Yeah, I did not underline it but $F$ may take value $+\infty$: so the expression written above would be the definition of $F$ if $u \in W^{1,1}$ (or $BV$ with some abuse), with the convention that $F=+\infty$ elsewhere... Thanks in any case for the comment!

Dear Prof. Iosif Pinelis, I do thank you for your answer. Your help in these days has been extremely valuable to me and I therefore thank you. +1, of course

Right, thanks for pointing it out. I am still confused a bit and I will think about it. Again thanks for helping.

By integration by parts you see that $\int u_i \frac{\partial}{\partial u_j} \tilde{h}(u) \, du$ is $\delta_{ij}$. Thus it seems to me that we have found a lower bound for $\mathcal J$, i.e. $\mathcal J \ge$ trace of $A$. I suspect I am wrong somewhere but do not see where. I apologize once again for the confusion and I thank for the help.

Sorry for my confusion. So let me try to explain better. We have $J(h) = \int \vert \langle Au, \nabla h(\vert u \vert) \rangle \vert du$. Do you agree on this? Let me define a function $\tilde h$ so that $h(\vert u \vert) = \tilde h(u)$. Then $J(h) = \int \vert \langle Au, \nabla \tilde h(u) \rangle \vert du \ge \vert \int \langle Au, \nabla \tilde h(u) \rangle du \vert$. Am I right? This last guy, by definition, is the sum over i,j of $A_{ji} \cdot \int u_i \cdot \frac{\partial}{\partial u_j}\tilde h(u) du$. Do you agree? Now the last integral can be computed by integration by parts.

Sorry, there is still something not clear. It is easy to see that one can re-write the functional $\mathcal J$ as $\int \vert \langle Au, \nabla \tilde{h}(u) \rangle \vert du \ge \vert \int \langle Au, \nabla \tilde{h}(u) \rangle du \vert = \vert \text{tr}\, A\vert$ since $\int u_i \partial_j \tilde{h}(u) du = \delta_{ij}$ (up to a sign, integration by parts). Where am I mistaken? Thanks again.

Oh nice, thanks for the useful answer! I admit I am a bit puzzled, as I expected the result to depend on the matrix $A$. However seems to be working, nice! Thanks again, +1!