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Piotr Hajlasz
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Suppose that $Ju_i$ does not converge to $Ju$ in $L^1$. Then for a subsequence you will have $\Vert Ju_i-Ju\Vert_1\geq\epsilon$. If you take a subequence you can assume that $\nabla u_i\to\nabla u$ a.e. and hence $Ju_i\to Ju$ a.e. which would imply convergence $Ju_i\to Ju$ on that sequence and that would contradict the inequality $\Vert Ju_i-Ju\Vert_1\geq\epsilon$.

To make thigs more straightforward, you can also use a finite covering of $\mathcal{M}$ so you can assume that $u,u_i\in W^{1,n}(B^n(0,1),\mathcal{N})$.

In fact your claim is true not only for Lipschitz maps but for any $W^{1,n}$.

Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184