# Doubts about a functional defined on a Sobolev space

I am given this definition of quasi convexity:

$g: \mathbb{R}^{N\times n} \rightarrow [0,+\infty)$ a $C^0$ function is said quasi convex if for every $A\in\mathbb{R}^{N\times n}$

$$f(A)\; m(\Omega)\leq \int_{\Omega} f(A+\nabla \phi)dx$$

for every open bounded set $\Omega\subset \mathbb{R}^n$ and $\forall \phi \in W^{1,\infty}_0(\Omega, \mathbb{R}^N)$.

The book I'm looking at consider next the following functional:

let be $f:\mathbb{R}^{N\times n} \rightarrow \mathbb{R}$ quasi convex and $\Omega\subset\mathbb{R}^n$ open and bounded it considers

$F:W^{1,\infty}(\Omega)\rightarrow [0,\infty)$ given by $F(u)=\int_{\Omega}f(\nabla u)dx.$

• Why is this functional well defined? How can I say that $F<+\infty$ ?Do I need the condition of quasi convex to claim that or only the fact that the gradient is bounded and f continuos?

• Can I generalize in this case the definition of $f$ and consider an $f$ with values in all $\mathbb{R}$? Does $F$ remai well defined?

Thank you!

If $f$ is quasiconvex it is continuous. If $u \in W^{1,\infty}(\Omega)$ then $\nabla u$ is bounded, hence $f(\nabla u)$ is bounded too. Since $\Omega$ is a bounded set it follows that $f(\nabla u) \in L^1(\Omega)$. The full assumption of quasiconvexity isn't required.
• Thank you very much! If in the initial definition of quasi convex I consider function with values in $\mathbb{R}$ nothing of what you said changes, is it right? Thank you! – Gggl. Sep 1 '16 at 20:36
• All that is needed for $F$ to be finite is that $f$ is bounded on bounded sets and $f(\nabla u)$ measurable for all $u \in W^{1,\infty}$. – Umberto P. Sep 1 '16 at 20:38