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We know that if $1\leq p < \infty$, the following embbedings of Sobolev spaces in $\mathbb{R}^k$ are true:

  • $\text { if } 1 \leq p<k, W^{1, p}\left(\mathbb{R}^{k}\right) \subset L^{q}\left(\mathbb{R}^{k}\right) \text { with } q \in\left[p, p_{*}\right] \text { and } p_{*}=\frac{kp}{k-p}$
  • $W^{1, k}\left(\mathbb{R}^{k}\right) \subset L^{q}\left(\mathbb{R}^{k}\right) \text { for every } q \in[k, \infty)$
  • $\text { if } p>N, W^{1, p}\left(\mathbb{R}^{k}\right) \subset L^{\infty}\left(\mathbb{R}^{k}\right)$.

Now, I am interested in proving that these emmbedings are not compacts. I am a little lost with this proof, I have found this related quiestion $W^{1,1}(\mathbb{R}^n)$ is not compactly embedded in $L^1(\mathbb{R}^n)$?, but I don't know if the proof which appears there can be generalised to answer my question.

mejopa
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    Essentially, You can always take a fixed very nice function $f$ and consider the sequence $f(x-(k,0,\dots,0))$ for $k \in N$. Then show that this sequence is bounded in $W^{k,p}$, nur has no convergent subsequence in $L^q$. – PhoemueX May 12 '20 at 10:56
  • Is the argument to prove that it has not convergent subsequences similar to which is used in the linked question? – mejopa May 12 '20 at 12:25

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