Theorem 15.30.3.B36BAA6B4AA0 The Hermitian form $H$ is not closed and satisfies $v_-(H)=0$. For every $0<\alpha <1$, the function $f=Rq_\alpha $ is a probability density in $L^1(\mathrm{d}V)$, belonging to no $L^ p(\mathrm{d}V)$ with $p>1$, and
\[ b_ H(f)=CV_0>0,\qquad \mathrm{MA}^{\mathrm{ext}}_ H(u_\alpha )=b_ H(f)f\, \mathrm{d}V. \]
The potential $u_\alpha $, pulled back to $X$, has supremum zero and finite mass $CV_0$. It belongs to $\mathcal E_{\mathrm{env}}(X,H,0)$, and $P_{\delta H}(u_\alpha )\not\equiv -\infty $ for every $0<\delta <1$.
Proof.
Step 1: the threshold. Write $a=\min _ XR>0$, $A=\max _ XR$, and $M_ s=\int _ Y\min (q_\alpha ,s)\kappa ^ k$ for $s>0$. By lemma 15.30.1, the truncations $\min (q_\alpha ,s)$ are continuous and strictly positive, and $M_ s\uparrow 1$ as $s\to \infty $. The truncations of $f$ satisfy
\[ R\min (q_\alpha ,j/A)\leq \min (Rq_\alpha ,j) \leq R\min (q_\alpha ,j/a). \]
Hence lemma 15.30.2 and the monotonicity of the scalar give
\[ \frac{CV_0}{M_{j/a}}\leq c_ H(\min (f,j)) \leq \frac{CV_0}{M_{j/A}}. \]
Both bounds tend to $CV_0$ as $j\to \infty $, so $b_ H(f)=CV_0$, whatever the reference volume $\mathrm{d}V$.
Step 2: the equation. For $u_{\alpha ,j}=\max (u_\alpha ,-j)$, the binomial expansion of bounded local products, with factor forms of degrees $m$ and $k$, gives
\[ \mathrm{MA}_ H(u_{\alpha ,j}) =C h_0^ m\wedge \mathrm{MA}_\kappa (u_{\alpha ,j}). \]
Multiplying by $\mathbf1_{\{ u_\alpha >-j\} }$ and letting $j\to \infty $, as in the definition (13.14.4) of the extended measure, we obtain
\[ \mathrm{MA}^{\mathrm{ext}}_ H(u_\alpha )=Ch_0^ m\wedge q_\alpha \kappa ^ k =CV_0\, q_\alpha \mu =CV_0f\, \mathrm{d}V, \]
a measure of mass $CV_0$. The bounds $a\leq R\leq A$ transfer the integrability properties of $q_\alpha $ to $f$.
Step 3: lower mass and nonclosedness. Pull back bounded $h_0$-plurisubharmonic functions whose masses on $X_0$ tend to zero. The same expansion shows that their $H$-masses are $C$ times their masses on $X_0$, so $v_-(H)=0$. If $h_0$ were closed, Stokes' theorem would give every bounded $h_0$-plurisubharmonic function the mass $V_0>0$, contradicting $v_-(h_0)=0$. Hence $h_0$, and therefore $H$, is not closed.
Step 4: the envelopes. Choose $\alpha <\beta <1$. For $b\geq 1$ and $\varepsilon >0$ with $\varepsilon \beta \leq 1$, the number
\[ K_ b=\sup _{s\geq 1}\{ b s^\alpha -\varepsilon s^\beta \} -b \]
is finite, since $\beta >\alpha $. The function $g_ b=-\varepsilon (1-\psi )^\beta -K_ b$ on $Y$ satisfies $\kappa +\mathrm{dd}^{\mathrm{c}}g_ b\geq (1-\varepsilon \beta )\kappa \geq 0$, by the chain-rule computation in the proof of lemma 15.30.1, applied to the increasing convex function $t\mapsto -\varepsilon (1-t)^\beta $, whose derivative is at most $\varepsilon \beta $ for $t\leq 0$. With $L=1-\psi \geq 1$, the inequality $g_ b\leq bu_\alpha $ reads $bL^\alpha -\varepsilon L^\beta -b\leq K_ b$, so it holds on $Y\setminus \{ o\} $, and also at $o$, where both sides are $-\infty $. Thus the pullback of $g_ b$ belongs to $\mathrm{PSH}(X,H)$ and lies below $bu_\alpha $. Since $b\geq 1$ is arbitrary, $u_\alpha \in \mathcal E_{\mathrm{env}}(X,H,0)$ by definition 16.28.61. For $0<\delta <1$, take $b=1$ and $\varepsilon \beta \leq \delta $. Then $g_1$ is $\delta H$-plurisubharmonic and lies below $u_\alpha $, so $P_{\delta H}(u_\alpha )\geq g_1$ is not identically $-\infty $.
$\square$
Pome
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