Let $\Omega$ be a domain in $\mathbb{C}^n$, and let $f$ and $g$ be two complex differential forms on $\Omega$: $$ f=\sum_{I, J} f_{I, J} d z^{I} \wedge d \bar{z}^{J}, $$ $$ g=\sum_{K, L} f_{K, L} d z^{K} \wedge d \bar{z}^{L}. $$ Define the pointwise inner product and norm as usual: $$ |f|^{2}:=\sum_{I, J}\left|f_{I, J}\right|^{2}, $$ $$ \langle f, g\rangle:=\int_{\Omega} f \bar{g} d \lambda, $$ where $d \lambda$ is the Lebesgue measure.
a) Assume the bidegree of $f\wedge g$ is not more than $(n,n)$. Then does the following hold pointwisely? $$|f\wedge g|^2\leq |f|^2|g|^2$$ (the calculation sees rather tedious and I am not clear whether it is right or not.)
b) This is similar with a). Assume $f$ is of $(p,q)$ degree and $g$ is of $(n-p,n-q)$ degree. Then how can we prove $$|\int_{\Omega} f\wedge g |^2 \leq \int_{\Omega}|f|^2 d \lambda \int_{\Omega}|g|^2 d \lambda?$$
Consider a real vector space V with an orthonormal base e_1,…,e_n. The (squared) norm |.|^2 on ΛV is defined by the requirement that the exterior products e_I, I⊆{1,…,n}, should be orthonormal as well. Example: n=6 and α=e_12+e_34+e_56. Then |α|^2=3, but α∧α=2e_1234+2e_1256+2e_3456, |α∧α|^2=4+4+4=12.
By the example, the inequality is not true without a constant. An easy estimate with a large constant can be found in Lemma 2.6 of [Matsumura, arxiv 1308.2033].