathbb R$. Let $\chi _\eta $ be the convolution of $t\mapsto \max (t,0)$ with a smooth even probability density supported in $[-\eta ,\eta ]$, and put
\[ \begin{split} q(\zeta )& =\varepsilon \log (|\zeta |^2/R^2),\qquad T(\zeta ,z')=v(z')-A-q(\zeta ),\\ h(\zeta ,z')& =q(\zeta )+\chi _\eta (T(\zeta ,z')) \end{split} \]
for $\zeta \ne 0$. Then $h$ extends to a smooth plurisubharmonic function on $\mathbb C\times U$. Set
\[ w_\eta (t)=n\chi _\eta '(t)^{n-1}\chi _\eta ''(t),\qquad r_ t(z')=R\exp \bigl ((v(z')-A-t)/(2\varepsilon )\bigr ). \]
The function $w_\eta $ is nonnegative, supported in $[-\eta ,\eta ]$, and has integral one. For every $f\in C_ c(\mathbb C\times U)$,
97724F2A7D19
5.27.30
\begin{equation} \begin{split} \int f(\mathrm{dd}^{\mathrm{c}}h)^ n =\varepsilon \int _ U\biggl [\int _{-\eta }^{\eta }w_\eta (t) \int _0^{2\pi }f(r_ t(z')\mathrm e^{\mathrm i\vartheta },z') \frac{\mathrm d\vartheta }{2\pi }\, \mathrm dt\biggr ] (\mathrm{dd}^{\mathrm{c}}v)^{n-1}. \end{split} \end{equation}
Proof. By convexity, $0\leq \chi _\eta '\leq 1$ and $\chi _\eta ''\geq 0$. The function $\chi _\eta $ vanishes on $(-\infty ,-\eta ]$ and equals $t$ on $[\eta ,\infty )$. Since $T\to +\infty $ as $\zeta \to 0$, we have $h=v-A$ near each point with $\zeta =0$, which gives the smooth extension. Off the hyperplane $\{ \zeta =0\} $, the function $q$ is pluriharmonic and
\[ \mathrm{dd}^{\mathrm{c}}h=\chi _\eta '(T)\mathrm{dd}^{\mathrm{c}}v +\chi _\eta ''(T)\, \mathrm dT\wedge \mathrm d^{\mathrm c}T\geq 0. \]
The second summand has rank at most one, and the first has rank at most $n-1$. Hence
\[ (\mathrm{dd}^{\mathrm{c}}h)^ n=n\chi _\eta '(T)^{n-1}\chi _\eta ''(T) \, \mathrm dT\wedge \mathrm d^{\mathrm c}T\wedge (\mathrm{dd}^{\mathrm{c}}v)^{n-1}. \]
The last factor has top degree in the variables $z'$, so the derivatives of $v$ in $\mathrm dT\wedge \mathrm d^{\mathrm c}T$ contribute nothing. Write $s=\log (|\zeta |^2/R^2)$. In polar coordinates, $\mathrm dq=\varepsilon \, \mathrm ds$ and $\mathrm d^{\mathrm c}q=\varepsilon \, \mathrm d\vartheta /(2\pi )$. At fixed $z'$, the change of variables from $s$ to $t=v-A-\varepsilon s$ gives (5.27.30). There is no contribution near $\zeta =0$, where $h=v-A$ and the form $\mathrm{dd}^{\mathrm{c}}h$ has rank at most $n-1$. Finally, $w_\eta $ is the derivative of $(\chi _\eta ')^ n$, whose values at $-\eta $ and $\eta $ are zero and one. $\square$
Proposition 5.27.31.026A64B7A5A2 Let $\Omega \subset \mathbb C^ n$, $n\geq 2$, be a bounded domain and $a\in \Omega $. For every $c>0$, there are smooth bounded plurisubharmonic functions $u_ j\leq 0$ on $\Omega $ such that
\[ \| u_ j\| _{L^ p(\Omega )}\longrightarrow 0\quad (1\leq p<\infty ), \qquad (\mathrm{dd}^{\mathrm{c}}u_ j)^ n\longrightarrow c\delta _ a \quad \text{vaguely}. \]
The same sequence tends to zero in local capacity. Here the norms are taken with respect to Lebesgue measure, and vague convergence means convergence against every compactly supported continuous test function.
