While investigating a form of the Euler-Lagrange equation for convex but nondifferentiable lagrangians, my professor makes the following two assumptions for a Generalized Riesz Theorem:
- $x\mapsto\|D(x)\|=\sup_{z\in D(x)}\|z\|$ is $L^\infty$;
- $x\mapsto m_{D(x)-v(x)}$ is measurable for every $v\in L^\infty$, $m_{\dots}$ being the projection of minimal distance of the origin onto the set $\dots$.
With these hypotheses, plus that $T$ is a functional on $L^1$ with:
$$T\xi\leq\int_\Omega I_{D(x)}^\ast(\xi(x))dx,$$
where $I_{D(x)}^\ast$ is the Legendre transform of the indicator of $D(x)$, and the indicator is the one from convex analysis, i.e.:
$$I_{D(x)}(p)=\begin{cases} 0 & p\in D(x) \\ +\infty & p\notin D(x) \end{cases},$$
not the characteristic function, which gives us:
$$I_{D(x)}^\ast(p)=\sup_{d\in D(x)}\langle d,p\rangle,$$
then $T$ is represented as an integral against a function.
Then he wants to apply this to $D(x)=\partial L(\nabla u(x))$ with $L$ a convex not necessarily differentiable Lagrangian, and $u$ is a solution to the minimization problem:
$$\min\int_\Omega[L(\nabla u(x))+u(x)]dx\qquad\text{on}\qquad u^0+W_0^{1,1}(\Omega),$$
$u^0$ being any boundary datum.
He will deal with the assumption on $T$, and also the $L^\infty$ assumption (actually, he assumes measurability, which seems unnecessary in the proof of said theorem, and gets boundedness by normalization), but he takes the other one (the one on the projections) for granted, and I cannot find them in Dacorogna.
The projection of minimal distance $m_{\partial L(\nabla u(x))}$ is well-defined since, as Dacorogna proves, $\partial L(\nabla u(x))$ is convex (and compact, as well). Hence, my question is:
How do I prove that, given $\Omega\subseteq\mathbb{R}^n$ an open set and $L:\mathbb{R}^n\to\mathbb{R}$ a convex but not necessarily differentiable function, for any $v:\mathbb{R}^n\to\mathbb{R}^n$ measurable, with $u$ as above, we have that $x\mapsto m_{\partial L(\nabla u(x))}$ is measurable, $m_K$ being the projection of minimal distance of the origin onto $K$?
Since he assumed this seemingly superfluous hypothesis, we have a bonus question:
Can I prove that $x\mapsto\|\partial L(\nabla u(x))\|$ is also measurable?
Update
The bonus question is now a needed point, since, though that measurability is not needed to apply Riesz, it is needed in the normalization process.