A Bounded Sequence in Sobolev Space contains a subsequence that converges to a limit in that Sobolev Space

Question

I have some questions about the last steps in a compactness theorem at in Evans and Gariepy. My questions first, the proof is below.

1. In the last step he just asserts that $\{f_k\}$ is bounded in $L^{p^*}$. How is this so?
2. In the last step he shows that there's a subsequence that converges in $L^q$ for $1 \leq q < p^*$. Then concludes $f \in W^{1,p}$. Why does this conclusion follow?
3. There's a comment at the end of the proof for $p = 1$ that I did not follow -- why is $f \in BV(U)$? Note the theorem in reference states that if $f_k \in BV(U)$ and $f_k \rightarrow f$ in $L^1_{loc}$ then $\|Df\|(U) \leq \liminf_{k \rightarrow \infty}\|Df_k\|(U)$ where $$\|Df\|(U)= \sup_{\phi \in C^1_c(U;\mathbb{R}^n),|\phi|\leq 1}\left\{\int_Uf\,(\nabla \cdot \phi) \right\}$$

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The statement and proof:

This is proved in a couple of steps:

1. Creating extension of sobolev functions, $f_k$, to $\mathbb{R}^n$ -- call these $\overline{f}_k$
2. Defining mollifiers for each extension: $\overline{f^{\epsilon}_k} = \eta_{\epsilon} * \overline{f}_k$
3. Proving $\{\overline{f^{\epsilon}_k}\}$ is a bounded, equicontinuous family.
4. Showing that this implies that there exists a subsequence $\{f_{k_j}\} \subset \{f_{k}\}$ s.t. $\limsup_{i,j \rightarrow \infty} \|f_{k_i} - f_{k_j}\| \leq \delta$
5. Let $p^* = np/(n-p)$. Finally,

Theorem 3 is below:

Answer 1

1. By assumption, $(f_k)$ is bounded in $W^{1,p}$. From the Sobolev embedding theorem we know that $W^{1,p}$ embeds continuously into $L^{p^\ast}$. Thus $(f_k)$ is also bounded in $L^{p^\ast}$.
2. Assume that $f_k\to f$ in $L^q$ and $(f_k)$ is bounded in $W^{1,p}$. In particular, $\partial_j f_k$ is bounded in $L^p$ for $1\leq j\leq n$. Since $L^p$ is reflexive for $1<p<\infty$, there is a subsequence $f_{k_l}$ and $g_1,\dots,g_n\in L^p$ such that $\partial_j f_{k_l}\to g_j$ weakly in $L^p$ (this is your Theorem 3) and $\|g_j\|_p\leq \liminf_{l\to\infty}\|\partial_j f_{k_l}\|_p$ (this holds for any weakly convergent sequence in a Banach space). This implies $$ \int \partial_j\phi f\,dx=\lim_{l\to\infty}\int \partial_j\phi f_{k_l}\,dx=-\lim_{l\to\infty}\int \phi\partial_j f_{k_l}\,dx=-\int \phi g_j\,dx $$ for all $\phi\in C_c^\infty$. Thus $f\in W^{1,p}$ and $$ \|f\|_{W^{1,p}}=\left(\|f\|_p^p+\sum_{j=1}^n \|\partial_j f\|_p^p\right)^{1/p}\leq \liminf_{l\to\infty}\|f_{k_l}\|_{W^{1,p}}. $$
3. By definition, $f\in \mathrm{BV}(U)$ if $f\in L^1(U)$ and $\|Df\|(U)<\infty$. Moreover, if $g\in W^{1,1}$, then $\|Dg\|(U)=\|\nabla g\|_1$. Hence in your case $$ \|Df\|(U)\leq \liminf_{l\to\infty}\|D f_{k_l}\|(U)=\liminf_{l\to\infty}\|\nabla f_{k_l}\|_1\leq \liminf_{l\to\infty}\|f_{k_l}\|_{W^{1,1}}<\infty. $$