Why can we assume a subsequence converges to the limit? This result in Evans discusses a sufficient condition for a functional (convexity in one of the coordinates) to be lower semicontinuous.
The proof is a bit long and requires some notation and my more problematic questions is about the end. So I am posting a picture of the proof.
For easy searching here is the statement to be proved:

Define $I[w] = \int_U L(Dw(x),w(x),x)\, dx$. Assume that $L$ is smooth and bounded below. In addition, the mapping $p \mapsto L(Dw(x),w(x),x)$ is convex ($p$ is an element of $Dw$), for each $w(x) \in \mathbb{R}, x \in U$. $I$ so defined is weakly lower semicontinuous on $W^{1,q}(U)$.

My questions:


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*At the start they pass to subsequence and suppose $l = \lim I[u_k]$. This is just done by selecting the subsequence that "creates" the $\liminf$ right?

*He uses convexity to prove weak LSC. I'm having trouble following how he concludes $(24)$ and $(25)$.


 A: For 1, you are right. 
For (23), the author uses the convexity inequality $f(y) \geq f(x)+(d_xf)(y-x)$ (except that you have additional parameters but they do not matter). 
The result in (24) does not depend on (23). To get the equality, I am assuming $U$ is bounded (at least, (22) makes no sense if $U$ has infinite measure). 
It follows because $G_{\epsilon} \subset E_{\epsilon}$, thus $u_k \rightarrow u$ uniformly on $G_{\epsilon}$. Therefore, because of the choice of function in $F_{\epsilon}$, if $k$ is large enough, then for all $x \in G_{\epsilon}$, $|Du(x)|+|u_k(x)| \leq \frac{2}{\epsilon}$. 
$L$ is smooth on $\{(p,q,r),\,|p|+|q| \leq 2\epsilon^{-1},\,|r| \leq \sup_{x \in U}\,|x|\}$ (which is compact) thus is Lipschitz (*), therefore there is a uniform convergence of $$x \in G_{\epsilon} \longmapsto L(Du(x),u_k(x),x)$$ to $$x \in G_{\epsilon} \longmapsto L(Du(x),u(x),x).$$ 
Thus ($G_{\epsilon}$ has finite measure) there is convergence of the integrals. 
Note that only $\lim\,\inf \, \ldots \geq \ldots $ in (24) matters and it follows from (22), (16) and Fatou’s lemma.
For (25), we use the same kind of estimate as in (*) to show that $$\|D_pL(Du,u_k,x)-D_pL(Du,u,x)\|_{\infty,x \in G_{\epsilon}} \rightarrow 0.$$
Since $U$ has finite measure and $Du_k$ converges weakly to $Du$ in $L^q(U)$, $$\sup_k\,\|Du-Du_k\|_{L^1(U,\mathbb{R}^n)} < \infty.$$
Therefore, $$\int_{G_{\epsilon}}{(D_pL(Du,u_k,x)-D_pL(Du,u,x)) \cdot (Du_k-Du)} \rightarrow 0.$$
It remains to prove that $$\int_{G_{\epsilon}}{L(Du,u,x)Du_k} \rightarrow \int_{G_{\epsilon}}{L(Du,u,x)Du}.$$
For that, you just note that $$f \in L^q(U,\mathbb{R}^n) \longmapsto \int_{G_{\epsilon}}{L(Du,u,x)f(x)dx}$$ is a continuous linear form (because on $G_{\epsilon}$, $L(Du,u,x)$ is bounded), and $Du_k \rightarrow Du$ weakly.
