Quasiconvexity via dominated convergence theorem in Dacorogna

Question

I am reading the proof of the Dacorogna formula for quasiconvex envelop on page 271 of the book Direct methods in the calculus of variations by Dacorogna (Theorem 6.9).

In the beginning of step 3 of the proof. We have already established $$\int_DQ'f(\xi+\nabla\psi(x))dx\geq Q'f(\xi)\cdot\textrm{meas}D$$ for all $\xi\in\mathbb R^{N\times n}$ and $\psi\in\mathcal A(D)$, where $\mathcal A(D)$ is the set $$\mathcal A(D)=\{\varphi\in W^{1,\infty}_0(D;\mathbb R^N)\cap\textrm{Aff}_{piec}(\overline D;\mathbb R^N):\textrm{supp }\varphi\subseteq D\}$$ The author then claims that the continuity of $Q'f$ and the fact that $\mathcal A(D)$ is dense in $W^{1,\infty}_0$ in any $W^{1,p}$ norm, $1\leq p<\infty$, implies the quasiconvexity of $Q'f$ by the dominated convergence theorem.

Now, given $\psi\in W^{1,\infty}_0$, we can find a sequence $\psi_n$ converging to $\psi$ in any $W^{1,p}$ norm. However, I don't see why this should imply $\nabla \psi_n$ converges to $\nabla\psi$ pointwise which is what I think we would need to apply the dominated convergence theorem. Also would the dominating function be $f$? We already know that $f\geq Q'f$ and $f$ is bounded on compact sets, but I'm not sure how to handle the possible domain difference for $f$ and $Q'f(\xi+\nabla\psi(x))$. I am not very fluent in Sobolev space theory. Any help is appreciated.

Answer 1

Note that you already know, at Step 3, that for all $\xi\in\mathbb R^{N\times n}$ and $\psi\in\mathcal A(D)$ $$\int_DQ'f(\xi+\nabla\psi(x))dx\geq Q'f(\xi)\cdot\textrm{meas}D$$

Which is almost quasiconvexity of $f$: you only need to go from the condition for all $\psi\in\mathcal A(D)$ to the condition for all $\psi\in W_0^{1, \infty}(D ; \mathbb{R}^n).$ Since $\mathcal A(D)$ is dense in $W_0^{1, \infty}(D ; \mathbb{R}^n)$, then, by continuity, the following quantities are the same (to see this, just recall that the infimum is just a limit and taking a limit of a continuous function on a dense set of $X$ is the same as taking it on all $X$):

$$\tilde{Q} f(\xi):=\inf \left\{\frac{1}{\text { meas } D} \int_{D} f(\xi+\nabla \varphi(x)) d x: \varphi \in W_{0}^{1, \infty}\left(D ; \mathbb{R}^{N}\right)\right\}$$ and $$ Q^{\prime} f(\xi):=\inf \left\{\frac{1}{\text { meas } D} \int_{D} f(\xi+\nabla \varphi(x)) d x: \varphi \in \mathcal{A}(D)\right\}$$

But now, the quantity $Q'f$ is under the integral, hence to be able to "take the limit", you need to appeal to a convergence theorem: the easiest is the Lebesgue one. Hence, by density you have $\psi_n \rightarrow \psi$ is Sobolev norm, hence you have convergence in $L^p$ of the gradients as well. From them, you can extract a subsequence converging pointwise almost everywhere (which is all that you need to apply Lebesgue dom. conv. thm, since your measure space is complete: see Remark 2). The domination comes from the fact that for all $\xi\in\mathbb R^{N\times n}$ you have that $Q'f$ is an infumum of a bounded quantity since $f$ is locally bounded: note that since you are converging to $\tilde{Q} f(\xi)$, which is an inf on a bigger set, then your sequence is decreasing. Hence its enough to take $Q'f$ as dominated function, which is bounded on $D$.