Functional continuity with respect to strong $W^{1,p}$-convergence

Question

Suppose $p\in [1,\infty), f\in C([a,b],\Bbb R^m,\Bbb R^m)$ such that $f$ satisfies the following growth bound: There exist $C_1,C_2 <\infty$ such that $|f(x,y,z)|\leq C_1|z|^p+C_2$ for all $x,y$.

Then, the functional $F(u)=\int f(x,u,Du)dx$ ist continous w.r.t. to the strong $W^{1,p}$-topology, i.e. $u_k \rightarrow u \implies F(u_k)\rightarrow F(u)$.

I know we can extract a pointwise convergent subsequence and then we can apply Egorov's theorem an a big set $E_\varepsilon \subset \Omega$ with $|\Omega \setminus E_\varepsilon| < \varepsilon$ to receive uniform convergence on that set, so we can shift taking Limits within the integral on that set. However I do not get how to use the growth bound to get to convergence on the "bad" complement too?

Answer 1

You are missing one important thing when extracting the pointwise converging subsequence: in addition you get a dominating function, that can be used to apply dominated convergence. The full theorem is:

Let $(x_n)$ converge to $x$ in $L^p$, $p<\infty$. Then there is a subsequence $(x_{n_k})$ and a dominating function $g\in L^p$ such that $x_{n_k} \to x$ and $|x_{n_k}|\le g$ almost everywhere.

This is Theorem 4.9 in Brezis: Functional analysis, Sobolev spaces, PDEs.

The claim is proven as part of the standard proof of completeness of $L^p$-spaces, but most authors do not mention the existence of the dominating function in the statement of the theorem, although the dominating function is constructed explicitly in the proof.

While Egorov theorem is very nice, most of the time the above result plus dominated convergence is sufficient.