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 $1 \leq q<\infty$.
It seems to me we can extend $(7)$ to
- $(8) \quad$ the injection $W^{1,p}(I) \subset L^q(I)$ is compact for all $1\le p \leq q<\infty$.
Could you confirm if $(8)$ is true and if my below attempt of proving it is fine?
We need the following results (in the same book), i.e.,
Notation (shift of function). We set $\left(\tau_h f\right)(x)=f(x+h)$ for $x \in \mathbb{R}^N$ and $h \in \mathbb{R}^N$.
Theorem 4.26 (Kolmogorov-M. Riesz-Fréchet). Let $\mathcal{F}$ be a bounded set in $L^p(\mathbb{R}^N)$ with $1 \leq p<\infty$. Assume that $\lim _{|h| \to 0} \|\tau_h f-f\|_{L^p (\mathbb R^N)}=0$ uniformly in $f \in \mathcal{F}$, i.e., for each $\varepsilon>0$ there is $\delta>0$ such that $\|\tau_h f-f\|_{L^p (\mathbb R^N)}<\varepsilon$ for all $f \in \mathcal{F}$ and $h \in \mathbb{R}^N$ with $|h|<\delta$. Then the closure of $\mathcal{F}_{\restriction \Omega}$ in $L^p(\Omega)$ is compact for any measurable set $\Omega \subset \mathbb{R}^N$ with finite measure. [Here $\mathcal{F}_{\restriction \Omega}$ denotes the restrictions to $\Omega$ of the functions in $\mathcal{F}$.]
Proposition 8.5 Let $u \in L^p(\mathbb{R})$ with $1<p<\infty$. The following statments are equivalent: (i) $u \in W^{1, p}(\mathbb{R})$. (ii) there exists a constant $C$ such that for all $h \in \mathbb{R}$ we have $\|\tau_h u-u\|_{L^p (\mathbb R)} \leq C|h|$. Moreover, we can choose $C=\|u'\|_{L^p (\mathbb R)}$ in (ii). The implication (i) $\implies$ (ii) is also valid when $p = 1$
Theorem 8.6 (extension operator). Let $1 \leq p \leq \infty$. There exists a bounded linear operator $P: W^{1, p}(I) \rightarrow W^{1, p}(\mathbb{R})$, called an extension operator, such that for each $u \in W^{1, p}(I)$ we have $P u_{\restriction I}=u, \|P u\|_{L^p(\mathbb{R})} \leq C\|u\|_{L^p(I)}$ and $\|P u\|_{W^{1, p}(\mathbb{R})} \leq C\|u\|_{W^{1, p}(I)}$ where $C$ depends only on $|I| \leq \infty$.
Let $\mathcal H$ be the closed unit ball of $W^{1,p} (I)$. We need to prove that $\mathcal H$ has a compact closure in $L^q (I)$. Let $P$ be the extension operator given by Theorem 8.6. Let $\mathcal F := \{Pu : u \in \mathcal H\} \subset W^{1,p} (\mathbb R)$. Then $\mathcal H = \mathcal F_{\restriction I}$ and $\mathcal F$ is bounded $W^{1, p}(\mathbb{R})$. In particular, $\mathcal F$ is bounded in $L^p (\mathbb R)$. By Theorem 8.8(5), $\mathcal F$ is bounded in $L^\infty (\mathbb R)$. By interpolation inequality, $\mathcal F$ is bounded in $L^q (\mathbb R)$. By Theorem 4.26, it remains to prove $\lim _{|h| \to 0} \|\tau_h f-f\|_{L^q (\mathbb R)}=0$ uniformly in $f \in \mathcal{F}$.
By Proposition 8.5, we have for each $f \in \mathcal F$ that $\|\tau_h f-f\|_{L^p (\mathbb R)} \leq C|h|$ where $C := \sup_{f \in \mathcal F} \|f'\|_{L^p (\mathbb R)} \le \sup_{f \in \mathcal F} \|f\|_{W^{1, p} (\mathbb R)} < \infty$. By interpolation inequality, we have for each $f \in \mathcal F$ that $$ \begin{align*} \|\tau_h f-f\|_{L^q (\mathbb R)}^q &\le \|\tau_h f-f\|_{L^p (\mathbb R)}^p \|\tau_h f-f\|_{L^\infty (\mathbb R)}^{q-p} \\ &\le C^p |h|^p (\|\tau_h f\|_{L^\infty (\mathbb R)} + \|f\|_{L^\infty (\mathbb R)})^{q-p} \\ &= C^p |h|^p (2 \|f\|_{L^\infty (\mathbb R)})^{q-p}. \end{align*} $$
The claim then follows.
