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Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \\ , \\ \\ \\ 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, and has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \, , \quad 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, and has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \\ , \\ \\ \\ 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, an has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \\ , \\ \\ \\ 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, and has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, $u\neq 0$ with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \big( \frac{x} {\epsilon} \big) \\ , 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, and has constant non-zero norm in $L^{p^*}$, and of course has no convergent subsequences there, since it converges a.e. to zero.

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \\ , \\ \\ \\ 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, an has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.