Proposition 18.28.45A86573651304

Proof. 1 As $v$ is bounded, $N^ k_{v,K}$ is clearly finite on $\mathrm{H}^0(X,L^ k\otimes T)$. In order to show that it is a norm, it suffices to show that for any $s\in \mathrm{H}^0(X,L^ k\otimes T)$, $N^ k_{v,K}(s)=0$ implies that $s=0$. In fact, we have $s|_ K=0$, hence $s=0$ by the connectedness of $X$ and the assumption that $K$ is non-pluripolar.

2 As $v$ is bounded, clearly $N^ k_{v,\mu }$ is finite and satisfies the triangle inequality. Non-degeneracy follows from the fact that $N^ k_{v,K}$ is a norm and (18.28.44). $\square$