Proposition 5.35.1DBEB8E0E0318

Proof. For $\| f\| _ m\leq 1$, the submean inequality gives

Taking the supremum over the unit ball of $\mathcal H_ m$ and then the logarithm proves the upper bound.

For the lower bound, suppose that $\psi (x)>-\infty $. The local Ohsawa–Takegoshi point-extension theorem gives a holomorphic $f$ with $f(x)=1$ and

where the constant is independent of the plurisubharmonic weight and of $m$. After normalizing $f$ in $\mathcal H_ m$, we see that the squared norm of point evaluation is at least $C^{-1}\mathrm{e}^{m\psi (x)}$, which proves the lower bound. At a pole of $\psi $ the lower bound holds trivially.

Choosing $r=1/m$ in the upper bound and using upper semicontinuity of $\psi $ proves pointwise convergence. On a fixed relatively compact set, the same estimates bound $\beta _ m$ above by a constant and below by $\psi -C_1/m$. Since $\psi $ is locally integrable, dominated convergence also gives convergence in $L^1_{\mathrm{loc}}$.

For the Lelong estimate, take suprema over $B(x,r)$ in the two inequalities, bounding the right-hand side of the upper one by a supremum over $B(x,2r)$. Divide by $\log r^2<0$ and let $r\downarrow 0$. After the division, the term $-(2n/m)\log r$ contributes $-n/m$, while the constant terms disappear. This proves the two inequalities in (5.35.3); since Lelong numbers are nonnegative, the lower bound may be replaced by its maximum with zero. $\square$