Flat norm

Remarks

Taking $S = 0$ shows $\mathbf{F}(T) \le \mathbf{M}(T)$. If $\mathbf{F}(T_j - T) \to 0$, then $T_j(\omega) \to T(\omega)$ for every smooth compactly supported $m$-form $\omega$, so flat convergence implies weak convergence. When $\mathbf{F}(T) < \infty$, there is a dual description of $\mathbf{F}(T)$ as the supremum of $T(\omega)$ over the forms $\omega$ such that $\omega$ and $d\omega$ have comass at most $1$ everywhere.

Example: in $\R$, let $a < b$ and $T = \delta_b - \delta_a$, a $0$-current. Then $T = \partial [\![a, b]\!]$, so $\mathbf{F}(T) \le b - a$, and $\mathbf{F}(T) \le \mathbf{M}(T) = 2$. Testing $T$ on a function $\varphi$ with $\abs{\varphi} \le 1$ and $\abs{\varphi'} \le 1$, chosen with $\varphi(b) - \varphi(a) = \min(b - a, 2)$, shows that $\mathbf{F}(T) = \min(b - a, 2)$. So $\delta_b - \delta_a \to 0$ in the flat norm as $b \to a$, while its mass stays $2$.

Non-example: the mass $\mathbf{M}$ is a norm on currents of finite mass, but it is not the flat norm, as the example shows.