Proposition 15.32.6C926F63790E3

Proof. Suppose $v_+(h)<+\infty $. A continuous solution for a bounded density $q$ has mass $c_ h(q)\int _ Xq\, \mathrm{d}V\leq v_+(h)$. Applying this to the truncations of $|p|^ n$ and letting the truncation level increase gives $b_ h(|p|^ n)\| p\| _{L^ n}^ n\leq v_+(h)$, which is the asserted inequality. Hence $\rho _ h$ and $\| \cdot \| _{\mathcal D_ h}$ are equivalent, and completeness follows from proposition 15.32.5.

Suppose instead that $v_+(h)=+\infty $. By theorem 12.15.4, theorem 12.2.3, the normalization scalars of smooth strictly positive probability densities are unbounded above. For each $j$, choose such a density $f_ j$ with

Then $\| p_ j\| _{L^ n}=j$ and $\rho _ h(p_ j)=jc_ h(f_ j)^{-1/n}\leq 2^{-j}$, so the partial sums $P_ N=\sum _{j\leq N}p_ j$ form a Cauchy sequence for $\rho _ h$.

Fix a Gauduchon form $\gamma $. Its coefficient $a_\gamma $ is bounded below by a positive constant, so (15.29.2) gives $\rho _ h(p)\geq c\| p\| _{L^1}$ with a constant $c>0$. Hence $\sum _ jp_ j$ converges in $L^1$ to a nonnegative function $P$. If $P_ N$ converged for $\rho _ h$ to an element of $\mathcal D_ h$, that element would also be the $L^1$ limit, hence equal to $P$ almost everywhere. But $P\geq p_ j$ gives $\| P\| _{L^ n}\geq j$ for every $j$, so $P\notin L^ n$. Therefore $\rho _ h$ is not complete. $\square$