Proof. Translate $a$ to the origin and write $z=(\zeta ,z')\in \mathbb C\times \mathbb C^ m$, $m=n-1$. It suffices to construct measures of mass one and then multiply the potentials by $c^{1/n}$. For $\delta >0$, put $H_\delta (z')=\log (|z'|^2+\delta ^2)$. With $s=|z'|^2$, the complex Hessian of $H_\delta $ has the eigenvalue $(s+\delta ^2)^{-1}$ with multiplicity $m-1$ in the directions orthogonal to $z'$ and the eigenvalue $\delta ^2(s+\delta ^2)^{-2}$ in the radial direction. Taking the determinant and integrating in polar coordinates gives
\[ (\mathrm{dd}^{\mathrm{c}}H_\delta )^ m= \frac{m!}{\pi ^ m}\frac{\delta ^2}{(|z'|^2+\delta ^2)^{m+1}} \, \mathrm d\lambda _{2m}, \qquad \int _{|z'|<b}(\mathrm{dd}^{\mathrm{c}}H_\delta )^ m =\left(\frac{b^2}{b^2+\delta ^2}\right)^ m, \]
where $\lambda _{2m}$ is Lebesgue measure. These probability measures tend to $\delta _0$ as $\delta \downarrow 0$, since their mass outside each ball about the origin tends to zero.
Choose $R>\max (1,\sup _\Omega |\zeta |)$ and a ball $P\subset \mathbb C^ m$ containing the projection of $\Omega $ to the $z'$ variables. Set
\[ \begin{gathered} \delta _ j=\varepsilon _ j=j^{-1},\qquad \eta _ j=j^{-2}, \qquad v_ j=\varepsilon _ j^{-1/m}H_{\delta _ j},\\ A_ j=\sup _ Pv_ j+2\varepsilon _ j\log (Rj). \end{gathered} \]
Apply lemma 5.27.29 on $\mathbb C\times \mathbb C^ m$ to these data, and let $u_ j=h_ j-\eta _ j$, where $h_ j$ is the resulting function. The convolution satisfies $0\leq \chi _{\eta _ j}(t)-\max (t,0)\leq \eta _ j$, and $q+\max (T,0)=\max (q,v_ j-A_ j)$. Both branches of this maximum are negative on $\Omega $, so
\[ \varepsilon _ j\log (|\zeta |^2/R^2)-\eta _ j \leq u_ j\leq 0. \]
Each $u_ j$ is smooth and bounded on $\Omega $, since the second branch $v_ j-A_ j$ is bounded below on the bounded set $P$. The logarithm has finite $L^ p(\Omega )$ norm for every finite $p$, by polar integration on a fixed polydisk containing $\Omega $. Hence $u_ j\to 0$ in all these norms.
Let $B'\Subset B\Subset \Omega $ be concentric balls, and let $w\in \mathrm{PSH}(B)$ with $0\leq w\leq 1$. Since $|u_ j|\leq \varepsilon _ j|\log (|\zeta |^2/R^2)|+\eta _ j$, the weighted Chern–Levine–Nirenberg estimate applied to $\log (|\zeta |^2/R^2)$, together with the range estimate (5.27.3) for the constant term, gives
\[ \int _{B'}|u_ j|(\mathrm{dd}^{\mathrm{c}}w)^ n\leq C_ B(\varepsilon _ j+\eta _ j), \]
where $C_ B$ depends on the fixed balls and the logarithm but not on $j$ or $w$. Restricting the integral to $B'\cap \{ |u_ j|>t\} $, dividing by $t>0$ and taking the supremum over $w$ shows that the capacity of $B'\cap \{ |u_ j|>t\} $ relative to $B$ tends to zero.
In (5.27.30), the factor $\varepsilon _ j(\mathrm{dd}^{\mathrm{c}}v_ j)^ m$ equals $(\mathrm{dd}^{\mathrm{c}}H_{\delta _ j})^ m$. For $z'\in P$ and $|t|\leq \eta _ j$, the radii satisfy
\[ r_ t(z')\leq j^{-1}\exp (\eta _ j/(2\varepsilon _ j)) \leq \mathrm e^{1/2}/j. \]
Extend $f\in C_ c(\Omega )$ by zero to $\mathbb C^ n$. Outside the projection of $\operatorname {supp}f$ to the $z'$ variables, both $f(r_ t\mathrm e^{\mathrm i\vartheta },z')$ and $f(0,z')$ vanish. On this projection, the uniform continuity of $f$ and the bound on the radii make their difference tend to zero uniformly. Since the weights $w_{\eta _ j}(t)\, \mathrm dt$ and $\mathrm d\vartheta /(2\pi )$ and the measures $(\mathrm{dd}^{\mathrm{c}}H_{\delta _ j})^ m$ all have total mass one, it follows that
\[ \int _\Omega f(\mathrm{dd}^{\mathrm{c}}u_ j)^ n -\int _{\mathbb C^ m}f(0,z')(\mathrm{dd}^{\mathrm{c}}H_{\delta _ j})^ m \longrightarrow 0. \]
The second integral tends to $f(0)$, which proves the asserted vague limit. $\square$
Each approximating measure has a smooth density and charges no pluripolar set, whereas the limit is a point mass. By lemma 5.27.28, the potentials therefore have no common lower bound on some compact set. Thus convergence in capacity is compatible with concentration of the Monge–Ampère measures once common local bounds are lost.
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