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Whenever $n<q,$ it is known that given a sequence $\{ u_{k} \}$ which is weakly convergent in $W^{1,q}(U)$ one has that the Jacobian determinants $\text{det} Du_{k}$ converge weakly in $L^{q/n}(U).$

Together with polyconvexity, this serves to prove weak lower semicontinuity of some functionals whose lagrangians fail to be convex.

Motivated by this fact (proving weak lower semicontinuity of some functionals) I wish to find out whether the above result can be improved. In other words, can one have $q<n$ and still have weak convergence of the jacobians? I know one cannot lower $q$ too much, but still one would expect to be able to lower it a bit.

The reason I think this might not be true is that in the proof one crucially needs Morrey's inequality, which of course only holds when $q<n,$ but hopefully there is some way around this difficulty (perhaps by imposing an extra condition on the matrix of cofactors?)

Thank you for your time

Whenever $n<q,$ it is known that given a sequence $\{ u_{k} \}$ which is weakly convergent in $W^{1,q}(U)$ one has that the Jacobian determinants $\text{det} Du_{k}$ converge weakly in $L^{q/n}(U).$

Together with polyconvexity, this serves to prove weak lower semicontinuity of some functionals whose lagrangians fail to be convex.

Motivated by this fact (proving weak lower semicontinuity of some functionals) I wish to find out whether the above result can be improved. In other words, can one have $q<n$ and still have weak convergence of the jacobians? I know one cannot lower $q$ too much, but still one would expect to be able to lower it a bit.

The reason I think this might not be true is that in the proof one crucially needs Morrey's inequality, which of course only holds when $n<q,$ but hopefully there is some way around this difficulty (perhaps by imposing an extra condition on the matrix of cofactors?)

Thank you for your time

Whenever $n<q,$ it is known that given a sequence $\{ u_{k} \}$ which is weakly convergent in $W^{1,q}(U)$ one has that the Jacobian determinants $\text{det} Du_{k}$ converge weakly in $L^{q/2}(U).$

Together with polyconvexity, this serves to prove weak lower semicontinuity of some functionals whose lagrangians fail to be convex.

Motivated by this fact (proving weak lower semicontinuity of some functionals) I wish to find out whether the above result can be improved. In other words, can one have $q<n$ and still have weak convergence of the jacobians? I know one cannot lower $q$ too much, but still one would expect to be able to lower it a bit.

The reason I think this might not be true is that in the proof one crucially needs Morrey's inequality, which of course only holds when $q<n,$ but hopefully there is some way around this difficulty (perhaps by imposing an extra condition on the matrix of cofactors?)

Thank you for your time

Whenever $n<q,$ it is known that given a sequence $\{ u_{k} \}$ which is weakly convergent in $W^{1,q}(U)$ one has that the Jacobian determinants $\text{det} Du_{k}$ converge weakly in $L^{q/n}(U).$

Together with polyconvexity, this serves to prove weak lower semicontinuity of some functionals whose lagrangians fail to be convex.

Motivated by this fact (proving weak lower semicontinuity of some functionals) I wish to find out whether the above result can be improved. In other words, can one have $q<n$ and still have weak convergence of the jacobians? I know one cannot lower $q$ too much, but still one would expect to be able to lower it a bit.

The reason I think this might not be true is that in the proof one crucially needs Morrey's inequality, which of course only holds when $q<n,$ but hopefully there is some way around this difficulty (perhaps by imposing an extra condition on the matrix of cofactors?)

Thank you for your time