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In Demailly's Complex Analytic and Differential Geometry page 139: enter image description here

He said the trivial (zero) extension of the positive current $T$ (on $X\setminus E$), which denoted by $\tilde T$ is always positive on $X$. It seems like quite obvious, but I came across some obstacle when verifying that claim.

Notice that a current $T\in\mathcal D^{'}_{p,p}(X)$ is said to be positive if $\langle T, u\rangle\geqslant 0$ for all test forms $u\in\mathcal D_{p,p}(X)$ that are strongly positive at each point.Another way of stating the definition is:

$T$ is positive if and only if $T \wedge u \in \mathcal D_{0,0}^{'}(X)$ is a positive measure for all strongly positive forms $u \in\mathcal C_{p,p} ^{\infty}(X)$.

This is so because a distribution $S\in\mathcal D^{'}(X)$ such that $S(f)\geqslant 0$ for every non-negtive function $f\in\mathcal D (X)$ is a positive measure.

Here is my thought:

First, select arbitrary $u \in\mathcal C_{p,p} ^{\infty}(X)$, we have $u \in\mathcal C_{p,p} ^{\infty}(X\setminus E)$, then we would like to show that

$$\langle \tilde T\wedge u,f\rangle=\langle T\wedge u,f\rangle\geqslant 0, \qquad f\in\mathcal D(X),$$

due to the construction of $\tilde T$. However, supp $(f)\cap (X\setminus E)$ isn't compact in $X\setminus E$. Then, how can I infer LHS is non-negative?

Any help and suggestion are appreciated. Thanks a lot!

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    $\begingroup$ This is where the finite mass condition comes in. It ensures that, with your notations, $\langle T \wedge u ,f \rangle$ is convergent. To simplify notation, say $T= \sum \lambda_I dz_I \wedge d \bar {z_I}$ and $u=dz_J\wedge d \bar{z_J}$. Then $\langle T\wedge u,f\rangle = \int_{X\setminus E}\lambda_{J^c} fdV_{\mathbb C^n}$ while we know that the functions $\lambda_I$ are non-negative on $X\setminus E$ and locally integrable near $E$. $\endgroup$
    – Henri
    Commented May 11, 2020 at 7:44
  • $\begingroup$ I meant that the $\lambda_I$'s are positive measures (they don't necessarily have a density wrt the Lebesgue measure) with locally finite mass near $E$, hence $\langle \lambda_I, f\rangle$ is well-defined. $\endgroup$
    – Henri
    Commented May 11, 2020 at 8:08
  • $\begingroup$ @Henri, thanks for your help. One more question maybe sounds stupid, why we need $\langle T\wedge u,f \rangle $ is convergent, as when consdering the local representation, it seems like non-negative already? $\endgroup$
    – Invariance
    Commented May 11, 2020 at 9:07
  • $\begingroup$ The quantity $\langle T \wedge u ,f \rangle $ has to be a number (not $+\infty$) if you want to make $\widetilde T$ into a current. $\endgroup$
    – Henri
    Commented May 11, 2020 at 9:35
  • $\begingroup$ @Henri, ok, now I get it, thanks for your patience and assistance again. $\endgroup$
    – Invariance
    Commented May 11, 2020 at 9:37

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