I am trying to understand why Quasiconvexity implies Rank-One convexity.
In a standard proof of this fact a sequence of functions is constructed, which converges weakly to zero in $W^{1,p}.$ in order to understand why this is useful, I would like to understand why my simpler argument is wrong.
Quasiconvexity of a function $f: \mathbb{R}^{m\times d} \rightarrow \mathbb{R}$ means that for any bounded domain $\Omega \subset \mathbb{R}^d:$ $$\int_{\Omega}f(C + \nabla \omega(y))dy \ge | \Omega | f(C), $$ for any $\omega \in W_0^{1,\infty}(\Omega)$.
Now let us consider two matrices $A, B$ that are rank one connected, through $A-B = \xi \otimes \eta.$ Wlog we assume that $\eta = e_1 \in \mathbb{R^d}$ (else we rotate everything). We want to prove that $$f (\theta A + (1-\theta)B) \le \theta f(A) + (1-\theta) f(B)$$ So let us define $C = \theta A + (1-\theta)B$ and $ \Omega = (0,1)^d$ and $$ \omega(y) = \begin{cases} (1-\theta) (x \cdot e_1) \xi, \text{ if } 0 \le x \cdot e_1 \le \theta \\ \\ -\theta (x \cdot e_1 -1) \xi, \text{ if } \theta < x \cdot e_1 \le 1 \end{cases} $$ Now we find that the gradient of $\omega$ is: $$\nabla \omega (y) = \begin{cases} (1-\theta)\xi \otimes e_1, \text{ if } 0 \le x \cdot e_1 \le \theta \\ \\ -\theta \xi \otimes e_1, \text{ if } \theta < x \cdot e_1 \le 1 \end{cases}$$ So eventually we find: $$\theta f(A) + (1-\theta) f(B) = \int_{\Omega}f(C + \nabla \omega(y))dy \ge f(C)$$ where the latter inequality holds by quasiconvexity.
A proof of this fact is given for example by Dacorogna in "Direct methods for calculus of variations," but it is actually more complicated.
Your proof is the natural first iteration of the argument. But it can't be the final form of the proof, because your function $\omega$ is not in $W^{1,\infty}_0(\Omega)$. It vanishes on two faces of the cube $(0,1)^d$ but not on the other faces.
One needs to multiply by a cutoff function $\eta$ to ensure zero boundary values of $\omega\eta$. Then of course $\nabla (\omega \eta) = \eta \nabla \omega + \omega \nabla \eta$, and one has to prevent the term with $\nabla \eta$ from messing up the inequality.