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Let $I$ be an open bounded interval of $\mathbb R$. I have recently proved in this thread that the injection $W^{1,p}(I) \subset L^q(I)$ is compact for all $1\le p \leq q<\infty$.

We fix $1\le r < p <\infty$. We have the injection $L^p (I) \subset L^r (I)$ is continuous. Combining this with the above result, we get that the injection $W^{1,p}(I) \subset L^q(I)$ is compact for any $p, q \in [1, \infty)$. This is quite surprising to me when I look at Rellich-Kondrachov theorem.

Could you confirm if my above reasoning is fine?

Akira
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    No, this is normal, in dimension $1$, $p^*$ is always infinite, so every injection is subcritical, except $p=1$, $q=\infty$. – LL 3.14 Nov 24 '23 at 14:39
  • @LL3.14 Thank you so much for your verification! – Akira Nov 24 '23 at 14:40
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    If you don't know the classical picture, here it is in dimension $d$ http://laurent-lafleche.perso.math.cnrs.fr/images/GagNir.svg. The inside of the triangle contains compact embedding of $L^r\cap W^{1,p}$ in the whole space – LL 3.14 Nov 24 '23 at 14:43
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    In bounded domains, as soon as you are in some space, you are also on all the spaces on the right ... – LL 3.14 Nov 24 '23 at 14:49

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