Let $M$ be a smooth oriented closed Riemannian manifold. Let $\delta$ be the adjoint of the exterior derivative. Let $\sigma$ be a differential $k$-form on $M$ with coefficients in $L^2(M)$. We say $\sigma$ is
$$ \text{ weakly closed if } \, \int_M \langle \sigma, \delta \alpha \rangle=0 \, \text{ for every $\alpha \in \Omega^{k+1}(M)$}, \tag{1}$$
$$ \text{ weakly co-closed if } \, \int_M \langle \sigma, d \alpha \rangle=0 \, \text{ for every $\alpha \in \Omega^{k-1}(M)$}, \tag{2}$$
$$ \text{ weakly harmonic if } \, \int_M \langle \sigma, \Delta \alpha \rangle=0 \, \text{ for every $\alpha \in \Omega^{k}(M)$}, \tag{3}$$
(where all the integrals are w.r.t the Riemannian volume form.)
Question: Suppose $\sigma$ is weakly harmonic. Is there a way to show $\sigma$ is weakly closed and co-closed, without using the regularity theorem (i.e. any such $\sigma$ must be smooth, hence strongly harmonic, hence closed and co-closed).
The usual way to show (classical) harmonic forms are closed and co-closed is to plug in $\alpha=\sigma$ in $(3)$:
$$ 0= \int_M \langle \sigma, \Delta \sigma \rangle=\int_M \langle \sigma, d\delta \sigma \rangle+\int_M \langle \sigma, \delta d \sigma \rangle=\int_M \langle \delta \sigma, \delta \sigma \rangle+\int_M \langle d\sigma, d \sigma \rangle .$$
This reasoning cannot be imitated directly in the weak setting.
Interestingly, it can be shown immediately from the definitions, that weakly closed and co-closed forms are weakly harmonic (without passing through regularity theory).