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Aug 13, 2020 at 17:55 vote accept Eduardo Longa
Aug 13, 2020 at 17:54 comment added Rohil Prasad Oh, I wasn't very careful about that. I think my argument can be fixed to show that, by picking some two-form $\eta$ supported disjointly from $\chi\omega_k$ such that $\beta(\eta) > 0$. Then, if we assume $\alpha(\chi\omega_k) > 0$ for every $k$ as well, you get $\lim_{k \to \infty} |(\alpha + \beta)(\chi\omega_k + \eta)|$ is equal to $\lim_{k \to \infty} |\alpha(\chi\omega_k)| + |\beta(\eta)|$. This is strictly greater than $M(\alpha)$, but bounded above by $M(\alpha + \beta)$ which gives you the strict inequality.
Aug 13, 2020 at 17:47 comment added Eduardo Longa Nice argument. In the particular case I described, the mass of $\alpha$ is stricly smaller than that of the sum, right?
Aug 13, 2020 at 17:37 history answered Rohil Prasad CC BY-SA 4.0