Is it true that $\frac{1}{meas(\Omega) }\int_{\Omega}\nabla u\in conv [\nabla u(\Omega)]$?

Question

Suppose $u\in W^{1,\infty}(\Omega) $. Then is it true that $\frac{1}{meas(\Omega) }\int_{\Omega}\nabla u\in conv [\nabla u(\Omega)]\ \ ? $

Here $\Omega \subset \mathbb R^n$ is open, bounded.

$\nabla u$ denotes the gradient of $u$. And $conv A$ means convex hull of a set $A$. Here $meas(\Omega) $ is Lebesgue measure of $\Omega$.

A function $u\in W^{1,\infty}(\Omega) $ means that $u\in L^{\infty}(\Omega) $ and first order weak derivatives of $u$ are also in $L^{\infty}(\Omega) $.

This is a part of a theorem, named Pyramidal construction , in Differential Inclusions . It has been used to prove the theorem so the above statement is true. But I can't prove it.

Any help is appreciated. Thank you.

Answer 1

This is true more generally: for any integrable vector-valued function $g:\Omega\to \mathbb{R}^m$ the mean value of $g$ is contained in the convex hull of its essential range.

First, recall that the closed convex hull of a set is the intersection of all closed half-spaces containing that set. A closed half-space is described as $\{x\in \mathbb{R}^m : \langle x, a\rangle \ge b\}$ for some $a\in\mathbb{R}^m$ and $b\in\mathbb{R}$. If the range of $g$ is contained in this half-space, then $$ \left< \frac{1}{\operatorname{meas}(\Omega)} \int_\Omega g, a \right> =\frac{1}{\operatorname{meas}(\Omega)} \int_\Omega \left< g, a \right> \ge \frac{1}{\operatorname{meas}(\Omega)} \int_\Omega b = b \tag1$$ hence the mean of $g$ is also in the same half-space.

Second, we can remove closed by observing that, unless the essential range of $g$ consists of one point, the mean of $g$ is contained in its relative interior (the interior with respect to the affine span). To this end, replace $V$ by the affine span of the essential range of $g$; this ensures the convex hull has nonempty interior. If the mean of $g$ lies on the boundary of the closed convex hull of the essential range, then there exists a supporting hyperplane at that point, which means having $a, b$ as above such that equality holds in (1). But then $\langle g, a\rangle =b$ a.e., which means the convex hull has empty interior, a contradiction.