This question comes from the *Complex Analytic and Differential Geometry* by *Demailly*. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if 
$$u= \sum_{s=1}^N \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$
where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.

**Here is my question:** Demailly claimed that the set of strongly positive forms is a closed set and he didn't give a proof. I don't think it is an obvious result: Consider an absolutely convergent infnite sum of $(q,q)$-forms:
$$\sum_{s=1}^{\infty} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$
It is an element in the closure of strongly positive cone, but I can't prove it is strongly positive unless $p=1,n-1$.

There is a similar (but not the same) question on math Stack Exchange: [question](https://math.stackexchange.com/questions/3795941/positive-forms-and-strongly-positive-forms-are-bidual). So I copy part of that question here for convenience.