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Proposition 15.32.54C06D1B94BD1

Proposition 15.32.5.4C06D1B94BD1 The real or complex vector space

\[ \mathcal D_ h=\{ p\in L^ n(\mathrm{d}V):\rho _ h(p)<+\infty \} \]

is complete for the norm $\| p\| _{\mathcal D_ h}=\| p\| _{L^ n}+\rho _ h(p)$. It is an order ideal in $L^ n$: if $q\in \mathcal D_ h$ and $|p|\leq |q|$, then $p\in \mathcal D_ h$. It contains $L^\infty $ and is therefore dense in $L^ n$ for the $L^ n$ norm.

Moreover, $\mathcal D_ h=L^ n$ if and only if $v_-(h)>0$, and in that case the norm $\| \cdot \| _{\mathcal D_ h}$ is equivalent to the $L^ n$ norm. If $h$ is Kähler, then

\[ \rho _ h(p)=\left(\int _ Xh^ n\right)^{-1/n}\| p\| _{L^ n}. \]

Proof. By proposition 15.32.4, $\mathcal D_ h$ is a vector space and an order ideal, and $\| \cdot \| _{\mathcal D_ h}$ is a norm on it. Bounded functions belong to $\mathcal D_ h$, because $b_ h(|p|^ n)=c_ h(|p|^ n)>0$ for bounded nonzero $p$.

We next show that $\rho _ h$ is lower semicontinuous on $L^ n$. If $p_ j\to p$ in $L^ n$, then $|p_ j|^ n\to |p|^ n$ in $L^1$, and the upper semicontinuity of $b_ h$ in theorem 15.32.1 gives

\[ \rho _ h(p)\leq \liminf _ j\rho _ h(p_ j). \]

This includes the case $b_ h(|p|^ n)=0$, in which the right side must be infinite; for $p=0$ the inequality is trivial. Now let $(p_ j)$ be a Cauchy sequence in $\mathcal D_ h$. It converges in $L^ n$ to some $p$, and lower semicontinuity gives $\rho _ h(p)\leq \liminf _ j\rho _ h(p_ j)<+\infty $, so $p\in \mathcal D_ h$. Applied to $p_ j-p_ k$ as $k\to \infty $, lower semicontinuity also gives

\[ \rho _ h(p_ j-p)\leq \liminf _{k\to \infty }\rho _ h(p_ j-p_ k), \]

which tends to zero as $j\to \infty $. Together with $L^ n$ convergence, this proves completeness.

If $v_-(h)>0$, proposition 15.28.7 gives $\rho _ h(p)\leq v_-(h)^{-1/n}\| p\| _{L^ n}$, so $\mathcal D_ h=L^ n$ with an equivalent norm. If $v_-(h)=0$, the density $f$ of theorem 15.27.3 has $b_ h(f)=0$ by theorem 15.28.4; thus $f^{1/n}$ belongs to $L^ n$ but not to $\mathcal D_ h$. The Kähler formula follows from the same proposition. $\square$

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