Let $I:=(a, b)$ be an open interval, possibly unbounded. I'm reading below theorem in Brezis' Functional Analysis:
Theorem 8.8. There exists a constant $C$ (depending only on $|I| \leq \infty)$ such that $$ (5) \quad \|u\|_{L^{\infty}(I)} \leq C\|u\|_{W^{1, p}(I)} \quad \forall u \in W^{1, p}(I), \quad \forall 1 \leq p \leq \infty . $$ In other words, $W^{1, p}(I) \subset L^{\infty}(I)$ with continuous injection for all $1 \leq p \leq \infty$. Further, if $I$ is bounded then
- $(6) \quad$ the injection $W^{1, p}(I) \subset C(\bar{I})$ is compact for all $1<p \leq \infty$,
- $(7) \quad$ the injection $W^{1,1}(I) \subset L^q(I)$ is compact for all $1 \leq q<\infty$.
Now we assume $I$ is bounded. It is clear that
- $8\quad$ the injection $C(\bar{I}) \subset L^p$ is continuous for all $p \in [1, \infty]$.
If $p \in (1, \infty]$ then $(6, 8)$ implies the injection $W^{1, p}(I) \subset L^p$ is compact. If $p=1$ then $(7)$ implies the injection $W^{1,1}(I) \subset L^1(I)$ is compact. In any case, $W^{1, p}(I) \subset L^{p}(I)$ with compact injection for all $p \in [1, \infty]$.
Could you confirm if my above reasoning is fine?
Thank you so much for your help!
