$\newcommand{\R}{\mathbb R}$ $\newcommand{\N}{\mathbb N}$ $\newcommand{\de}{\delta}$ $\newcommand{\sig}{\sigma}$ $\newcommand{\Average}[1]{\left\langle#1\right\rangle} $ $\newcommand{\IP}[2]{\Average{#1,#2}}$

Let $n,d \in \mathbb{N}$, and let $\Omega \subseteq \R^n$ be open.
Let $f \in W^{1,k}(\Omega,\R^d)$. Let $\omega \in \Omega^k(\R^d)$ be a smooth **closed** $k$-form, such that $\omega$ and its derivative $T\omega$ are both uniformly bounded globally; Is it true that $f^*\omega$ is weakly closed? i.e. does
$$\int_{\Omega} \IP{f^* \omega}{\de \sig}=0$$
hold for every compactly-supported $k+1$-form $\sigma \in \Omega^{k+1}(\Omega)$?

I have read that this should be true, but I am only able to show this in two special cases:

- The form $\omega$ is constant.
- $f$ is continuous.

Does this hold for non-continuous Sobolev maps in general?

Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n \in C^{\infty}(\Omega,\R^d)$ satisfy $f_n \to f$ in $W^{1,k}(\Omega,\R^d)$. Then $$ \int_{\Omega} \IP{f^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{d f_n^* \omega}{ \sig}=0. $$ Now, we need to justify the passage to the limit $\int_{\Omega} \IP{f^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^* \omega}{\de \sig}$.

If $\omega$ is constant, that is $\omega_q= \alpha$ independently of $q \in \R^d$, where $\alpha$ is a fixed element in $\bigwedge^k (\R^d)^* $, then we have $$ |f^* \omega-f_n^* \omega| \le |\alpha| \, \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|_{op}, $$ thus $$ \left|\int_{\Omega} \IP{f^* \omega}{\de \sig}- \IP{ f_n^* \omega}{\de \sig}\right| \le |\alpha| \|\de \sig\|_{\sup} \int_{\Omega} \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|. $$ and The RHS tends to zero since Sobolev approximation lifts to exterior powers.

When $\omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is $$ \begin{split} & |f^* \omega-f_n^* \omega|(p)= \\ &\left|\omega_{f(p)} \circ \bigwedge^{k} df_p -\omega_{f_n(p)} \circ \bigwedge^{k} (df_n)_p \right| \le \\ &\left|\big(\omega_{f(p)}-\omega_{f_n(p)}) \circ \bigwedge^{k} df_p +\omega_{f_n(p)} \circ \big( \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p) \right| \le \\ &|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \end{split} $$

So, we we need to estimate $\omega_{f(p)}-\omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:

$$ \begin{split} &|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\ &|T\omega|_{sup} \, \cdot \, |f(p)-f_n(p)| \, \cdot \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\ &|T\omega|_{sup} \, \cdot \, |f-f_n|_{sup} \, \cdot \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \stackrel{L^1}{\to} 0, %&|\alpha \circ \brk{\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p} | \le |\alpha| \cdot |\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p|_{op}. \end{split} $$

So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?